Chapter 7

Laplace Transform for Continuous-Time Signals and Systems

Chapter Objectives

• Learn the Laplace transform as a more generalized version of the Fourier transform studied in Chapter 4.
• Understand the convergence characteristics of the Laplace transform and the concept of region of convergence.
• Explore the properties of the Laplace transform.
• Understand the use of the Laplace transform for modeling CTLTI systems. Learn the s-domain system function and its use for solving signal-system interaction problems.
• Learn techniques for obtaining simulation diagrams for CTLTI systems based on the s-domain system function.
• Learn the use of the unilateral Laplace transform for solving differential equations with specified initial conditions.

7.1 Introduction

In earlier chapters we have studied Fourier analysis techniques for understanding characteristics of continuous-time signals and systems. Chapter 4 focused on representing continuous-time signals in the frequency domain through the use of Fourier series and Fourier transform. We have also adapted frequency domain analysis techniques for use with linear and time-invariant systems in the form of a system function.

Some limitations exist in the use of Fourier series and Fourier transform for analyzing signals and systems. A signal must be absolute integrable to have a representation based on Fourier series or Fourier transform. Consider, for example, a ramp signal in the form x (t) = tu (t). Such a signal cannot be analyzed using the Fourier transform since it is not absolute integrable. Similarly, a system function based on the Fourier transform is only available for systems for which the impulse response is absolute integrable. Consequently, Fourier transform techniques are usable only with stable systems.

Laplace transform, named after the French mathematician Pierre Simon Laplace (1749-1827), overcomes these limitations. It can be seen as an extension, or a generalization, of the Fourier transform. In contrast, the Fourier transform is a special case, or a limited view, of the Laplace transform. A signal that does not have a Fourier transform may still have a Laplace transform for a certain range of the transform variable. Similarly, an unstable system that does not have a system function based on the Fourier transform may still have a system function based on the Laplace transform.

The Fourier transform of a signal, if it exists, can be obtained from its Laplace transform while the reverse is not generally true. In addition to analysis of signals and systems, block diagram and signal flow graph structures for simulating continuous-time systems can be developed using Laplace transform techniques. The unilateral variant of the Laplace transform can be used for solving differential equations subject to specified initial conditions.

We begin with the basic definition of the Laplace transform and its application to some simple signals. The significance of the issue of convergence of the Laplace transform will become apparent throughout this discussion. Section 7.2 is dedicated to the convergence properties of the Laplace transform, and to the concept of region of convergence. We cover the fundamental properties of the Laplace transform in Section 7.3. These properties will prove useful in working with the Laplace transform for analyzing signals and systems, for understanding system characteristics, and for working with signal-system interaction problems. Proofs of significant Laplace transform properties are given not just for the sake of providing proofs, but also to provide further insight and experience on working with transforms. Techniques for computing the inverse Laplace transform are presented in Section 7.4. Use of the Laplace transform for the analysis of linear and time-invariant systems is discussed in Section 7.5, and the s-domain system function concept is introduced. Derivations of block diagram structures for the simulation of continuous-time systems based on the s-domain system function are covered in Section 7.6. In Section 7.7 the unilateral variant of the Laplace transform is introduced, and it is shown that it is useful in solving differential equations with specified initial conditions.

Notationally, the relationship between the signal x (t) and the transform X (s) can be expressed in the form

$$\begin{array}{ccc}X\left(s\right)=ℒ\left\{x\left(t\right)\right\}& & \left(7.2\right)\end{array}$$

or in the form

$$\begin{array}{ccc}x\left(t\right)\stackrel{ℒ}{\leftrightarrow }X\left(s\right)& & \left(7.3\right)\end{array}$$

Since the parameter s is a complex variable, we will represent it graphically as a point in the complex s-plane. It is customary to draw the s-plane with σ as its real axis and ω as its imaginary axis. Based on this convention the complex variable s is written in Cartesian form as

$$\begin{array}{ccc}s=\sigma +j\omega & & \left(7.4\right)\end{array}$$

A specific value s1 = σ1 + jω1 of the variable s can be shown as a point in the s-plane as illustrated in Fig. 7.1 with σ1 equal to the horizontal displacement of the point from the origin and ω1 as the vertical displacement from the origin. If the Laplace transform converges at the point s = s1, its value can be computed by evaluating the integral in Eqn. (7.1) at that point.

Using the Cartesian form of the variable s given by Eqn. (7.4), the transform of the signal x(t) becomes

$$\begin{array}{ccc}X\left(s\right){|}_{s=\sigma +j\omega }={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left(t\right)e^{-\left(\sigma +j\omega \right)t}dt}& & \left(7.5\right)\end{array}$$

At this point it would be interesting to let the value of σ, the real part of s, be fixed at σ = σ1. Thus, we have s = σ1 + jω. If the imaginary part ω is allowed to vary in the range −∞ < ω < ∞, the trajectory of the points s in the complex plane would be a vertical line with a horizontal displacement of σ1 from the origin, that is, a vertical line that passes through the point s = σ1 + j0. This is illustrated in Fig. 7.2.

If the transform in Eqn. (7.5) is evaluated at points on the vertical line represented by the trajectory s = σ1 + jω, we get

$$\begin{array}{ccccc}X\left({\sigma }_1,\omega \right)=X\left(s\right)|{}_{s={\sigma }_1+j\omega }& =& {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left(t\right)e^{-\left({\sigma }_1+j\omega \right)t}dt}& & \\ & =& {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left(t\right)e^{-{\sigma }_{1}t}{e}^{-j\omega t}dt}& & \left(7.6\right)\end{array}$$

Comparison of the result in Eqn. (7.6) with the definition of the Fourier transform in Eqn. (4.127) in Chapter 4 suggests that the Laplace transform X (s) evaluated on the trajectory s = σ1 + ω is identical to the Fourier transform of the signal $\left[x\left(t\right)e^{-{\sigma }_{1}t}\right]$.

$$\begin{array}{ccc}X\left(s\right){|}_{s={\sigma }_1+j\omega }=ℱ\left\{x\left(t\right)e^{-{\sigma }_{1}t}\right\}& & \left(7.7\right)\end{array}$$

This important conclusion can be summarized as follows:

If we choose σ1 = 0 then the vertical line in Fig. 7.2 coincides with the jω axis of the s-plane, and the Laplace transform evaluated on that trajectory becomes

$$X\left(s\right){|}_{s=0+j\omega }={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}x\left(t\right)e^{-j\omega t}dt=ℱ\left\{x\left(t\right)\right\}}$$

An easy way to visualize the relationship between the Laplace transform and the Fourier transform is the following: Imagine that the Laplace transform evaluated at every point in the s-plane results in a three-dimensional surface. The transform X (s) is complex-valued in general; therefore, it may be difficult to visualize it as a surface unless we are willing to accept the notion of a complex-valued surface. Alternatively, we may split the complex function X (s) into its magnitude and phase, each defining a surface. In any case, if we were to take the surface represented by the Laplace transform and cut it the through the length of the jω axis, the profile of the cross-section would be the same as the Fourier transform of the signal. This relationship is illustrated graphically in Fig. 7.3.

Consider a particular transform in the form

$$\begin{array}{ccc}X\left(s\right)=\frac{s+0.5}{{\left(s+0.5\right)}^2+4{\text{π}}^2}& & \left(7.8\right)\end{array}$$

which is complex-valued. We will choose to graph only its magnitude |X (s)| as a three-dimensional surface which is shown in Fig. 7.3(a). Also shown in the figure is the set of values of |X (s)| computed at points on the jω axis of the s-plane. In Fig. 7.3(b) the magnitude of the Fourier transform X (ω) of the same signal is graphed as a function of the radian frequency ω. Notice how the jω axis of the s-plane becomes the frequency axis for the Fourier transform.

For the same transform under consideration, Fig. 7.4 shows the values on the trajectory s = 0.5 + jω. Recall that the values of |X (s)| on this trajectory are the same as the magnitude of the Fourier transform of the modified signal [x (t)e−0.5t].

The Laplace transform as defined by Eqn. (7.1) is referred to as the bilateral (two-sided) Laplace transform. A variant of the Laplace transform known as the unilateral (one-sided) Laplace transform will be introduced in Section 7.7 as an alternative analysis tool. In this text, when we refer to Laplace transform without the qualifier word “bilateral” or “unilateral”, we will always imply the more general bilateral Laplace transform as defined in Eqn. (7.1).

Example 7.1: Laplace transform of the unit impulse

Find the Laplace transform of the unit-impulse signal

$$x\left(t\right)=\delta \left(t\right)$$

Solution: Applying the definition of the Laplace transform given by Eqn. (7.1) we get

$$\begin{array}{ccc}X\left(s\right)=ℒ\left\{\delta \left(t\right)\right\}={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\delta \left(t\right)e^{-st}dt=1}& & \left(7.9\right)\end{array}$$

using the sifting property of the unit-impulse function. Therefore, the Laplace transform of the unit-impulse signal is constant and equal to unity. Since it does not depend on the value of s, it converges at every point in the s-plane with no exceptions.

Example 7.2: Laplace Transform of a time-shifted unit impulse

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

$$x\left(t\right)=\delta \left(t-\tau \right)$$

Solution: Applying the Laplace transform definition we obtain

$$\begin{array}{ccc}X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\delta \left(t-\tau \right)e^{-st}dt=e^{-s\tau }}& & \left(7.10\right)\end{array}$$

Again we have used the sifting property of the unit-impulse function. If s = σ + jω, then Eqn. (7.10) becomes

$$X\left(s\right){|}_{s=\sigma +j\omega }=e^{-\sigma \tau }{e}^{-j\omega \tau }$$

The transform obtained in Eqn. (7.10) converges as long as σ = Re {s} > −∞.

Example 7.3: Laplace transform of the unit-step signal

Find the Laplace transform of the unit-step signal

$$x\left(t\right)=u\left(t\right)$$

Solution: Substituting x (t) = u (t) into the Laplace transform definition we obtain

$$X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}u\left(t\right)e^{-st}dt}$$

Since u (t) = 1 for t > 0 and u (t) = 0 for t < 0, the lower limit of the integral can be changed to t = 0 and the unit-step term can be dropped to yield

$$\begin{array}{ccc}X\left(s\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}dt=-\frac{1}{s}{e}^{-st}{|}_0^{\infty }}& & \left(7.11\right)\end{array}$$

To evaluate the integral of Eqn. (7.11) for the specified limits we will use the Cartesian form of the complex variable s. Substituting s = σ + jω

$$X\left(s\right){\left|{}_{s=\sigma +j\omega }=-\frac{1}{\sigma +j\omega }{e}^{-\left(\sigma +j\omega \right)t}\right|}_0^{\infty }$$

It is obvious that, for the exponential term eσt to converge as t → ∞, we need σ > 0:

$$\text{If }\sigma >0\Rightarrow {e^{-\sigma t}|}_{t\to \infty }=0$$

The transform becomes

$$\begin{array}{ccc}X\left(s\right)=-\frac{1}{\sigma +j\omega }\left[0-1\right]=\frac{1}{\sigma +j\omega }=\frac{1}{s},& \mathrm{Re}\left\{s\right\}>0& \left(7.12\right)\end{array}$$

The transform expression found in Eqn. (7.12) is valid only for points in the right half of the s-plane. This region is shown shaded in Fig. 7.5. Note that the transform does not converge at points on the jω axis. It converges at any point to the right of the jω axis regardless of how close to the axis it might be.

