Chapter 8

z-Transform for Discrete-Time Signals and Systems

Chapter Objectives

• Learn the z-transform as a more generalized version of the discrete-time Fourier transform (DTFT) studied in Chapter 5.
• Understand the convergence characteristics of the z-transform and the concept of region of convergence.
• Explore the properties of the z-transform.
• Understand the use of the z-transform for modeling DTLTI systems. Learn the z-domain system function and its use for solving signal-system interaction problems.
• Learn techniques for obtaining block diagrams for DTLTI systems based on the z-domain system function.
• Learn the use of the unilateral z-transform for solving difference equations with specified initial conditions.

8.1 Introduction

In this chapter we will consider the z-transform which, for discrete-time signals, plays the same role that the Laplace transform plays for continuous-time signals. In the process, we will observe that the development of the z-transform techniques for discrete-time signals and systems parallels the development of the Laplace transform in Chapter 7.

It was established in Chapter 7 that the Laplace transform is an extension of the Fourier transform for continuous-time signals such that the latter can be thought of as a special case, or a limited view, of the former. The relationship between the z-transform and the DTFT is similar. In Chapter 5 we have discussed the use of the DTFT for analyzing discrete-time signals. The use of the DTFT was adapted to the analysis of DTLTI systems through the concept of the system function. The z-transform is an extension of the DTFT that can be used for the same purposes.

Consider, for example, a discrete-time ramp signal x [n] = nu [n]. Its DTFT does not exist since the signal is not absolute summable. We will, however, be able to use z-transform techniques for analyzing the ramp signal. Similarly, the DTFT-based system function is only usable for stable systems since the impulse response of a system must be absolute summable for its DTFT-based system function to exist. The z-transform, on the other hand, can be used with systems that are unstable as well. The DTFT may or may not exist for a particular signal or system while the z-transform will generally exist subject to some constraints.

We will see that the DTFT is a restricted cross-section of the much more general z-transform. The DTFT of a signal, if it exists, can be obtained from its z-transform while the reverse is not generally true. In addition to analysis of signals and systems, implementation structures for discrete-time systems can be developed using z-transform based techniques. Using the unilateral variant of the z-transform we will be able to solve difference equations with non-zero initial conditions.

We begin our discussion with the basic definition of the z-transform and its application to some simple signals. In the process, the significance of the issue of convergence of the transform is highlighted. In Section 8.2 the convergence properties of the z-transform, and specifically the region of convergence concept, are studied. Section 8.3 covers the fundamental properties of the transform that are useful in computing the transforms of signals with more complicated definitions. Proofs of the significant properties of the z-transform are presented to provide further insight and experience on working with transforms. Methods for computing the inverse z-transform are discussed in Section 8.4. Application of the z-transform to the analysis of linear and time-invariant systems is the subject of Section 8.5 where the z-domain system function concept is introduced. Derivation of implementation structures for discrete-time systems based on z-domain system functions is covered in Section 8.6. In Section 8.7 we discuss the unilateral variant of the z-transform that is useful in solving difference equations with specified initial conditions.

If we were to expand the summation in Eqn. (8.1) we would get an expression in the form

$$\begin{array}{ccc}X (z) =\cdots +x[-2] z^2 + x[-1] z^1 + z[0] + x[1] z^{-1} + x[2] z^{-2}+...& & (8.2)\end{array}$$

which suggests that the transform is a polynomial that contains terms with both positive and negative powers of z. Since the independent variable z is complex, the resulting transform X (z) is also complex-valued in general. Notationally, the transform relationship of Eqn. (8.1) can be expressed in the compact form

$$\begin{array}{ccc}X\left(z\right)=Ƶ\left\{x\left[n\right]\right\}& & \left(8.3\right)\end{array}$$

which is read “X(z) is the z-transform of x[n]”. An alternative way to express the same relationship is through the use of the notation

$$\begin{array}{cccc}x[n]& \underleftrightarrow{Ƶ} X(z)& & (8.4)\end{array}$$

The derivation of the relationship between the z-transform and the DTFT is straightforward. Let the complex variable z be expressed in polar form as

$$\begin{array}{ccc}z=r{e}^{j\Omega }& & \left(8.5\right)\end{array}$$

Since z is a complex variable, we will represent it with a point in the complex z-plane as shown in Fig. 8.1. The parameter r indicates the distance of the point z from the origin. The parameter Ω is the angle, in radians, of the line drawn from the origin to the point z, measured counter-clockwise starting from the positive real axis.

Obviously, any point in the complex z-plane can be expressed by properly choosing the parameters r and Ω. Substituting Eqn. (8.5) into the z-transform definition in Eqn. (8.1) we have

$$\begin{array}{llll}X\left(r,\Omega \right)\hfill & ={X\left(z\right)|}_{z=r{e}^{j\Omega }}\hfill & \hfill & \hfill \\ \hfill & ={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]{\left(r{\text{ e}}^{j\Omega }\right)}^{-n}\hfill & \hfill & \hfill \\ \hfill & ={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]r^{-n}{e}^{-j\Omega n}\hfill & \hfill & \left(8.6\right)\hfill \end{array}$$

The result is a function of parameters r and Ω as well as the signal x[n]. We observe from Eqn. (8.6) that X (r, Ω) represents the DTFT of the signal x[n] multiplied by an exponential signal r−n:

$$\begin{array}{ccc}X\left(r,\Omega \right)={X\left(z\right)|}_{z=r{\text{ e}}^{j\Omega }}=ℱ\left\{x\left[n\right]r^{-n}\right\}& & \left(8.7\right)\end{array}$$

If we choose a fixed value of r = r1 in Eqn. (8.5) and allow the angle Ω to vary from 0 to 2π radians, the resulting trajectory of the complex variable z in the z-plane would be a circle with its center at the origin and with its radius equal to r1 as shown in Fig. 8.2.

This allows us to make an important observation:

If the parameter r is chosen to be equal to unity, then we have z = ejΩ and

$$\begin{array}{ccc}{X\left(z\right)|}_{z=e^{j\Omega }}={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]z^{-n}=ℱ\left\{x\left[n\right]\right\}& & \left(8.8\right)\end{array}$$

This is a significant observation. The z-transform of a signal x[n] evaluated for z = ejΩ produces the same result as the DTFT of the signal. The trajectory defined by z = ejΩ in the z-plane is a circle with unit radius, centered at the origin. This circle is referred to as the unit circle of the z-plane. Thus we can state an important conclusion:

An easy way to visualize the relationship between the z-transform and the DTFT is the following: Imagine that the unit circle of the z-plane is made of a piece of string. Suppose the z-transform is computed at every point on the string. Let each point on the string be identified by the angle Ω so that the range −π ≤ Ω < π covers the entire piece. If we now remove the piece of string from the z-plane and straighten it up, it becomes the Ω axis for the DTFT. The values of the z-transform computed at points on the string would be the values of one period of the DTFT.

Next we would like to develop a graphical representation of the z-transform in order to illustrate its relationship with the DTFT. Suppose the z-transform of some signal x[n] is given by

$$\begin{array}{ccc}X\left(z\right)=\frac{z\left(z-0.7686\right)}{z^2-1.5371z+0.9025}& & \left(8.9\right)\end{array}$$

In later sections of this chapter we will study techniques for computing the z-transform of a signal. At this point, however, our interest is in the graphical representation of the z-transform, the DTFT, and the relationship between them. Using the substitution z = r ejΩ, the transform in Eqn. (8.9) becomes

$$\begin{array}{ccc}X\left(r,\Omega \right)=\frac{r{e}^{j\Omega }\left(r{e}^{j\Omega }-0.7686\right)}{r^2{e}^{-j2\Omega }-1.5371r{e}^{j\Omega }+0.9025}& & \left(8.10\right)\end{array}$$

Using Eqn. (8.10) the transform at a particular point in the z-plane can be computed using r, its distance from the origin, and Ω, the angle it makes with the positive real axis. We would like to graph this transform; however, there are two issues that must first be addressed:

1. The z-transform is complex-valued. We must graph it either through its polar representation (magnitude and phase graphed separately) or its Cartesian representation (real and imaginary parts graphed separately). In this case we will choose to graph only its magnitude |X (z)| as a three-dimensional surface. The magnitude of the transform can be computed by

   $$\begin{array}{ccc}|X\left(r,\Omega \right)|={\left[X\left(r,\Omega \right)X^{*}\left(r,\Omega \right)\right]}^{1/2}& & \left(8.11\right)\end{array}$$

   where X* (r, Ω) is the complex conjugate of the transform X (r, Ω), and is computed as

   $$\begin{array}{ccc}{X}^{*}\left(r,\Omega \right)=\frac{r{e}^{j\Omega }\left(r{e}^{j\Omega }-0.7686\right)}{r^2{e}^{-j2\Omega }-1.5371r{e}^{j\Omega }+0.9025}& & \left(8.12\right)\end{array}$$

2. Depending on the signal x[n], the transform expression given by Eqn. (8.9) is valid either for r < 0.95 only, or for r > 0.95 only. This notion will elaborated upon when we discuss the “region of convergence” concept later in this chapter. In graphically illustrating the relationship between the z-transform and the DTFT at this point, we will assume that the sample transform in Eqn. (8.9) is valid for r > 0.95. The three-dimensional magnitude surface will be graphed for all values of r; however, we will keep in mind that it is only meaningful for values of r greater than 0.95.

The magnitude |X (z)| is shown as a mesh in part (a) of Fig. 8.3. The unit circle is shown in the (x, y) plane. Also shown in part (a) of the figure is the set of values of |X (z)| computed at points on the unit-circle. In part (b) of Fig. 8.3 the DTFT X (Ω) of the same signal is graphed. Notice how the unit circle of the z-plane is equivalent to the horizontal axis for the DTFT.

The z-transform defined by Eqn. (8.1) is sometimes referred to as the bilateral (two-sided) z-transform. A simplified variant of the transform termed the unilateral (one-sided) z-transform is introduced as an alternative analysis tool. The bilateral z-transform is useful for understanding signal characteristics, signal-system interaction, and fundamental characteristics of systems such as causality and stability. The unilateral z-transform is used for solving a linear constant-coefficient difference equation with specified initial conditions. We will briefly discuss the unilateral z-transform in Section 8.7, however, when we refer to z-transform without the qualifier word “bilateral” or “unilateral”, we will always imply the more general bilateral z-transform as defined in Eqn. (8.1).

Example 8.1: A simple z-transform example

Find the z-transform of the finite-length signal

$$\begin{array}{ll}x\left[n\right]& =\left\{3.7,1.3,-1.5,3.4,5.2\right\}\hfill \\ & \stackrel{\uparrow }{n=0}\hfill \end{array}$$

Solution: For the specified signal non-trivial samples occur for the index range n = 0,..., 4. Therefore, the z-transform is

$$\begin{array}{llll}X\left(z\right)\hfill & ={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]z^{-n}\hfill & \hfill & \hfill \\ \hfill & =x\left[0\right]+x\left[1\right]{\text{ z}}^{-1}+x\left[2\right]z^{-2}+x\left[3\right]z^{-3}+x\left[4\right]z^{-4}\hfill & \hfill & \hfill \\ \hfill & =3.7+1.3z^{-1}-1.5z^{-2}+3.4z^{-3}+5.2z^{-4}\hfill & \hfill & \left(8.13\right)\hfill \end{array}$$

The result is a polynomial of z−1. The samples of the signal x[n] become the coefficients of the corresponding powers of z−1. Essentially, information about time-domain sample amplitudes of the signal is contained in the coefficients of the polynomial X(z). The z-transform, therefore, is just an alternative way representing the signal x[n].

The value of the transform at a specific point in the complex plane can be evaluated from the polynomial in Eqn. (8.13). For example, if we wanted to know the value of the transform at z1 = 1 + j2 we would compute it as

$$\begin{array}{lll}X\left(1+j2\right)\hfill & =3.7+1.3{\left(1+j2\right)}^{-1}-1.5{\left(1+j2\right)}^{-2}+3.4{\left(1+j2\right)}^{-3}+5.2{\left(1+j2\right)}^{-4}\hfill & \hfill \\ \hfill & =3.7+\frac{1.3}{\left(1+j2\right)}-\frac{1.5}{{\left(1+j2\right)}^2}+\frac{3.4}{{\left(1+j2\right)}^3}+\frac{5.2}{{\left(1+j2\right)}^4}\hfill & \hfill \\ \hfill & =3.7826-j0.0259\hfill & \hfill \end{array}$$

The z-transform of a signal does not necessarily exist at every value of z. For example, the transform result we have found above could not be evaluated at the point z = 0 + j0 since the transform includes negative powers of z. In this case we conclude that the transform converges at all points in the complex z-plane with the exception of the origin.

Software resources:

ex_8_1.m

Example 8.2: z-Transform of a non-causal signal

Find the z-transform of the signal

$$x\left[n\right]=\left\{3.7,1.3,-\underset{\stackrel{\uparrow }{n=0}}{1.5},3.4,5.2\right\}$$

Solution: This is essentially the same signal the z-transform of which we have computed in Example 8.1 with one difference: It has been advanced by two samples so it starts with index n = −2. Applying the z-transform definition given by Eqn. (8.1) we obtain

$$\begin{array}{ll}X\left(z\right)\hfill & ={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]z^{-n}\hfill \\ \hfill & =x\left[-2\right]z^2+x\left[-1\right]z^1+x\left[0\right]+x\left[1\right]z^{-1}+x\left[2\right]z^{-2}\hfill \\ \hfill & =3.7z^2+1.3z^1-1.5+3.4z^{-1}+5.2z^{-2}\hfill \end{array}$$

As in the previous example, the transform fails to converge to a finite value at the origin of the z-plane. In addition, the transform does not converge for infinitely large values of |z| because of the z1 and z2 terms included in X (z). It converges at every point in the z-plane with the two exceptions, namely the origin and infinity.

Example 8.3: z-Transform of the unit-impulse

Find the z-transform of the unit-impulse signal

$$x\left[n\right]=\delta \left[n\right]=\{\begin{array}{cc}1,& n=0\\ 0,& n\ne 0\end{array}$$

Solution: Since the only non-trivial sample of the signal occurs at index n = 0, the z-transform is

$$X\left(z\right)=Ƶ\left\{\delta \left[n\right]\right\}={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]z^{-n}=x\left[0\right]z^0=1$$

The transform of the unit-impulse signal is constant and equal to unity. Since it does not contain any positive or negative powers of z, it converges at every point in the z-plane with no exceptions.

Example 8.4: z-Transform of a time shifted unit-impulse

Find the z-transform of the time shifted unit-impulse signal

$$x\left[n\right]=\delta \left[n-k\right]=\{\begin{array}{cc}1,& n=k\\ 0,& n\ne k\end{array}$$

where k ≠ 0 is a positive or negative integer.

Solution: Applying the z-transform definition we obtain

$$X\left(z\right)=Ƶ\left\{\delta \left[n-k\right]\right\}={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[n\right]z^{-n}=z^{-k}$$

The transform converges at every point in the z-plane with one exception:

1. If k > 0 then the transform does not converge at the origin z = 0 + j0.
2. If k < 0 then the transform does not converge at points with infinite radius, that is, at points for which |z| → ∞.

It becomes apparent from the examples above that the z-transform must be considered in conjunction with the criteria for the convergence of the resulting polynomial X (z). The collection of all points in the z-plane for which the transform converges is called the region of convergence (ROC). We conclude from Examples 8.1 and 8.2 that the region of convergence for a finite-length signal is the entire z-plane with the possible exception of z = 0 + j0, or |z| → ∞ or both.

1. The origin z = 0 + j0 is excluded from the region of convergence if any negative powers of z appear in the polynomial X (z). Negative powers of z are associated with samples of the signal x[n] with positive indices. Therefore, if the signal x[n] has any non-zero valued samples with positive indices, the z-transform does not converge at the origin of the z-plane.
2. Values of z with infinite radius must be excluded from the region of convergence if the polynomial X (z) includes positive powers of z. Positive powers of z are associated with samples of x[n] that have negative indices. Therefore, if the signal x[n] has any non-zero valued samples with negative indices, that is, if the signal is non-causal, its z-transform does not converge at |z| → ∞.

For finite-length signals, the ROC can be determined based on the sample indices of the leftmost and the rightmost significant samples of the signal. If the index of the leftmost significant sample is negative, then the transform does not converge for |z| → ∞. If the index of the rightmost significant sample is positive, then the transform does not converge at the origin z = 0 + j0.

How would we determine the region of convergence for a signal that is not finite-length? Recall that in Eqn. (8.7) we have represented the z-transform of a signal x[n] as equivalent to the DTFT of the modified signal x[n]r−n. As a result, the convergence of the z-transform of x[n] can be linked to the convergence of the DTFT of the modified signal x[n]r−n. Using Dirichlet conditions discussed in Chapter 5, this requires that x[n]r−n be absolute summable, that is,

$${\displaystyle \sum _{n=-\infty }^{\infty }}\begin{array}{ccc}\left|x\left[n\right]r^{-\mathrm{n}}\right|<\infty & & \left(8.14\right)\end{array}$$

The condition stated in Eqn. (8.14) highlights the versatility of the z-transform over the DTFT. Recall that, for the DTFT of a signal x[n] to exist, the signal has to be absolute summable. Therefore, the existence of the DTFT is a binary question; the answer to it is either yes or no. In contrast, if x[n] is not absolute summable, we may still be able to find values of r for which the modified signal x[n]r−n is absolute summable. Therefore, the z-transform of the signal x[n] may exist for some values of r. The question of existence for the z-transform is not a binary one; it is a question of which values of r allow the transform to converge.

The region of convergence concept will be discussed in detail in Section 8.2. The next few examples further highlight the need for a detailed discussion of the region of convergence.

Example 8.5:z-transform of the unit-step signal

Find the z-transform of the unit-step signal

$$x\left[n\right]=u\left[n\right]=\{\begin{array}{ll}1,\hfill & n\ge 0\hfill \\ 0,\hfill & n<0\hfill \end{array}$$

Solution: We will again apply the z-transform definition given by Eqn. (8.1).

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}u\left[n\right]z^{-n}& & \left(8.15\right)\end{array}$$

Since u[n] = 0 for n < 0, the lower limit of the summation in Eqn. (8.15) can be changed to n = 0 without affecting the result:

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}u\left[n\right]z^{-n}& & \left(8.16\right)\end{array}$$

Furthermore, u[n] = 1 for n ≥ 0, so dropping the u[n] term from the summation in Eqn. (8.16) has no effect either.

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}{z}^{-n}& & \left(8.17\right)\end{array}$$

Eqn. (8.17) represents the sum of an infinite-length geometric series, and can be computed in closed form using the formula1 derived in Appendix C to yield

$$X\left(z\right)=\frac{1}{1-z^{-1}}=\frac{z}{z-1}$$

which converges only for values of z for which

$$\begin{array}{ccc}\left|z^{-1}\right|<1& \Rightarrow & \left|z\right|>1\end{array}$$

The ROC for the transform of the unit-step signal is the collection of points outside a circle with radius equal to unity. This region is shown shaded in Fig. 8.5. Note that the unit-circle itself is not part of the ROC since the transform does not converge for values of z on the circle.

Example 8.6: z-Transform of a causal exponential signal

Find the z-transform of the signal

$$x\left[n\right]=a^{n}u\left[n\right]=\{\begin{array}{cc}{a}^n,& n\ge 0\\ 0,& n<0\end{array}$$

where a is any real or complex constant.

Solution: The signal x[n] is causal since x[n] = 0 for n < 0. Applying the z-transform definition given by Eqn. (8.1) we obtain

$$\begin{array}{ll}X\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}{a}^{n}u\left[n\right]z^{-n}\hfill & \left(8.18\right)\hfill \end{array}$$

Let us change the lower limit of the summation to n = 0 and drop the factor u[n]. Eqn. (8.18) becomes

$$X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}{a}^n{z}^{-n}={\displaystyle \sum _{n=0}^{\infty }}{\left(a{z}^{-1}\right)}^n$$

This is the sum of an infinite-length geometric series, and can be put into a closed form as

$$X\left(x\right)=\frac{1}{1-a{z}^{-1}}=\frac{z}{z-a}$$

which is valid only for values of z for which

$$\begin{array}{ccccc}\left|a{z}^{-1}\right|<1& \Rightarrow & \frac{\left|a\right|}{\left|z\right|}<1& \Rightarrow & \left|z\right|>\left|a\right|\end{array}$$

Four possible forms of the signal x[n] are shown in Fig. 8.6a,b,c,d corresponding to possible real values of the parameter a. It should be noted that there are other possibilities; the parameter a could also have a complex value. The ROC for X (z) is the region outside a circle with radius equal to |a|. This is shown in Fig. 8.7 for |a| < 1 and |a| > 1.

For the case |a| < 1 the unit circle is part of the ROC whereas, for the case |a| > 1, it is not. Recall from the previous discussion that the DTFT is equal to the z-transform evaluated on the unit circle. Consequently, the DTFT of the signal exists only if the ROC includes the unit circle. We conclude that, for the DTFT of the signal x[n] to exist, we need |a| < 1. This is consistent with the existence conditions discussed in Chapter 5.

Example 8.7 z-Transform of an anti-causal exponential signal

Find the z-transform of the signal

$$x\left[n\right]=-a^{n}u\left[-n-1\right]=\{\begin{array}{cc}-a^n,& n<0\\ 0,& n\ge 0\end{array}$$

where a is any real or complex constant.

Solution: In this case the signal x[n] is anti-causal since it is equal to zero for n ≥ 0. Applying the z-transform definition given by Eqn. (8.1) we obtain

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}-a^{n}u\left[-n-1\right]z^{-n}& & \left(8.19\right)\end{array}$$

Since

$$u\left[-n-1\right]=\{\begin{array}{cc}1,& n<0\\ 0,& n\ge 0\end{array}$$

changing the upper limit of the summation to n = −1 and dropping the factor u[−n − 1] would have no effect on the result. Eqn. (8.19) simplifies to

$$\begin{array}{ccc}X\left(z\right)=-{\displaystyle \sum _{n=-\infty }^{-1}}{a}^n{z}^{-\mathrm{n}}& & \left(8.20\right)\end{array}$$

Lower and upper summation limits are both negative. This can be fixed by employing a variable change m = −n, and the result found in Eqn. (8.20) can be written as

$$X\left(z\right)=-{\displaystyle \sum _{m=\infty }^1}{a}^{-m}{z}^m=-{\displaystyle \sum _{m=1}^{\infty }}{\left(a^{-1}z\right)}^m$$

which has the familiar sum of geometric series, however, the lower limit of the summation is m = 1 rather than m = 0 as required for the use of the closed-form formula. Let us apply another variable change to the summation, this time in the form k = m − 1, to get the lower limit of the summation to start at k = 0:

$$\begin{array}{llll}X\left(z\right)\hfill & =-{{\displaystyle \sum _{k=0}^{\infty }}\left(a^{-1}z\right)}^{k+1}\hfill & \hfill & \hfill \\ \hfill & =-a^{-1}z{\displaystyle \sum _{k=0}^{\infty }}{\left(a^{-1}z\right)}^k\hfill & \hfill & \left(8.21\right)\hfill \end{array}$$

Now the summation in Eqn. (8.21) can be put into a closed form to yield

$$\begin{array}{ccc}X\left(z\right)=-a^{-1}z\left(\frac{1}{1-a^{-1}z}\right)=\frac{z}{z-a},& \left|a^{-1}z\right|<1& \left(8.22\right)\end{array}$$

Notice how the closed-form expression found for the signal x[n] = −an u[−n − 1] in Eqn. (8.22) is identical to the one found in Example 8.6 for a different signal, namely x[n] = an u[n]. The transform expressions are identical; however, the regions in which those expressions are valid are different. The closed-form formula found in Eqn. (8.21) is valid only for values of z for which

$$\begin{array}{lllll}\left|a^{-1}z\right|<1\hfill & \Rightarrow \hfill & \frac{\left|z\right|}{\left|a\right|}<1\hfill & \Rightarrow \hfill & \left|z\right|<\left|a\right|\hfill \end{array}$$

Four possible forms of the signal x[n] are shown in Fig. 8.8 corresponding to different ranges of the real-valued parameter a (keep in mind that a could also be complex). The ROC for the transform X (z) found in this case is the region inside a circle with radius equal to |a|. This is shown in Fig. 8.9 for |a| < 1 and |a| > 1. For the case |a| > 1 the unit circle is part of the ROC whereas, for the case |a| < 1, it is not. Thus, the DTFT of the signal x[n] exists only when |a| > 1.

Examples 8.6 and 8.7 demonstrate a very important concept: It is possible for two different signals to have the same transform expression for the z-transform X (z). In order for us to uniquely identify which signal among the two led to a particular transform, the region of convergence must be specified along with the transform. The following two transform pairs are fundamental in determining to which of the possible signals a given transform corresponds:

Sometimes we need to deal with the inverse problem of determining the signal x[n] that has a particular z-transform X (z). Given a transform

$$X\left(z\right)=\frac{z}{z-a}$$

we need to know the region of convergence to determine if the signal x[n] is the one in Eqn. (8.23) or the one in Eqn. (8.24). This will be very important when we work with inverse z-transform in Section 8.4 later in this chapter. In order to avoid ambiguity, we will adopt the convention that the region of convergence is an integral part of the z-transform, and must be specified explicitly, or must be implied by means of another property of the underlying signal or system, every time a transform is given.

In each of the Examples 8.5, 8.6 and 8.7 we have obtained rational functions of z for the transform X (z). In the general case, a rational transform X (z) is expressed in the form

$$\begin{array}{ccc}X\left(z\right)=K\frac{B\left(z\right)}{A\left(z\right)}& & \left(8.25\right)\end{array}$$

where the numerator B (z) and the denominator A (z) are polynomials of z. The parameter K is a constant gain factor. Let the numerator polynomial be written in factored form as

$$\begin{array}{ccc}B\left(z\right)=\left(z-z_1\right)\left(z-z_2\right)...,\left(z-z_M\right)& & \left(8.26\right)\end{array}$$

and let p1, p2,..., pN be the roots of the denominator polynomial so that

$$\begin{array}{cc}A\left(z\right)=\left(z-p_1\right)\left(z-p_2\right)...,\left(z-p_N\right)& \left(8.27\right)\end{array}$$

The transform can be written as

$$\begin{array}{ccc}X\left(z\right)=K\frac{\left(z-z_1\right)\left(z-z_2\right)...,\left(z-z_M\right)}{\left(z-p_1\right)\left(z-p_2\right)...,\left(z-p_N\right)}& & \left(8.28\right)\end{array}$$

In Eqns. (8.26) and (8.27) the parameters M and N are the numerator order and the denominator order respectively. The larger of M and N is the order of the transform X (z). The roots of the numerator polynomial are referred to as the zeros of the transform X (z). In contrast, the roots of the denominator polynomial are the poles. The transform does not converge at a pole, therefore, the ROC cannot contain any poles. In the next section we will see that the boundaries of the ROC are determined by the poles of the transform.

Example 8.8: z-Transform of a discrete-time pulse signal

Determine the z-transform of the discrete-time pulse signal

$$x\left[n\right]=\{\begin{array}{ll}1,\hfill & 0\le n\le N-1\hfill \\ 0,\hfill & n<0\text{ or }n>N-1\hfill \end{array}$$

shown in Fig. 8.10.

Solution: The transform in question is computed as

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=0}^{N-1}}\left(1\right)z^{-n}=1+z^{-1}+...+z^{-N}=\frac{1-z^{-N}}{1-z^{-1}}& & \left(8.29\right)\end{array}$$

In computing the closed-form result in Eqn. (8.29) we have used the finite-length geometric series sum formula derived in Appendix C. Since x[n] is finite-length and causal, the transform converges at every point in the z-plane except at the origin z = 0 + j0. Thus we have

$$\begin{array}{cc}\text{ROC}:& \left|z\right|>0\end{array}$$

Often we find it more convenient to write the transform using non-negative powers of z. Multiplying both the numerator and the denominator polynomials in Eqn. (8.29) by b zN leads to

$$\begin{array}{ccc}X\left(z\right)=\frac{z^N-1}{z^{N-1}\left(z-1\right)}& & \left(8.30\right)\end{array}$$

which is in the general factored form given by Eqn. (8.28). The appearance of the closed-form result in Eqn. (8.30) may be confusing in the way it relates to the region of convergence. It seems as though X (z) might have a pole at z = 1. We need to realize, however, that the numerator polynomial is also equal to zero at z = 1:

$${B\left(z\right)|}_{z=1}={\left(z^N-1\right)|}_{z=1}=0$$

As a result, the numerator polynomial has a (z − 1) factor that effectively cancels the pole at z = 1. We will now go through the exercise of determining the zeros and poles of the transform X (z). The roots of the numerator polynomial are found by solving the equation

$$\begin{array}{lllll}B\left(z\right)=0\hfill & \Rightarrow \hfill & z^N=1\hfill & \hfill & \left(8.31\right)\hfill \end{array}$$

In order to determine the roots of the numerator polynomial that are not immediately obvious, we will use the fact that ej2πk = 1 for all integer k, and write Eqn. (8.31) as

$$\begin{array}{ccc}{z}^N=e^{j2\pi k}& & \left(8.32\right)\end{array}$$

Taking the N-th root of both sides leads to the set of roots

$$\begin{array}{ccc}{z}_k=e^{j2\pi k/N},k=0,...,N-1& & \left(8.33\right)\end{array}$$

The numerator polynomial of X (z) has N roots that are all on the unit circle of the z-plane. They are equally-spaced at angles that are integer multiples of 2π/N as shown in Fig. 8.11(a). Note that in the figure we have used N = 10.