Example 7.4: Laplace transform of a time-shifted unit-step signal

Find the Laplace transform of the time-shifted unit-step signal

$$x\left(t\right)=u\left(t-\tau \right)$$

Solution: Using x (t) = u (t − τ) in the Laplace transform definition we get

$$X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}u\left(t-\tau \right)e^{-st}dt}$$

Since u (t − τ) = 1 for t > τ and u (t − τ) = 0 for t < τ, the lower limit of the integral can be changed to t = τ and the unit-step function can be dropped without affecting the result.

$${X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{-st}dt}=-\frac{1}{s}{e}^{-st}|}_{\tau }^{\infty }$$

As in Example 7.3, the integral can be evaluated only for σ = Re {s} > 0, and its value is

$$\begin{array}{cccc}X\left(s\right)=\frac{1}{s}\left[0-e^{-st}\right]=\frac{e^{-st}}{s},& \mathrm{Re}\left\{s\right\}>0& & \left(7.13\right)\end{array}$$

Convergence conditions for the the transform of u (t − τ) are the same as those found in Example 7.3 for the transform of u (t).

We observe from the last two examples that, when we find the Laplace transform X (s) of a signal, we also need to specify the region in which the transform is valid. The collection of all points in the s-plane for which the Laplace transform converges is called the region of convergence (ROC).

Recall that in Eqn. (7.7) we have represented the Laplace transform of a signal x (t) as equivalent to the Fourier transform of the modified signal [x (t) e−σt]. Consequently, the conditions for the convergence of the Laplace transform of x (t) are identical to the conditions for the convergence of the Fourier transform of [x (t) e−σt]. Using Dirichlet conditions for the latter, we need the signal [x (t) e−σt] to be absolute integrable for the transform to exist, that is,

$$\begin{array}{ccc}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left|x\left(t\right)e^{-\sigma t}\right|dt<\infty }& & \left(7.14\right)\end{array}$$

The convergence condition stated in Eqn. (7.14) highlights the versatility of the Laplace transform over the Fourier transform. The Fourier transform of a signal x (t) exists is the signal is absolute integrable, and does not exist otherwise. Therefore, the existence of the Fourier transform is a binary question; the answer to it is either yes or no. In contrast, if the signal x (t) is not absolute integrable, we may still be able to find values of σ for which the modified signal [x (t) e−σt] is absolute integrable. Therefore, the Laplace transform of the signal x (t) may exist for some values of σ. The question of existence for the Laplace transform is not a binary one; it is a question of which values of σ allow the transform to converge.

The next two examples will further highlight the significance of the region of convergence for the Laplace transform. The fundamental characteristics of the region of convergence will be discussed in detail in Section 7.2.

Example 7.5: Laplace transform of a causal exponential signal

Find the Laplace transform of the signal

$$x\left(t\right)=e^{at}u\left(t\right)$$

where a is any real or complex constant.

Solution: The signal x (t) is causal since x (t) = 0 for t < 0. Applying the Laplace transform definition to x (t) results in

$$X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{at}u\left(t\right)e^{-st}dt}$$

Changing the lower limit of the integral to t = 0 and dropping the factor u (t) we get

$$\begin{array}{ccc}X(s)=\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{e}^{at}{e}^{-st}dt=\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{(a-s)t}dt=\frac{1}{a-s}{e}^{(a-s)t}{|}_0^{\infty }& & \left(7.15\right)\end{array}$$

The parameter a may be real or complex-valued. Let us assume that it is in the form a = ar + jai to keep the results general. For the purpose of evaluating the integral in Eqn. (7.15) we will substitute s = σ + jω and write the transform as

$$\begin{array}{ccccc}X\left(s\right)& =& {\frac{1}{a_r+j{a}_i-\sigma -j\omega }{e}^{\left(a_r+j{a}_i-\sigma -j\omega \right)t}|}_0^{\infty }& & \\ & =& {\frac{1}{a_r+j{a}_i-\sigma -j\omega }{e}^{\left(a_r-\sigma \right)t}{e}^{j\left(a_i-\omega \right)t}|}_0^{\infty }& & \left(7.16\right)\end{array}$$

For the result of Eqn. (7.16) to converge at the upper limit, we need

$$a_r-\sigma <0\Rightarrow \sigma >a_r$$

or equivalently

$$\begin{array}{ccc}\mathrm{Re}\left\{s\right\}>\mathrm{Re}\left\{a\right\}& & \left(7.17\right)\end{array}$$

which establishes the ROC for the transform. With the condition in Eqn. (7.17) satisfied, the exponential term in Eqn. (7.16) becomes zero as t → ∞, and we obtain

$$\begin{array}{cccc}X\left(s\right)=\frac{1}{a-s}\left[0-1\right]=\frac{1}{s-a},& \mathrm{Re}\left\{s\right\}>\mathrm{Re}\left\{a\right\}& & \left(7.18\right)\end{array}$$

The ROC is the region to the right of the vertical line that goes through the point s = ar + jai as shown in Fig. 7.6.

It is interesting to look at the shape of the signal x (t) in conjunction with the ROC.

1. Fig. 7.7 shows the signal for the case where a is real-valued. The signal x (t) is a decaying exponential for a < 0, and a growing exponential for a > 0.
2. Real and imaginary parts of the signal x (t) are shown in Figs. 7.8 and 7.9 for the case of parameter a being complex-valued. Fig. 7.8 illustrates the possibility of Re {a} < 0. Real and imaginary parts of x (t) exhibit oscillatory behavior with exponential damping. In contrast, if Re {a} > 0, both real and imaginary parts of x (t) grow exponentially as shown in Fig. 7.9.

Since the ROC is to the right of a vertical line, for the case Re {a} < 0 the jω axis is part of the ROC whereas, for the case Re {a} > 0, it is not.

Recall from the previous discussion that the Fourier transform of the signal x (t) is equal to the Laplace transform evaluated on the jω axis of the s-plane. Consequently, the Fourier transform of the signal exists only if the region of convergence includes the jω axis. We conclude that, for the Fourier transform of the signal x (t) to exist, we need Re {a} < 0. Signals in Figs. 7.7(a) and 7.8 have Fourier transforms whereas signals in Figs. 7.7(b) and 7.9 do not. This is consistent with the existence conditions discussed in Chapter 4.

Software resources:

ex_7_5a.m

ex_7_5b. m

Example 7.6: Laplace transform of an anti-causal exponential signal

Find the Laplace transform of the signal

$$x\left(t\right)=-e^{at}u\left(-t\right)$$

where a is any real or complex constant.

Solution: In this case the signal x (t) is anti-causal since it is equal to zero for t > 0. Substituting x (t) into the Laplace transform definition we get

$$X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}-e^{at}u\left(-t\right)e^{-st}dt}$$

Since u (−t) = 1 for t < 0 and u (−t) = 0 for t > 0, changing the upper limit of the integral to t = 0 and dropping the factor u (−t) would have no effect on the result. Therefore

$$\begin{array}{ccc}{X\left(s\right)={\displaystyle \underset{-\infty }{\overset{0}{\int }}-e^{at}{e}^{-st}}dt={\displaystyle \underset{-\infty }{\overset{0}{\int }}-e^{\left(a-s\right)t}dt=-\frac{1}{a-s}{e}^{\left(a-s\right)t}}|}_{-\infty }^0& & \left(7.19\right)\end{array}$$

The parameter a may be real or complex valued. We will assume that it is in the general form a = ar + jai. For the purpose of evaluating the integral in Eqn. (7.19) let us substitute s = σ + jω and write the transform as

$$\begin{array}{ccccc}X\left(s\right)& =& {-\frac{1}{a_r+j{a}_i-\sigma -j\omega }{e}^{\left(a_r+j{a}_i-\sigma -j\omega \right)t}|}_{-\infty }^0& & \\ & =& {-\frac{1}{a_r+j{a}_i-\sigma -j\omega }{e}^{\left(a_r-\sigma \right)t}{e}^{j\left(a_i-\omega \right)t}|}_{-\infty }^0& & \left(7.20\right)\end{array}$$

In contrast with Example 7.5, the critical end of the integral with respect to convergence is its lower limit. To evaluate the result of Eqn. (7.20) at the lower limit, we need

$$a_r-\sigma >0\Rightarrow \sigma <a_r$$

or equivalently

$$\begin{array}{ccc}\mathrm{Re}\left\{s\right\}<\mathrm{Re}\left\{a\right\}& & \left(7.21\right)\end{array}$$

which establishes the ROC for the transform. With the condition in Eqn. (7.21) satisfied, the exponential term in Eqn. (7.20) becomes zero as t → −∞, and we obtain

$$\begin{array}{cc}X\left(s\right)=\frac{1}{a-s}\left[1-0\right]=\frac{1}{s-a},& \mathrm{Re}\left\{s\right\}<\mathrm{Re}\left\{a\right\}\end{array}$$

The ROC is the region to the left of the vertical line that goes through the point s = ar + jai as shown in Fig. 7.10.

As in Example 7.5 we will look at the shape of the signal x (t) in conjunction with the ROC.

1. Fig. 7.11 shows the signal for the case where a is real-valued. For a < 0, the signal grows exponentially as t → −∞. In contrast, for a > 0 the signal decays exponentially.

2. For complex a, real and imaginary parts of x (t) exhibit oscillatory behavior. For Re {a} < 0 the signal grows exponentially as shown in Fig. 7.12. In contrast, oscillations are damped exponentially for Re {a} > 0. This is shown in Fig. 7.13.

   x (t) = —eat u (—t) , a < 0 and real x (t) = —eat u (—t) , a > 0 and real

For the case Re {a} > 0 the jω axis of the s-plane is part of the ROC. For the case Re {a} < 0, however, it is not. Thus, the Fourier transform of the signal x (t) exists only when Re {a} > 0.

Software resources:

ex_7_6a.m

ex_7_6b.m

Examples 7.5 and 7.6 demonstrate a fundamental concept for the Laplace transform: It is possible for two different signals to have the same transform expression for X (s). In both of the examples above we have found the transform

$$X\left(s\right)=\frac{1}{s-a}$$

What separates the two results apart is the ROC. In order for us to uniquely identify which signal among the two led to a particular transform, the ROC must be specified along with the transform. The following two transform pairs will be fundamental in determining to which signal a given transform corresponds:

Sometimes we need to solve the inverse problem of finding the signal x (t) with a given Laplace transform X (s). Given a transform

$$X\left(s\right)=\frac{1}{s-a}$$

we need to know the ROC to determine if the signal x (t) is the one in Eqn. (7.22) or the one in Eqn. (7.23). This will be very important when we work with inverse Laplace transform in Section 7.4 later in this chapter. In order to avoid ambiguity, we will adopt the convention that the ROC is an integral part of the Laplace transform. It must be specified explicitly, or implied by means of another property of the signal or the system under consideration every time a transform is given.

In each of the Examples 7.3, 7.5 and 7.6 we have obtained the Laplace transform of the specified signal in the form of a rational function of s, that is, a ratio of two polynomials in s. In the general case, a rational transform X (s) is expressed in the form

$$X\left(s\right)=K\frac{B\left(s\right)}{A\left(s\right)}$$

where the numerator B (s) and the denominator A (s) are polynomials of s. Let the numerator polynomial be written in factored form as

$$\begin{array}{ccc}B\left(s\right)=\left(s-z_1\right)\left(s-z_2\right)\mathrm{...}\left(s-z_M\right)& & \left(7.25\right)\end{array}$$

where z1, z2,..., zM are its roots. Similarly, let p1, p2,..., pN be the roots of the denominator polynomial so that

$$\begin{array}{ccc}A\left(s\right)=\left(s-p_1\right)\left(s-p_2\right)\mathrm{...}\left(s-p_N\right)& & \left(7.26\right)\end{array}$$

The transform can be written as

$$\begin{array}{ccc}X\left(s\right)=K\frac{\left(s-z_1\right)\left(s-z_2\right)\mathrm{...}\left(s-z_M\right)}{\left(s-p_1\right)\left(s-p_2\right)\mathrm{...}\left(s-p_N\right)}& & \left(7.27\right)\end{array}$$

In Eqns. (7.26) and (7.27) 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 (s). The roots of the numerator polynomial are referred to as the zeros of X (s). In contrast, the roots of the denominator polynomial are the poles of X (s). The transform does not converge at a pole; therefore, the ROC cannot contain any poles. In the next section we will see that the poles of the transform also determine the boundaries of the ROC.

Example 7.7: Laplace transform of a pulse signal

Determine the Laplace transform of the pulse signal

$$x\left(t\right)=\prod \left(\frac{t-\tau /2}{\tau }\right)$$

which is shown in Fig. 7.14.

Solution: The Laplace transform of the signal x (t) is computed as

$${X\left(s\right)={\displaystyle \underset{0}{\overset{\tau }{\int }}\left(1\right)e^{-st}dt=\frac{e^{-st}}{-s}}|}_0^{\tau }=\frac{1-e^{-s\tau }}{s}$$

At a first glance we may be tempted to think that the transform X (s) might not converge at s = 0 since the denominator of X (s) becomes equal to zero at s = 0. We must realize, however, that the numerator of X (s) is also equal to zero at s = 0. Using L’Hospital’s rule on X (s) we obtain its value at s = 0 as

$$X\left(s\right){\left|{}_{s=0}=\frac{\tau e^{-s\tau }}{1}\right|}_{s=0}=\tau$$

confirming the fact that X (s) does indeed converge at s = 0. As a matter of fact, X (s) converges everywhere on the s-plane with the exception of s = −∞±jω. In order to observe this, let us write the Taylor series representation of the exponential term e−sτ:

$$\begin{array}{ccc}{e}^{-s\tau }=1-s\tau +\frac{s^2{\tau }^2}{2!}-\frac{s^3{\tau }^3}{3!}+\frac{s^4{\tau }^4}{4!}-\frac{s^5{\tau }^5}{5!}+...& & \left(7.28\right)\end{array}$$

The numerator of X (s) can be written as

$$1-e^{-s\tau }=s\tau -\frac{s^2{\tau }^2}{2!}+\frac{s^3{\tau }^3}{3!}-\frac{s^4{\tau }^4}{4!}+\frac{s^5{\tau }^5}{5!}-...$$

and the transform X (s) can be written in series form

$$X\left(s\right)=\frac{1-e^{-s\tau }}{s}=\tau -\frac{s{\tau }^2}{2!}+\frac{s^2{\tau }^3}{3!}-\frac{s^3{\tau }^4}{4!}+\frac{s^4{\tau }^5}{5!}-...$$

Thus, the transform X (s) has an infinite number of zeros and no finite poles. Its ROC is

$$\mathrm{Re}\left\{s\right\}>-\infty$$

It will be interesting to determine the zeros of the transform. The roots of the numerator are found by solving the equation

$$\begin{array}{ccc}1-e^{-st}=0,\Rightarrow e^{-s\tau }=1& & \left(7.29\right)\end{array}$$

The value S = 0 is a solution for Eqn. (7.29), however, there must be other solutions as well. Realizing that ej2πk = 1 for all integer values of k, Eqn. (7.29) can be written as

$$e^{-s\tau }=e^{-j2\text{π}k}$$

which leads to the set of solutions

$$\begin{array}{cc}s=\frac{j2\text{π}k}{\tau },& k\text{integer}\end{array}$$

The numerator of X (s) has an infinite number of roots, all on the jω-axis of the s-plane. The roots are uniformly spaced, and occur at integer multiples of j2π/τ. The denominator polynomial of X (s) is just equal to s, and has only one root at s = 0. This single root of the denominator cancels the root of the numerator at s = 0, so that there is neither a zero nor a pole at the origin of the s-plane. We are left with zeros at

$$\begin{array}{cc}{z}_k=\frac{j2\text{π}k}{\tau },& k\text{integer}\end{array},\text{and}k\ne 0$$

Numerator and denominator roots are shown in Fig. 7.15. The pole-zero diagram for X (s) is shown in Fig. 7.16.

To get a sense of what the transform looks like in the s-plane, we will compute and graph the magnitude |X (s)| of the transform as a three-dimensional surface (recall that the transform X (s) is complex-valued, so we can only graph part of it as a surface, that is, we can graph magnitude, phase, real part or imaginary part). Let s = σ + jω, and write the transform as

$$\begin{array}{ccc}{X\left(s\right)|}_{s=\sigma +j\omega }=\frac{1-e^{-\sigma \tau }{e}^{-j\omega \tau }}{\sigma +j\omega }& & \left(7.30\right)\end{array}$$

The squared magnitude of a complex function is computed by multiplying it with its own complex conjugate; therefore

$$\begin{array}{lllll}{\left|X\left(s\right)\right|}^2\hfill & =\hfill & X\left(s\right)X^{*}\left(s\right)\hfill & \hfill & \hfill \\ \hfill & =\hfill & \left(\frac{1-e^{-\sigma \tau }{e}^{-j\omega \tau }}{\sigma +j\omega }\right){\left(\frac{1-e^{-\sigma \tau }{e}^{-j\omega \tau }}{\sigma +j\omega }\right)}^{*}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \left(\frac{1-e^{-\sigma \tau }{e}^{-j\omega \tau }}{\sigma +j\omega }\right)\left(\frac{1-e^{-\sigma \tau }{e}^{-j\omega \tau }}{\sigma -j\omega }\right)\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{1-2e^{-\sigma \tau }\cos \left(\omega \tau \right)+e^{-2\sigma \tau }}{{\sigma }^2+{\omega }^2}\hfill & \hfill & \left(7.31\right)\hfill \end{array}$$

and the magnitude of the transform is

$$\left|X\left(s\right)\right|={\left[\frac{1-2e^{-\sigma \tau }\cos \left(\omega \tau \right)+e^{-2\sigma \tau }}{{\sigma }^2+{\omega }^2}\right]}^{1/2}$$

which is shown in Fig. 7.17(a) as a surface graph. Note how the zeros equally spaced on the jω-axis cause the magnitude surface to dip down. In addition to the surface graph, magnitude values computed at points on the jω-axis of the s-plane are marked on the surface in Fig. 7.17(a). For comparison, the magnitude of the Fourier transform of the signal x (t) computed for the range of angular frequency values −10π/τ < ω < 10π/τ is shown in Fig. 7.17(b). It should be compared to the values marked on the surface graph that correspond to the magnitude of the Laplace transform evaluated on the jω-axis.

Software resources:

ex_7_7a.m

ex_7_7b.m

Example 7.8: Laplace Transform of complex exponential signal

Find the Laplace transform of the signal

$$x\left(t\right)=e^{j{\omega }_{0}t}u\left(t\right)$$

The parameter ω0 is real and positive, and is in radians.

Solution: Applying the Laplace transform definition directly to the signal x (t) leads to

$$X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{j{\omega }_{0}t}u\left(t\right)e^{-st}dt}$$

Since x (t) is causal, the lower limit of the summation can be set to t = 0, and the u (t) term can be dropped to yield

$$\begin{array}{ccc}{X\left(s\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{e}^{j{\omega }_{0}t}{e}^{-st}dt={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{\left(j{\omega }_0-s\right)t}dt}=\frac{e^{\left(j{\omega }_0-s\right)t}}{j{\omega }_0-s}}|}_0^{\infty }& & \left(7.32\right)\end{array}$$

The upper limit is critical in Eqn. (7.32). To be able to evaluate the expression at s → ∞ we need

$$\mathrm{Re}\left\{j{\omega }_0-s\right\}<0$$

and therefore

$$\mathrm{Re}\left\{s\right\}>0$$

Provided that this condition is satisfied, the transform in Eqn. (7.32) is computed as

$$\begin{array}{cc}X\left(s\right)=\frac{1}{j{\omega }_0-s}\left[0-1\right]=\frac{1}{s-j{\omega }_0},& \text{ROC}:\end{array}\mathrm{Re}\left\{s\right\}>0$$

The transform X (s) has one pole at s = jω0, and its ROC is the region to the right of the vertical line that passes through its pole. This is illustrated in Fig. 7.18.

7.2 Characteristics of the Region of Convergence

Through the examples we have worked on so far, we have observed that the Laplace transform of a continuous-time signal always needs to be considered in conjunction with its region of convergence, that is, the region in the s-plane in which the transform converges. In this section we will summarize and justify the fundamental characteristics of the region of convergence.

Let the signal x (t) be equal to zero outside the interval t1 < t < t2. The Laplace transform is

$$\begin{array}{ccc}X\left(s\right)={\displaystyle \underset{t_1}{\overset{t_2}{\int }}x\left(t\right)e^{-st}dt}& & \left(7.33\right)\end{array}$$

If the signal x (t) is absolute integrable, then the transform given by Eqn. (7.33) converges on the vertical line s = jω. Since the limits of the integral are finite, it also converges for all other values of s.