The roots of the denominator polynomial are found by solving

$$\begin{array}{lllll}A\left(z\right)=0\hfill & \Rightarrow \hfill & z^{N-1}\left(z-1\right)=0\hfill & \hfill & \left(8.34\right)\hfill \end{array}$$

There are N − 1 solutions at pk = 0 and one solution at z = 1 as shown in Fig. 8.11(b).

The pole-zero diagram for the transform X (z) can now be constructed. The factors (z − 1) in numerator and denominator polynomials cancel each other, therefore there is neither a zero nor a poleat z = 1. We are left with zeros at

$$\begin{array}{cc}{z}_k=e^{j2\pi k/N},& k=1,...,N-1\end{array}$$

and a total of N − 1 poles all at

$$\begin{array}{cc}{p}_k=0,& k=1,...,N-1\end{array}$$

as shown in Fig. 8.12.

In light of what we have discovered regarding the the poles of X (z) it makes sense that the ROC is the entire z-plane with the exception of a singular point at the origin. Since all N − 1 poles of the transform are at the origin, that is the only point where the transform does not converge.

The magnitude of the z-transform computed in Eqn. (8.30) is shown in Fig. 8.13(a) as a surface graph for N = 10. Since the origin of the z-plane is excluded from the ROC, the magnitude is not computed at the origin. Also, very large magnitude values in close proximity of the origin are clipped to make the graph fit into a reasonable scale. Note how the zeros equally spaced around the unit-circle cause the magnitude surface to dip down. In addition to the surface graph, Fig. 8.13(a) also shows the unit circle of the z-plane. Magnitude values computed at points on the unit circle are marked on the surface. Fig. 8.13(b) shows the magnitude of the DTFT computed for the range of angular frequency −π ≤ Ω < π. It should be compared to the values marked on the surface graph that correspond to the magnitude of the z-transform evaluated on the unit circle.

To provide a slightly different perspective and to help with visualization, imagine that we take a knife and carefully cut through the surface in Fig. 8.13(a) along the perimeter of the unit-circle of the z-plane, as if the surface is a cake that we would like to fit into a cylindrical box. The profile of the cutout would match the DTFT shown in Fig. 8.13(b). This is illustrated in Fig. 8.14.

Example 8.9: z-Transform of complex exponential signal

Find the z-transform of the signal

$$x\left[n\right]=e^{j{\Omega }_{0}n}u\left[n\right]$$

The parameter Ω0 is real and positive, and is in radians.

Solution: Applying the z-transform definition directly to the signal x[n] leads to

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}{e}^{j{\Omega }_{0}n}u\left[n\right]z^{-\mathrm{n}}& & \left(8.35\right)\end{array}$$

Since x[n] is causal, the lower limit of the summation can be set to n = 0, and the u[n] term can be dropped to yield

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}{e}^{j{\Omega }_{0}n}{z}^{-n}& & \left(8.36\right)\end{array}$$

The expression in Eqn. (8.36) is the sum of an infinite-length geometric series, and can be put into closed form as

$$\begin{array}{ll}X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}{\left(e^{j{\Omega }_0}{z}^{-1}\right)}^n=\frac{1}{1-e^{j{\Omega }_0}{z}^{-1}}\hfill & \left(8.37\right)\hfill \end{array}$$

The region of convergence is obtained from the convergence condition for the geometric series:

$$\begin{array}{lllll}\left|e^{j{\Omega }_0}{z}^{-1}\right|<1\hfill & \Rightarrow \hfill & \left|z\right|>1\hfill & \hfill & \left(8.38\right)\hfill \end{array}$$

The transform in Eqn. (8.37) can also be written using non-negative powers of z as

$$\begin{array}{ccc}X\left(z\right)=\frac{z}{z-e^{j{\Omega }_0}}& & \left(8.39\right)\end{array}$$

It has a zero at the origin and a pole at z = ejΩ0. The zero and the pole of the transform as well as its ROC are shown in Fig. 8.15.

Example 8.10: A more involved z-transform example

Determine the z-transform of the signal x[n] defined by

$$\begin{array}{ll}x\left[n\right]=\{\begin{array}{ll}{\left(1/2\right)}^n,\hfill & n\ge 0\text{ and even}\hfill \\ -{\left(1/3\right)}^n,\hfill & n>0\text{ and odd}\hfill \\ 0,\hfill & n<0\hfill \end{array}\hfill & \left(8.40\right)\hfill \end{array}$$

Solution: This example is somewhat tricky because of the particular way x[n] is defined. Using the definition of the z-transform we can write

$$\begin{array}{ll}X\left(z\right)={\displaystyle \sum _{\begin{array}{c}n=0\\ n\text{ even}\end{array}}^{\infty }}{\left(\frac{1}{2}\right)}^n{z}^{-n}-{\displaystyle \sum _{\begin{array}{c}n=1\\ n\text{ odd}\end{array}}^{\infty }}{\left(\frac{1}{3}\right)}^n{z}^{-n}\hfill & \left(8.41\right)\hfill \end{array}$$

The first summation in Eqn. (8.41) will be carried out for only the even values of the index n, and the second summation is for only the odd values of it. To facilitate the goal of limiting the summations to even and odd index values, we will use two variable changes: For the first summation, let n = 2m where m is the new integer index. This substitution will ensure that n is always an even value. Similarly, for the second summation, we will use the variable change n = 2m + 1 to ensure that the resulting value of n is always odd. Incorporating these variable changes into the summations, Eqn. (8.41) can be written as

$$\begin{array}{ll}X\left(z\right)={\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{2}\right)}^{2m}{z}^{-2m}-{\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{3}\right)}^{2m+1}{z}^{-\left(2m+1\right)}\hfill & \left(8.42\right)\hfill \end{array}$$

or in the equivalent form

$$\begin{array}{llll}X\left(z\right)\hfill & ={\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{4}\right)}^m{\left(z^2\right)}^{-m}-\frac{1}{3}{z}^{-1}{\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{9}\right)}^m{\left(z^2\right)}^{-m}\hfill & \hfill & \hfill \\ \hfill & ={\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{4}{z}^{-2}\right)}^m-\frac{1}{3}{z}^{-1}{\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{9}{z}^{-2}\right)}^m\hfill & \hfill & \left(8.43\right)\hfill \end{array}$$

The two summations in Eqn. (8.43) can be thought of as two transforms X1 (z) and X2 (z) so that

$$X\left(z\right)=X_1\left(z\right)-X_2\left(z\right)$$

The closed form expression for X1 (z) and its associated ROC can be obtained as

$$\begin{array}{lllll}{X}_1\left(z\right)={\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{4}{z}^{-2}\right)}^m=\frac{z^2}{z^2-\frac{1}{4}},\hfill & \text{ROC}:\hfill & \left|z^2\right|>\frac{1}{4}\hfill & \hfill & \left(8.44\right)\hfill \end{array}$$

Similarly, X2 (z) is

$$\begin{array}{llll}{X}_2\left(z\right)=\frac{1}{3}{z}^{-1}{\displaystyle \sum _{m=0}^{\infty }}{\left(\frac{1}{9}{z}^{-2}\right)}^m=\frac{\frac{1}{3}z}{z^2-\frac{1}{9}},\hfill & \text{ROC}:\hfill & \left|z^2\right|>\frac{1}{9}\hfill & \left(8.45\right)\hfill \end{array}$$

Combining the closed form expressions for X 1 (z)and X 2 (z) as given by Eqns. (8.44) and (8.45) under a common denominator, we find the transform X (z) and its ROC as

$$\begin{array}{cccc}X\left(z\right)=\frac{z^4-\frac{1}{3}{z}^3-\frac{1}{9}{z}^2+\frac{1}{12}z}{\left(z^2-\frac{1}{4}\right)\left(z^2-\frac{1}{9}\right)},& \text{ROC}:& \left|z\right|>\frac{1}{2}& \left(8.46\right)\end{array}$$

The resulting transform has two real zeros at z = 0 and z = −0.4158 as well as a complex conjugate pair of zeros at z = 0.3746 ± j0.2451. Its poles are at $z=\pm \frac{1}{2}$ and $z=\pm \frac{1}{3}$. The pole-zero diagram and the ROC for X (z) are shown in Fig. 8.16.

Example 8.11: Transform of a downsampled signal

The downsampling operation was briefly discussed in Chapter 6. Consider a signal g[n] obtained by downsampling another signal x[n] by a factor of 2:

$$g\left[n\right]=x\left[2n\right]$$

Determine the z-transform of the downsampled signal g[n] in terms of the transform of the original signal x[n].

Solution: Using the z-transform definition, G (z) is

$$\begin{array}{ccc}G\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}g\left[n\right]z^{-n}& & \left(8.47\right)\end{array}$$

Let us substitute g[n] = x[2n] into Eqn. (8.47) to obtain

$$\begin{array}{ccc}G\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}x\left[2n\right]z^{-n}& & \left(8.48\right)\end{array}$$

To resolve the summation in Eqn. (8.48) we will use the variable change m = 2n and write the summation in terms of the new index m:

$$\begin{array}{ll}G\left(z\right)={\displaystyle \sum _{\begin{array}{c}m=-\mathrm{\infty }\\ m\text{ even}\end{array}}^{\infty }}x\left[m\right]z^{-m/2}\hfill & \left(8.49\right)\hfill \end{array}$$

Naturally, the summation should only have the terms for which m is even. This restriction on the values of the summation index makes it difficult for us to relate Eqn. (8.49) to the transform X (z) of the original signal. To overcome this hurdle we will use a simple trick. Consider a discrete-time signal w[n] with the following definition:

$$w\left[m\right]=\{\begin{array}{cc}1,& m\text{ even}\\ 0,& m\text{ odd}\end{array}$$

The transform in Eqn. (8.49) can be written as

$$\begin{array}{ccc}G\left(z\right)={\displaystyle \sum _{m=-\infty }^{\infty }}x\left[m\right]w\left[m\right]z^{-m/2}& & \left(8.50\right)\end{array}$$

Notice how we removed the restriction on m in Eqn. (8.50) since the factor w[m] ensures that odd-indexed terms do not contribute to the sum. One method of creating the signal w[m] is through

$$\begin{array}{ll}\omega \left[m\right]=\frac{1+{\left(-1\right)}^m}{2}\hfill & \left(8.51\right)\hfill \end{array}$$

Substituting Eqn. (8.51) into Eqn. (8.50) we obtain

$$\begin{array}{llll}G\left(z\right)\hfill & ={\displaystyle \sum _{m=-\infty }^{\infty }}x\left[m\right]\left[\frac{1+{\left(-1\right)}^m}{2}\right]z^{-m/2}\hfill & \hfill & \hfill \\ \hfill & =\frac{1}{2}{\displaystyle \sum _{m=-\infty }^{\infty }}x\left[m\right]z^{-m/2}+\frac{1}{2}{\displaystyle \sum _{m=-\infty }^{\infty }}x\left[m\right]{\left(-1\right)}^m{z}^{-m/2}\hfill & \hfill & \left(8.52\right)\hfill \end{array}$$

Recognizing that $z^{-m/2}={\left(\sqrt{z}\right)}^{-m}$ and ${\left(-1\right)}^m{z}^{-m/2}={\left(-\sqrt{z}\right)}^{-m}$, the result in Eqn. (8.52) can be rewritten as

$$\begin{array}{llll}G\left(z\right)\hfill & =\frac{1}{2}{\displaystyle \sum _{m=-\infty }^{\infty }}x\left[m\right]{\left(\sqrt{z}\right)}^{-m}+\frac{1}{2}{\displaystyle \sum _{m=-\infty }^{\infty }}x\left[m\right]{\left(-\sqrt{z}\right)}^{-m}\hfill & \hfill & \hfill \\ \hfill & =\frac{1}{2}X\left(\sqrt{2}\right)+\frac{1}{2}X\left(-\sqrt{2}\right)\hfill & \hfill & \left(8.53\right)\hfill \end{array}$$

Let the ROC for the original transform X (z) be

$$r_1<\left|z\right|<r_2$$

The ROC for $X\left(\sqrt{z}\right)$ and $X\left(-\sqrt{z}\right)$ terms in Eqn. (8.53) is

$$\begin{array}{ccc}{r}_1<\left|\pm \sqrt{z}\right|<r_2& \Rightarrow & r_1^2<\left|z\right|\end{array}<r_2^2$$

and therefore the ROC for G (z) is also

$$r_1^2<\left|z\right|<r_2^2$$

Example 8.12: Transform of a downsampled signal revisited

Consider the causal exponential signal

$$x\left[n\right]={\left(0.9\right)}^{n}u\left[n\right]$$

Let a new signal g[n] be defined as g[n] = x[2n]. Find the z-transform of g[n].

Solution: The transform of the signal x[n] is

$$\begin{array}{ccc}X\left(z\right)=\frac{z}{z-0.9},& \text{ROC}:& \left|z\right|>0.9\end{array}$$

Applying the result found in Eqn. (8.53) of Example 8.11, the transform G (z) of the down-sampled signal g[n] is found as

$$\begin{array}{ccc}G\left(z\right)=\frac{\frac{1}{2}\sqrt{z}}{\sqrt{z}-0.9}+\frac{-\frac{1}{2}\sqrt{z}}{-\sqrt{z}-0.9}& & \left(8.54\right)\end{array}$$

Combining the two terms of Eqn. (8.54) under a common denominator we get

$$\begin{array}{lllll}G\left(z\right)=\frac{z}{z-{\left(0.9\right)}^2},\hfill & \text{ROC}:\hfill & \left|z\right|>\left(0.9\right)\hfill & \hfill & \left(8.55\right)\hfill \end{array}$$

It is easy to verify the validity of the result in Eqn. (8.55) if we apply the downsampling operation in the time domain and then find the z-transform of the resulting signal g[n] directly. The time domain expression for the signal g[n] is

$$g\left[n\right]=x\left[2n\right]={\left({0.9}^2\right)}^{n}u\left[2n\right]={\left(0.81\right)}^{n}u\left[n\right]$$

and its z-transform is

$$\begin{array}{ccc}G\left(z\right)=\frac{z}{z-0.81},& \text{ROC}:& \left|z\right|>0.81\end{array}$$

8.2 Characteristics of the Region of Convergence

The examples we have worked on so far made it clear that the z-transform of a discrete-time signal always needs to be considered in conjunction with its region of convergence, that is, the collection of points in the z-plane for which the transform converges. In this section we will summarize and justify the fundamental characteristics of the region of convergence.

This property is easy to justify when we recall that, for z = r ejΩ, the values of the z-transform are identical to the values of the DTFT of the signal x[n]r−n, as derived in Eqn. (8.7). Therefore, the convergence of the z-transform for z =r ejΩ is equivalent to the convergence of the DTFT of the signal x[n]r−n which requires that

$$\begin{array}{ll}{\displaystyle \sum _{n=-\infty }^{\infty }}\left|x\left[n\right]r^{-n}\right|<\infty \hfill & \left(8.56\right)\hfill \end{array}$$

Thus, the ROC depends only on r and not on Ω, explaining the circular nature of the region. Following are the possibilities for the ROC:

$$\begin{array}{lll}r<r_1\hfill & :\hfill & \text{Inside a circle}\hfill \\ r>r_2\hfill & :\hfill & \text{Outside a circle}\hfill \\ r_1<r<r_2\hfill & :\hfill & \text{Between two circles}\hfill \end{array}$$

By definition, poles of X (z) are values of z that make the value of the transform infinitely large. For rational z-transforms, poles are the roots of the denominator polynomial. Since the transform does not converge at a pole, the ROC must naturally exclude all poles of the transform.

Since a causal signal does not have any significant samples for n < 0, its z-transform can be written as

$$X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}x\left[n\right]z^{-n}$$

Writing z as z = rjΩ and remembering that the convergence of the z-transform is equivalent to the signal x[n]r−n being absolute summable leads to the convergence condition

$$\begin{array}{ccc}X\left(z\right)={\displaystyle \sum _{n=0}^{\infty }}|x\left[n\right]r^{-n}|<\infty & & \left(8.57\right)\end{array}$$

which can be expressed in the equivalent form

$$\begin{array}{ccc}{\displaystyle \sum _{n=0}^{\infty }}\left|x\left[n\right]\right|{\left(\frac{1}{\left|r\right|}\right)}^n<\infty & & \left(8.58\right)\end{array}$$

If we can find a value of r for which Eqn. (8.58) is satisfied, then it is obvious that any larger value of r will satisfy Eqn. (8.58) as well. All we need to do is find the radius r1 of the bounding circle, and the ROC is the region outside that circle.

A special case of a causal signal is one that is both causal and finite-length. Let N1 be the largest value of the index for which the signal x[n] has a non-zero value, that is

$$x\left[n\right]=0\text{ for }n<0\text{ or }n>N_1$$

In this case the absolute summability condition in Eqn. (8.58) can be written as

$$\begin{array}{ccc}{{\displaystyle \sum _{n=0}^{N_1}}\left|x\left[n\right]\right|\left(\frac{1}{\left|r\right|}\right)}^n<\infty & & \left(8.59\right)\end{array}$$

The condition in Eqn. (8.59) is satisfied for all values of r with the exception of r = 0. Our initial assessment that the ROC is the outside of a circle is still valid if we are willing to consider a circle with radius of r1 = 0, that is, one that shrinks down to a single point at the origin. The ROC for a finite-length causal signal is therefore

$$\begin{array}{ccc}\left|z\right|>0& & \left(8.60\right)\end{array}$$

The justification of this property will be similar to that of the previous one. An anti-causal signal does not have any significant samples for n ≥ 0, and its z-transform can be written as

$$X\left(x\right)={\displaystyle \sum _{n=-\infty }^{-1}}x\left[n\right]z^{-n}$$

Using z = rjΩ the condition for the convergence of the z-transform can be expressed through the equivalent condition for the absolute summability of the signal x[n]r−n as

$$\begin{array}{ccc}{\displaystyle \sum _{n=-\infty }^{-1}}\left|x\left[n\right]r^{-n}\right|<\infty & & \left(8.61\right)\end{array}$$

which can be expressed in the equivalent form

$$\begin{array}{ll}{\displaystyle \sum _{n=-\infty }^1}\left|x\left[n\right]\right|·{\left|r\right|}^{-n}<\infty \hfill & \left(8.62\right)\hfill \end{array}$$

Let us apply the variable change n = −m to the summation in Eqn. (8.62) to write it as

$$\begin{array}{ll}{\displaystyle \sum _{m=1}^{\infty }}\left|x\left[-m\right]\right|\cdot {\left|r\right|}^m<\infty \hfill & \left(8.63\right)\hfill \end{array}$$

If we can find a value of r for which Eqn. (8.63) is satisfied, then it is obvious that any smaller value of r will satisfy Eqn. (8.63) as well. If r2 is the radius of the bounding circle, then the ROC is the region inside that circle.

A special case of an anti-causal signal is one that is both anti-causal and finite-length. Let −N2 be the most negative value of the index for which the signal x[n] has a non-zero value, that is

$$\begin{array}{lll}x\left[n\right]=0\hfill & \text{for}\hfill & n\ge 0\text{ or }n<-N_2\hfill \end{array}$$

In this case the absolute summability condition in Eqn. (8.62) can be written as

$$\begin{array}{lllll}{\displaystyle \sum _{n=-N_2}^{-1}}\left|x\left[n\right]\right|.{\left|r\right|}^{-n}<\infty \hfill & \Rightarrow \hfill & {\displaystyle \sum _{m=1}^{N_2}}\left|x\left[-m\right]\right|.{\left|r\right|}^m<\infty \hfill & \hfill & \left(8.64\right)\hfill \end{array}$$

The condition in Eqn. (8.64) is satisfied for all values of r that are finite; the only exception would be r → ∞. If we simply take the bounding circle to be one with infinite radius r2 = ∞ then our conclusion that the ROC is the inside of a circle is still valid, The ROC for a finite-length anti-causal signal is therefore

$$\begin{array}{ccc}\left|z\right|<\infty & & \left(8.65\right)\end{array}$$

Any signal x[n] can be written as the sum of a causal signal and an anti-causal signal. Let the two signals xR[n] and xL[n] be defined in terms of the signal x[n] as

$$\begin{array}{ll}{x}_R\left[n\right]=\{\begin{array}{ll}x\left[n\right],\hfill & n\ge 0\hfill \\ 0,\hfill & n<0\hfill \end{array}\hfill & \left(8.66\right)\hfill \end{array}$$

and

$$\begin{array}{ll}{x}_L\left[n\right]=\{\begin{array}{ll}x\left[n\right],\hfill & n<0\hfill \\ 0,\hfill & n\ge 0\hfill \end{array}\hfill & \left(8.67\right)\hfill \end{array}$$

so that xR[n] is causal, and xL[n] is anti-causal, and they add up to x[n]:

$$\begin{array}{ccc}x\left[n\right]=x_R=\left[n\right]+x_L\left[n\right]& & \left(8.68\right)\end{array}$$

Let the z-transforms of these two signals be

$$\begin{array}{lll}{X}_R\left(z\right)=Ƶ\left\{x_R\left[n\right]\right\}\hfill & \text{ROC}:\hfill & \left|z\right|>r_1\hfill \\ X_L\left(z\right)=Ƶ\left\{x_L\left[n\right]\right\}\hfill & \text{ROC}:\hfill & \left|z\right|<r_2\hfill \end{array}$$

The z-transform of the signal x [n] is

$$\begin{array}{ccc}X\left(x\right)=X_R\left(z\right)+X_L\left(z\right)& & \left(8.69\right)\end{array}$$

The ROC for X (z) is at least the overlap of the two regions, that is,

$$\begin{array}{ccc}{r}_1<\left|z\right|<r_2& & \left(8.70\right)\end{array}$$

provided that r2 > r1 (otherwise there may be no overlap, and the z-transform may not exist at any point in the z-plane).

In some cases the ROC may actually be larger than the overlap in Eqn. (8.70) if the addition of the two transforms in Eqn. (8.69) results in the cancellation of a pole that sets the boundary for either XR (z) or XL (z).

As a special case, if the causal term xR[n] is of finite-length, the inner circle of the ROC may shrink down to a single point at the origin, resulting in r1 = 0. Similarly, if the anti-causal term xL[n] is of finite-length, the outer circle of the ROC may grow to have infinite radius, resulting in r2 → ∞.

Example 8.13: z-Transform of two-sided exponential signal

Find the z-transform of the signal

$$x\left[n\right]=a^{\left|n\right|}$$

Solution: The specified signal exists for all values of the index n, and can be written as

$$\begin{array}{llll}x\left[n\right]\hfill & =\{\begin{array}{cc}{a}^n,& n\ge 0\\ a^{-n},& n<0\end{array}\hfill & \hfill & \hfill \\ \hfill & =a^{n}u\left[n\right]+a^{-n}u\left[-n-1\right]\hfill & \hfill & \left(8.71\right)\hfill \end{array}$$

or, equivalently as

$$\begin{array}{ccc}x\left[n\right]=a^{n}u\left[n\right]+{\left(1/a\right)}^{n}u\left[-n-1\right]& & \left(8.72\right)\end{array}$$

We will think of x[n] as the sum of a causal signal xR[n] and an anti-causal signal xL[n] in the form

$$\begin{array}{ccc}x\left[n\right]=x_R\left[n\right]+x_L\left[n\right]& & \left(8.73\right)\end{array}$$

with the two components given by

$$\begin{array}{ll}{x}_R\left[n\right]\hfill & =a^{n}u\left[n\right]\hfill \\ x_L\left[n\right]\hfill & ={\left(1/a\right)}^{n}u\left[-n-1\right]\hfill \end{array}$$

When we discuss the properties of the z-transform later in Section 8.3 we will show that the z-transform is linear, and therefore, the transform of the sum of two signals is equal to the sum of their respective transforms. Therefore X (z) can be written as

$$\begin{array}{llll}X\left(z\right)\hfill & =X_R\left(z\right)+X_L\left(z\right)\hfill & \hfill & \hfill \\ \hfill & =Ƶ\left\{a^{n}u\left[n\right]\right\}+Ƶ\left\{{\left(1/a\right)}^{n}u\left[-n-1\right]\right\}\hfill & \hfill & \left(8.74\right)\hfill \end{array}$$

The individual transforms that make up X (z) in Eqn. (8.74) can be determined by adapting the results obtained earlier in Eqns. (8.23) and (8.23) as:

$$\begin{array}{cccc}Ƶ\left\{a^{n}u\left[n\right]\right\}=\frac{z}{z-a},& \text{ROC}:\left|z\right|>\left|a\right|& & \left(8.75\right)\end{array}$$

$$\begin{array}{cccc}Ƶ\left\{{\left(1/a\right)}^{n}u\left[-n-1\right]\right\}=-\frac{z}{z-1/a},& \text{ROC}:\left|z\right|<\frac{1}{\left|a\right|}& & \left(8.76\right)\end{array}$$

The regions of convergence for XR (z) and XL (z) are shown in Fig. 8.18.

The transform X (z) can now be computed as

$$X\left(z\right)=\frac{z}{z-a}-\frac{z}{z-1/a}=\frac{\left(a^2-1\right)z}{a{z}^2-\left(a^2+1\right)z+a}$$

The transform has a zero at z = 0, and poles at z = a and z = 1/a. For it to converge, both XR (z) and XL (z) must converge. Therefore, the ROC for X (z) is the overlap of the ROCs of XR (z) and XL (z) if such an over lap exists:

$$\begin{array}{ccc}\text{ROC}:\left|a\right|<\left|z\right|<\frac{1}{\left|a\right|}& & \left(8.77\right)\end{array}$$

If |a| < 1, then |a| < 1/|a|, and the two regions shown in Fig. 8.18(a) and (b) do indeed overlap, creating a ring-shaped region between two circles with radii |a| and 1/|a|. This is shown in Fig. 8.19.

If |a| ≥ 1, then |a| ≥ 1/|a|, and therefore the inequality in Eqn. (8.77) cannot be satisfied for any value of z. In that case, the transform found in Eqn. (8.77) does not converge at any point in the z-plane.

8.3 Properties of the z-Transform

In this section we focus on some of the important properties of the z-transform that will help us later in using the z-transform effectively for the analysis and design of discrete-time systems. The proofs of the z-transform properties are also be given. The motivation behind discussing the proofs of various properties of the z-transform is twofold:

1. The techniques used in proving various properties of the z-transform are also useful in working with z-transform problems in general.
2. The proofs provide further insight into the z-transform, and allow us to later identify opportunities for the effective use of z-transform properties in solving problems.

8.3.1 Linearity

Example 8.14: Using the linearity property of the z-transform

Determine the z-transform of the signal

$$x\left[n\right]=3{\left(\frac{1}{2}\right)}^{n}u\left[n\right]-5{\left(\frac{1}{3}\right)}^{n}u\left[n\right]$$

Solution: A general expression for the z-transform of the causal exponential signal was found in Example 8.6 as

$$\begin{array}{cc}Ƶ\left\{a^{n}u\left[n\right]\right\}=\frac{z}{z-a},& \text{ROC}:\left|z\right|>\left|a\right|\end{array}$$

Applying this result to the exponential terms in x [n] we get

$$\begin{array}{cccc}Ƶ\left\{{\left(\frac{1}{2}\right)}^{n}u\left[n\right]\right\}=\frac{z}{z-\frac{1}{2}},& \text{ROC}:\left|z\right|>\frac{1}{2}& & \left(8.81\right)\end{array}$$

and

$$\begin{array}{lll}Ƶ\left\{{\left(\frac{1}{3}\right)}^{n}u\left[n\right]\right\}=\frac{z}{z-\frac{1}{3}},\hfill & \text{ROC}:\left|z\right|>\frac{1}{3}\hfill & \left(8.82\right)\hfill \end{array}$$

Combining the results in Eqns. (8.81) and (8.82) using the linearity property we arrive at the desired result:

$$\begin{array}{llll}X\left(z\right)\hfill & =3\left(\frac{z}{z-\frac{1}{2}}\right)-5\left(\frac{z}{z-\frac{1}{3}}\right)\hfill & \hfill & \hfill \\ \hfill & =\frac{-2z\left(z-\frac{3}{4}\right)}{\left(z-\frac{1}{2}\right)\left(z-\frac{1}{3}\right)}\hfill & \hfill & \left(8.83\right)\hfill \end{array}$$

The ROC for X (z) is the overlap of the two regions in Eqns. (8.81) and (8.82), namely

$$\left|z\right|>\frac{1}{2}$$

The ROC for the transform X (z) is shown in Fig. 8.20 along with poles and zeros of the transform.

Example 8.15: z-Transform of a cosine signal

Find the z-transform of the signal

$$x\left[n\right]=cos\left({\Omega }_{0}n\right)u\left[n\right]$$

Solution: Using Euler’s formula for the cos (Ω0n) term, the signal x[n] can be written as

$$\begin{array}{ll}cos\left({\Omega }_{0}n\right)u\left[n\right]=\frac{1}{2}{e}^{j{\Omega }_{0}n}u\left[n\right]+\frac{1}{2}{e}^{-j{\Omega }_{0}n}u\left[n\right]\hfill & \left(8.84\right)\hfill \end{array}$$

and its z-transform is

$$\begin{array}{ll}Ƶ\left\{cos\left({\Omega }_{0}n\right)u\left[n\right]\right\}=Ƶ\left\{\frac{1}{2}{e}^{j{\Omega }_{0}n}u\left[n\right]+\frac{1}{2}{e}^{-{\Omega }_{0}n}u\left[n\right]\right\}\hfill & \left(8.85\right)\hfill \end{array}$$

Applying the linearity property of the z-transform, Eqn. (8.85) becomes

$$\begin{array}{lll}Ƶ\left\{cos\left({\Omega }_{0}n\right)u\left[n\right]\right\}\hfill & =\frac{1}{2}Ƶ\left\{e^{j{\Omega }_{0}n}u\left[n\right]\right\}+\frac{1}{2}Ƶ\left\{e^{-j{\Omega }_{0}n}u\left[n\right]\right\}\hfill & \hfill \\ \hfill & =\frac{1/2}{1-e^{j{\Omega }_0}{z}^{-1}}+\frac{1/2}{1-e^{-j{\Omega }_0}{z}^{-1}}\hfill & \left(8.86\right)\hfill \end{array}$$

Combining the terms of Eqn. (8.86) under a common denominator we have

$$\begin{array}{ll}X\left(z\right)=\frac{1-cos\left({\Omega }_0\right)z^{-1}}{1-2cos\left({\Omega }_0\right)z^{-1}+z^{-2}}\hfill & \left(8.87\right)\hfill \end{array}$$

Let us multiply both the numerator and the denominator by z2 to eliminate negative powers of z:

$$\begin{array}{ll}X\left(z\right)=\frac{z\left[z-cos\left({\Omega }_0\right)\right]}{z^2-2cos\left({\Omega }_0\right)z+1}\hfill & \left(8.88\right)\hfill \end{array}$$

Poles of X (z) are both on the unit circle of the z-plane at z = e±jΩ0. Since x[n] is causal, the ROC is the outside of the unit circle. The unit circle itself is not included in the ROC. This is illustrated in Fig. 8.21.