This property is easy to justify. We have established in Eqn. (7.7) of the previous section that, for s = σ + jω, the values of the Laplace transform of the signal x (t) are identical to the values of the Fourier transform of the signal x (t) e−σt. Therefore, the convergence of the Laplace transform for s = σ + jω is equivalent to the convergence of the Fourier transform of the signal x (t) e−σt, and it requires that

$$\begin{array}{ccc}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left|x\left(t\right)e^{-\sigma t}\right|dt<\infty }& & \left(7.34\right)\end{array}$$

Thus, the ROC depends only on σ and not on ω, explaining why the region is in the form of a vertical strip. Following are the possibilities for the ROC:

$$\begin{array}{rrr}\hfill \mathrm{Re}\left\{s\right\}>{\sigma }_1& \hfill :& \hfill \text{To the right of a vertical line}\\ \hfill \mathrm{Re}\left\{s\right\}<{\sigma }_2& \hfill :& \hfill \text{To the left of a vertical line}\\ \hfill {\sigma }_1<\mathrm{Re}\left\{s\right\}<{\sigma }_2& \hfill :& \hfill \text{Between two vertical lines}\end{array}$$

By definition, poles of X (s) are values of s that make the transform infinitely large. For rational Laplace 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 equals zero for all t < 0, its Laplace transform can be written as

$$X\left(s\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}x\left(t\right)e^{-st}dt}$$

Using s = σ + jω and remembering that the convergence of the Laplace transform is equivalent to the signal x (t) e−σt being absolute integrable leads to the convergence condition

$$\begin{array}{ccc}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left|x\left(t\right)e^{-\sigma t}\right|}dt<\infty & & \left(7.35\right)\end{array}$$

If we can find a value of σ for which Eqn. (7.35) is satisfied, then any larger value of σ will satisfy Eqn. (7.35) as well. All we need to do is find the value σ = σ1 for the boundary, and the ROC will be the region to the right of the vertical line that passes through the point s = σ1.

The justification of this property will be similar to that of the previous one. Since an anti-causal signal is equal to zero for all t > 0, its Laplace transform can be written as

$$X\left(s\right)={\displaystyle \underset{-\infty }{\overset{0}{\int }}x\left(t\right)e^{-st}dt}$$

Using s = σ + jω the condition for the convergence of the Laplace transform can be expressed through the equivalent condition for the absolute integrability of the signal x (t) e−σt as

$$\begin{array}{ccc}{\displaystyle \underset{-\infty }{\overset{0}{\int }}\left|x\left(t\right)e^{-\sigma t}\right|dt<\infty }& & \left(7.36\right)\end{array}$$

If we can find a value of σ for which Eqn. (7.36) is satisfied, then any smaller value of σ will satisfy Eqn. (7.36) as well. Once a value σ = σ2 is found for the boundary, the ROC will be the region to the left of the vertical line that passes through the point s = σ2.

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

$$x_R\left(t\right)=x\left(t\right)u\left(t\right)=\{\begin{array}{cc}x\left(t\right),& t>0\\ 0,& t<0\end{array}$$

and

$$x_L\left(t\right)=x\left(t\right)u\left(-t\right)=\{\begin{array}{cc}x\left(t\right),& t<0\\ 0,& t>0\end{array}$$

so that xR (t) is causal, xL (t) is anti-causal, and

$$x\left(t\right)=x_R\left(t\right)+x_L\left(t\right)$$

Let the Laplace transforms of these two signals be

$$\begin{array}{cc}{X}_R\left(s\right)=ℒ\left\{x_R\left(t\right)\right\},& \text{ROC:}\text{Re}\left\{s\right\}>{\sigma }_1\\ X_L\left(s\right)=ℒ\left\{x_L\left(t\right)\right\},& \text{ROC:}\text{Re}\left\{s\right\}>{\sigma }_2\end{array}$$

The Laplace transform of the signal x (t) is

$$\begin{array}{ccc}X\left(s\right)=X_R\left(s\right)+X_L\left(s\right)& & \left(7.37\right)\end{array}$$

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

$$\begin{array}{ccc}{\sigma }_1<\mathrm{Re}\left\{s\right\}<{\sigma }_2& & \left(7.38\right)\end{array}$$

provided that σ2 > σ1 (otherwise there may be no overlap, and the Laplace transform may not exist at any point in the s-plane).

In some cases the ROC may actually be larger than the overlap in Eqn. (7.38) if the addition of the two transforms in Eqn. (7.37) results in the cancellation of a pole that sets one of the boundaries.

Example 7.9: Laplace transform of two-sided exponential signal

Find the Laplace transform of the signal

$$x\left(t\right)=e^{-\left|t\right|}$$

Solution: The specified signal exists for all t, and can be written as

$$x\left(t\right)=e^{-t}u\left(t\right)+e^{t}u\left(-t\right)=\{\begin{array}{cc}{e}^{-t},& t>0\\ e^t,& t<0\end{array}$$

Thus, x (t) is the sum of a causal signal xR (t) and an anti-causal signal xL (t):

$$x\left(t\right)=x_R\left(t\right)+x_L\left(t\right)$$

Two components of x (t) are

$$\begin{array}{c}{x}_R\left(t\right)=e^{-t}u\left(t\right)\\ x_L\left(t\right)=e^{t}u\left(-t\right)\end{array}$$

When we discuss the properties of the Laplace transform later in Section 7.3 we will show that the Laplace transform is linear. Consequently, the transform of the sum of two signals is equal to the sum of their respective transforms, and X (s) can be written as

$$\begin{array}{ccccc}X\left(s\right)& =& X_R\left(s\right)+X_L\left(s\right)& & \\ & =& ℒ\left\{e^{-t}u\left(t\right)\right\}+ℒ\left\{e^{t}u\left(-t\right)\right\}& & \left(7.39\right)\end{array}$$

The individual transforms that make up X (s) in Eqn. (7.39) can be determined by adapting the results obtained earlier in Eqns. (7.22) and (7.23) as:

$$\begin{array}{cccc}ℒ\left\{e^{-t}u\left(t\right)\right\}=\frac{1}{s+1},& \text{ROC}:\mathrm{Re}\left\{s\right\}>-1& & \left(7.40\right)\end{array}$$

$$\begin{array}{cccc}ℒ\left\{e^{t}u\left(-t\right)\right\}=-\frac{1}{s-1},& \text{ROC}:\mathrm{Re}\left\{s\right\}<1& & \left(7.41\right)\end{array}$$

The ROC for XR (s) and XL (s) are shown in Fig. 7.20.

The transform X (s) can be computed as

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

X (s) has a two poles at s = ±1. For this transform to converge both XR (s) and XL (s) must converge. Therefore, the ROC for X (s) is the overlap of the ROCs of XR (s) and XL (s)

and is shown in Fig. 7.21.

Software resources:

ex_7_9.m

7.3 Properties of the Laplace Transform

The important properties of the Laplace transform will be summarized in this section, and the proof of each will be given. The use of these properties greatly simplifies the application of the Laplace transform to problems involving the analysis of continuous-time signals and systems.

7.3.1 Linearity

Laplace transform is linear. For any two transforms x1 (t) and x2 (t) with their respective transforms

$$\begin{array}{c}{x}_1\left(t\right)\stackrel{ℒ}{\leftrightarrow }{X}_1\left(s\right)\\ x_2\left(t\right)\stackrel{ℒ}{\leftrightarrow }{X}_2\left(s\right)\end{array}$$

and any two constants α1 and α2, it can be shown that the following relationship holds:

The Laplace transform of a weighted sum of two signals is equal to the same weighted sum of their respective transforms X1 (s) and X2 (s). The ROC for the resulting transform is at least the overlap of the two individual transforms, if such an overlap exists. The ROC may be greater than the overlap of the two regions if the addition of the two transforms results in the cancellation of a pole.

The linearity property proven above for two signals can be generalized to any arbitrary number of signals. The Laplace transform of a weighted sum of any number of signals is equal to the same weighted sum of their respective transforms.

Example 7.10: Using the linearity property of the Laplace transform

Determine the Laplace transform of the signal

$$x\left(t\right)=2e^{-t}u\left(t\right)+5e^{-2t}u\left(t\right)$$

Solution: A general expression for the Laplace transform of the causal exponential signal was found in Example 7.5 as

$$\begin{array}{cc}ℒ\left\{e^{\alpha t}u\left(t\right)\right\}=\frac{1}{s-a},& \text{ROC:}\text{Re}\end{array}\left\{s\right\}>\mathrm{Re}\left\{a\right\}$$

Applying this result to the exponential terms in x (t) we get

$$\begin{array}{ccc}\begin{array}{cc}ℒ\left\{e^{-t}u\left(t\right)\right\}=\frac{1}{s+1},& \text{ROC:}\text{Re}\end{array}\left\{s\right\}>-1& & \left(7.45\right)\end{array}$$

and

$$\begin{array}{ccc}\begin{array}{cc}ℒ\left\{e^{-2t}u\left(t\right)\right\}=\frac{1}{s+2},& \text{ROC:}\text{Re}\end{array}\left\{s\right\}>-2& & \left(7.46\right)\end{array}$$

Combining the results in Eqns. (7.45) and (7.46) and using the linearity property we arrive at the desired result:

$$X\left(s\right)=2\left(\frac{1}{s+1}\right)+5\left(\frac{1}{s+2}\right)=\frac{7s+9}{s^2+3s+2}$$

The ROC for X (s) is the overlap of the two regions in Eqns. (7.45) and (7.46), namely

$$\mathrm{Re}\left\{s\right\}>-1$$

Fig. 7.22 illustrates the poles and the ROCs of the individual terms X1 (s) and X2 (s) as well as the ROC of the transform X (s) obtained as the overlap of the two.

Example 7.11: Laplace transform of a cosine signal

Find the Laplace transform of the signal

$$x\left(t\right)=\cos \left({\omega }_{0}t\right)u\left(t\right)$$

Solution: Using Euler’s formula for the cos (ω0t) term, the signal x (t) can be written as

$$\cos \left({\omega }_{0}t\right)u\left(t\right)=\frac{1}{2}{e}^{j{\omega }_{0}t}u\left(t\right)+\frac{1}{2}{e}^{-j{\omega }_{0}t}u\left(t\right)$$

and its Laplace transform is

$$\begin{array}{ccc}ℒ\left\{\cos \left({\omega }_{0}t\right)u\left(t\right)\right\}=ℒ\left\{\frac{1}{2}{e}^{j{\omega }_{0}t}u\left(t\right)+\frac{1}{2}{e}^{-j{\omega }_{0}t}u\left(t\right)\right\}& & \left(7.47\right)\end{array}$$

Using the linearity property of the Laplace transform Eqn. (7.47) can be written as

$$\begin{array}{lllll}ℒ\left\{\cos \left({\omega }_{0}t\right)u\left(t\right)\right\}\hfill & =\hfill & \frac{1}{2}ℒ\left\{e^{j{\omega }_{0}t}u\left(t\right)\right\}+\frac{1}{2}ℒ\left\{e^{-j{\omega }_{0}t}u\left(t\right)\right\}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{1/2}{s-j{\omega }_0}+\frac{1/2}{s+j{\omega }_0}\hfill & \hfill & \left(7.48\right)\hfill \end{array}$$

Combining the terms of Eqn. (7.48) under a common denominator we have

$$\begin{array}{ccc}X\left(s\right)=\frac{s}{s^2+{\omega }_0^2}& & \left(7.49\right)\end{array}$$

The transform X (s) has a zero at s = 0 and a pair of complex conjugate poles at s = ±jω0. The ROC is the region to the right of the jω-axis of the s-plane, as illustrated in Fig. 7.23. The jω-axis itself is not included in the ROC since there are poles on it.

Example 7.12: Laplace transform of a sine signal

Find the Laplace transform of the signal

$$x\left(t\right)=\sin \left({\omega }_{0}t\right)u\left(t\right)$$

Solution: This problem is quite similar to that in Example 7.11. Using Euler’s formula for the signal x (t) we get

$$\sin \left({\omega }_{0}t\right)u\left(t\right)=\frac{1}{2j}{e}^{j{\omega }_{0}t}u\left(t\right)-\frac{1}{2j}{e}^{-j{\omega }_{0}t}u\left(t\right)$$

and its Laplace transform is

$$\begin{array}{lllll}ℒ\left\{\sin \left({\omega }_{0}t\right)u\left(t\right)\right\}\hfill & =\hfill & ℒ\left\{\frac{1}{2j}{e}^{j{\omega }_{0}t}u\left(t\right)-\frac{1}{2j}{e}^{-j{\omega }_{0}t}u\left(t\right)\right\}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{1}{2j}ℒ\left\{e^{j{\omega }_{0}t}u\left(t\right)\right\}-\frac{1}{2j}ℒ\left\{e^{-j{\omega }_{0}t}u\left(t\right)\right\}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{1}{2j}\left(\frac{1}{s-j{\omega }_0}\right)-\frac{1}{2j}\left(\frac{1}{s+j{\omega }_0}\right)\hfill & \hfill & \left(7.50\right)\hfill \end{array}$$

Combining the terms of Eqn. (7.50) under a common denominator we have

$$\begin{array}{ccc}X\left(s\right)=\frac{{\omega }_0}{s^2+{\omega }_0^2}& & \left(7.51\right)\end{array}$$

The transform X (s) has a pair of complex conjugate poles at s = ±jω0. The ROC of the transform is Re {s} > 0 as in the previous example.

7.3.2 Time shifting

Time shifting the signal x (t) in the time domain by τ corresponds to multiplication of the transform X (s) by e−Sτ.

Example 7.13: Laplace transform of a pulse signal revisited

The Laplace transform of the pulse signal

$$x\left(t\right)=\prod \left(\frac{t-\tau /2}{\tau }\right)$$

was determined earlier in Example 7.7. Find the same transform through the use of linearity and time-shifting properties.

Solution: The signal x (t) can be expressed as the difference of a unit-step signal and a time-shifted unit-step signal in the form

$$x\left(t\right)=u\left(t\right)-u\left(t-\tau \right)$$

Using the linearity of the Laplace transform, X (s) can be expressed as

$$\begin{array}{lll}X\left(s\right)\hfill & =\hfill & ℒ\left\{u\left(t\right)-u\left(t-\tau \right)\right\}\hfill \\ \hfill & =\hfill & ℒ\left\{u\left(t\right)\right\}-ℒ\left\{u\left(t-\tau \right)\right\}\hfill \end{array}$$

The Laplace transform of the unit-step function was found in Example 7.3 to be

$$\begin{array}{cccc}ℒ\left\{u\left(t\right)\right\}=\frac{1}{s},& \text{ROC:}\text{Re}\left\{s\right\}>0& & \left(7.55\right)\end{array}$$

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

$$\begin{array}{cccc}ℒ\left\{u\left(t-\tau \right)\right\}=\frac{e^{-s\tau }}{s},& \text{ROC:}\text{Re}\left\{s\right\}>0& & \left(7.56\right)\end{array}$$

Subtracting the transform in Eqn. (7.56) from the transform in Eqn. (7.55) yields

$$X\left(s\right)=\frac{1-e^{-s\tau }}{s}$$

which matches the earlier result found in Example 7.7. Since x (t) is a finite-duration signal, the ROC is the entire s-plane with the exception of s = −∞. This is one example of the possibility mentioned earlier regarding the boundaries. The ROC for X (s) is larger than the overlap of the two regions given by Eqns. (7.55) and (7.56). The individual transforms ℒ{u (t)} and ℒ{u (t − τ)} each have a pole at s = 0, and therefore the individual ROCs are to the right of the jω-axis. When the two terms are added to construct X (s), however, the pole at s = 0 is canceled, resulting in the ROC for X (s) being larger than the overlap of the individual regions.

Example 7.14: Laplace transform of a truncated sine function

Determine the Laplace transform of the signal

$$x\left(s\right)=\{\begin{array}{cc}\sin \left(\text{π}t\right),& 0<t<1\\ 0,& \text{otherwise}\end{array}$$

which is graphed in Fig. 7.24.

Solution: We will solve this problem first by direct application of the Laplace transform definition, and then through the use of linearity and shifting properties of the Laplace transform. Applying the definition of the Laplace transform we get

$$\begin{array}{ccc}X\left(s\right)={\displaystyle \underset{0}{\overset{1}{\int }}\sin \left(\text{π}t\right)e^{-s\tau }dt}& & \left(7.57\right)\end{array}$$

Writing sin(πt) using Euler’s formula leads to

$$\sin \left(\text{π}t\right)=\frac{1}{2j}{e}^{j\pi t}-\frac{1}{2j}{e}^{-j\pi t}$$

and substituting this result into Eqn. (7.57) yields

$$X\left(s\right)=\frac{1}{2j}{\displaystyle {\int }_0^1{e}^{-j\pi t}}{e}^{-st}dt-\frac{1}{2j}{\displaystyle {\int }_0^1{e}^{-j\pi t}{e}^{-st}dt}$$

which can be evaluated as

$${X\left(s\right)=\frac{1}{2j}\frac{e^{\left(j\text{π}-s\right)t}}{\left(j\text{π}-s\right)}|}_0^1+{\frac{1}{2j}\frac{e^{-\left(j\text{π+}s\right)t}}{\left(j\text{π}+s\right)}|}_0^1=\frac{\text{π}\left(1+e^{-s}\right)}{s^2+{\text{π}}^2}$$

In Eqn. (7.58) we have used the fact that e±jπ = 1. Since the signal x (t) is of finite duration, the ROC of the transform is the entire s-plane with just the exception of points where Re {s} → −∞.

The solution can be simplified by clever use of the properties of the Laplace transform. The first step is to recognize that the signal x (t) can be expressed as the sum of a causal sinusoidal signal and its time-shifted version as shown in Fig. 7.25.

Mathematically we have

$$\begin{array}{ccc}x\left(t\right)=\sin \left(\text{π}t\right)u\left(t\right)+\sin \left(\text{π}\left[t-1\right]\right)u\left(t-1\right)& & \left(7.58\right)\end{array}$$

Recall that the Laplace transform of a causal sinusoidal signal was found in Example 7.12 as

$$\begin{array}{cc}ℒ\left\{\sin \left({\omega }_{0}t\right)u\left(t\right)\right\}=\frac{{\omega }_0}{s^2+{\omega }_0^2},& \text{ROC:}\text{Re}\left\{s\right\}>0\end{array}$$

Using this result with ω0 = π along with the linearity of the Laplace transform and the time-shifting property, we obtain the transform as

$$\begin{array}{lllll}X\left(s\right)\hfill & =\hfill & ℒ\left\{\sin \left(\text{π}t\right)u\left(t\right)\right\}+ℒ\left\{\sin \left(\text{π}\left[t-1\right]\right)u\left(t-1\right)\right\}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\text{π}}{s^2+{\text{π}}^2}+e^{-s}\frac{\text{π}}{s^2+{\text{π}}^2}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\text{π}\left(1+e^{-s}\right)}{s^2+{\text{π}}^2}\hfill & \hfill & \left(7.59\right)\hfill \end{array}$$

which matches the result found earlier.

Software resources:

ex_7_14.m

7.3.3 Shifting in the s-domain

Let the ROC for original transform X (s) be

$${\sigma }_1<\mathrm{Re}\left\{s\right\}<{\sigma }_2$$

For X (s − s0) to converge, we need

$${\sigma }_1<\mathrm{Re}\left\{s-s_0\right\}<{\sigma }_2$$

Therefore, the ROC for X (s − s0) is

$${\sigma }_1+\mathrm{Re}\left\{s_0\right\}<\mathrm{Re}\left\{s\right\}<{\sigma }_2+\mathrm{Re}\left\{s_0\right\}$$

The ROC for X (s − s0) is a shifted version of the ROC for X (s), shifted horizontally by an amount equal to the real part of the parameter s0. This is illustrated in Fig. 7.26.