Example 8.16: z-Transform of a sine signal

Find the z-transform of the signal

$$x\left[n\right]=sin\left({\Omega }_{0}n\right)u\left[n\right]$$

Solution: This problem is quite similar to that in Example 8.15. Using Euler’s formula for the signal x[n] we get

$$\begin{array}{ccc}sin\left({\Omega }_{0}n\right)u\left[n\right]=\frac{1}{2j}{e}^{j{\Omega }_{0}n}u\left[n\right]-\frac{1}{2}{e}^{-j{\Omega }_{0}n}u\left[n\right]& & \left(8.89\right)\end{array}$$

and its z-transform is

$$\begin{array}{lll}Ƶ\left\{sin\left({\Omega }_{0}n\right)u\left[n\right]\right\}\hfill & =Ƶ\left\{\frac{1}{2j}{\text{e}}^{j{\Omega }_{0}n}u\left[n\right]-\frac{1}{2}{e}^{-j{\Omega }_{0}n}u\left[n\right]\right\}\hfill & \hfill \\ \hfill & =\frac{1}{2j}Ƶ\left\{e^{j{\Omega }_{0}n}u\left[n\right]\right\}-\frac{1}{2j}\left\{e^{-j{\Omega }_{0}n}u\left[n\right]\right\}\hfill & \hfill \\ \hfill & =\frac{1/2j}{1-e^{j{\Omega }_{0}n}}-\frac{1/2j}{1-e^{-j{\Omega }_0}{z}^{-1}}\hfill & \left(8.90\right)\hfill \end{array}$$

Combining the terms of Eqn. (8.90) under a common denominator we have

$$\begin{array}{ccc}X\left(z\right)=\frac{sin\left({\Omega }_0\right)z^{-1}}{1-2cos\left({\Omega }_0\right)z^{-1}+z^{-2}}& & \left(8.91\right)\end{array}$$

or, using non-negative powers of z,

$$\begin{array}{ccc}X\left(z\right)=\frac{sin\left({\Omega }_0\right)z}{z^2-2\text{ cos}\left({\Omega }_0\right)z+1}& & \left(8.92\right)\end{array}$$

The ROC of the transform is |z| > 1 as in the previous example.

8.3.2 Time shifting

Thus, shifting the signal x[n] in the time domain by k samples corresponds to multiplication of the transform X (z) by z−k. The ability to express the z-transform of a time-shifted version of a signal in terms of the z-transform of the original signal will be very useful in working with difference equations.

The region of convergence for the resulting transform z−k X (z) is the same as that of X (z) with some possible exceptions:

1. If time-shifting the signal for k < 0 samples (left shift) causes some negative indexed samples to appear in x [n − k], then points at |z| → ∞ need to be excluded from the ROC.
2. If time-shifting the signal for k > 0 samples (right shift) causes some positive indexed samples to appear in x[n − k], then the origin z = 0 + j0 of the z-plane needs to be excluded from the ROC.

Example 8.17: z-Transform of discrete-time pulse signal revisited

The z-transform of the discrete-time pulse signal

$$x\left[n\right]=\{\begin{array}{ll}1,\hfill & 0\le n\le N-1\hfill \\ 0,\hfill & n<0\text{ or }n>N-1\hfill \end{array}$$

was determined earlier in Example 8.8. Find the same transform through the use of linearity and time-shifting properties.

Solution: The signal x[n] can be expressed as the difference of a unit-step signal and a time-shifted unit-step signal, that is,

$$\begin{array}{ccc}x\left[n\right]=u\left[n\right]-u\left[n-N\right]& & \left(8.96\right)\end{array}$$

We know from the linearity property of the z-transform that

$$\begin{array}{ccc}X\left(z\right)=Ƶ\left\{u\left[n\right]\right\}-Ƶ\left\{u\left[n-N\right]\right\}& & \left(8.97\right)\end{array}$$

The z-transform of the unit-step function was found in Example 8.5 as

$$\begin{array}{cccc}Ƶ\left\{u\left[n\right]\right\}-\frac{z}{z-1},& \text{ROC}:\left|z\right|>1& & \left(8.98\right)\end{array}$$

Using the time-shifting property of the z-transform, we find the transform of the shifted unit-step signal as

$$\begin{array}{cccc}Ƶ\left\{u\left[n-N\right]\right\}=z^{-N}\frac{z}{z-1},& \text{ROC}:\left|z\right|>1& & \left(8.99\right)\end{array}$$

Adding the two transforms in Eqns. (8.98) and (8.99) yields

$$\begin{array}{ccc}X\left(z\right)=\frac{z}{z-1}-z^{-N}\frac{z}{z-1}=\left(1-z^{-N}\right)\frac{z}{z-1}& & \left(8.100\right)\end{array}$$

The result in Eqn. (8.100) can be simplified to

$$\begin{array}{ccc}X\left(z\right)=\frac{z^N-1}{z^{N-1}\left(z-1\right)}& & \left(8.101\right)\end{array}$$

which matches the earlier result found in Example 8.8. Since x[n] is a finite-length causal signal, the transform converges everywhere except at the origin. Thus, the region of convergence is

$$\left|z\right|>0$$

This is one example of the possibility mentioned earlier regarding the region of convergence: The ROC for X (z) is larger than the overlap of the two regions given by Eqns. (8.98) and (8.99). The individual transforms Ƶ{u[n]} and Ƶ{u[n − N]} each have a pole at z = 1 causing the individual ROCs to be the outside of a circle with unit radius. When the two terms are added to construct X (z), however, the pole at z = 1 is canceled, resulting in the region of convergence for X (z) being larger than the overlap of the individual regions.

8.3.3 Time reversal

Since we have replaced z by z−1 in the transform, the ROC must be adjusted for this change. Let the ROC of the original transform X (z) be

$$r_1<\left|z\right|<r_2$$

The ROC for X(z−1) is

$$\begin{array}{ccc}{r}_1<\left|\frac{1}{2}\right|<r_2& \Rightarrow & \frac{1}{r_2}<\left|z\right|\end{array}<\frac{1}{r_1}$$

Example 8.18: z-Transform of anti-causal exponential signal revisited

The z-transform of the anti-causal exponential signal

$$x\left[n\right]=-a^{n}u\left[-n-1\right]$$

was found in Example 8.7 as

$$\begin{array}{cc}X\left(z\right)=\frac{z}{z-a},& \text{ROC}:\end{array}\left|z\right|<\left|a\right|$$

Obtain the same result from the z-transform of the causal exponential signal using the time-reversal and time-shifting properties.

Solution: We will start with the known transform relationship and try to convert it into the relationship that we seek, applying appropriate properties of the z-transform at each step. The causal exponential signal and its z-transform are

$$\begin{array}{cccc}{b}^{n}u\left[n\right]\stackrel{Ƶ}{\leftrightarrow }\frac{z}{z-a}& \text{ROC}:\left|z\right|>\left|a\right|& & \left(8.106\right)\end{array}$$

where we have used the parameter b instead of a. Let us begin by applying the time-reversal operation to the signal on the left, and adjusting the transform on the right according to the time-reversal property of the z-transform:

$$\begin{array}{cccc}{b}^{-n}u\left[-n\right]\stackrel{Ƶ}{\leftrightarrow }\frac{z^{-1}}{z^{-1}-b},& \text{ROC}:\left|z\right|<\frac{1}{\left|b\right|}& & \left(8.107\right)\end{array}$$

The transform relationship in Eqn. (8.107) can be written in the equivalent form

$$\begin{array}{cccc}{\left(\frac{1}{b}\right)}^{n}u\left[-n\right]\stackrel{Ƶ}{\leftrightarrow }\frac{-1/b}{z-1/b},& \text{ROC}:\left|z\right|<\frac{1}{\left|b\right|}& & \left(8.108\right)\end{array}$$

We will now apply a time shift to the signal by one sample to the left through the substitution n → n + 1 which causes the transform to be multiplied by z, resulting in the relationship

$$\begin{array}{cccc}{\left(\frac{1}{b}\right)}^{n+1}u\left[-n-1\right]\stackrel{Ƶ}{\leftrightarrow }\frac{(-1/b)z}{z-1/b},& \text{ROC}:\left|z\right|<\frac{1}{\left|b\right|}& & \left(8.109\right)\end{array}$$

Multiplying both sides of the relationship by −b leads to

$$\begin{array}{cccc}-{\left(\frac{1}{b}\right)}^{n}u\left[-n-1\right]\stackrel{Ƶ}{\leftrightarrow }\frac{z}{z-1/b},& \text{ROC}:\left|z\right|<\frac{1}{\left|b\right|}& & \left(8.110\right)\end{array}$$

Finally, choosing b = 1/a, we obtain the desired result:

$$\begin{array}{cccc}-a^{n}u\left[-n-1\right]\stackrel{Ƶ}{\leftrightarrow }\frac{z}{z-a},& \text{ROC}:\left|z\right|<\frac{1}{\left|b\right|}& & \left(8.111\right)\end{array}$$

8.3.4 Multiplication by an exponential signal

The ROC must be adjusted for the new transform variable (z/a). Let the ROC of the original transform X (z) be

$$r_1<\left|z\right|<r_2$$

The ROC for X (z/a) is

$$\begin{array}{ccc}{r}_1<\left|\frac{z}{a}\right|<r_2& \Rightarrow & \left|a\right|\end{array}{r}_1<\left|z\right|<\left|a\right|r_2$$

Example 8.19: Multiplication by an exponential signal

Determine the z-transform of the signal

$$x\left(n\right)=a^n\text{ cos}\left({\Omega }_0\right)u\left[n\right]$$

Assume a is real.

Solution: Let the signal x1[n] be defined as

$$x_1\left[n\right]=cos\left({\Omega }_{0}n\right)u\left[n\right]$$

so that

$$x\left[n\right]=a^n{x}_1\left[n\right]$$

The z-transform of the signal x1[n] was found in Example 8.15 to be

$$X_1\left(z\right)=Ƶ\left\{cos\left({\Omega }_{0}n\right)u\left[n\right]\right\}=\frac{z\left[z-cos\left({\Omega }_0\right)\right]}{z^2-2cos\left({\Omega }_0\right)z+1}$$

Multiplication of the time-domain signal by an causes the transform to be evaluated for z → z/a, resulting in the transform relationship

$$\begin{array}{llll}X\left(z\right)\hfill & =X_1\left(z/a\right)\hfill & \hfill & \hfill \\ \hfill & =\frac{\left(z/a\right)\left[\left(z/a\right)-cos\left({\Omega }_0\right)\right]}{{\left(z/a\right)}^2-2cos\left({\Omega }_0\right)\left(z/a\right)+1}\hfill & \hfill & \hfill \\ \hfill & =\frac{z\left[z-acos\left({\Omega }_0\right)\right]}{x^2-2acos\left({\Omega }_0\right)z+a^2}\hfill & \hfill & \left(8.114\right)\hfill \end{array}$$

The transform X (z) has two poles at z = ae±jΩ0. Its ROC is

$$\begin{array}{ccc}\text{ROC}:\le \left|z\right|>\left|a\right|& & \left(8.115\right)\end{array}$$

The pole-zero diagram and the ROC are shown in Fig. 8.22.

8.3.5 Differentiation in the z-domain

Example 8.20: Using the differentiation property

Determine the z-transform of the signal

$$x\left[n\right]=n{a}^{n}u\left[n\right]$$

Solution: In Example 8.6 we found the z-transform of a causal exponential signal to be

$$\begin{array}{cc}Ƶ\left\{a^{n}u\left[n\right]\right\}=\frac{z}{z-a},& \text{ROC}:\left|z\right|>\left|a\right|\end{array}$$

Applying the differentiation property given by Eqn. (8.116) we get

$$\begin{array}{llll}X\left(z\right)\hfill & =\left(-z\right)\frac{d}{dz}\left[\frac{z}{z-a}\right]\hfill & \hfill & \hfill \\ \hfill & =\left(-z\right)\frac{\left(-a\right)}{{\left(z-a\right)}^2}=\frac{a\mathrm{z}}{{\left(z-a\right)}^2}\hfill & \hfill & \left(8.121\right)\hfill \end{array}$$

and the ROC is

$$\left|z\right|>\left|a\right|$$

Example 8.21: z-Transform of a unit-ramp signal

Determine the z-transform of the unit-ramp signal

$$x\left[n\right]=nu\left[n\right]$$

Solution: The solution is straightforward if we use the result obtained in Example 8.20. The signal n an u [n] becomes the unit-ramp signal we are considering if the parameter a is set equal to unity. Setting a = 1 in the transform found in Example 8.20 we have

$$\begin{array}{ccc}X\left(z\right)={\frac{a\mathrm{z}}{{\left(z-a\right)}^2}|}_{a=1}=\frac{z}{{\left(z-1\right)}^2}& & \left(8.122\right)\end{array}$$

with the ROC

$$\left|z\right|>1$$

Example 8.22: More on the use of the differentiation property

Determine the z-transform of the signal

$$x\left[n\right]=n\left(n+2\right)u\left[n\right]$$

Solution: The signal x[n] can be written as

$$\begin{array}{ccc}x\left[n\right]=n^{2}u\left[n\right]+2nu\left[n\right]& & \left(8.123\right)\end{array}$$

The z-transform of the unit-ramp function was found in Example 8.21 through the use of the differentiation property as

$$\begin{array}{cccc}Ƶ\left\{nu\left[n\right]\right\}=\frac{z}{{\left(z-1\right)}^2},& \text{ROC}:\left|z\right|>1& & \left(8.124\right)\end{array}$$

We will now apply the differentiation property to the transform pair in Eqn. (8.124):

$$\begin{array}{lllll}Ƶ\left\{n^{2}u\left[n\right]\right\}\hfill & =\left(-z\right)\frac{d}{dz}\left[\frac{z}{{\left(z-1\right)}^2}\right]\hfill & \hfill & \hfill & \hfill \\ \hfill & =\frac{z\left(z+1\right)}{{\left(z-z\right)}^3},\hfill & \text{ROC}:\left|z\right|>1\hfill & \hfill & \left(8.125\right)\hfill \end{array}$$

Finally, combining the results in Eqns. (8.124) and (8.125) we construct the z-transform of x[n] as

$$\begin{array}{lllll}X\left(z\right)\hfill & =Ƶ\left\{n^{2}u\left[n\right]\right\}+2Ƶ\left\{nu\left[n\right]\right\}\hfill & \hfill & \hfill & \hfill \\ \hfill & =\frac{z\left(z+1\right)}{{\left(z-1\right)}^3}+2\frac{z}{{\left(z-1\right)}^2}\hfill & \hfill & \hfill & \hfill \\ \hfill & =\frac{3z\left(z-1/3\right)}{\left(z-3\right)},\hfill & \text{ROC}:\left|z\right|>1\hfill & \hfill & \left(8.126\right)\hfill \end{array}$$

The transform found in Eqn. (8.126) has three poles at z = 1. Its zeros are at z = 0 and $z=\frac{1}{3}$. The pole-zero diagram and the ROC are shown in Fig. 8.23.

8.3.6 Convolution property

Convolution property of the z-transform is fundamental in its application to DTLTI systems. It forms the basis of the z-domain system function concept that will be explored in detail in Section 8.5.

As before, the ROC for the resulting transform is at least the overlap of the two individual transforms, if such an overlap exists. It may be greater than the overlap of the two regions if the multiplication of the two transforms X1 (z) and X2 (z) results in the cancellation of a pole that sets a boundary for either X1 (z) or X2 (z).

Convolution property of the z-transform is very useful in the sense that it provides an alternative to actually computing the convolution of two signals in the time-domain. Instead, we can obtain x[n] = x1[n] * x2[n] using the procedure outlined below:

Example 8.23: Using the convolution property

Consider two signals x1[n] and x2[n] given by

$$\underset{n=0}{\overset{x_1\left[n\right]-\left\{4,3,2,1\right\}}{\uparrow }}$$

and

$$\underset{n=0}{\overset{x_2\left[n\right]=\left\{3,7,4\right\} }{\uparrow }}$$

Let x[n] be the convolution of these two signals, i.e.,

$$x\left[n\right]=x_1\left[n\right]*x_2\left[n\right]$$

Determine x[n] using z-transform techniques.

Solution: Convolution of these two signals was computed in Example 3.19 of Chapter 1 using the time-domain method. In this exercise we will use the z-transform to obtain the same result. Individual transforms of the signals x1[n] and x2[n] are

$$X_1\left(z\right)=4+3z^{-1}2z^{-2}+z^{-3}$$

and

$$X_2\left(z\right)=3+7z^{-1}+4z^{-2}$$

Applying the convolution property, the transform of x[n] can be written as the product of the two transforms:

$$\begin{array}{llll}X\left(z\right)\hfill & =X_1\left(z\right)X_2\left(z\right)\hfill & \hfill & \hfill \\ \hfill & =\left(4+3z^{-1}+2z^{-2}+z^{-3}\right)\left(3+7z^{-1}+4z^{-2}\right)\hfill & \hfill & \hfill \\ \hfill & =12+37z^{-1}+43z^{-2}+29z^{-3}+15z^{-4}+4z^{-5}\hfill & \hfill & \left(8.132\right)\hfill \end{array}$$

Comparing the result obtained in Eqn. (8.132) with the definition of the z-transform we conclude that

$$\begin{array}{llllll}x\left[0\right]=12\hfill & x\left[1\right]=37\hfill & x\left[2\right]=43\hfill & x\left[3\right]=29\hfill & x\left[4\right]=15\hfill & x\left[5\right]=4\hfill \end{array}$$

Equivalently, the signal x[n] can be written as

$$\underset{n=0}{\overset{ x\left[n\right]=12,37,43,29,15,4}{\uparrow }}$$

The solution of the convolution problem above using z-transform techniques leads to an interesting result: Convolution operation can be used for multiplying two polynomials. Suppose we need to multiply two polynomials

$$A\left(\nu \right)=ak{\nu }^k+a{k}_{-1}{\nu }^{k-1}+...+a_1\nu +a_0$$

and

$$B\left[\nu \right]=b_m{\nu }^m+b_{m-1}{\nu }^{m-1}+...+b_{1}v+b_0$$

and we have a convolution function available in software. To find the polynomial C (ν) = A (ν) B (ν), we can first create a discrete-time signal with the coefficients of each polynomial:

$$\underset{n=0}{\overset{ x_a\left[n\right]=\left\{a_0,a_1,...,a_{k-1},ak\right\}}{\uparrow }}$$

and

$$\underset{n=0}{\overset{ x_b\left[n\right]=\left\{b_0,b_1,...,b_{m-1},bm\right\}}{\uparrow }}$$

It is important to ensure that the polynomial coefficients are listed in ascending order of powers of ν in the signals xa[n] and xb[n], and any missing coefficients are accounted for in the form of zero-valued samples. If we now compute the convolution of the two signals as

$$x_c\left[n\right]=x_a\left[n\right]*x_b\left[n\right]$$

the resulting signal xc[n] holds the coefficients of the product polynomial C (ν), also in ascending order of powers of the independent variable. This will be demonstrated in MATLAB Exercise 8.4.

Example 8.24: Finding the output signal of a DTLTI system using inverse z-transform

Consider a DTLTI system described by the impulse response

$$h\left[n\right]={\left(0.9\right)}^{n}u\left[n\right]$$

driven by the input signal

$$x\left[n\right]=u\left[n\right]-u\left[n-7\right]$$

Compute the output signal y[n] through the use of z-transform techniques. Recall that this problem was solved in Example 3.21 of Chapter 3 using the convolution operation.

Solution: The z-transform of the impulse response is

$$\begin{array}{cc}H\left(z\right)=Ƶ\left\{h\left[n\right]\right\}=\frac{z}{z-0.9},& \text{ROC}:\end{array}\left|z\right|>0.9$$

The z-transform of the input signal x[n] is

$$\begin{array}{ll}X\left(z\right)\hfill & =\sum _{n=0}^6{z}^{-n}\hfill \\ \hfill & =1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5}+z^{-6}\hfill \\ \hfill & =\frac{z^7-1}{z^6\left(z-1\right)}\hfill \end{array}$$

The finite-length geometric series form of the transform X (z) will prove to be more convenient for use in this case. The z-transform of the output signal is found by multiplying the z-transforms of the input signal and the impulse response:

$$\begin{array}{ll}Y\left(z\right)\hfill & =X\left(z\right)H\left(z\right)\hfill \\ \hfill & =\left(1+z^{-1}+z^{-3}+z^{-4}+z^{-5}+z^{-6}\right)H\left(z\right)\hfill \\ \hfill & =H\left(z\right)+z^{-1}H\left(z\right)+z^{2}H\left(z\right)+z^{-3}H\left(z\right)+z^{-4}H\left(z\right)+z^{-5}H\left(z\right)+z^{-6}H\left(z\right)\hfill \end{array}$$

The output signal y[n] can now be found as the inverse z-transform of Y (z) with the use of linearity and time-shifting properties of the z-transform.

$$\begin{array}{lll}y\left[n\right]\hfill & =h\left[n\right]+h\left[n-1\right]+h\left[n-2\right]+h\left[n-3\right]+h\left[n-4\right]\hfill & \hfill \\ \hfill & +h\left[n-5\right]+h\left[n-6\right]\hfill & \hfill \\ \hfill & ={\left(0.9\right)}^{n}u\left[n\right]+{\left(0.9\right)}^{n-1}u\left[n-1\right]+{\left(0.9\right)}^{n-2}u\left[n-2\right]\hfill & \hfill \\ \hfill & +{\left(0.9\right)}^{n-3}u\left[n-3\right]+{\left(0.9\right)}^{n-4}u\left[n-4\right]\hfill & \hfill \\ \hfill & +{\left(0.9\right)}^{n-5}u\left[n-5\right]+{\left(0.9\right)}^{n-6}u\left[n-6\right]\hfill & \left(8.133\right)\hfill \end{array}$$

It can be shown that the result obtained in Eqn. (8.133) is identical to the one found in Example 3.21 when the two results are compared on a sample-by-sample basis.

Software resources:

ex_8_24.m

8.3.7 Initial value

Initial value property of the z-transform applies to causal signals only.

It is interesting to note that the limit in Eqn. (8.134) exists only for a causal signal, provided that x[0] is finite. Consequently, the convergence of limz→∞ [X(z)] can be used as a test of the causality of the signal. For a causal signal x[n], the ROC of the z-transform must include infinitely large values of z. If the transform X(z) is a rational function of z, the numerator order must not be greater than the denominator order.

Example 8.25: Using the initial value property

The z-transform of a causal signal x[n] is given by

$$\begin{array}{cc}X(z)=\frac{3z^3+2z+5}{2z^3-7z^2+z-4}& (8.136)\end{array}$$

Determine the initial value x[0] of the signal.

Solution: Using Eqn. (8.134) we get

$$\underset{z\to \infty }{lim}[X(z)]=\underset{z\to \infty }{lim}[\frac{3z^3+2z+5}{2z^3-7z^2+z-4}]=\underset{z\to \infty }{lim}[\frac{3z^3}{2z^3}]=\frac{3}{2}$$

The result can be easily justified by computing the inverse z-transform of X(z) using either partial fraction expansion or long division. These techniques will be explored in Section 8.4.

8.3.8 Correlation property

Cross-correlation of two discrete-time signals x[n] and y [n] is defined as

$$\begin{array}{cc}{r}_{xy}[m]={\displaystyle \sum _{n=-\infty }^{\infty }}x[n]y[n-m]& (8.137)\end{array}$$

The argument for the ROC of the transform uses the same reasoning we have employed earlier: The ROC for the resulting transform Rxy (z) is atleast the overlap of the two individual transforms, if such an overlap exists. It may be greater than the overlap of the two regions if the multiplication of the two transforms X(z) and Y(z−1) results in the cancellation of a pole that sets a boundary for either X(z) or Y(z−1).

As a special case of Eqn. (8.146), the transform of the autocorrelation of a signal x[n] can be found as

$$R_{xx}(z)=X(z)X(z^{-1})$$

Example 8.26: Using the correlation property

Determine the cross-correlation of the two signals

$$x\left[n\right]=a^{n}u\left[n\right]$$

and

$$y\left[n\right]=u\left[n\right]-u\left[n-3\right]$$

using the correlation property of the z-transform.

Solution: The z-transform of x[n] is

$$\begin{array}{ccc}X(z)=\frac{z}{z-a},& \text{R}\text{O}\text{C}:\left|z\right|>\left|a\right|& (8.146)\end{array}$$

Next we will determine the z-transform of y [n]:

$$\begin{array}{ccc}Y(z)=1+z^{-1}+z^{-2}& \text{R}\text{O}\text{C}:\left|z\right|>0& (8.147)\end{array}$$

Substituting z → z −1 in this last result we obtain

$$\begin{array}{ccc}Y(z^{-1})=1+z+z^2& \text{R}\text{O}\text{C}:\left|z\right|<\infty & (8.148)\end{array}$$

which is the z-transform of the time-reversed signal y [−n]. We have modified the ROC in Eqn. (8.148) to account for the fact that the transform variable z has been reciprocated. Now we obtain Rxy (z) by multiplying the two results in Eqns. (8.146) and (8.148):

$$\begin{array}{llll}\begin{array}{ll}{R}_{xy}(z)\hfill & =X(z)Y(z^{-1})\hfill \\ \hfill & =\frac{z}{z-a}(1+z+z^2)\hfill \\ \hfill & =\frac{z}{z-a}+z\frac{z}{z-a}+z^2\frac{z}{z-a},\hfill \end{array}\hfill & \text{R}\text{O}\text{C}:\left|a\right|<\left|z\right|<\infty \hfill & \hfill & (8.149)\hfill \end{array}$$

The cross-correlation rxy[n] is the inverse z-transform of the result in Eqn. (8.149) which, using the time-shifting property of the z-transform, can be found as

$$\begin{array}{cc}{r}_{xy}[n]=a^{n}u[n]+a^{n+1}u[n+1]+a^{n+2}u[n+2]& (8.150)\end{array}$$

Software resources:

ex_8_26.m

Example 8.27: Finding auto-correlation of a signal

Using the correlation property of the z-transform, determine the auto-correlation of the signal

$$\begin{array}{cc}x[n]=a^{n}u[n]& (8.151)\end{array}$$

Solution: The z-transform of the signal is

$$\begin{array}{ccc}x(z)=\frac{z}{z-a},& \text{R}\text{O}\text{C}:\left|z\right|>\left|a\right|& (8.152)\end{array}$$

Substituting z −1 for z in Eqn. (8.152) we obtain

$$\begin{array}{ccc}x(z^{-1})=\frac{z^{-1}}{z^{-1}-a}=\frac{-1/a}{z-1/a},& \text{R}\text{O}\text{C}:\left|z\right|<\frac{1}{\left|a\right|}& (8.153)\end{array}$$

The z-transform of the auto-correlation rxx[n] is

$$\begin{array}{llll}{R}_{xx}(z)\hfill & =X(z)X(Z^{-1})\hfill & \hfill & \hfill \\ \hfill & =\frac{-Z}{a{z}^2-(a^2+1)z+a},\hfill & \text{R}\text{O}\text{C}:\left|a\right|<\left|z\right|<\frac{1}{\left|a\right|}\hfill & (8.154)\hfill \end{array}$$

provided that |a| < 1. Recall that in Example 8.13 we found the z-transform of the two-sided exponential signal to be

$$Ƶ―\left\{a^{\left|n\right|}\right\}=\frac{(a^2-1)z}{a{z}^2-(a^2+1)z+a}$$

Comparing this earlier result to what we have in Eqn. (8.154) leads to the conclusion

$$\begin{array}{cc}{r}_{xx}[n]=(\frac{1}{1-a^2})a^{\left|n\right|}& (8.155)\end{array}$$

8.3.9 Summation property

Example 8.28: z-Transform of a unit-ramp signal revisited

The transform of the unit-ramp signal

$$x[n]=nu[n]$$

was found in Example 8.21 through the use of the differentiation property. In this example we will use it as an opportunity to apply the summation property of the z-transform. The ramp signal x[n] can be expressed as the running sum of a time-shifted unit-step signal as

$$\begin{array}{cc}x[n]={\displaystyle \sum _{k=-\infty }^n}u[n-1]& (8.166)\end{array}$$

It can easily be verified that the definition in Eqn. (8.166) produces x[n]= 0 for n ≤ 0, and x[1] = 1,x [2] = 2, and so on, consistent with the ramp signal. The z-transform of the time-shifted unit-step signal is

$$\begin{array}{cc}Ƶ―\{u[n-1]\}=z^{-1} Ƶ―\{u[n]\}=z^{-1}(\frac{z}{z-1})=\frac{1}{z-1}& (8.167)\end{array}$$

Using the summation property we find the the transform X(z) as

$$X(z)=\frac{z}{z-1}Ƶ―\{u[n-1]\}=(\frac{z}{z-1})(\frac{1}{z-1})=\frac{z}{{(z-1)}^2}$$

which matches the answer found in Example 8.20.