Example 7.15: Laplace transform of exponentially damped sinusoidal signal

Determine the Laplace transform of the signal

$$x\left(t\right)=e^{-2t}\cos \left(3t\right)u\left(t\right)$$

Solution: The transform can be found easily by the use of the s-domain shifting property. Let the signal x1 (t) be defined as

$$x_1\left(t\right)=\cos \left(3t\right)u\left(t\right)$$

so that

$$x\left(t\right)=e^{-2t}{x}_1\left(t\right)$$

The Laplace transform of a causal cosine signal was derived in Example 7.11. Using the result found in that example with ω0 = 3 yields

$$\begin{array}{cc}{X}_1\left(s\right)=ℒ\left\{\cos \left(3t\right)u\left(t\right)\right\}=\frac{s}{s^2+9},& \text{ROC}:\end{array}\mathrm{Re}\left\{s\right\}>0$$

We are now ready to apply the s-domain shifting property.

$${X\left(s\right)=X_1\left(s+2\right)=\frac{s}{s^2+9}|}_{s\to s+2}=\frac{s+2}{{\left(s+2\right)}^2+9}$$

The resulting transform has a zero at s = −2 and a pair of complex conjugate poles at s = −2 ± j3 as shown in Fig. 7.27. The boundary for the ROC is set by the two poles, therefore the ROC is

$$\begin{array}{cc}\text{ROC:}& \mathrm{Re}\left\{s\right\}>-2\end{array}$$

7.3.4 Scaling in time and s-domains

The ROC of the result still needs to be related to the ROC for the original transform X (s). Let the latter be

$${\sigma }_1<\mathrm{Re}\left\{s\right\}<{\sigma }_2$$

For the term X (s/a) in Eqn. (7.66) to converge, we need

$$\begin{array}{ccc}{\sigma }_1<\mathrm{Re}\left\{\frac{s}{a}\right\}<{\sigma }_2& & \left(7.67\right)\end{array}$$

Since the parameter a is real-valued, the ROC in Eqn. (7.67) can be written as

$$\begin{array}{ccc}{\sigma }_1<\frac{\mathrm{Re}\left\{s\right\}}{a}<{\sigma }_2& & \left(7.68\right)\end{array}$$

Depending on the sign of the parameter a, two possibilities need to be considered for the ROC of ℒ{x (at)}:

1. $\begin{array}{cc}\text{If}a>0:& a{\sigma }_1<\mathrm{Re}\left\{s\right\}<a{\sigma }_2\end{array}$
2. $\begin{array}{cc}\text{If}a>0:& a{\sigma }_2<\mathrm{Re}\left\{s\right\}<a{\sigma }_1\end{array}$

Example 7.16: Using the scaling property of the Laplace transform

Use the scaling property to find the Laplace transform of the signal

$$x\left(t\right)=e^{2t}u\left(-t\right)$$

from the knowledge of the transform pair

$$\begin{array}{cccc}g\left(t\right)=e^{-2t}u\left(t\right)& \Rightarrow & G\left(s\right)=\frac{1}{s+2},& \text{ROC:}\text{Re}\left\{s\right\}>-2\end{array}$$

Solution: It is obvious from a comparison of the signals x (t) and g (t) that

$$x\left(t\right)=g\left(-t\right)$$

so that the scaling property given by Eqn. (7.62) can be used with the scale factor a = −1 to yield

$$X\left(s\right)=\frac{1}{\left|-1\right|}G\left(\frac{s}{-1}\right)=G\left(-s\right)=\frac{1}{-s+2}$$

and the ROC is

$$\begin{array}{ccc}\mathrm{Re}\left\{-s\right\}>-2& \Rightarrow & \mathrm{Re}\left\{s\right\}<2\end{array}$$

The result found is consistent with Eqn. (7.23).

7.3.5 Differentiation in the time domain

The ROC for the transform of dx (t) /dt is at least equal to the ROC of the original transform. If the original transform X (s) has a single pole at s = 0 that sets the boundary of its ROC, then the cancellation of that pole due to multiplication by s causes the ROC of the new transform sX (s) to be larger.

Example 7.17: Using the time-domain differentiation property of the Laplace transform

Use the time-domain differentiation property to find the Laplace transform of the signal

$$x\left(t\right)=\cos \left(3t\right)u\left(t\right)$$

from the knowledge of the transform pair

$$\begin{array}{cccc}g\left(t\right)=\sin \left(3t\right)u\left(t\right)& \Rightarrow & G\left(s\right)=\frac{3}{s^2+9},& \text{ROC}:\mathrm{Re}\left\{s\right\}>0\end{array}$$

Solution: Let us differentiate the signal g (t):

$$\begin{array}{ccc}\frac{dg\left(t\right)}{dt}=3\cos \left(3t\right)u\left(t\right)& & \left(7.73\right)\end{array}$$

Based on the result in Eqn. (7.73), the signal x (t) can be written as

$$x\left(t\right)=\left(\frac{1}{3}\right)\frac{dg\left(t\right)}{dt}$$

and its Laplace transform is

$$X\left(s\right)=\left(\frac{1}{3}\right)sG\left(s\right)=\frac{s}{s^2+9}$$

as expected. The ROC of X (s) is

$$\mathrm{Re}\left\{s\right\}>0$$

7.3.6 Differentiation in the s-domain

The ROC for the transform of tx (t) is the same as the ROC for the original transform X (s).

Example 7.18: Using the s-domain differentiation property of the Laplace transform

Determine the Laplace transform of the unit-ramp signal

$$r\left(t\right)=tu\left(t\right)$$

Solution: The Laplace transform of the unit-step function is

$$\begin{array}{cc}ℒ\left\{u\left(t\right)\right\}=\frac{1}{s},& \text{ROC:}\end{array}\mathrm{Re}\left\{s\right\}>0$$

Using the s-domain differentiation property, the Laplace transform of the unit-ramp signal is

$$\begin{array}{ccc}R\left(s\right)=ℒ\left\{tu\left(t\right)\right\}=-\frac{d}{ds}\left[\frac{1}{s}\right]=\frac{1}{s^2}& & \left(7.77\right)\end{array}$$

and the ROC of the transform is

$$\mathrm{Re}\left\{s\right\}>0$$

Example 7.19: Transform of a signal using multiple ramp functions

Determine the Laplace transform of the signal x (t) shown in Fig. 7.28.

Solution: The signal x (t) can be expressed in terms of a unit-ramp function and its delayed version as

$$\begin{array}{lll}x\left(t\right)\hfill & =\hfill & r\left(t\right)-r\left(t-1\right)\hfill \\ \hfill & =\hfill & tu\left(t\right)-\left(t-1\right)u\left(t-1\right)\hfill \end{array}$$

The transform of the unit-ramp function was determined in Example 7.18. Using the result in Eqn. (7.77) along with the time-shifting property we obtain

$$X\left(s\right)=\frac{1}{s^2}-e^{-s}\frac{1}{s^2}=\frac{1-e^{-s}}{s^2}$$

with the ROC

$$\mathrm{Re}\left\{s\right\}>0$$

Example 7.20: Transform of an exponentially damped ramp function

Determine the Laplace transform of the signal

$$x\left(t\right)=t{e}^{-2t}u\left(t\right)$$

which is shown in Fig. 7.29.

Solution: The transform of the causal exponential signal e−2t u (t) is

$$\begin{array}{cc}ℒ\left\{e^{-2t}u\left(t\right)\right\}=\frac{1}{s+2},& \text{ROC:}\text{Re}\left\{s\right\}>-2\end{array}$$

Using the s-domain differentiation property

$$X\left(s\right)=ℒ\left\{t{e}^{-2t}u\left(t\right)\right\}=-\frac{d}{ds}\left[\frac{1}{s+2}\right]=\frac{1}{{\left(s+2\right)}^2}$$

with the ROC

$$\mathrm{Re}\left\{s\right\}>-2$$

7.3.7 Convolution property

For any two signals x1 (t) and x2 (t) with their respective transforms

$$x_1\left(t\right)\stackrel{ℒ}{\leftrightarrow }{X}_1\left(s\right)$$

and

$$x_2\left(t\right)\stackrel{ℒ}{\leftrightarrow }{X}_2\left(s\right)$$

it can be shown that the following transform relationship holds:

Convolution property of the Laplace transform is fundamental in its application to CTLTI systems.

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

Convolution property of the Laplace transform is very useful in the sense that it provides an alternative to computing the convolution of two signals directly in the time-domain. Instead, the convolution result x (t) = x1 (t) * x2 (t) can be obtained using the procedure outlined below:

Example 7.21: Using the convolution property of the Laplace transform

Consider two signals x1 (t) and x2 (t) given by

$$x_1\left(t\right)=e^{-t}u\left(t\right)$$

and

$$x_2\left(t\right)=\delta \left(t\right)-e^{-2t}u\left(t\right)$$

Determine

$$x\left(t\right)=x_1\left(t\right)*x_2\left(t\right)$$

using Laplace transform techniques.

Solution: Individual transforms of the signals x1 (t) and x2 (t) are

$$\begin{array}{cccc}{X}_1\left(s\right)=\frac{1}{s+1},& \text{ROC:}\text{Re}\left\{s\right\}>-1& & \left(7.84\right)\end{array}$$

and

$$\begin{array}{cccc}{X}_2\left(s\right)=1-\frac{1}{s+2}=\frac{s+1}{s+2},& \text{ROC:}\text{Re}\left\{s\right\}>-2& & \left(7.85\right)\end{array}$$

Applying the convolution property, the transform of x (t) is the product of the two transforms:

$$X\left(s\right)=X_1\left(s\right)X_2\left(s\right)=\left(\frac{1}{s+1}\right)\left(\frac{s+1}{s+2}\right)=\frac{1}{s+2}$$

The overlap of the two ROCs in Eqns. (7.84) and (7.85) would be Re {s} > −1; however, the pole at s = −1 is cancelled in the process of multiplying the two transforms. Consequently, the ROC for X (s) is

$$\mathrm{Re}\left\{s\right\}>-2$$

as shown in Fig. 7.30.

The signal x (t) is the inverse Laplace transform of X (s):

$$x\left(t\right)={ℒ}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}u\left(t\right)$$

7.3.8 Integration property

The ROC may need to be adjusted. Let the ROC of the original transform X (s) be

$${\sigma }_1<\mathrm{Re}\left\{s\right\}<{\sigma }_2$$

The ROC for the Laplace transform of the unit-step function is

$$\mathrm{Re}\left\{s\right\}>0$$

Therefore, the ROC of the transform W (s) must be at least the overlap of these regions. It may be larger than the overlap if X (s) has a zero at s = 0 to counter the pole at s = 0 introduced by the transform of the unit-step function.

7.4 Inverse Laplace Transform

Consider a transform pair

$$x\left(t\right)\stackrel{ℒ}{\leftrightarrow }X\left(s\right)$$

If X (s) is given, the signal x (t) can be found using the inverse Laplace transform

$$\begin{array}{ccc}x\left(t\right)={ℒ}^{-1}\left\{X\left(s\right)\right\}=\frac{1}{2\text{πj}}{\displaystyle \underset{C}{\int }X\left(s\right)e^{st}ds}& & \left(7.93\right)\end{array}$$

The contour C is a vertical line within the ROC of the transform as shown in Fig. 7.31.

Even though the contour integral in Eqn. (7.93) will not be used in this text for the actual computation of the inverse Laplace transform, it will be explored a bit further to provide insight.

For a rational function X (s) it is usually easier to compute the inverse Laplace transform through the use of partial fraction expansion (PFE).

7.4.1 Partial fraction expansion with simple poles

Consider a rational transform in the form

$$\begin{array}{ccc}x\left(s\right)=\frac{B\left(s\right)}{\left(s-p_1\right)\left(s-p_2\right)...\left(s-p_N\right)}& & \left(7.97\right)\end{array}$$

where the poles p1,p2,... ,p N are distinct. Furthermore, let the order of the numerator polynomial B (s) be less than N, the order of the denominator polynomial. The transform X (s) can be expanded into partial fractions in the form

$$\begin{array}{ccc}X\left(s\right)=\frac{k_1}{s-p_1}+\frac{k_2}{s-p_2}+...+\frac{k_N}{s-p_N}& & \left(7.98\right)\end{array}$$

The coefficients k1,k2,...,k N are called the residues of the partial fraction expansion. They can be computed by (see Appendix E)

$$\begin{array}{cccc}{k_i=\left(s-p_i\right)X\left(s\right)|}_{s=p_i},& i=1,2,\mathrm{...},N& & \left(7.99\right)\end{array}$$

Once the residues are determined, the inverse transform of each term in the partial fraction expansion is determined using either Eqn. (7.22) or Eqn. (7.23), depending on the placement of each pole relative to the ROC.

$$\begin{array}{cc}{ℒ}^{-1}\left\{\frac{1}{s-p_i}\right\}=\{\begin{array}{ll}{e}^{p_{i}t}u\left(t\right),\hfill & \text{if the ROC is to the right of }{p}_i\hfill \\ -e^{p_{i}t}u\left(-t\right),\hfill & \text{if the ROC is to the left of }{p}_i\hfill \end{array}& \left(7.100\right)\end{array}$$

This will be illustrated in the next several examples.

Example 7.22: Inverse Laplace transform using PFE

A causal signal x (t) has the Laplace transform

$$X\left(s\right)=\frac{s+1}{s\left(s+2\right)}$$

Determine x (t) using partial fraction expansion.

Solution: Since x (t) is specified to be causal, the ROC of the transform must be to the right of a vertical line. X (s) has poles at s = −2 and s = 0, therefore the ROC is

$$\mathrm{Re}\left\{s\right\}>0$$

Partial fraction expansion of X (s) is in the form

$$X\left(s\right)=\frac{k_1}{s}+\frac{k_2}{s+2}$$

Residues are found by the application of residue formulas:

$${k_1=sX\left(s\right)|}_{s=0}={\frac{s+1}{s+2}|}_{s=0}=\frac{1}{2}$$

$${k_2=\left(s+2\right)X\left(s\right)|}_{s=-2}={\frac{s+1}{s}|}_{s=-2}=\frac{1}{2}$$

Using the values found, X (s) is

$$\begin{array}{ccc}X\left(s\right)=\frac{1/2}{s}+\frac{1/2}{s+2}& & \left(7.101\right)\end{array}$$

Both terms of X (s) in Eqn. (7.101) correspond to causal terms in the time domain. Use of the transform pair given by Eqn. (7.22) for both terms yields

$$x\left(t\right)=\frac{1}{2}u\left(t\right)+\frac{1}{2}{e}^{-2t}u\left(t\right)$$

Software resources:

ex_7_22a.m

ex_7_22b.m

Example 7.23: Using PFE with complex poles

The Laplace transform of a signal x (t) is

$$X\left(s\right)=\frac{s+1}{s\left(s^2+9\right)}$$

with the ROC specified as

$$\mathrm{Re}\left\{s\right\}>0$$

Determine x (t).

Solution: The transform X (s) can be written in factored form as

$$X\left(s\right)=\frac{s+1}{s\left(s+j3\right)\left(s-j3\right)}$$

and expanded into partial fractions as

$$\begin{array}{ccc}X\left(s\right)=\frac{k_1}{s}+\frac{k_2}{s+j3}+\frac{k_3}{s-j3}& & \left(7.102\right)\end{array}$$

The residues are determined using the residue formulas. The residues associated with complex poles at ±j3 will be complex-valued, however, the method used for computing them is the same.

$$\begin{array}{c}{k_1=sX\left(s\right)|}_{s=0}={\frac{s+1}{\left(s+j3\right)\left(s-j3\right)}|}_{s=0}=\frac{1}{9}\\ {k_2=\left(s+j3\right)X\left(s\right)|}_{s=-j3}={\frac{s+1}{s\left(s-j3\right)}|}_{s=-j3}=-\frac{1}{18}+j\frac{1}{6}\end{array}$$

The residue k3 can be computed using the same method, however, there is a shortcut available. Since the transform X (s) is a rational function with real coefficients, residues of complex conjugate poles must be complex conjugates of each other. Therefore

$$k_3=k_2^{*}=-\frac{1}{18}-j\frac{1}{6}$$

Based on the specified ROC, the signal x (t) is causal. Using Eqn. (7.22) for inverting each term of the partial fraction expansion we get

$$\begin{array}{lll}x\left(t\right)\hfill & =\hfill & k_{1}u\left(t\right)+k_2{e}^{-j3t}u\left(t\right)+k_3{e}^{j3t}u\left(t\right)\hfill \\ \hfill & =\hfill & \frac{1}{9}u\left(t\right)+\left(-\frac{1}{18}+j\frac{1}{6}\right)e^{-j3t}u\left(t\right)+\left(-\frac{1}{18}-j\frac{1}{6}\right)e^{j3t}u\left(t\right)\hfill \end{array}$$

which can be simplified as

$$\begin{array}{lll}x\left(t\right)\hfill & =\hfill & \frac{1}{9}u\left(t\right)-\frac{1}{18}\left[e^{-j3t}+e^{j3t}\right]u\left(t\right)+j\frac{1}{6}\left[e^{-j3t}-e^{j3t}\right]u\left(t\right)\hfill \\ \hfill & =\hfill & \frac{1}{9}u\left(t\right)-\frac{1}{9}\cos \left(3t\right)u\left(t\right)+\frac{1}{3}\sin \left(3t\right)u\left(t\right)\hfill \end{array}$$

An alternative approach would be to combine the two partial fractions with complex poles into a second-order term and write Eqn. (7.102) as

$$X\left(s\right)=\frac{1/9}{s}+\frac{-\frac{1}{9}s+1}{s^2+9}$$

Afterwards the transform pairs

$$\begin{array}{cc}\cos \left(3t\right)u\left(t\right)\stackrel{ℒ}{\leftrightarrow }\frac{s}{s^2+9}\text{and}& \cos \left(3t\right)u\left(t\right)\stackrel{ℒ}{\leftrightarrow }\frac{3}{s^2+9}\end{array}$$

can be used for arriving at the same result.

Software resources:

ex_7_23a.m

ex_7_23b.m

Example 7.24: Using PFE in conjunction with the ROC

The Laplace transform of a signal x (t) is

$$X\left(s\right)=\frac{5\left(s-1\right)}{\left(s+1\right)\left(s+2\right)\left(s-2\right)\left(s-3\right)}$$

with the ROC specified as

$$-1<\mathrm{Re}\left\{s\right\}<2$$

Determine x (t).

Solution: We will first find the partial fraction expansion of X (s). The ROC will be taken into consideration in the next step. Partial fraction expansion of X (s) is in the form

$$X\left(s\right)=\frac{k_1}{s+1}+\frac{k_2}{s+2}+\frac{k_3}{s-2}+\frac{k_4}{s-3}$$

The residues are

$${k_1=\left(s+1\right)X\left(s\right)|}_{s=-1}={\frac{5\left(s-1\right)}{\left(s+2\right)\left(s-2\right)\left(s-3\right)}|}_{s=-1}=-0.8333$$

$${k_2=\left(s+2\right)X\left(s\right)|}_{s=-2}={\frac{5\left(s-1\right)}{\left(s+1\right)\left(s-2\right)\left(s-3\right)}|}_{s=-2}=0.75$$

$${k_3=\left(s-2\right)X\left(s\right)|}_{s=2}={\frac{5\left(s-1\right)}{\left(s+1\right)\left(s+2\right)\left(s-3\right)}|}_{s=2}=-0.4167$$

$${k_4=\left(s-3\right)X\left(s\right)|}_{s=}={\frac{5\left(s-1\right)}{\left(s+1\right)\left(s+2\right)\left(s-2\right)}|}_{s=3}=-0.5$$

The completed form of the partial fraction expansion is

$$X\left(s\right)=-\frac{0.8333}{s+1}+\frac{0.75}{s+2}-\frac{0.4167}{s-2}+\frac{0.5}{s-3}$$

Next we need to pay attention to the ROC which is a vertical strip in the s-plane between the poles at s = −1 and s = 2 as shown in Fig. 7.32.

The shape of the ROC indicates a two-sided signal. Accordingly, we will write the transform as the sum of two transforms belonging to causal and anti-causal components of x (t):

$$X\left(s\right)=X_R\left(s\right)+X_L\left(s\right)$$

The poles at s = −1 and s = −2 are associated with xR (t), the causal component of x (t). Therefore XR (s) can be written as

$$\begin{array}{cc}{X}_R\left(s\right)=-\frac{0.8333}{s+1}+\frac{0.75}{s+2}=\frac{-0.0833\left(s+11\right)}{\left(s+1\right)\left(s+2\right)},& \text{ROC:}\text{Re}\left\{s\right\}>-1\end{array}$$

Conversely the poles at s = 2 and s = 3 are associated with xL (t), the anti-causal component of x (t).

$$\begin{array}{cc}{X}_L\left(s\right)=-\frac{0.4167}{s-2}+\frac{0.5}{s-3}=\frac{0.0833\left(s+3\right)}{\left(s-2\right)\left(s-3\right)};,& \text{ROC:}\text{Re}\left\{s\right\}<2\end{array}$$

The individual ROCs of the transforms XR (s) and XL (s) are shown in Fig. 7.33(a) and (b).

The signal xR (t) is found by using Eqn. (7.22) on the terms of XR (s):

$$x_R\left(t\right)=-0.8333e^{-t}u\left(t\right)+0.75e^{-2t}u\left(t\right)$$

The signal xL (t) is found by using Eqn. (7.23) on the terms of XL (s):

$$x_L\left(t\right)=0.4167e^{2t}u\left(-t\right)-0.5e^{3t}u\left(-t\right)$$

The signal x (t) is the sum of these two components:

$$\begin{array}{lll}x\left(t\right)\hfill & =\hfill & x_R\left(t\right)+x_L\left(t\right)\hfill \\ \hfill & =\hfill & -0.8333e^{-t}u\left(t\right)+0.75e^{-2t}u\left(t\right)+0.4167e^{2t}u\left(-t\right)-0.5e^{3t}u\left(-t\right)\hfill \end{array}$$

Software resources:

ex_7_24a.m

ex_7_24b.m

If the order of the numerator polynomial B (s) in Eqn. (7.97) is equal to or greater than the order of the denominator polynomial, special precautions need to be taken before partial fraction expansion can be used. In this case X (s) must be written as

$$\begin{array}{ccc}X(s)=C(s)+\frac{\overline{B}(s)}{(s-p_1)(s-p_2)...(s-p_N)}& & (7.103)\end{array}$$

where C (s) is a polynomial of S, and the order of the new numerator polynomial $\overline{B}\left(s\right)$ is N − 1.