8.4 Inverse z-Transform

Inverse z-transform is the problem of finding x[n] from the knowledge of X(z). Often we need to determine the signal x[n] that has a specified z-transform. There are three basic techniques for computing the inverse z-transform:

1. Direct evaluation of the inversion integral

2. Partial fraction expansion technique for a rational transform

3. Expansion of the rational transform into a power series through long division

We will focus our attention on the last two methods. The inversion integral method will be briefly mentioned, but will not be considered further due to its complexity. We will find that methods 2 and 3 will be sufficient for most problems encountered in the analysis of signals and linear systems.

8.4.1 Inversion integral

Let X(r, Ω) be the function obtained by evaluating the z-transform of a signal x[n] for z =re jΩ, that is, at points on a circle with radius equal to r. We have established in Section 8.1 that X(r, Ω) is the same as the DTFT of the signal x[n]r−n, that is

$$\begin{array}{cc}X(r,\Omega )=X(z){|}_{z=re{j}^{\Omega }}=F\{x[n]r^{-n}\}& (8.168)\end{array}$$

Consequently, the inverse DTFT equation given by Eqn. (5.70) should yield the signal x[n]r−n:

$$\begin{array}{cc}x[n]r^{-n}=F^{-1}\{X(r,\Omega )\}=\frac{1}{2\pi }{{\displaystyle \int }}_{-\pi }^{\pi }X(r,\Omega )e^{j\Omega }d\Omega & (8.169)\end{array}$$

Multiplying both sides of Eqn. (8.169) with r n we obtain

$$\begin{array}{lll}x[n]\hfill & =r^n\frac{1}{2\pi }\underset{-x}{\overset{x}{{\displaystyle \int }}}X(r,\Omega )e^{j\Omega }d\Omega \hfill & \hfill \\ \hfill & =\frac{1}{2\pi }\underset{-x}{\overset{x}{{\displaystyle \int }}}X(r,\Omega ){(r{e}^{j\Omega })}^{n}d\Omega \hfill & (8.170)\hfill \end{array}$$

Eqns. (8.168) through (8.170) provide us with a method for finding x[n] fromits z-transform:

The three-step procedure outlined above can be reduced to a contour integral. Since z =re jΩ and r is a constant in the evaluation of the integral in Eqn. (8.172), we can write

$$\begin{array}{cc}dz=jr{e}^{j\Omega }d\Omega =jzd\Omega & (8.173)\end{array}$$

and therefore

$$\begin{array}{cc}d\Omega =\frac{1}{jz}dz& (8.174)\end{array}$$

In the next step we will substitute Eqn. (8.174) into Eqn. (8.172), change re jΩ to z, and change X(r, Ω) to X(z) to arrive at the result

$$\begin{array}{cc}x[n]=\frac{1}{2\pi j}\oint X(z)z^{n-1}dz& (8.175)\end{array}$$

where we have used the contour integral symbol ∮ to indicate that the integral should be evaluated by traveling counter-clockwise on a closed contour in the z-plane within the ROC of the transform. The values of the transform X(z) on the closed contour are multiplied by zn−1 and integrated. Integration can start at an arbitrary point on the contour, but it must end at the same point. Note the similarity of the inversion integral to that given by Eqn. (7.93) for the computation of the inverse Laplace transform.

In general, direct evaluation of the contour integral is difficult when X(z) is a rational function of z. An indirect method of evaluating the integral in Eqn. (8.175) is to rely on the Cauchy residue theorem named after Augustin-Louis Cauchy (1789-1857). In this text we will not consider the inversion integral further for computing the inverse z-transform since more practical methods exist to accomplish the same task.

8.4.2 Partial fraction expansion

It was established in Examples 8.6 and 8.7 that the z-transform of a causal exponential signal is

$$\begin{array}{ccc}Ƶ―\{a^{n}u[n]\}=\frac{z}{z-a},& \text{R}\text{O}\text{C}:\left|z\right|>\left|a\right|& (8.176)\end{array}$$

and the z-transform of an anti-causal exponential signal is

$$\begin{array}{ccc}Ƶ―\{-a^{n}u[-n-1]\}=\frac{z}{z-a},& \text{R}\text{O}\text{C}:\left|z\right|<\left|a\right|& (8.177)\end{array}$$

The two signals in Eqns. (8.176) and (8.177) lead to the same rational function for the transform, albeit with different ROCs. These two transform pairs can be used as the basis for determining the inverse z-transform of rational functions expressed using partial fractions. Consider a transform X(z) given with its denominator factored out as

$$\begin{array}{cc}X(z)=\frac{B(z)}{(z-z_1)(z-z_2)\mathrm{...}(z-z_N)}& (8.178)\end{array}$$

Provided that the order of the numerator polynomial B (z) is not greater than N, expanding the transform into partial fractions in the form

$$\begin{array}{cc}X(z)=\frac{k_{1}z}{z-z_1}+\frac{k_{2}z}{z-z_2}+\mathrm{...}+\frac{k_{N}z}{z-z_N}& (8.179)\end{array}$$

would allow us to use the standard forms in Eqns. (8.176) and (8.177) for finding the inverse transform of X(z). Let individual terms in the partial fraction expansion be

$$\begin{array}{cc}{X}_i(z)=\frac{k_{i}z}{z-z_i}\text{for}i=1,\mathrm{...},N& (8.180)\end{array}$$

so that

$$\begin{array}{cc}X(z)=X_1(z)+X_2(z)+\mathrm{...}{X}_N(z)& (8.181)\end{array}$$

The inverse transform is therefore computed as

$$\begin{array}{cc}x[n]=x_1[n]+x_2[n]+\mathrm{...}+x_N[n]& (8.182)\end{array}$$

where each contributing term x i[n] represents the inverse transform of the corresponding term on the right side of Eqn. (8.181), that is,

$$\begin{array}{cc}{x}_i\left[n\right]={Ƶ}^{-1}\left\{X_i\left(z\right)\right\}={Ƶ}^{-1}\left\{\frac{k_{i}z}{z-z_i}\right\},\text{ for }i=1,\mathrm{...},N& \left(8.183\right)\end{array}$$

In order to find the correct terms xi[n] for Eqn. (8.182) we need to determine for each Xi (z) whether it is the transfom of a causal signal as in Eqn. (8.176) or an anti-causal signal as in Eqn. (8.177). These decisions must be made by looking at the ROC for X(z) and reasoning what the contribution from the ROC of each individual term Xi (z) must be in order to get the overlap that we have. Since each term in the partial fraction expansion has only one pole, we will adopt the following simple rules:

These two rules are illustrated in Fig. 8.24. The next two examples will serve to clarify the process explained above.

Example 8.29: Expanding a rational z-transform into partial fractions

Express the z-transform

$$X(z)=\frac{(z-1)(z+2)}{(z-\frac{1}{2})(z-2)}$$

using partial fractions.

Solution: The first step is to divide X(z) by z to obtain

$$\begin{array}{cc}\frac{X(z)}{z}=\frac{(z-1)(z+2)}{z(z-\frac{1}{2})(z-2)}& (8.186)\end{array}$$

The function X(z)/z has three single poles at z = 0, ½, and 2, and can be expressed as

$$\begin{array}{cc}\frac{X(z)}{z}=\frac{k_1}{z}+\frac{k_2}{z-\frac{1}{2}}+\frac{k_3}{z-2}& (8.187)\end{array}$$

Multiplying both sides of Eqn. (8.187) by z allows us to express X(z) using familiar terms:

$$\begin{array}{cc}X(z)=k_1+\frac{k_{2}z}{z-\frac{1}{2}}+\frac{k_{3}z}{z-2}& (8.188)\end{array}$$

The transition from Eqn. (8.187) to Eqn. (8.188) explains the motivation for dividing X(z) by z in Eqn. (8.186) before expanding the result into partial fractions in Eqn. (8.187). Had we not divided X(z) by z in Eqn. (8.187), we would not have obtained partial fractions in the standard form z/ (z − pi)) in Eqn. (8.188).

The residues in Eqn. (8.187) can be found by using the residue formulas derived in Appendix E:

$$k_1=z\frac{X(z)}{z}{|}_{z=0}=\frac{(z-1)(z+2)}{(z-\frac{1}{2})(z-2)}{|}_{z=0}=\frac{(0-1)(0+2)}{(0-\frac{1}{2})(0-2)}=-2$$

$$k_2=(z-\frac{1}{2})\frac{X(z)}{z}{|}_{z=\frac{1}{2}}=\frac{(z-1)(z+2)}{(z-2)}{|}_{z=0}=\frac{(\frac{1}{2}-1)(\frac{1}{2}+2)}{\frac{1}{2}(\frac{1}{2}-2)}=\frac{5}{3}$$

$$k_3=(z-2)\frac{X(z)}{z}{|}_{z=-2}=\frac{(z-1)(z+2)}{z(z-\frac{1}{2})}{|}_{z=0}=\frac{(2-1)(2+2)}{2(2-\frac{1}{2})}=\frac{4}{3}$$

Using the residues computed, the transform X(z) is

$$X(z)=-2+\frac{\frac{5}{3}z}{z-\frac{1}{2}}+\frac{\frac{4}{3}z}{z-2}$$

Example 8.30: Finding the inverse z-transform using partial fractions

Consider again the transform X(z) expanded into partial fractions in Example 8.29. The region of convergence was not specified in that example, and therefore, we have only determined the partial fraction expansion for X(z).

1. How many possibilities are there for the ROC?

2. For each possible choice of ROC, determine the inverse transform.

Solution: A pole-zero diagram for X(z) is shown in Fig. 8.25. We know from previous discussion that

1. There can be no poles in the ROC.
2. The boundaries of ROC must be determined by poles.

We will begin by defining the three components of X(z) in the partial fraction expansion given by Eqn. (8.189). Let

$$X_1(z)=-2$$

$$X_2(z)=\frac{\frac{5}{3}z}{z-\frac{1}{2}}$$

and

$$X_3(z)=\frac{\frac{4}{3}z}{z-2}$$

The first component,X 1 (z), is a constant, and its ROC is always the entire z-plane. Therefore it has no effect on the overall ROC for X(z). The inverse transform of X1 (z) is

$$x_1\left[n\right]={Ƶ}^{-1}\left\{-2\right\}=-2\delta \left[n\right]$$

The ROC of X(z) will be determined based on the individual ROCs of the terms X2 (z) and X3 (z). The term X2 (z) has a zero at z =0, and a pole at z = ½. Its region of convergence is either the inside or the outside of the circle with a radius of ½. Similarly, the term X3 (z) has a zero at z =0, and a pole at z = 2. Its region of convergence is either the inside or the outside of the circle with a radius of 2. Applying the rules (a) and (b) mentioned above in conjunction with the pole-zero diagram shown in Fig. 8.25 we obtain the following possibilities for the ROC of X(z):

Possibility 1: $\text{ROC }:\left|\text{z}\right|<\frac{1}{2}$

In this case both X2 (z) and X3 (z) must correspond to anti-causal signals, so that the overlap of individual ROCs yields the region chosen. Thus we need

$${\text{ROC for x}}_2(z):\left|\text{z}\right|<\frac{1}{2}$$

$${\text{ROC for x}}_3(z):\left|\text{z}\right|<2$$

The resulting ROC for X(z) is the overlap of the two individual ROCs, and is therefore the inside of the circle with radius ½. This is illustrated in Fig. 8.26.

The inverse transforms of X2 (z) and X3 (z) are determined using the anti-causal signal given by Eqn. (8.177):

$$\begin{array}{cc}{x}_2\left[n\right]=Z^{-1}\left\{\frac{\frac{5}{3}z}{z-\frac{1}{2}}\right\}=-\frac{5}{3}{\left(\frac{1}{2}\right)}^{n}u\left[-n-1\right]& \left(8.189\right)\end{array}$$

$$\mathrm{ }\begin{array}{lll}{x}_3\mathrm{ }[n]=\mathrm{ }{Ƶ}^{-\mathrm{ }1}\mathrm{ }\left\{\frac{\frac{4}{3}\mathrm{ }\mathrm{ }z}{z\mathrm{ }-\mathrm{ }2}\right\}\mathrm{ }=\mathrm{ }-\mathrm{ }\frac{4}{3}\mathrm{ }{(2)}^n\mathrm{ }u[-n\mathrm{ }-\mathrm{ }1]\hfill & \hfill & (8.190)\hfill \end{array}$$

Combining Eqns. (8.189), (8.189) and (8.190) we find the signal x[n] to be

$$\mathrm{ }\begin{array}{lll}x[n]\mathrm{ }=\mathrm{ }-2\mathrm{ }\delta \mathrm{ }\left[n\right]\mathrm{ }-\mathrm{ }\frac{5}{3}\mathrm{ }{\left(\frac{1}{2}\right)}^n\mathrm{ }u[-n\mathrm{ }-\mathrm{ }1]\mathrm{ }-\mathrm{ }\frac{4}{3}\mathrm{ }{\left(2\right)}^n\mathrm{ }u\left[-n\mathrm{ }-\mathrm{ }1\right]\hfill & \hfill & \left(8.191\right)\hfill \end{array}$$

Numerical evaluation of x [n] in Eqn. (8.191) for a few values of the index n results in

$$x\left[n\right]\mathrm{ }=\mathrm{ }\{\mathrm{...},\mathrm{ }-53.375,\mathrm{ }-26.75,\mathrm{ }-13.5,\mathrm{ }-7,\mathrm{ }-4,\mathrm{ }-\underset{n=0}{\overset{2}{\uparrow }}\mathrm{ }\}$$

Possibility 2: ROC: |z| > 2

This ROC is only possible as the overlap of individual ROCs if both X2 (z) and X3 (z) correspond to causal signals, that is,

$$\text{R}\text{O}\text{C}\mathrm{ }\text{for}\mathrm{ }{X}_2\mathrm{ }(z)\mathrm{ }:\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }>\mathrm{ }\frac{1}{2}$$

$$\text{R}\text{O}\text{C}\mathrm{ }\text{for}\mathrm{ }{X}_3\mathrm{ }(z)\mathrm{ }:\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }>\mathrm{ }2$$

The resulting ROC for X (z) in this case is the overlap of the two individual ROCs which is the outside of the circle with a radius of 2. This is illustrated in Fig. 8.27.

In this case, the inverse transforms of both X2 (z) and X3 (z) are causal signals, and they are computed from Eqn. (8.176) as

$$\begin{array}{lll}{x}_2\mathrm{ }[n]\mathrm{ }=\mathrm{ }{Ƶ}^{-1\mathrm{ }}\mathrm{ }\left\{\frac{\frac{5}{3}\mathrm{ }z}{z\mathrm{ }-\mathrm{ }\frac{1}{2}}\right\}\mathrm{ }=\mathrm{ }\frac{5}{3}\mathrm{ }{\left(\frac{1}{2}\right)}^n\mathrm{ }u[n]\hfill & \hfill & \left(8.192\right)\hfill \end{array}$$

$$\begin{array}{lll}{x}_3\mathrm{ }[n]\mathrm{ }=\mathrm{ }{Ƶ}^{-1}\mathrm{ }\left\{\frac{\frac{4}{3}\mathrm{ }z}{z\mathrm{ }-\mathrm{ }2}\right\}\mathrm{ }=\mathrm{ }\frac{4}{3}\mathrm{ }{\left(2\right)}^n\mathrm{ }u\left[n\right]\hfill & \hfill & \left(8.193\right)\hfill \end{array}$$

Using Eqns. (8.189), (8.192) and (8.193) the inverse transform for X (z) is

$$\begin{array}{lll}x[n]\mathrm{ }=\mathrm{ }-2\mathrm{ }\delta \left[n\right]\mathrm{ }+ \mathrm{ }\frac{5}{3}\mathrm{ }{\left(\frac{1}{2}\right)}^n\mathrm{ }u[n]\mathrm{ }+ \mathrm{ }\frac{4}{3}\mathrm{ }{(2)}^n\mathrm{ }u[n]\hfill & \hfill & (8.194)\hfill \end{array}$$

In this case the first few samples of x[n] are computed as

$$x[n]\mathrm{ }=\mathrm{ }\{ \underset{n=0}{\overset{1}{\uparrow }},\mathrm{ }3.5,\mathrm{ }5.75,\mathrm{ }8.2083,\mathrm{ }10.7708,\mathrm{ }13.3854,\mathrm{...}\}$$

Possibility 3: $\text{R}\text{O}\text{C}:\mathrm{ }\frac{1}{2}\mathrm{ }<\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }<\mathrm{ }2\mathrm{ }\mathrm{ }\mathrm{ }$

For this ROC to be the overlap of the two individual ROCs, X2 (z) must correspond to a causal signal while X3 (z) corresponds to an anti-causal signal. We need

$$\text{R}\text{O}\text{C}\mathrm{ }\text{for}\mathrm{ }{X}_2\mathrm{ }(z)\mathrm{ }:\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }>\mathrm{ }\frac{1}{2}$$

$$\text{R}\text{O}\text{C}\mathrm{ }\text{for}\mathrm{ }{X}_3\mathrm{ }(z)\mathrm{ }:\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }<\mathrm{ }2$$

Thus the ROC for X2 (z) is the outside of a circle, and the ROC for X3 (z) is the inside of a circle. The overlap of the two ROCs is the region between the two circles. Fig. 8.28 illustrates this possibility.

Inverting each component accordingly, we obtain

$$\begin{array}{lll}{x}_2[n]\mathrm{ }=\mathrm{ }{Ƶ}^{-1}\mathrm{ }\left\{\frac{\frac{5}{3}\mathrm{ }z}{z\mathrm{ }-\mathrm{ }\frac{1}{2}}\right\}\mathrm{ }=\mathrm{ }\frac{5}{3}\mathrm{ }{\left(\frac{1}{2}\right)}^n\mathrm{ }u[n]\hfill & \hfill & \left(8.195\right)\hfill \end{array}$$

and

$$\begin{array}{lll}{x}_3[n]\mathrm{ }=\mathrm{ }{Ƶ}^{-1}\mathrm{ }\left\{\frac{\frac{4}{3}\mathrm{ }z}{z\mathrm{ }-\mathrm{ }2}\right\}\mathrm{ }=\mathrm{ }-\frac{4}{3}\mathrm{ }{(2)}^n\mathrm{ }u[-n\mathrm{ }-1]\hfill & \hfill & (8.196)\hfill \end{array}$$

Using Eqns. (8.189), (8.195) and (8.196) the inverse transform for X (z) can be constructed as

$$\begin{array}{lll}x[n]\mathrm{ }=\mathrm{ }-2\delta [n]\mathrm{ }+ \mathrm{ }\frac{5}{3}\mathrm{ }{\left(\frac{1}{2}\right)}^n\mathrm{ }u[n]\mathrm{ }-\mathrm{ }\frac{4}{3}\mathrm{ }{(2)}^n\mathrm{ }u[-n\mathrm{ }-1]\hfill & \hfill & \left(8.197\right)\hfill \end{array}$$

In this case the first few samples of x[n] are computed as

$$x[n]\mathrm{ }=\mathrm{ }\{\mathrm{...},\mathrm{ }-0.1667,\mathrm{ }-0.3333,\mathrm{ }-0.6667,\mathrm{ }\underset{n=0}{\overset{0.3333}{\uparrow }},\mathrm{ }0.8333,\mathrm{ }0.4167,\mathrm{ }0.2083,\mathrm{...}\}$$

8.4.3 Long division

An alternative method for computing the inverse z-transform is the long division technique. Recall that the definition of the z-transform given by Eqns. (8.1) and (8.2) is essentially in the form of a power series involving powers of z and z−1. The long division idea is based on converting a rational transform X (z) back into its power series form, and associating the coefficients of the power series with the sample amplitudes of the signal x [n]. In contrast with the partial fraction expansion method discussed in the previous section, the long division method does not produce an analytical solution for the signal x [n]. Instead, it allows us to obtain the signal one sample at a time. Its main advantage over the partial fraction expansion method is that it is suitable for use on a computer.

Consider a rational transform in the general form

$$\begin{array}{lll}X(z)\mathrm{ }= \mathrm{ } \frac{B(Z)}{A(Z)}\mathrm{ }=\mathrm{ } \frac{b_M{z}^M\mathrm{ }+\mathrm{ }\mathrm{ }{b}_{M-1}{z}^{M-1}\mathrm{ }+\mathrm{...}+\mathrm{ } b_{1}z\mathrm{ }+\mathrm{ }{b}_0\mathrm{ }}{a_N{z}^N\mathrm{ }+\mathrm{ }{a}_{N-1}{z}^{N-1}\mathrm{ }+\mathrm{...}+\mathrm{ } a_{1}z\mathrm{ } +\mathrm{ }{a}_0}\mathrm{ }\hfill & \hfill & \left(8.198\right)\hfill \end{array}$$

where the M-th order denominator polynomial B (z) is

$$\begin{array}{lll}B(z)\mathrm{ }=\mathrm{ }{b}_M{z}^M\mathrm{ }+\mathrm{ }{b}_{M-1}{z}^{M-1}\mathrm{ }+\mathrm{...}+\mathrm{ }{b}_{1}z\mathrm{ }+\mathrm{ }{b}_0\hfill & \hfill & \left(8.199\right)\hfill \end{array}$$

and the N-th order numerator polynomial A (z) is

$$\begin{array}{lll}A(z)\mathrm{ }=\mathrm{ }{a}_N{z}^N\mathrm{ }+\mathrm{ }{a}_{N-1}{z}^{N-1}\mathrm{ }+\mathrm{...}+\mathrm{ }{a}_{1}z\mathrm{ }+\mathrm{ }{a}_0\hfill & \hfill & \left(8.200\right)\hfill \end{array}$$

Initially we will focus on the case where the signal x [n] that led to the transform X (z) is causal. This implies the following:

1. The ROC is in the form |z| > r1.
2. The transform X (z) converges at |z| → ∞.
3. The order of the numerator is less then or equal to the order of the denominator, that is, M ≤ N.

Dividing the numerator polynomial by the denominator polynomial, the transform X (z) can be written in the alternative form

$$\begin{array}{lll}X(z)\mathrm{ }=\mathrm{ }\left(\frac{b_M}{a_N}\right)\mathrm{ }{z}^{-(N-M)}\mathrm{ }+\mathrm{ }\frac{\overline{B}(z)}{A(z)}\hfill & \hfill & (8.201)\hfill \end{array}$$

where (bM /aN)z −(N−M) is the quotient of the division. The remainder is a polynomial $\overline{B}\mathrm{ }(z)$ in the form

$$\begin{array}{lll}\overline{B}(z)\mathrm{ }=\mathrm{ }{\overline{b}}_{M\mathrm{ }-\mathrm{ }1}{z}^{M\mathrm{ }-\mathrm{ }1}\mathrm{ }+\mathrm{...}+\mathrm{ }{\overline{b}}_{M\mathrm{ }-\mathrm{ }N}{z}^{M\mathrm{ }-\mathrm{ }N}\mathrm{ }\mathrm{ }\hfill & \hfill & \left(8.202\right)\hfill \end{array}$$

Associating the expression in Eqn. (8.202) with the power series form of the z-transform we recognize that

$$\begin{array}{lll}x[N\mathrm{ }-\mathrm{ }M]\mathrm{ }=\mathrm{ }\frac{b_M}{a_N}\hfill & \hfill & \left(8.203\right)\hfill \end{array}$$

providing us with one sample of the signal x [n]. We can now take the function

$$\begin{array}{ccc}\overline{X}\left(z\right)\mathrm{ }=\mathrm{ }\frac{\overline{B}\mathrm{ }\left(z\right)}{A\mathrm{ }\left(z\right)}& & \left(8.204\right)\end{array}$$

and repeat the process, obtaining

$$\begin{array}{lll}\overline{X}\mathrm{ }(z)\mathrm{ }=\mathrm{ }\left(\frac{{\overline{b}}_M\mathrm{ }-\mathrm{ }1}{a_N}\right)\mathrm{ }{z}^{-(N\mathrm{ }- M\mathrm{ }+\mathrm{ }1)}\mathrm{ }+\mathrm{ }\frac{\hat{B}\mathrm{ }(z)}{A(z)}\hfill & \hfill & (8.205)\hfill \end{array}$$

which produces another sample of the signal x [n] as

$$\begin{array}{lll}x[N\mathrm{ }-\mathrm{ }M\mathrm{ }+\mathrm{ }1]\mathrm{ }=\mathrm{ }\frac{{\overline{b}}_M-1}{a_N}\hfill & \hfill & (8.206)\hfill \end{array}$$

and a new remainder polynomial

$$\begin{array}{lll}\hat{B}(z)\mathrm{ }=\mathrm{ }{\hat{b}}_{M\mathrm{ }-\mathrm{ }2}{z}^{M\mathrm{ }-2}\mathrm{ }+\mathrm{...}+{\hat{b}}_{M-N-1}{z}^{M-N-1}\hfill & \hfill & \left(8.207\right)\hfill \end{array}$$

The next example will illustrate this process.

Example 8.31: Using long division with right-sided signal

Use the long division method to determine the first few samples of the signal x [n] of Example 8.22 from its z-transform which was determined to be

$$\begin{array}{lll}X(z)\mathrm{ }=\mathrm{ } \frac{3z^2\mathrm{ }-\mathrm{ }z}{{(z\mathrm{ }-\mathrm{ }1)}^3}\mathrm{ },\hfill & \text{R}\text{O}\text{C}:\hfill & \left|z\right|\mathrm{ }>\mathrm{ }1\hfill \end{array}$$

Solution: By multiplying out the denominator, the transform X (z) can be written as

$$\begin{array}{lll}X(z)\mathrm{ }=\mathrm{ } \frac{3z^2\mathrm{ }-\mathrm{ }z}{z^3\mathrm{ }-\mathrm{ }3z^{2\mathrm{ }}+\mathrm{ }3z\mathrm{ }-\mathrm{ }1}\mathrm{ },\hfill & \text{R}\text{O}\text{C}:\hfill & \left|z\right|\mathrm{ }>\mathrm{ }1\hfill \end{array}$$

The numerator polynomial is

$$B(z)\mathrm{ }=\mathrm{ }3z^2\mathrm{ }-\mathrm{ }z$$

and the denominator polynomial is

$$A(z)\mathrm{ }=\mathrm{ }{z}^3\mathrm{ }-\mathrm{ }3z^2\mathrm{ }+\mathrm{ }3z\mathrm{ }-\mathrm{ }1$$

Dividing the numerator polynomial by the denominator polynomial we obtain an alternative form of X (z) as

$$\begin{array}{lll}X(z)\mathrm{ }=\mathrm{ }3z^{-1}\mathrm{ }+\mathrm{ }\frac{8z\mathrm{ }-\mathrm{ }9\mathrm{ }+\mathrm{ }3z^{-1}}{z^3\mathrm{ }-\mathrm{ }3z^2\mathrm{ }+\mathrm{ }3z\mathrm{ }-\mathrm{ }1}\mathrm{ }=\mathrm{ }3z^{-1\mathrm{ }}+\mathrm{ }\overline{X}(z)\hfill & \hfill & (8.208)\hfill \end{array}\mathrm{ }$$

Thus we have extracted one term of the power series representation of X (z), and it indicates that x [1] = 3. Repeating the process with $\overline{X}$ (z), we obtain

$$\begin{array}{ll}\overline{X}\left(z\right)=8z^{-2}+\frac{15-21z^{-1}+8z^{-2}}{z^3-3z^2+3z-}=8z^{-2}+\hat{X}\left(z\right)\hfill & \left(8.209\right)\hfill \end{array}$$

indicating that x [2] = 8. The results in Eqns. (8.208) and (8.209) can be combined, yielding

$$X(z)\mathrm{ }=\mathrm{ }3z^{-1}\mathrm{ }+\mathrm{ }\mathrm{ }8z^{-2}\mathrm{ }+\mathrm{ }\frac{15\mathrm{ }-\mathrm{ }21z^{-1}\mathrm{ }+\mathrm{ }8z^{-2}}{z^3\mathrm{ }-\mathrm{ }3z^2\mathrm{ }+\mathrm{ }3z\mathrm{ }-\mathrm{ }1}$$

Thus, each iteration through the long division operation produces one more sample of the signal x [n]. We are now ready to set up the long division:

Using the quotient and the remainder of the division, the transform X (z) can be written as

$$\begin{array}{lll}X(z)\mathrm{ }=\mathrm{ }3z^{-1}\mathrm{ }+\mathrm{ }\mathrm{ }8z^{-2}\mathrm{ }+\mathrm{ }15z^{-3\mathrm{ }}+\mathrm{ }24z^{-4}\mathrm{ }+\mathrm{ }\frac{24z^{-1}\mathrm{ }-\mathrm{ }37z^{-2}\mathrm{ }+\mathrm{ }15z^{-3}}{z^3\mathrm{ }-\mathrm{ }3z^2\mathrm{ }+\mathrm{ }3z\mathrm{ }-\mathrm{ }1}\hfill & \hfill & (8.210)\hfill \end{array}$$

Comparing the result in Eqn. (8.210) with the definition of the z-transform in Eqn. (8.1) we conclude that the signal x [n] is in the form

$$x[n]\mathrm{ }=\mathrm{ }\{\underset{n=1}{\overset{3}{\uparrow }},\mathrm{ }8,\mathrm{ }15,\mathrm{ }24,\mathrm{...}\}$$

The sample amplitudes obtained should be in agreement with those obtained by directly evaluating the values of the signal x[n] = n(n + 2) u[n] that led to the transform in question.

In Example 8.31 the use of the long division technique produced a causal signal as the inverse transform of the specified function X (z). That was fine since the ROC specified for the transform also supported this conclusion. What if we have a transform and associated ROC that indicate an anti-causal or a non-causal signal as the inverse transform? How would we need to modify the long division technique to produce the correct result in such a case?

Consider again a rational transform X (z) with one change from Eqn. (8.198): This time we will order the terms of numerator and denominatior polynomials in ascending powers of z. The result is

$$\begin{array}{lll}X(z)\mathrm{ }= \mathrm{ } \frac{B(Z)}{A(Z)}\mathrm{ }=\mathrm{ } \frac{b_0\mathrm{ }+b_{1}z\mathrm{ }\mathrm{ }+\mathrm{...}+\mathrm{ }\mathrm{ }{b}_{M-1}{z}^{M-1}\mathrm{ }+\mathrm{ } b_M{z}^M\mathrm{ }}{a_0\mathrm{ }+a_{1}z\mathrm{ }\mathrm{ }+\mathrm{...}+\mathrm{ }\mathrm{ }{a}_{N-1}{z}^{N-1}\mathrm{ }+\mathrm{ } a_N{z}^N}\mathrm{ }\hfill & \hfill & (8.211)\hfill \end{array}$$

If we now divide the numerator polynomial by the denominator polynomial, we can write X (z) as

$$\begin{array}{lll}X(z)\mathrm{ }= \mathrm{ } \left(\frac{b_0}{a_0}\right)\mathrm{ }+\mathrm{ } \frac{\overline{B}(z)\mathrm{ }}{A(z)}\mathrm{ }\hfill & \hfill & (8.212)\hfill \end{array}$$

The term (b0/a0) is the quotient of the division. The remainder is a polynomial $\tilde{B}\mathrm{ }(z)$ in the form

$$\begin{array}{lll}B(z)\mathrm{ }=\mathrm{ }\overline{b_1}\mathrm{ }z\mathrm{ }+\mathrm{ }\overline{b_2}\mathrm{ }{z}^2\mathrm{ }+\mathrm{ }\mathrm{...}\hfill & \hfill & (8.213)\hfill \end{array}$$

Associating the expression in Eqn. (8.202) with the power series form of the z-transform we recognize that

$$\begin{array}{lll}x[0]\mathrm{ }=\mathrm{ }\frac{b_0}{a_0}\hfill & \hfill & (8.214)\hfill \end{array}$$

We can now take the function

$$\begin{array}{ccc}\overline{X}(z)\mathrm{ }=\mathrm{ }\frac{\tilde{B}(z)}{A(z)}& & (8.215)\end{array}$$

and repeat the process, obtaining

$$\begin{array}{ccc}\overline{X}(z)\mathrm{ }=\mathrm{ }\left(\frac{\overline{b_1}}{a_0}\right)\mathrm{ }z\mathrm{ }+\mathrm{ }\frac{\hat{B}\mathrm{ }(z)}{A\mathrm{ }(z)}& & (8.216)\end{array}$$

which produces another sample of the signal x [n] as

$$\begin{array}{ccc}x[-1]\mathrm{ }=\mathrm{ }\frac{\overline{b_1}}{a_0}& & (8.217)\end{array}$$

and a new remainder polynomial

$$\begin{array}{ccc}\hat{B}(z)\mathrm{ }=\mathrm{ }\widehat{b_2}{z}^2\mathrm{ }+\mathrm{ }{\hat{b}}_3{z}^3\mathrm{ }+\mathrm{ }.\mathrm{ }.\mathrm{ }.& & \left(8.218\right)\end{array}$$

The next example will illustrate the use of long division with different types of ROCs.

Example 8.32: Using long division with specified ROC

In Example 8.29 we have used the partial fraction expansion method to determine the inverse z-transform of the rational function

$$X(z)\mathrm{ }=\mathrm{ }\frac{(z\mathrm{ }-\mathrm{ }1)\mathrm{ }(z\mathrm{ }+\mathrm{ }2)}{(z\mathrm{ }-\mathrm{ }\frac{1}{2})\mathrm{ }(z\mathrm{ }-\mathrm{ }2)}$$

for all possible choices of the ROC. Verify the results of that example by determining the first few samples of the inverse z-transform through the use of the long division method for each possible choice of the ROC.

Solution: Multiplying out the factors of the numerator and the denominator of X (z) we obtain

$$\begin{array}{ccc}X(z)\mathrm{ }=\mathrm{ }\frac{z^2\mathrm{ }+\mathrm{ }z\mathrm{ }-\mathrm{ }2}{z^2\mathrm{ }-\mathrm{ }2.5z\mathrm{ }+\mathrm{ }1}& & \left(8.219\right)\end{array}$$

As we have discussed in Example 8.30, there are three possible choices for the ROC:

Possibility 1: $\text{R}\text{O}\text{C}:\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }<\mathrm{ }\frac{1}{2}\mathrm{ }\mathrm{ }$

In this case the inverse transform x [n] must be anti-causal. In order to obtain an anti-causal solution from the long division, we will rewrite X (z) with its numerator and denominator polynomials arranged in ascending powers of z :

$$X(z)\mathrm{ }=\mathrm{ }\frac{-2\mathrm{ }+\mathrm{ }z\mathrm{ }+\mathrm{ }{z}^2}{1\mathrm{ }-\mathrm{ }2.5z\mathrm{ }+\mathrm{ }{z}^2}$$

and set up the long division:

Using the quotient and the remainder of the division, the transform X (z) can be written as

$$\begin{array}{ccc}X(z)\mathrm{ }=\mathrm{ }\mathrm{ }-2\mathrm{ }-4x\mathrm{ }-7z^2\mathrm{ }-13.5z^3\mathrm{ }+\mathrm{ }\frac{-26.75z^4\mathrm{ }+\mathrm{ }13.5z^5}{1\mathrm{ }-\mathrm{ }2.5z\mathrm{ }+\mathrm{ }{z}^2}& & \left(8.220\right)\end{array}$$

Comparing the result in Eqn. (8.220) with the z-transform definition in Eqn. (8.1) we conclude that the signal x [n] must be in the form

$$x[n]\mathrm{ }=\mathrm{ }\{\mathrm{...},\mathrm{ }-13.5,\mathrm{ }-7,\mathrm{ }-4,\mathrm{ }\underset{n=0}{\overset{-2}{\uparrow }}\}$$

consistent with the answer found in Example 8.30.

Possibility 2: ROC: |z| > 2

In this case the signal x [n] is causal. To obtain a causal answer we will use the original form of X (z) in Eqn. (8.219) with numerator and denominator polynomials arranged in descending orders of z. The long division is set up as follows:

Using the results obtained so far, the transform X (z) is expressed as

$$X\left(z\right)\mathrm{ }=\mathrm{ }1\mathrm{ }+\mathrm{ }3.5z^{-1}\mathrm{ }+\mathrm{ }5.75z^{-2}\mathrm{ }+\mathrm{ }\frac{10.875z^{-1}\mathrm{ }-\mathrm{ }5.75z^2}{z^2\mathrm{ }-\mathrm{ }2.5z\mathrm{ }+\mathrm{ }1}$$

The first few samples of the signal x [n] are

$$x\left[n\right]\mathrm{ }=\mathrm{ }\{\underset{n=0}{\overset{1}{\uparrow }},\mathrm{ }3.5,\mathrm{ }5.75,\mathrm{...}\}\mathrm{ }$$

identical to the result that was found in Example 8.30.

Possibility 3: $\text{R}\text{O}\text{C}:\mathrm{ }\frac{1}{2}\mathrm{ }<\mathrm{ }\mathrm{ }\left|z\right|\mathrm{ }<\mathrm{ }2\mathrm{ }\mathrm{ }\mathrm{ }$

In this case the inverse transform x[n] has a causal component and an anti-causal component, so it can be written in the form

$$x\left[n\right]\mathrm{ }=\mathrm{ }{x}_R\left[n\right]\mathrm{ }+\mathrm{ }{x}_L\mathrm{ }\left[n\right]$$

Accordingly, we need to partition the transform X (z) into two parts: One that corresponds to the causal component xR[n], and one that corresponds to the anti-causal component xL[n]. Recall that in Example 8.29 the transform X (z) was expressed through partial fractions as

$$X\left(z\right)\mathrm{ }=\mathrm{ }{X}_1\mathrm{ }\left(z\right)\mathrm{ }+\mathrm{ }{X}_2\mathrm{ }(z)\mathrm{ }+\mathrm{ }{X}_3\mathrm{ }\left(z\right)$$

with

$$\begin{array}{ccc}{X}_{1\mathrm{ }}(z)\mathrm{ }=\mathrm{ }-2,& \text{R}\text{O}\text{C}:& \text{all}\mathrm{ }z\\ X_2(z)\mathrm{ }=\mathrm{ }\frac{\frac{5}{3}\mathrm{ }z\mathrm{ }}{z\mathrm{ }-\mathrm{ }\frac{1}{2}},& \text{R}\text{O}\text{C}:& \left|z\right|\mathrm{ }>\mathrm{ }\frac{1}{2}\end{array}$$

and

$$\begin{array}{ccc}{X}_3(z)\mathrm{ }=\mathrm{ }\frac{\frac{4}{3}\mathrm{ }z\mathrm{ }}{z\mathrm{ }-\mathrm{ }2},& \text{R}\text{O}\text{C}:& \left|z\right|\mathrm{ }<\mathrm{ }2\end{array}$$

It is clear that X2 (z) should be included in XR (z), and X3 (z) should be included in XL (z). The constant term, X1 (z) = −2, could be included with either function; we will choose to include it with X R (z). Consequently we have

$$\begin{array}{ccc}{X}_R\left(z\right)\mathrm{ }=\mathrm{ }{X}_1\left(z\right)\mathrm{ }+\mathrm{ }{X}_2\mathrm{ }\left(z\right)\mathrm{ }=\mathrm{ }-2\mathrm{ }+\mathrm{ }\frac{\frac{5}{3}\mathrm{ }z}{z\mathrm{ }-\mathrm{ }\frac{1}{2}}\mathrm{ }=\mathrm{ }\frac{-\frac{1}{3}\mathrm{ }z\mathrm{ }+\mathrm{ }1}{z\mathrm{ }-\mathrm{ }\frac{1}{2}},& \text{R}\text{O}\text{C}:& \left|z\right|\mathrm{ }>\mathrm{ }\frac{1}{2}\end{array}$$

and

$$\begin{array}{ccc}{X}_L\left(z\right)\mathrm{ }=\mathrm{ }{X}_3\left(z\right)\mathrm{ }\mathrm{ }=\mathrm{ }\mathrm{ }\frac{\frac{4}{3}\mathrm{ }z}{z\mathrm{ }-\mathrm{ }2}\mathrm{ },& \text{R}\text{O}\text{C}:& \left|z\right|\mathrm{ }<\mathrm{ }2\end{array}$$

Let us begin with xR[n]. The long division for the causal component is set up using the numerator and the denominator of XR (z) arranged in descending powers of z:

We conclude that xR[0] = −1/3, xR[1] = 5/6 and xR[2] = 5/12. Next we will set up the long division for xL[z] with numerator and denominator polynomials arranged in ascending powers of z:

Thus, the first few samples of the anti-causal component of x[n] are xL[−1] = −2/3, xL[−2] = −1/3 and xL[−3] = −1/6. Combining the results of the two long divisions we obtain x[n] as

$$x\left[n\right]\mathrm{ }=\mathrm{ }\{\mathrm{...},\mathrm{ }-\frac{1}{6},\mathrm{ }-\frac{1}{3},\mathrm{ }-\frac{2}{3},\mathrm{ }-\frac{1}{\underset{n=0}{\overset{3}{\uparrow }}},\mathrm{ }\frac{5}{6},\mathrm{ }\frac{5}{12},\mathrm{...}\}$$

8.5 Using the z-Transform with DTLTI Systems

In Chapter 3 we have demonstrated that the output signal of a DTLTI system is related to its input signal and its impulse response through the convolution sum. Specifically, if the impulse response of a DTLTI system is h [n], and if the signal x [n] is applied to the system as input, the output signal y [n] is computed as

$$y\left[n\right]\mathrm{ }=\mathrm{ }x\left[n\right]\mathrm{ }*\mathrm{ }h\left[n\right]\mathrm{ }=\mathrm{ }\sum _{k=-\infty }^{\infty }x\left[k\right]\mathrm{ }h\left[n\mathrm{ }-\mathrm{ }k\right]$$

Based on the convolution property of the z-transform introduced by Eqn. (8.127) in Section 8.3.6, the z-transform of the output signal is equal to the product of the z-transforms of the input signal and the impulse response, that is,

$$\begin{array}{ccc}Y\left(z\right)\mathrm{ }=\mathrm{ }X\left(z\right)H\left(z\right)\mathrm{ }& & \left(8.221\right)\end{array}$$

If the input signal x [n] and the impulse response h [n] are specified, we could find the z-transforms X (z) and H (z), multiply them to obtain Y (z), and then determine y [n] by means of inverse z-transform. This process provides us an alternative to computing the output by direct application of the convolution sum, and was demonstrated in Examples 8.23 and 8.24.

Alternatively, if the input and the output signals x [n] and y [n] are specified, their z-transforms can be computed, and then the z-transform of the impulse response can be determined from them. Solving Eqn. (8.221) for H (z) we get

$$\begin{array}{ccc}H\left(z\right)\mathrm{ }=\mathrm{ }\frac{Y\mathrm{ }\left(z\right)}{X\left(z\right)}& & \left(8.222\right)\end{array}$$

The function H (z) is the z-domain system function of the system. We already know that a DTLTI system can be described fully and uniquely by means of its impulse response h [n]. Since the system function H (z) is just the z-transform of the impulse response h [n], it also represents a complete description of the DTLTI system.

8.5.1 Relating the system function to the difference equation

In Section 3.4 of Chapter 3 we have established the fact that a DTLTI system can be described by means of a constant-coefficient linear difference equation in the standard form

$$\begin{array}{ccc}\sum _{k=0}^N{a}_k\mathrm{ }y\left[n\mathrm{ }-\mathrm{ }k\right]\mathrm{ }=\mathrm{ }\sum _{k=0}^M{b}_k\mathrm{ }x\left[n\mathrm{ }-\mathrm{ }k\right]& & \left(8.223\right)\end{array}$$

Therefore it must be possible to obtain the other two forms of description of the DTLTI system, namely the system function and the impulse response, from the knowledge of its difference equation. If we take the z-transform of both sides of Eqn. (8.223) the equality would still be valid:

$$\begin{array}{ccc}Ƶ\left\{\sum _{k=0}^N{a}_k\mathrm{ }y\left[n\mathrm{ }-\mathrm{ }k\right]\right\}\mathrm{ }=\mathrm{ }Ƶ\mathrm{ }\left\{\sum _{k=0}^M{b}_k\mathrm{ }x\left[n\mathrm{ }-\mathrm{ }k\right]\right\}& & \left(8.224\right)\end{array}$$

Using the linearity property of the z-transform, Eqn. (8.224) can be written as

$$\begin{array}{ccc}\sum _{k=0}^N{a}_k\mathrm{ }Ƶ\left\{y[n\mathrm{ }-\mathrm{ }k]\right\}\mathrm{ }=\mathrm{ }\sum _{k=0}^M{b}_kƵ\mathrm{ }\left\{x[n\mathrm{ }-\mathrm{ }k]\right\}& & \left(8.225\right)\end{array}$$

and using the time-shifting property of the z-transform leads to

$$\begin{array}{ccc}\sum _{k=0}^N{a}_k{z}^{-k}Y\left(z\right)\mathrm{ }=\mathrm{ }\sum _{k=0}^M{b}_k{z}^{-k}\mathrm{ }X\left(z\right)& & \left(8.226\right)\end{array}$$

The transforms X (z) and Y (z) do not depend on the summation index k, and can therefore be factored out of the summations in Eqn. (8.226) resulting in

$$Y\left(z\right)\begin{array}{ccc}\sum _{k=0}^N{a}_k{z}^{-k}\mathrm{ }=\mathrm{ }X\mathrm{ }\left(z\right)\mathrm{ }\sum _{k=0}^M{b}_k{z}^{-k}& & \left(8.227\right)\end{array}$$

Finally, the system function can be obtained from Eqn. (8.227) as

$$H\mathrm{ }(z)\mathrm{ }=\mathrm{ }\frac{Y(z)}{X(z)}\mathrm{ }=\mathrm{ }\frac{{\displaystyle \sum _{k=0}^M}{b}_k{z}^{-k}}{{\displaystyle \sum _{K=0}^N}{a}_k{z}^{-k}}\begin{array}{ccc}\mathrm{ }\mathrm{ }& & (8.228)\end{array}\mathrm{ }$$

At this point we will make two important observations:

1. In the above development leading up to the z-domain system function we have relied on the convolution operation in Eqn. (8.221). In Eqn. (8.228) we have used the convolution property of the z-transform. We know from Chapter 3 that the convolution operation is only applicable to problems involving linear and time-invariant systems. Therefore the system function concept is meaningful only for systems that are both linear and time-invariant. This notion was introduced in earlier discussions involving system functions as well.
2. Furthermore, it was justified in Section 3.4 of Chapter 3 that a constant-coefficient difference equation corresponds to a linear and time-invariant system only if all initial conditions are set to zero.

If we need to use z-transform based techniques to solve a difference equation subject to non-zero initial conditions, we can do that through the use of the unilateral z-transform, but not through the use of the system function. The unilateral z-transform and its use for solving difference equations will be discussed in Section 8.7.

Example 8.33: Finding the system function from the difference equation

Let a DTLTI system be defined by the difference equation

$$y[n]\mathrm{ }\mathrm{ }-\mathrm{ }0.4y[n\mathrm{ }-\mathrm{ }1]\mathrm{ }\mathrm{ }+\mathrm{ }0.89y[n\mathrm{ }-\mathrm{ }2]\mathrm{ }=\mathrm{ }x[n]\mathrm{ }-\mathrm{ }x[n\mathrm{ }-\mathrm{ }1]$$

Find the system function H (z) for this system.

Solution: We will assume zero initial conditions and take the z-transforms of both sides of the difference equation to get

$$Y\left(z\right)\mathrm{ }-\mathrm{ }0.4z{\mathrm{ }}^{-1}\mathrm{ }Y\mathrm{ }\left(z\right)\mathrm{ }+\mathrm{ }0.89z^{-2}Y\mathrm{ }\left(z\right)\mathrm{ }=\mathrm{ }X\mathrm{ }\left(z\right)\mathrm{ }-\mathrm{ }{z}^{-1}\mathrm{ }X\left(z\right)$$

from which the system function is obtained as

$$H\left(z\right)\mathrm{ }=\mathrm{ }\frac{1\mathrm{ }-\mathrm{ }{z}^{-1}}{1\mathrm{ }-\mathrm{ }0.4z^{-1}\mathrm{ }+\mathrm{ }0.89z^{-2}}$$

or, using non-negative powers of z

$$H\left(z\right)\mathrm{ }=\mathrm{ }\frac{z(z\mathrm{ }-\mathrm{ }1)}{z^2\mathrm{ }-\mathrm{ }0.4z\mathrm{ }+\mathrm{ }0.89}$$

We will make another important observation based on the result obtained in Example 8.33: The characteristic equation for the system considered in Example 8.33 is

$$z^{2\mathrm{ }}-\mathrm{ }0.4z\mathrm{ }+\mathrm{ }0.89\mathrm{ }=\mathrm{ }0$$

and the solutions of the characteristic equation are the modes of the system as defined in Section 3.5.1 of Chapter 3. When we find the system function H (z) from the difference equation we see that its denominator polynomial is identical to the characteristic polynomial. The roots of the denominator polynomial are the poles of the system function in the z-domain, and consequently, they are identical to the modes of the difference equation of the system.

Recall that in Section 3.5.1 we have reached some conclusions about the relationship between the modes of the difference equation and the natural response of the corresponding system. The same conclusions would apply to the poles of the system function. Specifically, if all poles pk of the system are real-valued and distinct, then the natural response is in the form

$$y\left[n\right]\mathrm{ }=\mathrm{ }\sum _{k\mathrm{ }=\mathrm{ }1}^N{c}_k{p}_k^n$$

Complex poles appear in conjugate pairs provided that all coefficients of the system function are real-valued. A pair of complex conjugate poles

$$\begin{array}{ccc}{p}_{1a}\mathrm{ }=\mathrm{ }{r}_1\mathrm{ }{e}^{j\Omega 1}\mathrm{ },& \text{and}& p_{1b}\mathrm{ }\mathrm{ }=\mathrm{ }{r}_1\mathrm{ }{e}^{-j\Omega 1}\end{array}$$

yields a response of the type

$$y_1\left[n\right]=d_1{r}_1^{n}cos\left({\Omega }_{1}n+{\theta }_1\right)$$

Finally, a pole of multiplicity m at z = p1 leads to a response in the form

$$y_1\left[n\right]=c_{11}{p}_1^n+c_{12}n{p}_1^n+...+c_m{n}^{m-1}{p}_1^n+\text{other terms}$$

Justifications for these relationships were given in Section 3.5.1 of Chapter 3 through the use of the modes of the difference equation, and will not be repeated here.

Sometimes we need to reverse the problem represented in Example 8.33 and find the difference equation from the knowledge of the system function. The next example will demonstrate this.

Example 8.34: Finding the difference equation from the system function

Find the difference equation of the DTLTI system defined by the system function

$$H\left(z\right)=\frac{z^2-5z+6}{z^3+2z^2-z-2}$$

Solution: Let us first write H(z) as a rational function of z− by multiplying both its numerator and denominator with z −3:

$$H\left(z\right)=\frac{z^{-1}-5z^{-2}+6z^{-3}}{1+2z^{-1}-z^{-2}-2z^{-3}}$$

We know that the system function is the ratio of the output transform to the input transform, that is,

$$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}$$

so that we can write

$$\left(1+2z^{-1}-z^{-2}-2z^{-3}\right)Y\left(z\right)=\left(z^{-1}-5z^{-2}+6z^{-3}\right)X\left(z\right)$$

or equivalently

$$y\left[n\right]+2y\left[n-1\right]-y\left[n-2\right]-2y\left[n-3\right]=x\left[n-1\right]-5x\left[n-2\right]+6x\left[n-3\right]$$

8.5.2 Response of a DTLTI system to complex exponential signal

An interesting interpretation of the system function concept is obtained when one considers the response of a DTLTI system to a complex exponential signal in the form

$$x\left[n\right]=z_0^n$$

where z0 represents a point in the z-plane within the ROC of the system function. Let h[n] be the impulse response of the DTLTI system under consideration. The output signal is determined through the use of the convolution sum

$$\begin{array}{lll}y\left[n\right]=h\left[n\right]*x\left[n\right]={\displaystyle \sum _{k=-\infty }^{\infty }}h\left[k\right]x\left[n-k\right]\hfill & \hfill & \left(8.229\right)\hfill \end{array}$$

Substituting $x\left[n-k\right]=z_0^{n-k}$ into Eqn. (8.229) and simplifying the resulting summation we obtain

$$\begin{array}{llll}y\left[n\right]\hfill & ={\displaystyle \sum _{k=-\infty }^{\infty }}h\left[k\right]z_0^{n-k}\hfill & \hfill & \hfill \\ \hfill & =z_0^n{\displaystyle \sum _{k=-\infty }^{\infty }}h\left[k\right]z_0^{n-k}\hfill & \hfill & \left(8.230\right)\hfill \end{array}$$

In Eqn. (8.230) the summation corresponds to the system function H(z) evaluated at the point z = z0, so the response of the DTLTI system to the input signal $x\left[n\right]=z_0^n$ is

$$\begin{array}{lll}y\left[n\right]=z_0^{n}H\left(z_0\right)\hfill & \hfill & \left(8.231\right)\hfill \end{array}$$

We conclude that the response of a DTLTI system to the exponential signal $x\left[n\right]=z_0^n$ is a scaled version of the input signal. The scale factor is the value of the system function at the point z = z0.

Example 8.35: Response of a DTLTI system to a complex exponential signal

Consider a DTLTI system with the system function

$$H(z)=\frac{z+1}{z^2-\frac{5}{6}z+\frac{1}{6}}$$

Find the response of the system to the exponential signal

$$x\left[n\right]={\left(0.9e^{j0.2\pi }\right)}^n$$

Solution: The input signal is complex-valued. Using Euler’s formula we can write x [n]as

$$x\left[n\right]={\left(0.9\right)}^{n}cos\left(0.2\pi n\right)+j{\left(0.9\right)}^{n}sin\left(0.2\pi n\right)$$

Real and imaginary parts of x [n] are shown in Fig. 8.29. The value of the system function at z = 0.9ej0.2π is

$$\begin{array}{ll}H\left(0.9e^{j0.2\pi }\right)\hfill & =\frac{0.9e^{j0.2\pi }+1}{{\left(0.9e^{j0.2\pi }\right)}^2-\frac{5}{6}\left(0.9e^{j0.2\pi }\right)+\frac{1}{6}}\hfill \\ \hfill & =-1.0627-j4.6323\hfill \\ \hfill & =4.7526e^{-j1.7963}\hfill \end{array}$$

The output signal is found using Eqn. (8.231) as

$$\begin{array}{ll}y[n]\hfill & ={(0.9e^{j0.2\pi })}^n(4.7526 e^{-j1.7963})\hfill \\ \hfill & =4.7526{(0.9)}^n{e}^{j(0.2\pi n-1.7963)}\hfill \end{array}$$

or, in Cartesian form

$$y\left[n\right]=4.7526{\left(0.9\right)}^{n}cos\left(0.2\pi n-1.7963\right)+j4.7526{\left(0.9\right)}^{n}sin\left(0.2\pi n-1.7963\right)$$

Real and imaginary parts of the output signal y[n]are shown in Fig. 8.30.

Software resources:

ex_8_35a.m

ex_8_35b.m

8.5.3 Response of a DTLTI system to exponentially damped sinusoid

Next we will consider an input signal in the form of an exponentially damped sinusoid such as

$$\begin{array}{lll}x\left[n\right]=r_0^{n}cos\left({\Omega }_{0}n\right)\hfill & \hfill & \left(8.232\right)\hfill \end{array}$$

In order to find the response of a DTLTI system to this signal, let us write x [n] using Euler’s formula:

$$\begin{array}{lll}x\left[n\right]=\frac{1}{2}{r}_0^n{e}^{j{\Omega }_{0n}}+\frac{1}{2}{r}_0^n{e}^{-j{\Omega }_{0n}}\hfill & \hfill & \left(8.233\right)\hfill \end{array}$$

Let z0 be defined as

$$z_0=r_0{e}^{j{\Omega }_0}$$

It follows that

$$z_0^{*}=r_0{e}^{-j{\Omega }_0}$$

and Eqn. (8.233) can be written as

$$\begin{array}{lll}x\left[n\right]=\frac{1}{2}{z}_0^n+\frac{1}{2}{\left(z_0^{*}\right)}^n\hfill & \hfill & \left(8.234\right)\hfill \end{array}$$

The response of the system is determined using the linearity of the system:

$$\begin{array}{llll}y\left[n\right]\hfill & =\text{sys}\left\{\frac{1}{2}{z}_0^n+\frac{1}{2}{\left(z_0^{*}\right)}^n\right\}\hfill & \hfill & \hfill \\ \hfill & =\frac{1}{2}\text{sys}\left\{z_0^n\right\}+\frac{1}{2}\text{sys}\left\{{\left(z_0^{*}\right)}^n\right\}\hfill & \hfill & \left(8.235\right)\hfill \end{array}$$

We already know that the response of the system to the term $z_0^n$ is

$$\begin{array}{lll}\text{sys}\left\{z_0^n\right\}=z_0^{n}H\left(z_0\right)\hfill & \hfill & \left(8.236\right)\hfill \end{array}$$

and its response to the term ${\left(z_n^{*}\right)}^n$ is

$$\begin{array}{lll}\text{sys}\left\{{\left(z_0^{*}\right)}^n\right\}={\left(z_0^{*}\right)}^{n}H\left(z_0^{*}\right)\hfill & \hfill & \left(8.237\right)\hfill \end{array}$$

so that the output signal y [n] can be written as

$$\begin{array}{lll}y\left[n\right]=\frac{1}{2}{z}_0^{n}H\left(z_o\right)+\frac{1}{2}{\left(z_0^{*}\right)}^{n}H\left(z_0^{*}\right)\hfill & \hfill & \left(8.238\right)\hfill \end{array}$$

The result obtained in Eqn. (8.238) can be further simplified. Let the value of the system function evaluated at the point z = z0 be written in polar complex form as

$$\begin{array}{lll}{H}_{\left(z0\right)}=H_0{e}^{j{\Theta }_0}\hfill & \hfill & \left(8.239\right)\hfill \end{array}$$

where H0 and Θ0 represent the magnitude and the phase of the system function at the point z = z 0 respectively:

$$\begin{array}{lll}{H}_0=\left|H_{\left(z0\right)}\right|\hfill & \hfill & \left(8.240\right)\hfill \end{array}$$

and

$$\begin{array}{lll}{\Theta }_0=\Theta H(z_0)\hfill & \hfill & (8.241)\hfill \end{array}$$

It can be shown (see Problem 8.34 at the end of this chapter) that, for a system with a real-valued impulse response, the value of the system function at the point $z=z_0^{*}$ is the complex conjugate of its value at the point z = z0,thatis,

$$\begin{array}{lll}H\left(z_0^{*}\right)={\left[H\left(z_0\right)\right]}^{*}=H_0{e}^{-j{\Theta }_0}\hfill & \hfill & \left(8.242\right)\hfill \end{array}$$

Using Eqns. (8.239) and (8.242) in Eqn. (8.238), the output signal y [n] becomes

$$\begin{array}{llll}y\left[n\right]\hfill & =\frac{1}{2}{z}_0^n{H}_0{e}^{j{\Theta }_0}+\frac{1}{2}{\left(z_0^{*}\right)}^n{H}_0{e}^{-j{\Theta }_0}\hfill & \hfill & \hfill \\ \hfill & =\frac{1}{2}{\left(r_0{e}^{j{\Omega }_0}\right)}^n{H}_0{e}^{j{\Theta }_0}+\frac{1}{2}{\left(r_0{e}^{-j{\Omega }_0}\right)}^n{H}_0{e}^{-j{\Theta }_0}\hfill & \hfill & \hfill \\ \hfill & =\frac{1}{2}{r}_0^n{H}_0\left[e^{j\left({\Omega }_{0}n+{\Theta }_0\right)+}{e}^{-j\left({\Omega }_{0}n+{\Theta }_0\right)}\right]\hfill & \hfill & \hfill \\ \hfill & =H_0{r}_0^{n}cos\left({\Omega }_{0}n+{\Theta }_0\right)\hfill & \hfill & \left(8.243\right)\hfill \end{array}$$

Comparison of the input signal in Eqn. (8.232) and the output signal Eqn. (8.243) reveals the following:

1. The amplitude of the signal is multiplied by the magnitude of the system function evaluated at the point z = r 0ejΩ.
2. The phase of the cosine function is incremented by an amount equal to the phase of the system function evaluated at the point z = r0ejΩ.

Example 8.36: Response of a DTLTI system to an exponentially damped sinusoid

Consider again the system function used in Example 8.35. Determine the response of the system to the input signal

$$x\left[n\right]={\left(0.9\right)}^{n}cos\left(0.2\pi n\right)$$

Solution: In Example 8.35 we have evaluated the system function at the point z = 0.9ej0.2π and have obtained

$$\begin{array}{lll}H\left(0.9e^{j0.2\pi }\right)=H_0{e}^{j{\Theta }_0}=4.7526e^{-j1.7963}\hfill & \hfill & \left(8.244\right)\hfill \end{array}$$

Using Eqn. (8.243) with H0 = 4.7526 and Θ0 = −1.7963 radians, we obtain the output signal as

$$y\left[n\right]=4.7526{\left(0.9\right)}^{n}cos\left(0.2\pi n-1.7963\right)$$

The input and the output signals are shown in Fig. 8.31.

Software resources:

ex_8_36.m

8.5.4 Graphical interpretation of the pole-zero plot

It was discussed earlier in this chapter that the complex variable z can be represented as a point in the z-plane. Alternatively, a complex number can be thought of as a vector in the complex plane (see Appendix A), drawn from the origin to the point of interest. Let z = rejΩ. Fig. 8.32 illustrates the use of a vector with norm (or length) equal to r and angle equal to Ω for representing z.

Using vector notation we will write

$$\begin{array}{lll}\overrightarrow{z}=\overrightarrow{r{e}^{j\Omega }}\hfill & \hfill & \left(8.245\right)\hfill \end{array}$$

The norm and the angle of the vector $\left|\overrightarrow{z}\right|$ are expressed as

$$\begin{array}{lll}\overrightarrow{\left|z\right|}=r\hfill & \hfill & \left(8.246\right)\hfill \end{array}$$

and

$$\measuredangle \overrightarrow{z}=\Omega$$

Fixing Ω and changing the value of r causes the tip of the vector to move either toward or away from the origin while the direction of the vector remains unchanged. Fixing r and changing the value of Ω causes the vector to rotate while its length remains unchanged. In this case the tip of the vector stays on a circle with radius r.

Consider a fixed point z 1 in the z-plane. It will be interesting to find the vector $\left(z\overrightarrow{-}{z}_1\right)$.

$$\begin{array}{llll}z\overrightarrow{-}{z}_1\hfill & =\overrightarrow{z}-\overrightarrow{z_1}\hfill & \hfill & \hfill \\ \hfill & =\overrightarrow{z}+\left(\overrightarrow{-z_1}\right)\hfill & \hfill & \left(8.248\right)\hfill \end{array}$$

Based on Eqn. (8.248) the vector $\left(z\overrightarrow{-}{z}_1\right)$ can be found by adding the two vectors and $\overrightarrow{z}$ using $\left(-\overrightarrow{z_1}\right)$ the parallelogram rule as shown in Fig. 8.33.

Vector representation of the factors that make up the system function H (z)is particularly useful for inferring the frequency response of a system from the distribution of its z-domain poles and zeros. It was shown earlier that the DTFT-based frequency response H (ω) of a DTLTI system can be obtained from the z-domain system function by evaluating H (z) at each point on the unit circle of the z-plane. Continuing with the first-order system function given by Eqn. (8.250) we obtain

$$\begin{array}{lll}{H(\Omega )=H\left(z\right)|}_{z=e^{j\Omega }}=\frac{e^{j\Omega }}{e^{j\Omega }-a}\hfill & \hfill & \left(8.255\right)\hfill \end{array}$$

In using Eqn. (8.255) we are assuming that |a| < 1 so that the ROC for the system function includes the unit circle, and therefore the frequency response H (Ω) of the system exists. Fig. 8.37 shows the two vectors

$$\begin{array}{lll}\overrightarrow{A}=\overrightarrow{e^{j\Omega }}\hfill & \hfill & \left(8.256\right)\hfill \end{array}$$

and

$$\begin{array}{lll}\overrightarrow{B}=\left(\overrightarrow{e^{j\Omega }}-a\right)\hfill & \hfill & \left(8.257\right)\hfill \end{array}$$

that are needed for evaluating H (Ω). The vectors originate from the zero and the pole and terminate at the point z = ejΩ on the unit circle.

As Ω is varied, the termination point moves on the unit circle, causing the lengths and the angles of the vectors to change, and therefore causing the frequency response H (Ω) to vary as a function of Ω.

Now consider a more general system function in the form

$$\begin{array}{lll}H\left(z\right)=K\frac{\left(z-z_1\right)\left(z-z_2\right)...\left(z-z_M\right)}{\left(z-p_1\right)\left(z-p_2\right)...\left(z-p_N\right)}\hfill & \hfill & \left(8.258\right)\hfill \end{array}$$

The system has M zeros at z 1, z2,...,zM and N poles at p1,p2,...,pN. Suppose we would like to evaluate the system function at the point z = za within the ROC. Vector representation of H(za)is

$$\begin{array}{ccc}\overrightarrow{H\left(z_a\right)}=k\frac{\left(z_a\overrightarrow{-}{z}_1\right)\left(z_a\overrightarrow{-}{z}_2\right)\cdots \left(z_a\overrightarrow{-}{z}_M\right)}{\left(z_a\overrightarrow{-}{p}_1\right)\left(z_a\overrightarrow{-}{p}_2\right)\cdots \left(z_a\overrightarrow{-}{p}_N\right)}& & \left(8.259\right)\end{array}$$

The magnitude of the system function is

$$\begin{array}{ccc}\left|H\left(z_a\right)\right|=K\frac{\left|\left(z_a\overrightarrow{-}{z}_1\right)\right|\left|\left(z_a\overrightarrow{-}{z}_2\right)\right|\cdots \left|\left(z_a\overrightarrow{-}{z}_M\right)\right|}{\left|\left(z_a\overrightarrow{-}{p}_1\right)\right|\left|\left(z_a\overrightarrow{-}{p}_2\right)\right|\cdots \left|\left(z_a\overrightarrow{-}{p}_N\right)\right|}& & \left(8.260\right)\end{array}$$

and its phase is

$$\begin{array}{ccc}\measuredangle H(z_a)=\measuredangle (z_a\overrightarrow{-}{z}_1)+\measuredangle (z_a\overrightarrow{-}{z}_2)+\cdots +\measuredangle (z_a\overrightarrow{-}{z}_M)-\measuredangle (z_a\overrightarrow{-}{p}_1)-\measuredangle (z_a\overrightarrow{-}{p}_1)-\cdots -\measuredangle (z_a\overrightarrow{-}{p}_N)& & \left(8.261\right)\end{array}$$

The vector-based graphical method described above is useful for understanding the correlation between pole-zero placement and system behavior. In order to build intuition in this regard, Fig. 8.38 illustrates several pole-zero layouts along with the magnitude spectrum that corresponds to each layout.

8.5.5 System function and causality

Causality in linear and time-invariant systems was discussed in Section 3.8 of Chapter 3. For a DTLTI system to be causal, its impulse response h [n] needs to be equal to zero for n 0. Thus, by changing the lower limit of the summation index to n = 0 in the definition of the z-transform, the z-domain system function for a causal DTLTI system can be written as

$$\begin{array}{lll}H\left(z\right)={\displaystyle \sum _{n=-\infty }^{\infty }}h\left[n\right]z^{-n}={\displaystyle \sum _{n=0}^{\infty }}h\left[n\right]z^{-n}\hfill & \hfill & \left(8.262\right)\hfill \end{array}$$

As discussed in Section 8.2 of this chapter, the ROC for the system function of a causal system is the outside of a circle in the z-plane. Consequently, the system function must also converge at |z| → ∞. Consider a system function in the form

$$H\left(z\right)=\frac{B\left(N\right)}{A\left(z\right)}=\frac{b_M{z}^M+b_{M-1}{z}^{M-1}+...b_{1}z+b_0}{a_N{z}^N+a_{N-1}{z}^{N-1}+...a_{1}z+a_0}$$

For the system described by H(z) tobecausal weneed

$$\begin{array}{lll}\underset{z\to \infty }{lim}\left[H\left(z\right)\right]=\underset{z\to \infty }{lim}\left[\frac{b_M}{a_N}{z}^{\left(M-N\right)}\right]<\infty \hfill & \hfill & \left(8.263\right)\hfill \end{array}$$

which requires that M − N ≤ 0 and consequently M ≤ N. Thus we arrive at an important conclusion:

Note that this condition is necessary for a system to be causal, but it is not sufficient. It is also possible for a non-causal system to have a system function with M ≤ N.

8.5.6 System function and stability

In Section 3.9 of Chapter 3 we concluded that for a DTLTI system to be stable its impulse response must be absolute summable, that is,

$${\displaystyle \sum _{n=-\infty }^{\infty }}\left|h\left[n\right]\right|<\infty$$

Furthermore, we have established in Section 5.3.3 of Chapter 5 that the DTFT of a signal exists if the signal is absolute summable. But the DTFT of the impulse response is equal to the z-domain transfer function evaluated on the unit circle of the z-plane, that is,

$${H(\Omega )=H\left(z\right)|}_{z=e^{j\Omega }}$$

provided that the unit circle of the z-plane is within the ROC.

What are the corresponding conditions that must be imposed on the locations of poles and zeros for stability? We will answer this question by taking three separate cases into account:

1. Causal system:

   The ROC for the system function of a causal system is the outside of a circle with radius r1, and is expressed in the form

   $$\left|z\right|>r_1$$

   For the ROC to include the unit circle, we need r1 < 1. Since the ROC can not have any poles in it, all the poles of the system function must be on or inside the circle with radius r1.

   For a causal system to be stable, the system function must not have any poles on or outside the unit circle of the z-plane.

2. Anti-causal system:

   If the system is anti-causal, its impulse response is equal to zero for n ≤ 0 and the ROC for the system function is the inside of a circle with radius r2, expressed in the form

   $$\left|z\right|<r_2$$

   For the ROC to include the unit circle, we need r2 > 1. Also, all the poles of the system function must reside on or outside the circle with radius r2 since there can be no poles within the ROC.

   For an anti-causal system to be stable, the system function must not have any poles on or inside the unit circle of the z-plane.

3. Neither causal nor anti-causal system:

   In this case the ROC for the system function, if it exists, is the region between two circles with radii r1 and r2, and is expressed in the form

   $$r_1<\left|z\right|<r_2$$

   The poles of the system function may be either

   1. On or inside the circle with radius r1
   2. On or outside the circle with radius r2
      and the ROC must include the unit circle.

Example 8.38: Impulse response of a stable system

A stable system is characterized by the system function

$$H\left(z\right)=\frac{z\left(z+1\right)}{\left(z-0.8\right)\left(z+1.2\right)\left(z-2\right)}$$

Determine the impulse response of the system.

Solution: The ROC for the system function is not directly stated; however, we are given enough information to deduce it. The poles of the system are at p1 = 0.8, p2 = − 1. 2 and p3 = 2. Since the system is known to be stable, its ROC must include the unit circle. The only possible choice is

$$0.8<\left|z\right|<1.2$$

Partial fraction expansion of H(z)is

$$H\left(z\right)=-\frac{0.75z}{z-0.8}-\frac{0.0312z}{z+1.2}+\frac{0.7813z}{z-2}$$

Based on the ROC determined above, the first partial fraction corresponds to a causal signal, and the other two correspond to anti-causal signals. The impulse response of the system is

$$h\left[n\right]=-0.75{\left(0.8\right)}^{n}u\left[n\right]+0.0312{\left(-1.2\right)}^{n}u\left[-n-1\right]-0.7813{\left(2\right)}^{n}u\left[-n-1\right]$$

which is shown in Fig. 8.39. We observe that h[n] tends to zero as the sample index is increased in both directions, consistent with the fact that h[n] must be absolute summable.

Software resources:

ex_8_38a.m

ex_8_38b.m

Example 8.39: Stability of a system described by a difference equation

A DTLTI system is characterized by the difference equation

$$y\left[n\right]=x\left[n-1\right]+3x\left[n-2\right]+2x\left[n-3\right]+2.3y\left[n-1\right]-2y\left[n-2\right]+1.2y\left[n-3\right]$$

Solution: As in the previous example, the ROC for the system function is not directly stated. On the other hand, the difference equation of the system is written with the current output sample y [n] on the left of the equal sign. All terms on the right side of the equal sign are past samples of the input and the output signals. The form of the difference equation suggests that the system is causal. When we find the system function we must choose the ROC in a way that includes infinitely large values of |z|.

Rearranging terms of the difference equation we obtain

$$y\left[n\right]=-2.3y\left[n-1\right]+2y\left[n-2\right]-1.2y\left[n-3\right]=x\left[n-1\right]+3x\left[n-2\right]+2x\left[n-3\right]$$

Taking the z-transform of each side yields

$$\left(1-2.3z^{-1}+2z^{-2}-1.2z^{-3}\right)Y\left(z\right)=\left(z^{-1}+3z^{-2}+2z^{-3}\right)X\left(z\right)$$

Finally, solving for the ratio of Y (z) and X(z) we find the system function as

$$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=\frac{z^{-1}+3z^{-2}+2z^{-3}}{1-2.3z^{-1}+2z^{-2}-1.2z^{-3}}$$

Let us multiply both the numerator and the denominator of H (z)by z3 so that we express it using powers of z :

$$H\left(z\right)=\frac{z^2+3x+2}{x^3-2.3z^2+2z-1.2}$$

Numerator and denominator polynomials can be factored to yield

$$H\left(z\right)=\frac{\left(z+1\right)\left(z+2\right)}{\left(z-0.4+j0.8\right)\left(z-0.4-j0.8\right)\left(z-1.5\right)}$$

The pole-zero diagram for the system function is shown in Fig. 8.40. There are two zeros, one at z = − 1 and one at z = − 2. The system function also has a pole at z = 1.5 and a conjugate pair of poles at z = 0.4 ± j 0. 8.

Since the system is known to be causal, its ROC must be the outside of a circle. The radius of the boundary circle is determined by the outermost pole of the system which happens to be at z = 1.5. Consequently, the ROC for the system function is

$$\left|z\right|>1.5$$

Since the ROC does not include the unit circle, the system is unstable. This can be verified by computing the first few samples of the impulse response h [n] from the system function using long division which yields

$$h\left[n\right]=\left\{\underset{\stackrel{\uparrow }{n=1}}{1},5.3,12.19,18.64,24.85,34.5,52.02,80.46,122.42,183.07,\mathrm{...}\right\}$$

The impulse response keeps growing with increasing values of the index n.

Software resources:

ex_8_39a.m

ex_8_39b.m

ex_8_39c.m

8.5.7 Allpass systems

A system the magnitude characteristic of which is constant across all frequencies is called an allpass system. For a system to be considered an allpass system we need

$$\begin{array}{lll}|H(\Omega )|=C(\text{constant}) -\pi \le \Omega \le \pi \hfill & \hfill & (8.264)\hfill \end{array}$$

Consider a first-order DTLTI system with a complex zero at

$$\begin{array}{lll}{z}_1=a=r{e}^{j{\Omega }_0}\hfill & \hfill & (8.265)\hfill \end{array}$$

and a complex pole at

$$\begin{array}{lll}{p}_1=\frac{1}{a^{*}}=\frac{1}{r{e}^{-j{\Omega }_0}}=\frac{1}{r}{e}^{j{\Omega }_0}\hfill & \hfill & (8.266)\hfill \end{array}$$

The resulting system function is

$$\begin{array}{lll}H(z)=\frac{z-z_1}{z-p_1}=\frac{z-r{e}^{j{\Omega }_0}}{z-(1/r)e^{j{\Omega }_0}}\hfill & \hfill & (8.267)\hfill \end{array}$$

The choice made for the zero and the pole ensures that the two lie on a line that goes through the origin. Their distances from the origin are r and 1/r respectively. Since it is desirable to have a system that is both causal and stable, we need to choose r>1 so that the resulting pole is inside the unit circle. Naturally this choice causes the zero to be outside the unit circle. This is illustrated in Fig. 8.41.

A pole and a zero that have the relationship shown in Fig. 8.41 are said to be mirror images of each other across the unit circle. To find H (Ω) we will evaluate the z-domain system function on the unit circle of the z-plane:

$$\begin{array}{lll}H(\Omega )=\frac{e^{j\Omega }-r{e}^{j{\Omega }_0}}{e^{j\Omega }-(1/r)e^{j{\Omega }_0}}\hfill & \hfill & (8.268)\hfill \end{array}$$

For convenience we will determine the squared magnitude of H (Ω):

$$\begin{array}{llll}{\left|H(\Omega )\right|}^2\hfill & =H(\Omega )H^{*(\Omega )}\hfill & \hfill & \hfill \\ \hfill & =\left(\frac{e^{j\Omega }-r{e}^{j{\Omega }_0}}{e^{j\Omega }-\left(1/r\right)e^{j{\Omega }_0}}\right)\left(\frac{e^{-j\Omega }-r{e}^{-j{\Omega }_0}}{e^{j\Omega }-\left(1/r\right)e^{-j{\Omega }_0}}\right)\hfill & \hfill & \hfill \\ \hfill & =\frac{1+r^2-2rcos\left(\Omega -{\Omega }_0\right)}{1+{\left(1/r\right)}^2-2\left(1/r\right)cos\left(\Omega -{\Omega }_0\right)}\hfill & \hfill & \hfill \\ \hfill & =r^2\left(\text{constant}\right)\hfill & \hfill & \left(8.269\right)\hfill \end{array}$$

The magnitude of the system function is |H (Ω)| = r. It can be shown (see Problem 8.40 at the end of this chapter) that the phase of the system function is

$$\begin{array}{lll}\measuredangle H(\Omega )={tan}^{-1}\left[\frac{\left(r^2-1\right)sin\left(\Omega -{\Omega }_0\right)}{2r-\left(r^2+1\right)cos\left(\Omega -{\Omega }_0\right)}\right]\hfill & \hfill & \left(8.270\right)\hfill \end{array}$$

The phase characteristic ∠H (Ω) is shown in Fig. 8.42 for the case of a pole and a zero both on the real axis, that is, for Ω0 = 0. The phase is shown for r = 2.5, r = 1.75 and r = 1.2.

Since the shape of the phase response can be controlled by the choice of r while keeping the magnitude response constant, an allpass system is also referred to as a phase-shifter. Increased versatility in controlling the phase response can be obtained by choosing the pole and the zero to be complex-valued. This requires Ω0 ≠ 0. Consider a first-order system with a pole at p1 =(1/r)ejΩ0 and a zero at z1 = re jΩ0. Again we choose r > 1 to obtain a system that is both causal and stable. Fig. 8.43 illustrates the phase response ∠H (Ω) is shown for Ω0 = π/6 and Ω0 = π/ 3. The parameter r is fixed at r = 1.2 in both cases.

If a system function with real-valued coefficients is desired, two first-order allpass sections with complex conjugate zeros and poles can be combined into a second-order allpass system. The resulting system function is

$$\begin{array}{lll}H\left(z\right)=\left(\frac{z-r{e}^{j{\Omega }_0}}{z-\left(1/r\right)e^{j{\Omega }_0}}\right)\left(\frac{z-r{e}^{-j{\Omega }_0}}{z-\left(1/r\right)e^{-j{\Omega }_0}}\right)\hfill & \hfill & \left(8.271\right)\hfill \end{array}$$

$$\begin{array}{lll}=\frac{z^2-2r\text{ cos }\left({\Omega }_0\right)z+r^2}{z^2-\left(2/r\right)cos\left({\Omega }_0\right)z+{\left(1/r\right)}^2}\hfill & \hfill & \left(8.272\right)\hfill \end{array}$$

Pole-zero diagram for a second-order allpass system with parameters r and Ω0 is shown in Fig. 8.44.

8.5.8 Inverse systems

The inverse of the system is another system which, when connected in cascade with the original system, forms an identity system. This relationship is depicted in Fig. 8.45.

The output signal of the original system is

$$\begin{array}{lll}y\left[n\right]=\text{Sys}\left\{x\left[n\right]\right\}\hfill & \hfill & \left(8.273\right)\hfill \end{array}$$

We require the inverse system to recover the original input signal x[n] from the output signal y [n], therefore

$$\begin{array}{lll}x\left[n\right]=\text{Sys}{}_i\left\{y\left[n\right]\right\}\hfill & \hfill & \left(8.274\right)\hfill \end{array}$$

Combining Eqns. (8.273) and (8.273) yields

$$\text{Sys}{}_i\begin{array}{lll}\left\{\text{sys}\left\{x\left[n\right]\right\}\right\}=x\left[n\right]\hfill & \hfill & \left(8.275\right)\hfill \end{array}$$

Let the original system under consideration be a DTLTI system with impulse response h[n], and let the inverse system be also a DTLTI system with impulse response hi[n] as shown in Fig. 8.46. For the output signal of the inverse system to be identical to the input signal of the original system we need the impulse response of the cascade combination to be equal to δ [n], that is,

$$\begin{array}{lll}{h}_{eq}\left[n\right]=h\left[n\right]*h_i\left[n\right]=\delta \left[n\right]\hfill & \hfill & \left(8.276\right)\hfill \end{array}$$

or, using the convolution summation

$$\begin{array}{lll}{h}_{eq}\left[n\right]={\displaystyle \sum _k}h\left[k\right]h_i\left[n-k\right]=\delta \left[n\right]\hfill & \hfill & \left(8.277\right)\hfill \end{array}$$

Corresponding relationship between the system functions is found by taking the z-transform of Eqn. (8.276):

$$\begin{array}{ccc}{H}_{eq}(z)=H(z)H_i(z)=1& & (8.278)\end{array}$$

Consequently, the system function of the inverse system is

$$\begin{array}{ccc}{H}_i(z)=\frac{1}{H(z)}& & (8.279)\end{array}$$

Two important characteristics of the inverse system are causality and stability. We will first focus on causality. Consider again a rational system function in the standard form

$$\begin{array}{ccc}H(z)=\frac{B(z)}{A(z)}=\frac{b_M{z}^M+b_{M-1}{z}^{M-1}+\mathrm{...}+b_{1}z+b_0}{a_N{z}^N+a_{N-1}{z}^{N-1}+\mathrm{...}+a_{1}z+a_0}& & (8.280)\end{array}$$

The system function for the inverse system is

$$H_i(z)=\frac{A(z)}{B(z)}=\frac{a_N{z}^N+a_{N-1}{z}^{N-1}+\mathrm{...}+a_{1}z+a_0}{b_M{z}^M+b_{M-1}{z}^{M-1}+\mathrm{...}+b_{1}z+b_0}$$

If the original system with system function H (z) is causal then M ≤ N as we have established in Section 8.5.5. By the same token, causality of the inverse system with system function H i (z) requires N ≤ M. Hence we need N = M if both the original system and its inverse are required to be causal.

To analyze the stability of the inverse system we will find it more convenient to write the system function H (z) in pole zero form. Using M = N we have

$$\begin{array}{ccc}H(z)=\frac{b_N(z-z_1)(z-z_2)\mathrm{...}(z-z_N)}{a_N(z-p_1)(z-p_2)\mathrm{...}(z-p_N)}& & (8.281)\end{array}$$

If the original system is both causal and stable, all its poles must be inside the unit circle of the z-plane (see Section 8.5.6), therefore

$$\begin{array}{cccc}\left|p_k\right|<1,& k=1,\mathrm{...},N& & (8.282)\end{array}$$

The system function of the inverse system, written in pole-zero form, is

$$\begin{array}{ccc}{H}_i(z)=\frac{a_N(z-p_1)(z-p_2)\mathrm{...}(z-p_N)}{b_N(z-z_1)(z-z_2)\mathrm{...}(z-z_N)}& & (8.283)\end{array}$$

For the inverse system to be stable, its poles must also lie inside the unit circle. The poles of the inverse system are the zeros of the original system. Therefore, for the inverse system to be stable, both zeros and poles of the original system must be inside the unit circle. In addition to Eqn. (8.282) we also need

$$\begin{array}{cccc}\left|z_k\right|<1,& k=1,\mathrm{...},N& & (8.284)\end{array}$$

Example 8.40: Inverse of a system described by a difference equation

A DTLTI system is described by a difference equation

$$y[n]=0.1y[n-1]+0.72y[n-2]+x[n]+0.5x[n-1]$$

Determine if a causal and stable inverse can be found for this system. If yes, find a difference equation for the inverse system.

Solution: The system function is found by taking the z-transform of the difference equation and solving for the ratio of the z-transforms of the output signal and the input signal:

$$Y[z][1-0.1z^{-1}-0.72z^{-2}]=X(z)[1+0.5z^{-1}]$$

$$H(z)=\frac{1+0.5z^{-1}}{1-0.1z^{-1}-0.72z^{-2}}=\frac{z(z+0.5)}{(z+0.8)(z-0.9)}$$

The zeros of the system function are at z1 = 0 and z2 = −0.5. Its two poles are at p1 = −0.8 and p2 = 0.9. Since all poles and zeros are inside the unit circle of the z-plane, H(z) is a minimum phase system. The system function for the inverse system is

$$H_i(z)=\frac{(z+0.8)(z-0.9)}{z(z+0.5)}=\frac{1-0.1z^{-1}-0.72z^{-2}}{1+0.5z^{-1}}$$

which is clearly causal and stable. Its corresponding difference equation is

$$y[n]=-0.5y[n-1]+x[n]-0.1x[n-1]-0.72x[n-2]$$

8.6 Implementation Structures for DTLTI Systems

Block diagram implementation of discrete-time systems was briefly discussed in Section 3.6 of Chapter 3. A method was presented for obtaining a block diagram from a linear constant-coefficient difference equation. Three types of elements were utilized; namely signal adder, constant-gain multiplier and one-sample delay. In this section we will build on the techniques presented in Section 3.6, this time utilizing the z-domain system function as the starting point.

8.6.1 Direct-form implementations

The general form of the z-domain system function for a DTLTI system is

$$\begin{array}{ccc}H(z)=\frac{Y(z)}{X(z)}=\frac{b_0+b_1{z}^{-1}+b_2{z}^{-2}\mathrm{...}+\mathrm{...}{b}_M{z}^{-M}}{1+a_1{z}^{-1}+a_2{z}^{-2}\mathrm{...}+\mathrm{...}{a}_N{z}^{-N}}& & (8.285)\end{array}$$

For the sake of convenience we chose to write H (z) in terms of negative powers of z. X (z) and Y (z) are the z-transforms of the input and the output signals respectively.

The method of obtaining a block diagram from a z-domain system function will be derived using a third-order system, but its generalization to higher-order system functions is quite straightforward. Consider a DTLTI system for which M = 3 and N = 3. The system function H (z) is

$$\begin{array}{ccc}H(z)=\frac{Y(z)}{X(z)}=\frac{b_0+b_1{z}^{-1}+b_2{z}^{-2}+b_3{z}^{-3}}{1+a_1{z}^{-1}+a_2{z}^{-2}+a_3{z}^{-3}}& & (8.286)\end{array}$$

Let us use an intermediate transform V (z) that corresponds to an intermediate signal v [n], and express the system function as

$$\begin{array}{ccc}H(z)=H_1(z)H_2(z)=\frac{Y(z)}{V(z)}\frac{V(z)}{X(z)}& & (8.287)\end{array}$$

We will elect to associate V (z)/X (z) with the numerator polynomial, that is,

$$\begin{array}{ccc}{H}_1(z)=\frac{V(z)}{X(s)}=b_0+b_1{z}^{-1}+b_2{z}^{-2}+b_3{z}^{-3}& & (8.288)\end{array}$$

To satisfy Eqn. (8.287), the ratio Y (z)/V (z) must be associated with the denominator polynomial.

$$\begin{array}{ccc}{H}_2(s)=\frac{Y(z)}{V(z)}=\frac{1}{1+a_1{z}^{-1}+a_2{z}^{-2}+a_3{z}^{-3}}& & (8.289)\end{array}$$

The relationships described by Eqns. (8.286) through (8.289) are illustrated in Fig. 8.47.

Solving Eqn. (8.288) for V (z) results in

$$\begin{array}{ccc}V(z)=b_{0}X(z)+b_1{z}^{-1}X(z)+b_2{z}^{-2}X(z)+b_3{z}^{-3}X(z)& & (8.290)\end{array}$$

Taking the inverse z-transform of each side of Eqn. (8.290) and remembering that multiplication of the transform by z−k corresponds to a delay of k samples in the time domain, we get

$$\begin{array}{ccc}v[n]=b_{0}x[n]+b_{1}x[n-1]+b_{2}x[n-2]+b_{3}x[n-3]& & (8.291)\end{array}$$

The relationship between X (z) and V (z) can be realized using the block diagram shown in Fig. 8.48(a). The second part of the system will be implemented by expressing the output transform Y (z) in terms of the intermediate transform V (z). From Eqn. (8.289) we write

$$\begin{array}{ccc}Y(z)=V(z)-a_1{z}^{-1}Y(z)-a_2{z}^{-2}Y(z)-a_3{z}^{-3}Y(z)& & (8.292)\end{array}$$

The corresponding time-domain relationship is

$$\begin{array}{ccc}y[n]=v[n]-a_{1}y[n-1]-a_{2}y[n-2]-a_{3}y[n-3]& & [8.293]\end{array}$$

The relationship between V (z) and Y (z) can be realized using the block diagram shown in Fig. 8.48(b).

The two block diagrams obtained for H1 (z) and H2 (z) can be connected in cascade as shown in Fig. 8.49 to implement the system function H (z). This is the direct-form I realization of the system.

The block diagram in Fig. 8.49 represents the functional relationship between X (z) and Y (z), the z-transforms of the input and the output signals. If a diagram using time-domain quantities is desired, the conversion is easy using the following steps:

1. Replace each transform X (z), Y (z) and V (z) with its time-domain version x [n], y [n] and v [n].
2. Replace each block with the factor z −1 with one-sample time delay.

The resulting diagram is shown in Fig. 8.50.

Example 8.41: Direct-form type-I implementation of a system

Consider a causal DTLTI system described by the z-domain system function

$$H(z)=\frac{z^3-7z+6}{z^4-z^3-0.34z^2+0.966z-0.2403}$$

Draw a direct-form I block diagram for implementing this system.

Solution: Let us multiply both the numerator and the denominator by z −4 to write the system function in the standard form of Eqn. (8.285):

$$H(z)=\frac{z^{-1}-7z^{-3}+6z^{-4}}{1-z^{-1}-0.34z^{-2}+0.966z^{-3}-0.2403z^{-4}}$$

We have

$$\begin{array}{lllll}M=4,\hfill & N=4\hfill & \hfill & \hfill & \hfill \\ b_0=0,\hfill & b_1=1,\hfill & b_2=0,\hfill & b_3=-7,\hfill & b_4=6,\hfill \\ a_0=1,\hfill & a_1=-1,\hfill & a_2=-0.34,\hfill & a_3=0.966,\hfill & a_4=-0.2403,\hfill \end{array}$$

The block diagram is shown in Fig. 8.51.

Let us consider the direct-form I diagram in Fig. 8.49 again. The subsystem with system function H1 (z) takes X (z) as input, and produces V (z). The subsystem with system function H2 (z) takes V (z), the output of the first subsystem, and produces Y (z) in response. Since each subsystem is linear,2 it does not matter which one comes first in a cascade connection. An alternative way of connecting the two subsystems is shown in Fig. 8.52, and it leads to the block diagram shown in Fig. 8.53.

The middle part of the diagram in Fig. 8.53 has two delay lines running parallel to each other, each set containing four nodes. These two sets of nodes hold identical values W (z), z−1 W (z), z −2 W (z) and z −3 W (z). Consequently, the two sets of nodes can be merged together to result in the diagram shown in Fig. 8.54. This is the direct-form II realization of the system.

The block diagram in Fig. 8.54 could easily have been obtained directly from the system function in Eqn. (8.286) by inspection, using the following set of rules:

Example 8.42: Direct-form II implementation of a system

Draw a direct-form II block diagram for implementing the system used in Example 8.41.

Solution: The diagram obtained by interchanging the order of the two subsystems in Fig. 8.51 and eliminating the nodes that become redundant is shown in Fig. 8.55.

8.6.2 Cascade and parallel forms

Instead of simulating a system with the direct-form block diagram discussed in the previous section, it is also possible to express the system function as either the product or the sum of lower order sections, and base the block diagram on cascade of parallel combination of smaller diagrams. Consider a system function of order M that can be expressed in the form

$$\begin{array}{ccc}\begin{array}{cc}H(z)& =H_1(z)H_2(z)\mathrm{...}{H}_M(z)\\ \\ \hfill & =\frac{W_1(z)}{X(z)}\frac{W_2(z)}{W_1(z)}\mathrm{...}\frac{Y(z)}{W_{M-1}(z)}\hfill \end{array}& & (8.294)\end{array}$$

One method of simulating this system would be to build a diagram for each of the subsections Hi (s) using the direct-form approach discussed previously, and then to connect those sections in cascade as shown in Fig. 8.56.

An easy method of sectioning a system function in the style of Eqn. (8.294) is to first find the poles and the zeros of the system function, and then use them for factoring numerator and denominator polynomials. Afterwards, each section may be constructed by using one of the poles. Each zero is incorporated into one of the sections, and some sections may have constant numerators. If some of the poles and zeros are complex-valued, we may choose to keep conjugate pairs together in second-order sections to avoid the need for complex gain factors in the diagram. The next example will illustrate this process.

Example 8.43: Cascade form block diagram

Develop a cascade form block diagram for the DTLTI system used in Example 8.41.

Solution: The system function specified in Example 8.41 has zeros at z1 = −3, z2 = 2, z3 = 1, and poles at p1 = −0.9, p2 = 0.3, p3, 4 = 0.8 ± j 0.5 (see MATLAB Exercise 8.9). It can be written in factored form as

$$H(z)=\frac{(z+3)(z-1)(z-1)}{(z+0.9)(z-0.3)(z-0.8-j0.5)(z-0.8+j0.5)}$$

The system function can be broken down into cascade sections in a number of ways. Let us choose to have two second-order sections by defining (see MATLAB Exercise 8.9)

$$H_1(z)=\frac{(z+3)(z-1)}{(z+0.9)(z-0.3)}=\frac{z^2+2z-3}{z^2+0.6z-0.27}$$

and

$$H_2(z)=\frac{z-2}{(z-0.8-j0.5)(z-0.8+j0.5)}=\frac{z-2}{z^2-1.6z+0.89}$$

so that

$$H(z)=H_1(z)H_2(z)$$

A cascade form block diagram is obtained by using direct-form II for each of H1 (z) and H2 (z), and connecting the resulting diagrams as shown in Fig. 8.57.

It is also possible to consolidate the neighboring adders in the middle although this would cause the intermediate signal W1 (z) to be lost. The resulting diagram is shown in Fig. 8.58.

An alternative to the cascade form simulation diagram is a parallel form diagram which is based on writing the system function as a sum of lower order functions:

$$\begin{array}{llll}H(s)\hfill & =\overline{H_1}(z)+\overline{H_2}(z)+\mathrm{...}+\overline{H_M}(z)\hfill & \hfill & \hfill \\ \hfill & =\frac{\overline{W_1}(z)}{X(z)}+\frac{\overline{W_2}(z)}{X(z)}+\mathrm{...}\frac{\overline{W_M}(z)}{X(z)}\hfill & \hfill & (8.295)\hfill \end{array}$$

A simulation diagram may be constructed by implementing each term in Eqn. (8.295) using the direct-form approach, and then connecting the resulting subsystems in a parallel configuration as shown in Fig. 8.59.

A rational system function H (z) may be sectioned in the form of Eqn. (8.295) using partial fraction expansion. If some of the poles and zeros are complex-valued, we may choose to keep conjugate pairs together in second-order sections to avoid the need for complex gain factors in the diagram. This process will be illustrated in the next example.

Example 8.44: Parallel form block diagram

Develop a cascade form block diagram for the DTLTI system used in Example 8.41.

Solution: The system function has poles at p1 = −0.9, p2 = 0.3and p3, 4 = 0.8 ± j 0.5. Let us expand it into partial fractions in the form

$$H(z)=\frac{k_1}{z+0.9}+\frac{k_2}{z-0.3}+\frac{k_3}{z-0.8-j0.5}+\frac{k_4}{z-0.8+j05}$$

Note that it is not necessary to divide H (z) by z before expanding it into partial fractions since we are not trying to compute the inverse transform. The residues of the partial fraction expansion are (see MATLAB Exercise 8.11)

$$\begin{array}{ll}{k}_1=-3.0709,\hfill & k_2=6.5450\hfill \\ k_3=-1.2371+j1.7480\hfill & k_4=-1.2371-j1.7480\hfill \end{array}$$

and H (s) is

$$H(z)=\frac{-3.0709}{z+0.9}+\frac{6.5450}{z-0.3}+\frac{-1.2371+j1.7480}{z-0.8-j0.5}+\frac{-1.2371-j1.7480}{z-0.8+j0.5}$$

Let us combine the first two terms to create a second-order section:

$$\overline{H_1}(z)=\frac{-3.0709}{z+0.9}+\frac{6.5450}{z-0.3}=\frac{3.474z+6.812}{z^2+0.6z-0.27}$$

The remaining two terms are combined into another second-order section, yielding

$$\overline{H_2}(z)=\frac{-1.2371+j1.7480}{z-0.8-j0.5}+\frac{-1.2371-j1.7480}{z-0.8+j0.5}=\frac{-2.474z+0.2314}{z^2-1.6z-0.89}$$

so that

$$H(z)=\overline{H_1}(z)+\overline{H_2}(z)$$

The parallel form simulation diagram is shown in Fig. 8.60.

8.7 Unilateral z-Transform

It was mentioned in earlier discussion that the z-transform as defined by Eqn. (8.1) is often referred to as the bilateral z-transform. An alternative version, known as the unilateral z-transform, is defined by

$$\begin{array}{ccc}{X}_u(z)={Ƶ}_u\{x[n]\}={\displaystyle \sum _{n=0}^{\infty }}x[n]z^{-n}& & (8.296)\end{array}$$

We use the subscript “u” to distinguish the unilateral z-transform from its bilateral counterpart. It is clear from a comparison of Eqn. (8.296) with Eqn. (8.1) that the only difference between the two definitions is the lower index of the summation. In Eqn. (8.296) we start the summation at n = 0 instead of n = ∞. If x [n] is a causal signal, that is, if x [n]= 0 for n < 0, then the unilateral transform X u (z) becomes identical to the bilateral transform X (z). On the other hand, for a signal that is x [n] is not necessarily causal, the unilateral transform is essentially the bilateral transform of x [n] multiplied by a unit-step function, that is,

$$\begin{array}{ll}{Ƶ}_u\{x[n]\}\hfill & ={Ƶ}_u\{x[n]u[n]\}\hfill \\ \hfill & =\underset{\infty }{\overset{n=-\infty }{{{\displaystyle \sum }}^{ }}}x[n]u[n]z^{-n}\hfill \end{array}$$

Because of the way X u (z) is defined, its region of convergence is always the outside of a circle, and does not have to be explicitly stated.

One property of the unilateral z-transform that differs from its counterpart for the bilateral z-transform is the time-shifting property. Consider the time-shifted signal x [n − 1]. In terms of the bilateral z-transform, we have proved the following relationship in earlier discussion:

$$\begin{array}{ccc}x[n-1]\underleftrightarrow{Ƶ}{z}^{-1}X(z)& & (8.297)\end{array}$$

The relationship in Eqn. (8.298) does not extend to the unilateral z-transform, that is, we cannot claim that Ƶ u{x [n − 1]} = z −1 Z u{x [n]}. The reason for this fundamental difference becomes obvious if we write the unilateral transform in open form:

$$\begin{array}{ccccc}{Ƶ}_u\{x[n]\}& =& {\displaystyle \sum _{n=0}^{\infty }}x[n]z^{-n}& & \\ & =& x[0]+x[1]z^{-1}+x[2]z^{-2}+x[3]z^{-3}\mathrm{...}& & (8.298)\end{array}$$

The transform of the time-shifted signal is

$$\begin{array}{ccccc}{Ƶ}_u\{x[n-1]\}& =& {\displaystyle {\displaystyle \sum }_{n=0}^{\infty }}x[n-1]z^{-n}& & \\ & =& x[-1]+x[0]z^{-1}+x[1]z^{-2}+x[3]z^{-3}\mathrm{...}& & (8.299)\end{array}$$

The sample x [−1] is not part of Z u{x [n]} but appears in Z u{x [n − 1]}. Let us rewrite the latter by separating x [−1] from the rest of the terms:

$$\begin{array}{ccccc}{Ƶ}_u\{x[n-1]\}& =& x[-1]+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}x[n-1]z^{-n}& & \\ & =& x[-1]+z^{-1}{\displaystyle {\displaystyle \sum }_{n=0}^{\infty }}x[n]z^{-n}& & \\ & =& x[-1]+z^{-1}\left[x[0]+x[1]z^{-1}+x[2]z^{-2}+x[3]z^{-3}+\mathrm{...}\right]& & (8.300)\end{array}$$

Consequently, the unilateral z-transform of x [n − 1] can be expressed in terms of the transform of x [n] as

$$\begin{array}{ccc}{Ƶ}_u\{x[n-1]\}=x[-1]+z^{-1}{X}_u(z)& & (8.301)\end{array}$$

Through similar reasoning it can be shown that, for the signal x [n − 2], we have

$$\begin{array}{ccc}{Ƶ}_u\{x[n-2]\}=x[-2]+x[-1]z^{-1}+z^{-2}{X}_u(z)& & (8.302)\end{array}$$

and in the general case of a signal time-shifted by k > 0 samples

$$\begin{array}{ccccc}{Ƶ}_u\{x[n-k]\}& =& x[-k]+x[-k+1]z^{-1}+\mathrm{...}+x[-1]z^{-k+1}+z^{-k}{X}_u(z)& & \\ & =& {\displaystyle \sum _{n=-k}^{-1}}x[n]z^{-n-k}+z^{-k}{X}_u(z)& & (8.303)\end{array}$$

If the signal is advanced rather than delayed in time, it can be shown that (see Problem 8.44 at the end of this chapter)

$$\begin{array}{ccc}{Ƶ}_u\{x[n+k]\}& =z^k{X}_u(z)-{\displaystyle \sum _{n=0}^{k-1}}x[n]z^{k-n}& (8.304)\end{array}$$

for k > 0. The unilateral z-transform is useful in the use of z-transform techniques for solving difference equations with specified initial conditions. The next couple of examples will illustrate this.

Example 8.45: Finding the natural response of a system through z-transform

The homogeneous difference equation for a system is

$$y[n]-\frac{5}{6}y[n-1]+\frac{1}{6}y[n-2]=0$$

Using z-transform techniques, determine the natural response of the system for the initial conditions

$$\begin{array}{ccc}y[-1]=19,& & y[-2]=53\end{array}$$

Recall that this problem was solved in Example 3.12 of Chapter 3 using time-domain solution techniques.

Solution: The first step will be to compute the unilateral z-transform of both sides of the homogeneous difference equation:

$$\begin{array}{ccc}{Ƶ}_u\left\{y[n]-\frac{5}{6}y[n-1]+\frac{1}{6}y[n-2]\right\}=0& & (8.305)\end{array}$$

Using the linearity of the z-transform we can write

$$\begin{array}{ccc}{Ƶ}_u\{y[n]\}-\frac{5}{6}{Ƶ}_u\{y[n-1]+\frac{1}{6}{Ƶ}_u\{y[n-2]\}=0& & (8.306)\end{array}$$

The unilateral z-transforms of the time-shifted versions of the output signal are

$$\begin{array}{ccccc}{Ƶ}_u\{y[n-1]\}& =& y[-1]+z^{-1}{Y}_u(z)& & \\ & =& 19+z^{-1}{Y}_u(z)& & (8.307)\end{array}$$

and

$$\begin{array}{ccccc}{Ƶ}_u\{y[n-2]\}& =& y[-2]+y[-1]z^{-1}+z^{-2}{Y}_u(z)& & \\ & =& 53+19z^{-1}+z^{-2}{Y}_u(z)& & (8.308)\end{array}$$

Substituting Eqns. (8.307) and (8.308) into Eqn. (8.306) we get

$$Y_u(z)-\frac{5}{6}[19+z^{-1}{Y}_u(z)]+\frac{1}{6}[53+19z^{-1}+z^{-2}{Y}_u(z)]=0$$

which can be solved for Y u (z) to yield

$$Y_u(z)=\frac{z(7z-\frac{19}{6})}{z^2-\frac{5}{6}z+\frac{1}{6}}=\frac{z(7z-\frac{19}{6})}{(z-\frac{1}{2})(z-\frac{1}{3})}$$

Now we need to find y [n] through inverse z-transform using partial fraction expansion. We will write Y u (z) as

$$Y_u(z)=\frac{k_{1}z}{z-\frac{1}{2}}+\frac{k_{2}z}{z-\frac{1}{3}}$$

The residues of the partial fraction expansion are

$$k_1=(z-\frac{1}{2})\frac{Y_u(z)}{z}{|}_{z=\frac{1}{2}}=\frac{(7z-\frac{19}{6})}{(z-\frac{1}{3})}{|}_{z=\frac{1}{2}}=\frac{7(\frac{1}{2})-\frac{19}{6}}{(\frac{1}{2}-\frac{1}{3})}=2$$

and

$$k_2=(z-\frac{1}{3}) \frac{Y_u(z)}{z}{|}_{z=\frac{1}{3}}=\frac{(7_z-\frac{19}{6})}{(z-\frac{1}{2})}{|}_{z=\frac{1}{3}}=\frac{7(\frac{1}{3}) - \frac{19}{6}}{(\frac{1}{3} - \frac{1}{2})}=5$$

Thus, the partial fraction expansion for Y u (z) is

$$Y_u(z)=\frac{2z}{z-\frac{1}{2}}+\frac{5z}{z-\frac{1}{3}}$$

resulting in the homogeneous solution

$$y_h[n]={Ƶ}_u^{-1}\{Y_u(z)\}=2{(\frac{1}{2})}^n u[n]+5 {(\frac{1}{3})}^n u[n]$$

This result is in perfect agreement with that found in Example 3.12.

Example 8.46: Finding the forced response of a system through z-transform

Consider a system defined by means of the difference equation

$$y\left[n\right]=0.9y\left[n-1\right]+0.1x\left[n\right]$$

Determine the response of this system for the input signal

$$x\left[n\right]=20cos\left(0.2\pi n\right)$$

if the initial value of the output is y [−1] = 2.5. Recall that this problem was solved in Example 3.15 of Chapter 3 using time-domain solution techniques.

Solution: Let us begin by computing the unilateral z-transform of both sides of the difference equation.

$$\begin{array}{llll}Ƶ\left\{y\left[n\right]\right\}\hfill & ={Ƶ}_u\left\{0.9y\left[n-1\right]\right\}+{Ƶ}_u\left\{0.1x\left[n\right]\right\}\hfill & \hfill & \hfill \\ \hfill & =0.9{Ƶ}_u\left\{y\left[n-1\right]\right\}+0.1{Ƶ}_u\left\{x\left[n\right]\right\}\hfill & \hfill & \left(8.309\right)\hfill \end{array}$$

In Example 8.15 we have found

$$\begin{array}{ccc}Ƶ\left\{cos\left({\Omega }_{0}n\right)u\left[n\right]\right\}=\frac{z\left[z-cos\left({\Omega }_0\right)\right]}{z^2-2cos\left({\Omega }_0\right)z+1}& & \left(8.310\right)\end{array}$$

Adapting that result to the transform of the input signal x [n] given, we get

$$\begin{array}{ccc}{Ƶ}_u\left\{x\left[n\right]\right\}={Ƶ}_u\left\{20cos\left(0.2\pi n\right)\right\}=\frac{20\left[z-cos\left(0.2\pi \right)\right]}{z^2-2cos\left(0.2\pi \right)z+1}& & \left(8.311\right)\end{array}$$

The unilateral z-transform of the y [n − 1] term is

$$\begin{array}{ccc}{Ƶ}_u\left\{y\left[n-1\right]\right\}=y\left[-1\right]+z^{-1}{Y}_u\left(u\right)=2.5+z^{-1}{Y}_u\left(z\right)& & \left(8.312\right)\end{array}$$

Substituting Eqns. (8.311) and (8.312) into Eqn. (8.309) we obtain

$$\begin{array}{cc}{Y}_u\left(z\right)& =0.9\left[2.5+z^{-1}{Y}_u\left(z\right)\right]+0.1\frac{20z\left[z-\text{ cos}\left(\text{0}\text{.2}\pi \right)\right]}{z^2-2cos\left(0.\text{2}\pi \right)z+1}\\ & =0.9z^{-1}{Y}_u\left(z\right)+2.25+\frac{2z\left[z-\text{ cos}\left(\text{0}\text{.2}\pi \right)\right]}{z^2-2cos\left(0.\text{2}\pi \right)z+1}\end{array}$$

from which the unilateral z-transform of the output signal can be obtained as

$$Y_u\left(z\right)=\frac{2z^2\left[z-cos\left(0.2\pi \right)\right]+2.25z\left(z-e^{j0.2\pi }\right)\left(z-e^{-j0.2\pi }\right)}{\left(z-\text{0}\text{.9}\right)\left(z-e^{j0.2\pi }\right)\left(z-e^{-j0.2\pi }\right)}$$

We are now ready to determine y [n] using partial fraction expansion. Let

$$Y_u\left(z\right)=\frac{k_{1}z}{z-0.9}+\frac{k_{2}z}{z-e^{j0.2\pi }}+\frac{k_{3}z}{z-e^{-j0.2\pi }}$$

The residues are

$$\begin{array}{ll}{k}_1\hfill & =\left(z-0.9\right){\frac{Y_u\left(z\right)}{z}|}_{z=0.9}\hfill \\ \hfill & {=\frac{2z\left[z-cos\left(0.2\pi \right)\right]+2.25\left(z-e^{j0.2\pi }\right)\left(z-e^{-j0.2\pi }\right)}{\left(z-e^{j02\pi }\right)\left(z-e^{-j0.2\pi }\right)}|}_{z=0.9}\hfill \\ \hfill & =\frac{2\left(0.9\right)\left[0.9-cos\left(0.2\pi \right)\right]+2.25\left(0.9-e^{j0.2\pi }\right)\left(0.9-e^{j0.2\pi }\right)}{\left(0.9-e^{j0.2\pi }\right)\left(0.9-e^{-j0.2\pi }\right)}\hfill \\ \hfill & =2.7129\hfill \end{array}$$

and

$$\begin{array}{ll}{k}_2\hfill & =(z-e^{j0.2\pi }) \frac{Y_u(z)}{z}{|}_{z=e^{j0.2\pi }}\hfill \\ \hfill & =\frac{2z[z- cos(0.2\pi )] +2.25 (z-e^{j0.2\pi }) (z-e^{-j0.2\pi })}{(z- 0.9) (z-e^{-j0.2\pi })}{|}_{z=e^{j0.2\pi }}\hfill \\ \hfill & =\frac{2e^{j0.2\pi }[e^{j0.2\pi }-cos(0.2\pi )]}{(e^{j0.2\pi }-0.9) (e^{j0.2\pi }-e^{-j0.2\pi })}\hfill \\ \hfill & =0.7685 - j1.4953\hfill \end{array}$$

The third residue, k3, can be computed in a similar manner, however, we will take a shortcut and recognize the fact that complex residues of a rational transform must occur in complex conjugate pairs provided that all coefficients of the denominator polynomial are real-valued. Consequently we have

$$k_3=k_2^{*}=0.7685+j1.4953$$

It will be more convenient to express k2 and k3 in polar complex form as

$$\begin{array}{cc}{k}_2=1.6813e^{-j1.096},& k_3=1.6813e^{j1.096}\end{array}$$

The forced response of the system is

$$\begin{array}{ll}y[n]\hfill & =2.7129{(0.9)}^{n}u[n]+1.6813e^{-j1.096}{e}^{j0.2\pi n}u[n]+1.6813e^{j1.096}{e}^{-j0.2\pi n}u[n]\hfill \\ \hfill & =2.7129{(0.9)}^{n}u[n]+1.6813[e^{j(0.2\pi n-1.096)}+e^{-j(0.2\pi n-1.096)}]u[n]\hfill \\ \hfill & =2.7129{(0.9)}^{n}u[n]+3.3626cos(0.2\pi n-1.096)u[n]\hfill \end{array}$$

or equivalently

$$y[n]=2.7129{\left(0.9\right)}^{n}u[n]+1.5371cos(0.2\pi n)u[n]+2.9907sin(0.2\pi n)u[n]$$

which is consistent with the answer found earlier in Example 3.15.

8.8 Further Reading

[1] U. Graf. Applied Laplace Transforms and Z-Transforms for Scientists and Engineers: A Computational Approach Using a Mathematica Package. Birkhäuser, 2004.

[2] D.G. Manolakis and V.K. Ingle. Applied Digital Signal Processing: Theory and Practice. Cambridge University Press, 2011.

[3] A.V. Oppenheim and R.W. Schafer. Discrete-Time Signal Processing. Prentice Hall, 2010.

[4] R.A. Schilling, R.J. Schilling, and P.D. Sandra L. Harris. Fundamentals of Digital Signal Processing Using MATLAB. Cengage Learning, 2010.

[5] L. Tan and J. Jiang. Digital Signal Processing: Fundamentals and Applications. Elsevier Science, 2013.

MATLAB Exercises

The first sample of the inverse transform is obtained using the following:

 >> num = [3 , -1 ,0];

 >> den = [1,-3,3,-1];

 >> [q,r] = ss_longdiv_util(num ,den)

 q =

  3

 r =

  8 -9 3

This result should be compared to Eqn. (8.208). The next step in the long division can be performed with the statement

 >> [q,r] = ss_longdiv_util(r,den)

 q =

  8

 r =

  15 -21 8

and matches Eqn. (8.209). This process can be repeated as many times as desired. The function ss_longdiv(..) listed below uses the function ss_longdiv_util(..) in a loop to compute n samples of the inverse transform.

 1 function [x] = ss_longdiv (num , den , n)

 2  [q,r] = ss_longdiv_util(num , den);

 3 x = q;

 4 for nn = 2:n ,

 5  [q,r] = ss_longdiv_util(r, den);

 6  x = [x , q];

 7 end;

 8 end

It can be tested using the problem in Example 8.31:

 >> x = ss_longdiv ([3 , -1 , 0] , [1 , -3 , 3 , -1] , 6)

 x =

 3 8  15 24 35 48

If a left-sided result is desired, the function ss_longdiv(..) can be still be used by simply reordering the coefficients in the vectors “num” and “den”. Refer to the transform in Example 8.32.

$$\begin{array}{ll}X\left(z\right)=\frac{z^2+z-2}{z^2-2.5z+1},\hfill & \text{ROC}\hfill \end{array}:\left|z\right|<\frac{1}{2}$$

The left-sided solution is obtained by

 >>  num = fliplr ([1 , 1 , -2]);

 >>  den = fliplr ([1 , -2.5 , 1]);

 >>  x = ss_longdiv (num , den , 6)

 x =

  -2.0000 -4.0000 -7.0000 -13.5000 -26.7500 -53.3750

Software resources:

ss_longdiv.m

ss_longdiv_util.m

Problems

1. 8.1. Using the definition of the z-transform, compute X (z) for the signals listed below. Write each transform using non-negative powers of z. In each case determine the poles and the zeros of the transform, and the region of convergence.

   1. $x\left[n\right]=\left\{\underset{\stackrel{\uparrow }{n=0}}{1},1,1\right\}$
   2. $x\left[n\right]=\left\{\underset{\stackrel{\uparrow }{n=0}}{1},1,1\right\}$
   3. $x\left[n\right]=\left\{1,1,\underset{\stackrel{\uparrow }{n=0}}{1},1,1\right\}$
   4. $x\left[n\right]=\left\{1,1,1,1,\underset{\stackrel{\uparrow }{n=0}}{1}\right\}$
2. 8.2. For each transform X (z) listed below, determine whether the DTFT of the corresponding signal x[n] exists. If it does, find it.
   1. $\begin{array}{lll}X\left(z\right)=\frac{z\left(z-1\right)}{\left(z+1\right)\left(z+2\right)},\hfill & \text{ROC}:\hfill & \left|z\right|\hfill \end{array}>2$
   2. $\begin{array}{lll}X\left(z\right)=\frac{z\left(z+2\right)}{\left(z+1/2\right)\left(z+3/2\right)},\hfill & \text{ROC}:\hfill & \frac{1}{2}<\left|z\right|<\frac{3}{2}\hfill \end{array}$
   3. $\begin{array}{lll}X\left(z\right)=\frac{z^2}{z^2+5z+6},\hfill & \text{ROC}:\hfill & \left|z\right|<2\hfill \end{array}$
   4. $\begin{array}{lll}X\left(z\right)=\frac{\left(z+1\right)\left(z-1\right)}{\left(z+2\right)\left(z-3\right)\left(z-4\right)},\hfill & \text{ROC}:\hfill & \left|z\right|<2\hfill \end{array}$
3. 8.3. Pole-zero diagrams for four transforms are shown in Fig. P.8.3. For each, determine the ROC if it is known that the DTFT of x[n] exists.

   Figure P. 8.3

   

4. 8.4. Determine the z-transforms of the signals given below. Indicate the ROC for each.
   1. $x\left[n\right]=\{\begin{array}{ll}n,\hfill & n=0,...,9\hfill \\ 0,\hfill & \text{otherwise}\hfill \end{array}$
   2. $x\left[n\right]=\{\begin{array}{ll}n,\hfill & n=0,...,9\hfill \\ 10,\hfill & n\ge 10\hfill \\ 0,\hfill & \text{otherwise}\hfill \end{array}$
   3. $x\left[n\right]=\{\begin{array}{ll}n,\hfill & n=0,...,9\hfill \\ -n+20,\hfill & n=10,...,19\hfill \\ 0,\hfill & \text{otherwise}\hfill \end{array}$
5. 8.5. Find the z-transform of the signal

   $$x\left[n\right]=\{\begin{array}{ll}1,\hfill & n\ge 0\text{ and even}\hfill \\ {\left(2/3\right)}^n,\hfill & n>0\text{ and odd}\hfill \\ 0,\hfill & n<0\hfill \end{array}$$

Also indicate the ROC.

1. 8.6. Consider the signal g[n] specified as

   $$g\left[n\right]={\left(0.9\right)}^{n}cos\left(0.3n\right)u\left[n\right]$$

   1. Determine the transform G(z) and its ROC.
   2. Let a new signal x[n] be obtained from g[n] through the relationship x[n] = g[2n]. Determine the transform X (z) and its ROC from the transform G (z) using the development outlined in Example 8.11.
   3. Determine the transform X (z) and its ROC by direct application of the z-transform definition to the signal x[n]. Compare to the result obtained in part (b).
2. 8.7. Determine the z-transform of each signal listed below, and indicate the ROC. Use linearity, time shifting and time reversal properties of the z-transform when needed.
   1. x[n] = δ[n + 2]
   2. x[n] = δ[n − 3]
   3. $x\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]$
   4. $x\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]+{\left(\frac{1}{3}\right)}^{n}u\left[n\right]$
   5. $x\left[n\right]={\left(\frac{1}{2}\right)}^{n-1}u\left[n-1\right]$
   6. $x\left[n\right]={\left(\frac{1}{2}\right)}^{n-1}u\left[n\right]$
   7. $x\left[n\right]={\left(\frac{1}{2}\right)}^{n+1}u\left[n\right]$
   8. x[n] = (3)n u[−n − 1]
   9. x[n] = (3)n u[−n +1]
3. 8.8. Consider the signal

   $$x\left[n\right]={\left(1/2\right)}^{\left|n\right|}$$

   1. Express x[n] as the sum of a causal signal xR[n] and an anti-causal signal xL[n]. Find an analytical expression for each component, and sketch each component.
   2. Determine the transforms XR(z) and XL(z) corresponding to the signals xR[n] and xL[n] respectively. Also determine the ROC for each transform.
   3. Use linearity of the z-transform to express X(z) and to determine its ROC.
4. 8.9. Use the time reversal property of the z-transform to determine X(z) for the signals listed below. Indicate the ROC for each.
   1. x[n] = u[−n]
   2. x[n] = u[−n − 1]
   3. x[n] = u[−n] −u[−n − 5]
   4. x[n] = n u[−n]
   5. x[n] = cos(Ω0n) u[−n]
5. 8.10. Use the differentiation property of the z-transform to find the transform of each signal below. Also indicate the ROC for each.
   1. x[n] = (n2 + 3n + 5) u[n]
   2. x[n] = (n2 − 5) u[−n − 1]
   3. x[n] = n cos (Ω0n) u[n]
   4. x[n] = (n + 1) sin (Ω0n) u[n]
6. 8.11. Transforms of several causal signals are listed below. Use the initial value property of the z transform to determine x[0] for each case.
   1. $X\left(z\right)=\frac{z^2+z+1}{z^2+3z+2}$
   2. $X\left(z\right)=\frac{z^2+z+1}{z^3+3z^2+3z+1}$
   3. $X\left(z\right)=\frac{z^{-1}+z^{-2}-z^{-3}}{1+0.7z^{-1}+1.2z^{-2}-1.5z^{-3}}$
7. 8.12.

   1. Use the correlation property of the z-transform to determine the autocorrelation function for

      $$x\left[n\right]=u\left[n\right]-u\left[n-2\right]$$

      Hint: Write X (z) in polynomial form.

   2. Generalize the result in part (a) to find the autocorrelation function for

      $$\begin{array}{ll}x\left[n\right]=u\left[n\right]-u\left[n-N\right],\hfill & N>0\hfill \end{array}$$

      using z-transform techniques.

8. 8.13. Use the correlation property of the z-transform to determine the cross correlation function for the signals

   $$x\left[n\right]=u\left[n\right]-u\left[n-3\right]$$

   and

   $$y\left[n\right]=u\left[n\right]-u\left[n-5\right]$$

   Hint: Write X (z) and Y (z) in polynomial form.

9. 8.14. Prove that the cross correlation of signals x[n] and y[n] is equal to the convolution of x[n] and y[−n], that is,

   $$r_{xy}\left[n\right]=x\left[n\right]*y\left[-n\right]$$

   1. Using direct application of the convolution sum
   2. Using correlation and time reversal properties of the z-transform
10. 8.15. Let x[n] = an u[n]. A signal w[n] is defined in terms of x[n] as

    $$w\left[n\right]={\displaystyle \sum _{k=-\infty }^n}x\left[k\right]$$

    1. Determine X (z). Afterwards determine W (z) from X (z) using the summation property of the z-transform.
    2. Determine w[n] from the transform W (z) using partial fraction expansion.
    3. Determine w[n] directly from the summation relationship using the geometric series formula and compare to the result found in part (b).

11. 8.16. Using the summation property of the z transform prove the well-known formula

    $${\displaystyle \sum _{k=0}^n}k=\frac{n\left(n+1\right)}{2}$$

    Hint: Use x [n] = n u [n] and $w\left[n\right]={\displaystyle \sum _{k=-\infty }^n}x\left[k\right]$.

12. 8.17. Consider the z-transforms listed below along with their ROCs. For each case determine the inverse transform x [n] using partial fraction expansion.

    1. $\begin{array}{cc}X(z)=\frac{z}{(z+1) (z+2)},& \text{ROC}:\end{array} |z|<1$
    2. $\begin{array}{cc}X(z)=\frac{z+1}{(z+1/2) (z+2/3)},& \text{ROC}:\end{array}|z|>\frac{2}{3}$
    3. $\begin{array}{cc}X(z)=\frac{z(z+1)}{(z-0.4) (z+0.7)},& \text{ROC}: |z|>0.7\end{array}$
    4. $\begin{array}{cc}X(z)=\frac{z(z+1)}{(z+3/4) (z-1/2) (z-3/2)},& \text{ROC}: \frac{3}{4}<|z|<\frac{3}{2}\end{array}$
    5. $\begin{array}{cc}X(z)=\frac{z(z+1)}{(z+3/4) (z-1/2) (z-3/2)},& \text{ROC}: \frac{1}{2}<|z|<\frac{3}{4}\end{array}$

13. 8.18. The transform X (z) is given by

    $$\frac{\left(z+1\right)\left(z-2\right)}{\left(z+1/2\right)\left(z-1\right)\left(z+2\right)}$$

    The ROC is not specified.

    1. Construct a pole-zero plot for X (z) and identify all possibilities for the ROC.
    2. Express X (z) using partial fractions and determine the residues.
    3. For each choice of the ROC, determine the inverse transform x [n].

14. 8.19. The transforms listed below have complex poles. Each is known to be the transform of a causal signal. Determine the inverse transform x [n] using partial fraction expansion.

    1. $X(z)=\frac{z^2+3z}{z^2- 1.4z + 0.85}$
    2. $X(z)=\frac{z^2}{z^2-1.6z +1}$
    3. $X\left(z\right)=\frac{z^2+3z}{z^2-\sqrt{3}z+\text{ 1}}$

15. 8.20. The transforms listed below have multiple poles. Express each transform using partial fractions, and determine the residues.

    1. $X\left(z\right)=\frac{z^2+3z+2}{z^2-2z+\text{ 1 }}$
    2. $X\left(z\right)=\frac{z\left(z+1\right)}{{\left(z+0.9\right)}^2\left(z-1\right)}$
    3. $X\left(z\right)=\frac{z^2+4z-7}{{\left(z+0.9\right)}^2{\left(z-1.2\right)}^2}$

16. 8.21. The difference equation that represents the remaining balance on a car loan was derived in Chapter 3 as

    $$y\left[n\right]=\left(1+c\right)y\left[n-1\right]-x\left[n\right]$$

    where c is the monthly interest rate,x [n] is the payment made in month n, and y [n] is the balance remaining.

    1. Treat the loan repayment system as a DTLTI system and find its system function H (z).

    2. Let A and B represent the initial amount borrowed and the monthly payment amount respectively. Sincewewould liketotreat thesystemas DTLTI we cannot use an initial value for the difference equation. Instead, we can view the borrowed amount as a negative payment made at n = 0, and monthly payments can be viewed as positive payments starting at n = 1. With this approach the input to the system would be

       $$x\left[n\right]=-A\delta \left[n\right]+Bu\left[n-1\right]$$

       Find the transform X (z).

    3. Find the transform Y (z) of the output signal that represents remaining balance each month.

    4. We would like to have the loan fully paid back after N payments, that is,y [N] = 0. Determine the amount of the monthly payment B in terms of parameters A, c and N.

17. 8.22. Consider again the transforms listed in Problem 8.19. Find the inverse of each transform using the long division method. In each case carry out the long division for 8 samples.

18. 8.23. Refer to the transform X (z) in Problem 8.18 for which the ROC is not given. For each possible choice of the ROC, determine the inverse transform using long division. Carry out the long division operation to compute significant samples of x [n]in the interval −5 ≤ n ≤ 5.

19. 8.24. Assume that the transforms listed below correspond to causal signals. Find the inverse of each transform for n = 0,..., 5 using long division.

    1. $X\left(z\right)=\frac{z}{\left(z+1\right)\left(z+2\right)}$
    2. $X\left(z\right)=\frac{z+1}{\left(z+1/2\right)\left(z+2/3\right)}$
    3. $X\left(z\right)=\frac{z\left(z+1\right)}{\left(z-0.4\right)\left(z+0.7\right)}$

20. 8.25. For each transform given below with its ROC, find the inverse transform using long division. For each, obtain x [n] in the range n = − 4,..., 4.

    1. $\begin{array}{cc}X(z)=\frac{z (z + 1)}{(z + 3/4) (z-1/2) (z - 3/2)},& \text{ROC}: \frac{3}{4}<|z|<\frac{3}{2}\end{array}$
    2. $\begin{array}{cc}X(z)=\frac{z (z + 1)}{(z + 3/4) (z - 1/2) (z-3/2)},& \text{ROC}: \frac{1}{2}<|z|<\frac{3}{4}\end{array}$

    Compare with the analytical results found in Problem 8.17 parts (d) and (e).

21. 8.26. Find the system function H (z) for each causal DTLTI system described below by means of a difference equation. Afterwards determine the impulse response of the system using partial fraction expansion.

    1. $y\left[n\right]=0.9y\left[n-1\right]+x\left[n\right]-x\left[n-1\right]$
    2. $y[n]=1.7y[n-1] -0.72y[n-2] +x[n] -2x[n-1]$
    3. $y\left[n\right]=1.7y\left[n-1\right]-\text{0}\text{.72y}\left[n-2\right]+x\left[n\right]+x\left[n-1\right]+x\left[n-2\right]$
    4. $y\left[n\right]=y\left[n-1\right]-0.11y\left[n-2\right]-0.07y\left[n-3\right]+x\left[n-1\right]$

22. 8.27. Several causal DTLTI systems are described below by means of their system functions. Find a difference equation for each system.

    1. $H\left(z\right)=\frac{z+1}{z^2+\text{ 5z }+6}$
    2. $H\left(z\right)=\frac{{\left(z-1\right)}^2}{\left(z+2/3\right)\left(z+1/2\right)\left(z-4/5\right)}$
    3. $H\left(z\right)=\frac{z^2+\text{ 1}}{z^3+1.2z^2-1.8}$

23. 8.28. A causal DTLTI system is described by the difference equation

    $$y[n]= -0.1y[n-1] + 0.56y[n-2]+x[n] -2x[n-1]$$

    1. Find the system function H (z).
    2. Determine the impulse response h [n] of the system.
    3. Using z-transform techniques, determine the unit-step response of the system.
    4. Using z-transform techniques, find the response of the system to a time-reversed unit-step signal x [n] = u [−n].

24. 8.29. Repeat Problem 8.28 for an anti-causal DTLTI system is described by the difference equation

    $$y\left[n\right]=-\frac{5}{6}y\left[n-1\right]-\frac{1}{6}y\left[n+2\right]+\frac{1}{6}x\left[n-1\right]+\frac{1}{6}x\left[n-2\right]$$

25. 8.30. A causal DTLTI system is characterized by the system function

    $$H\left(z\right)=\frac{0.04z}{z-0.96}$$

    1. Using z-transform techniques, determine the response of the system to the causal sinusoidal signal

       $$x\left[n\right]=sin\left(0.01n\right)u\left[n\right]$$

    2. Determine the response of the system to the non-causal sinusoidal signal

       $$x[n]=sin(0.01n)$$

       This is the steady-state response of the system. Refer to the discussion in Section 8.5.3.

    3. Compare the responses in parts (a) and (b). Approximately how many samples does it take for the response to the causal sinusoidal signal in part (a) to be almost equal to the steady-state response found in part (b)?

26. 8.31. The impulse response of a DTLTI system is

    $$h\left[n\right]={\left(0.8\right)}^n\text{ cos }\left(0.2\pi n\right)u\left[n\right]$$

    1. Determine the system function H (z) and indicate its ROC.
    2. Draw the pole-zero plot for the system function and determine its stability.
    3. Write a difference equation for this system.
    4. Determine the unit-step response of the system using z-transform techniques.

27. 8.32. Input and output signal pairs are listed below for several DTLTI systems. For each case, determine the system function H (z) along with its ROC. Also indicate if the system considered is stable and/or causal.

    1. $\begin{array}{cc}x\left[n\right]={\left(1/2\right)}^{n}u\left[n\right],& y\left[n\right]=3{\left(1/2\right)}^{n}u\left[n\right]+2{\left(3/4\right)}^{4}u\left[n\right]\end{array}$
    2. $\begin{array}{cc}x\left[n\right]={\left(1/2\right)}^{n}u\left[n\right],& y\left[n\right]={\left(1/2\right)}^{n+3}u\left[n+2\right]+\left(1/2\right)+{\left(1/2\right)}^{n+2}u\left[n+1\right]\end{array}$
    3. $\begin{array}{cc}x\left[n\right]=u\left[n\right],& y\left[n\right]=\left(n-1\right)u\left[n-1\right]\end{array}$
    4. $\begin{array}{cc}x\left[n\right]=1.25\delta \left[n\right]-0.25{\left(0.8\right)}^{n}u\left[n\right],& y\left[n\right]={\left(0.8\right)}^{n}u\left[n\right]\end{array}$

28. 8.33. Consider the feedback control system shown in Fig. P.8.33.

    Figure P. 8.33

    

    1. Determine the overall system function H (z) in terms of the system functions H1 (z) and H2 (z) of the two subsystems.
    2. Let the impulse responses of the two subsystems be h1[n] = u[n] and h2[n] = K δ [n − 1] where K is a constant gain parameter. Determine the system functions H1 (z) and H2 (z). Afterwards determine the overall system function H (z).
    3. Determine the range of K for which the system is stable.
    4. Let K = − 3/2. If the input signal is a unit step, find the signal e [n] and the output signal y [n].

29. 8.34. The impulse response h [n] of a DTLTI system is real-valued. Let the system function evaluated at some point z = z0 be expressed in the form

    $$H\left(z_0\right)=H_0{e}^{j{\Theta }_0}$$

    where

    $$H_0=\left|H\left(z_0\right)\right|$$

    and

    $${\Theta }_0=\measuredangle H\left(z_0\right)$$

    Show that the value of the system function at the point $z=z_0^{*}$ is the complex conjugate of its value at the point z = z0, that is,

    $$H\left(z_0^{*}\right)={\left[H\left(z_0\right)\right]}^{*}=H_0{e}^{-j{\Theta }_0}$$

    Hint: Apply the z-transform definition to h [n], evaluate the resulting expression at $z=z_0^{*}$, and manipulate the result to complete the proof.

30. 8.35. A DTLTI system has the system function

    $$H\left(z\right)=\frac{z^2+3z}{z^2-1.4z+0.85}$$

    Find the steady-state response of the system to each of the following signals:

    1. x[n] = (0.8)n ej0.4πn
    2. x[n] = (0.9)n cos (0.3πn)

31. 8.36. A DTLTI system has the system function

    $$H(z)=\frac{z-0.4}{z(z^2-1.4z + 0.85)}$$

    1. Draw a pole-zero diagram to show poles and zeros of the system function as well as the unit circle of the z-plane.
    2. Using the graphical method discussed in Section 8.5.4 determine the magnitude and the phase of the system function at the angular frequency Ω = π/6.
    3. Repeatpart (b) at Ω = π/4.

32. 8.37. For each function H (z) given below, determine if it could be the system function of a causal and stable system. In each case provide the reasoning behind your answer.

    1. $H\left(z\right)=\frac{\left(z-2\right)\left(z+2\right)}{\left(z+1/2\right)}$
    2. $H\left(z\right)=\frac{{\left(z+1\right)}^2}{\left(z-1/2\right)\left(z+1/3\right)}$
    3. $H(z)=\frac{z(z+1)}{(z+1/2) (z+3/2)}$

33. 8.38. Several causal DTLTI systems are described below by their difference equations. For each system, determine the system function H (z). Draw a pole-zero diagram for the system and indicate the ROC for the system function. Determine if the system is stable.

    1. $y\left[n\right]=1.5y\left[n-1\right]-0.54y\left[n-2\right]+x\left[n\right]+3x\left[n-1\right]$
    2. $y\left[n\right]=-0.64y\left[n-2\right]+2x\left[n\right]$
    3. $y\left[n\right]=0.25y\left[n-1\right]-0.125y\left[n-2\right]-0.5y\left[n-3\right]+x\left[n\right]$
    4. $y\left[n\right]=0.25y\left[n-1\right]-0.5y\left[n-2\right]-0.75y\left[n-3\right]+x\left[n\right]+x\left[n-1\right]$

34. 8.39. Consider again the car loan repayment system discussed in Problem 8.21.

    1. Show that the loan repayment system is unstable for any practical setting. You may find it helpful to draw a pole-zero diagram.
    2. Find two examples of bounded input signals |x[n]| < ∞ that result in unbounded output signals.
    3. Can a bounded input signal be found that produces an output signal with a non-zero constant steady-state value? What type of a payment scheme does this represent?

35. 8.40. A first-order allpass filter section is characterized by the system function

    $$H\left(z\right)=\frac{z-r{e}^{j{\Omega }_0}}{z-\left(1/r\right)e^{j{\Omega }_0}}$$

    Show that the phase characteristic of the system is

    $$\measuredangle H(\Omega )={tan}^{-1}\left[\frac{\left(r^2-1\right)sin\left(\Omega -{\Omega }_0\right)}{2r-\left(r^2+1\right)cos\left(\Omega -{\Omega }_0\right)}\right]$$

    Hint: First find H (Ω) by substituting z = ejΩ. Afterwards express H (Ω) in Cartesian form as

    $$H(\Omega )=H_r(\Omega )+j{H}_i(\Omega )$$

36. 8.41. Develop a direct-form II block diagram for each system specified below by means of a system function. Assume that each system is causal and initially relaxed.

    1. $H\left(z\right)=\frac{z+2}{z-1/2}$
    2. $H\left(z\right)=\frac{z^2+1}{z^3+0.8z^2-2.2z+0.6}$
    3. $H\left(z\right)=\frac{z^3+2z^2-\text{ 3z }+4}{2z^3+0.8z^2+1.8z+3.2}$

37. 8.42. Develop a cascade-form block diagram for each system specified below by means of a system function. Assume that each system is causal and initially relaxed. Use first- and second-order cascade sections and ensure that all coefficients are real.

    1. $H\left(z\right)=\frac{z+1}{\left(z+1/2\right)\left(z+2/3\right)}$
    2. $H\left(z\right)=\frac{z\left(z+1\right)}{\left(z-0.4\right)\left(z+0.7\right)}$
    3. $H\left(z\right)=\frac{z\left(z+1\right)}{\left(z+0.6\right)\left(z^2-1.4z+0.85\right)}$

38. 8.43. Develop a parallel-form block diagram for each system specified in Problem 8.42. Assume that each system is causal and initially relaxed. Use first and second-order parallel sections and ensure that all coefficients are real.

39. 8.44. Let Xu (z) be the unilateral z-transform of x [n].

    1. Using the definition of the unilateral z-transform show that the transform of x [n + 1] is

       $${Ƶ}_u\left\{x\left[n+1\right]\right\}=z{X}_u\left(z\right)-zx\left[0\right]$$

    2. Show that the transform of x [n + 2] is

       $${Ƶ}_u\{x[n+2]\}=z^2{X}_u(z)-z^{2}x\left[0\right]-zx\left[1\right]$$

    3. Generalize the results of parts (a) and (b), and show that for k > 0

       $${Ƶ}_u\left\{x\left[n-k\right]\right\}=z^k{X}_u\left(z\right)-{\displaystyle \sum _{n=0}^{k-1}}x\left[n\right]z^{k-n}$$

40. 8.45. Using the unilateral z-transform determine the natural response of each system with the homogeneous difference equation and initial conditions given below:

    1. $\begin{array}{cccc}y\left[n\right]-1.4y\left[n-1\right]+0.85y\left[n-2\right]=0,& y\left[-1\right]=5,& \text{and}& y\left[-2\right]=7\end{array}$
    2. $\begin{array}{cccc}y[n] -1.64y[n-1]+0.64y[n-2]=0,& y[-1]=2,& \text{and}& y[-2]=-3\end{array}$

41. 8.46. Consider the difference equations given below with the specified input signals and initial conditions. Using the unilateral z-transform determine the solution y [n] for each.

    1. $\begin{array}{ccc}y\left[n\right]-2y\left[n-1\right]=x\left[n\right],& x\left[n\right]=cos\left(0.2\pi n\right)u\left[n\right],& y\left[-1\right]=5\end{array}$
    2. $\begin{array}{ccc}y\left[n\right]+0.6y\left[n-1\right]=x\left[n\right]+x\left[n-1\right],& x\left[n\right]=u\left[n\right],& y\left[-1\right]=-3\end{array}$
    3. $\begin{array}{cccc}y\left[n\right]-0.2y\left[n-1\right]-0.48y\left[n-2\right]=x\left[n\right],& x\left[n\right]=u\left[n\right],& y\left[-1\right]=-1& y\left[-2\right]=2\end{array}$

42. 8.47. The Fibonacci sequence is a well-known sequence of integers. Fibonacci numbers start with the pattern

    $$0,1,1,2,3,4,5,8,13,21,34,55,89,...$$

    They follow the recursion relationship

    $$F_{n+2}=F_{n+1}+F_n$$

    with F0 = 0 and F1 = 1. In simplest terms, each number in the sequence is the sum of the two numbers before it. In this problem we will use difference equations and the unilateral z-transform to study some properties of the Fibonacci sequence.

    1. Construct a homogeneous difference equation to represent the recurrence relationship of the Fibonacci sequence. Select y[n] = Fn+2 with initial conditions y [−1] = F1 = 1 and y [−2] = F0 = 0.

    2. Find the solution y[n] of the homogeneous difference equation using the unilateral z-transform.

    3. It can be shown that, for large n, the ratio of consecutive Fibonacci numbers converges to a constant known as the golden ratio, that is,

       $$\underset{n\to \infty }{lim} \frac{y[n+1]}{y[n]}=\phi$$

       Using the solution y [n] found in part (b) determine the value of φ.

MATLAB Problems

1. 8.48. The signal x[n] is given by

   $$x\left[n\right]={\left(0.8\right)}^n\left(u\left[n\right]-u\left[n-8\right]\right)$$

   1. Find X (z). Determine its poles and zeros. Manually construct a pole-zero diagram.
   2. Write a MATLAB script to evaluate the magnitude of X (z) at a grid of complex points in the z-plane. Use the function meshgrid(..) to generate the grid of complex points within the ranges −1.5 < Re {z} < 1.5 and − 1.5 < Im {z} < 1.5 with increments of 0.05 in each direction.
   3. Use the function mesh(..) to produce a three-dimensional mesh plot of |X (z)|.
   4. Evaluate the z-transform for z = ejΩ and use the function plot3(..) to plot it over the three-dimensional mesh plot.

2. 8.49.

   1. Repeat Problem 8.48 with the signal

      $$x\left[n\right]={\left(0.6\right)}^n\left(u\left[n\right]-u\left[n-8\right]\right)$$

   2. Repeat Problem 8.48 with the signal

      $$x\left[n\right]={\left(0.4\right)}^4\left(u\left[n\right]-u\left[n-8\right]\right)$$

3. 8.50. Refer to the system in Problem 8.30.

   1. Construct a system object for H (z) using the function zpk(..).

   2. Compute the response of the system to the input signal

      $$x[n]=sin(0.01n) u[n]$$

      for n = 0,..., 49 using the function dlsim(..). Compare the first few samples of the response to the result of hand calculations in Problem 8.30.

4. 8.51. Refer to the system in Problem 8.31.

   1. Write a MATLAB script to find the impulse response of the system by iteratively solving the difference equation found in Problem 8.31 part (c). Compute h[n] for n = 0,..., 10. Does it match the expected answer?
   2. Modify the script written in part (a) so that the unit-step response of the system is computed by iteratively solving the difference equation. Compare the result to that found in Problem 8.31 part (d).
   3. Construct a system object for H (z) using the function tf(..). Afterwards compute the impulse response and the unit-step response of the system using functions dimpulse(..) and dstep(..) respectively. Compare to the answers obtained in previous parts.

5. 8.52. Refer to the DTLTI system in Problem 8.35.

   1. Write a MATLAB script to compute and graph the magnitude and the phase of the frequency spectrum for the system. Your script should also mark the points critical for the two input signals used in Problem 8.35.
   2. Graph the steady-state response to each of the two input signals specified.

6. 8.53. Develop a script to compute the phase response of a first-order allpass filter with a real pole at p = r + j0 for parameter values r = 0.2, 0.4, 0.6, 0.8. Graph the four phase responses on the same coordinate system for comparison.

7. 8.54. Develop a script to compute the phase response of a second-order allpass filter with a complex conjugate poles at

   $$p_{1,2}=r{e}^{j\theta }$$

   1. In your script set θ = π/ 6 radians. Compute the phase response for r = 0.4, 0.6, 0.8, and graph the results superimposed for comparison.
   2. Set r = 0.8. Compute the phase response for θ = π/ 6, π/4, π/3 radians, and graph the results superimposed for comparison.

8. 8.55. Refer to Problem 8.45.

   1. For each difference equation given, write a script to find the solution for sample indices n = 0,..., 10 using the iterative solution method with the specified initial conditions.
   2. For each difference equation given, write a script to find the analytical solution using symbolic mathematics capabilities of MATLAB.

MATLAB Project

1. 8.56. In Problem 1.51 of Chapter 1 the generation of DTMF signals was explored using MATLAB's trigonometric functions. In this project we will explore alternative means to sinusoidal tone generation.

   In some applications we may wish to use inexpensive processors that may either have no support for trigonometric functions or may be too slow in computing them. An alternative method of computing the samples of a sinusoidal signal is to use a discrete-time system the impulse response of which is sinusoidal, and to apply an impulse signal to its input. Recall that the z-transform of a discrete-time sine signal was found in Example 8.16 as

   $$Ƶ\left\{sin\left({\Omega }_{i}n\right)u\left[n\right]\right\}=\frac{sin\left({\Omega }_i\right)z}{z^2-2cos\left({\Omega }_i\right)z+1}$$

   which could be used as the basis of a second-order system as shown in Fig. P.8.56(a). DTMF signals that correspond to digits on a keypad can be generated by using two systems of this type in parallel as shown in Fig. P.8.56(b) by adjusting the coefficients properly.

   Figure P. 8.56

   

   1. Develop a MATLAB function ss_dtmf2(..) to generate samples of the DTMF signal for a specified digit for a specified duration. The syntax of the function should be as follows:

       x = ss_dtmf2 (digit,fs,n)

      The first argument “digit” is the digit for which the DTMF signal is to be generated. Let values digit = 0 through digit = 9 represent the corresponding keys on the keypad. Map the remaining two keys “*”and “#” to values digit = 10 and digit = 11 respectively. Finally, the value digit = 12 should represent a pause, that is, a silent period. The arguments “fs” and “n” are the sampling rate (in Hz) and the number of samples to be produced respectively. Internally you may want to use the function ss_iir2(..) that was developed in MATLAB Exercise 8.12.

   2. Develop a function named ss_dtmf(..) with the syntax

       x = ss_dtmf (number ,fs ,nd ,np)

      The arguments for the function ss_dtmf(..) are defined as follows:

      number: The phone number to be dialed, entered as a vector. For example, to dial the number 555-1212, the vector “number” would be entered as

       number = [5 ,5 ,5 ,1 ,2 ,1 ,2]

      fs: The sampling rate (in Hz) used in computing the amplitudes of the DTMF signal.

      nd: Number of samples for each digit.

      np: Number of samples for each pause between consecutive digits.

      The function ss_dtmf(..) should use the function ss_dtmf2(..) to produce the signals for each digit (and the pauses between digits) and append them together to create the signal x[n].

   3. Write a script to test the function ss_dtmf(..) with the number 555-1212. Use a sampling rate of fs = 8000 Hz. The duration of each digit should be 200 milliseconds with 80 millisecond pauses between digits.
   4. Play back the resulting signal x[n]using the sound(..) function.

1 $\begin{array}{cc}{\Sigma }_{n=0}^{\infty }{\alpha }^n=1/\left(1-\alpha \right),& \left|\alpha \right|<1.\end{array}$

2 Remember that, for linearity, each subsystem must be initially relaxed. All initial values must be zero. In Fig. 8.50, for example, we need x [−1] = x [−2] = x [−3] = y [−1] = y [−2] = y [−3] = 0.