Example 7.25: Using PFE when numerator order is not less than denominator order

The Laplace transform of a causal signal is

$$X\left(s\right)=\frac{s\left(s+1\right)}{\left(s+2\right)\left(s+3\right)}$$

Determine x (t).

Solution: Since the numerator order is equal to the denominator order, X (s) cannot be expanded into partial fractions directly. Let us first multiply numerator and denominator

$$X\left(s\right)=\frac{s^2+s}{s^2+5s+6}$$

which can be written as

$$X\left(s\right)=1+\frac{-4s-6}{s^2+5s+6}$$

Let X1 (s) be defined as

$$X_1\left(s\right)=\frac{-4s-6}{s^2+5s+6}=\frac{-4s-6}{\left(s+2\right)\left(s+3\right)}$$

so that X (s) = 1 + X1 (s). Expanding X1 (s) into partial fractions yields

$$X_1\left(s\right)=\frac{2}{s+2}-\frac{6}{s+3}$$

Consequently we have

$$X_1\left(t\right)=2e^{-2t}u\left(t\right)-6e^{-3t}u\left(t\right)$$

and

$$\begin{array}{lllll}x\left(t\right)\hfill & =\hfill & \delta \left(t\right)+x_1\left(t\right)\hfill & \hfill & \hfill \\ \hfill & =\hfill & \delta \left(t\right)+2e^{-2t}u\left(t\right)-6e^{-3t}u\left(t\right)\hfill & \hfill & \left(7.104\right)\hfill \end{array}$$

Software resources:

ex_7_25a.m

ex_7_25b.m

7.4.2 Partial fraction expansion with multiple poles

If the transform X (s) has repeated roots, residue calculations become a bit more complicated. Consider a rational transform in the form

$$\begin{array}{ccc}X\left(s\right)=\frac{B\left(s\right)}{{\left(s-p_1\right)}^r\left(s-p_2\right)...\left(s-p_N\right)}& & \left(7.105\right)\end{array}$$

The pole of multiplicity r at s = p1 requires r terms in the partial fraction expansion:

$$\begin{array}{ccc}X\left(s\right)=\frac{k_{1,1}}{s-p_1}+\frac{k_{1,2}}{{\left(s-p_1\right)}^2}+...+\frac{k_{1,r}}{{\left(s-p_1\right)}^r}+\frac{k_2}{s-p_2}+...+\frac{k_N}{s-p_N}& & \left(7.106\right)\end{array}$$

The residues of single poles at p2,...,pN are still computed using the residue formulas discussed earlier in the previous section. The residue k1,r is also easy to compute:

$$\begin{array}{ccc}{k_{1,r}={\left(s-p_1\right)}^{r}X\left(s\right)|}_{s=p_1}& & \left(7.107\right)\end{array}$$

The residue k1,r−1 for the partial fraction with (s − p1)r − 1 in its denominator requires more work, and is computed as

$$\begin{array}{ccc}{k_{1,r-1}=\frac{1}{1!}\frac{d}{ds}\left[{\left(s-p_1\right)}^{r}X\left(s\right)\right]|}_{s=p_1}& & \left(7.108\right)\end{array}$$

The residue k1, r−2 is

$$\begin{array}{ccc}{k_{1,r-2}=\frac{1}{2!}\frac{d^2}{d{s}^2}\left[{\left(s-p_1\right)}^{r}X\left(s\right)\right]|}_{s=p_1}& & \left(7.109\right)\end{array}$$

and so on. See Appendix E for details.

Example 7.26: Multiple-order poles

A causal signal x (t) has the Laplace transform

$$X\left(s\right)=\frac{s-1}{{\left(s-1\right)}^3\left(s+2\right)}$$

Determine x (t) using partial fraction expansion.

Solution: The partial fraction expansion for X (s) is in the form

$$X\left(s\right)=\frac{k_{1,1}}{s+1}+\frac{k_{1,2}}{{\left(s+1\right)}^2}+\frac{k_{1,3}}{{\left(s+1\right)}^3}+\frac{k_2}{s+2}$$

The residue k2 for the single pole at s = 2 is easily determined using the residue formula:

$${k_2=\left(s+2\right)X\left(s\right)|}_{s=-2}{=\frac{s-1}{{\left(s+1\right)}^3}|}_{s=-2}=3$$

The residues of the third-order pole at s = −1 are found using Eqns. (7.107) through

$$\begin{array}{c}{k_{1,3}={\left(s+1\right)}^{3}X\left(s\right)|}_{s=-1}{=\frac{s-1}{s+2}|}_{s=-1}=-2\\ {k_{1,2}=\frac{d}{ds}\left[{\left(s+1\right)}^{3}X\left(s\right)\right]|}_{s=-1}{=\frac{3}{{\left(s+2\right)}^2}|}_{s=-1}=3\\ {k_{1,1}=\frac{1}{2}\frac{d^2}{d{s}^2}\left[{\left(s+1\right)}^{3}X\left(s\right)\right]|}_{s=-1}={\frac{-3}{{\left(s+2\right)}^3}|}_{s=-1}=-3\end{array}$$

and the partial fraction expansion of X (s) is

$$X\left(s\right)=-\frac{3}{s+1}+\frac{3}{{\left(s+1\right)}^2}-\frac{2}{{\left(s+1\right)}^3}+\frac{3}{s+2}$$

Using the transform pairs

$$\begin{array}{c}{e}^{-t}u\left(t\right)\stackrel{ℒ}{\leftrightarrow }\frac{1}{s+1}\\ t{e}^{-t}u\left(t\right)\stackrel{ℒ}{\leftrightarrow }-\frac{d}{ds}\left[\frac{1}{s+1}\right]=\frac{1}{{\left(s+1\right)}^2}\end{array}$$

and

$$t^2{e}^{-t}u\left(t\right)\stackrel{ℒ}{\leftrightarrow }-\frac{d}{ds}\left[\frac{1}{{\left(s+2\right)}^2}\right]=\frac{2}{{\left(s+1\right)}^3}$$

to invert the terms of the partial fraction expansion, we arrive at the solution

$$x\left(t\right)=-3e^{-t}u\left(t\right)+3t{e}^{-t}u\left(t\right)-t^2{e}^{-t}u\left(t\right)+3e^{-2t}u\left(t\right)$$

Software resources:

ex_7_26.m

7.5 Using the Laplace Transform with CTLTI Systems

We have shown in Chapter 2 that the output signal of a CTLTI system can be computed from its input signal and its impulse response through the use of the convolution integral. If the impulse response of a CTLTI system is h (t), and if the signal x (t) is applied to the system as input, the output signal y (t) is found as

$$y\left(t\right)=x\left(t\right)*h\left(t\right)={\displaystyle {\int }_{-\infty }^{\infty }x\left(\lambda \right)h\left(t-\lambda \right)d\lambda }$$

Based on the convolution property of the Laplace transform introduced in Eqn. (7.78) in Section 7.3.7, the transform of the convolution of two signals is equal to the product of their individual transforms. Therefore, the transform of the output signal is

$$\begin{array}{ccc}Y\left(s\right)=X\left(s\right)H\left(s\right)& & \left(7.110\right)\end{array}$$

Solving Eqn. (7.110) for H (s) we get

$$\begin{array}{ccc}H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}& & \left(7.111\right)\end{array}$$

The function H (s) is the s-domain system function of the system under consideration.

We already know that a CTLTI system can be completely and uniquely described by means of its impulse response h (t). Since the system function H (s) is just the Laplace transform of the impulse response h (t), it also represents a complete description of the CTLTI system.

7.5.1 Relating the system function to the differential equation

As discussed in Section 2.4 of Chapter 2, the input-output relationship of a CTLTI system can be modeled by a constant-coefficient linear differential equation given in the standard form

$$\begin{array}{ccc}{\displaystyle \sum _{k=0}^N{a}_k}\frac{d^{k}y\left(t\right)}{d{t}^k}={\displaystyle \sum _{k=0}^M{b}_k}\frac{d^{k}x\left(t\right)}{d{t}^k}& & \left(7.112\right)\end{array}$$

The order of the CTLTI system is the larger of M and N. Up to this point we have considered three different forms of modeling for a CTLTI system, namely the differential equation, the impulse response and the system function. (A fourth method, state-space modeling, will be introduced in Chapter 9.) It must be possible to obtain any of the three models from the knowledge of any other. In this section we will focus on the problem of determining the system function from the differential equation. If we take the Laplace transform of both sides of Eqn. (7.112) the equality would still be valid:

$$\begin{array}{ccc}ℒ\left\{{\displaystyle \sum _{k=0}^N{a}_k}\frac{d^{k}y\left(t\right)}{d{t}^k}\right\}=ℒ\left\{{\displaystyle \sum _{k=0}^M{b}_k}\frac{d^{k}x\left(t\right)}{d{t}^k}\right\}& & \left(7.113\right)\end{array}$$

Laplace transform is linear; therefore, the transform of a summation is equal to the sum of individual terms, allowing us to write Eqn. (7.113) as

$$\begin{array}{ccc}{\displaystyle \sum _{k=0}^Nℒ\left\{a_k\frac{d^{k}y\left(t\right)}{d{t}^k}\right\}={\displaystyle \sum _{k=0}^Mℒ}\left\{b_k\frac{d^{k}x\left(t\right)}{d{t}^k}\right\}}& & \left(7.114\right)\end{array}$$

and subsequently as

$$\begin{array}{ccc}{\displaystyle \sum _{k=0}^N{a}_k}ℒ\left\{\frac{d^{k}y\left(t\right)}{d{t}^k}\right\}={\displaystyle \sum _{k=0}^M{b}_k}ℒ\left\{\frac{d^{k}x\left(t\right)}{d{t}^k}\right\}& & \left(7.115\right)\end{array}$$

Using the time-domain differentiation property of the Laplace transform, given by Eqn. (7.69), in Eqn. (7.115) yields

$$\begin{array}{ccc}{\displaystyle \sum _{k=0}^N{a}_k}{s}^{k}Y\left(s\right)={\displaystyle \sum _{k=0}^M{b}_k{s}^k}X\left(s\right)& & \left(7.116\right)\end{array}$$

The transforms X (s) and Y (s) do not depend on the summation indices k, and can therefore be factored out of the summations in Eqn. (7.116) resulting in

$$\begin{array}{ccc}Y\left(s\right){\displaystyle \sum _{k=0}^N{a}_k{s}^k}=X\left(s\right){\displaystyle \sum _{k=0}^M{b}_k{s}^k}& & \left(7.117\right)\end{array}$$

The system function can now be obtained from Eqn. (7.117) as

$$\begin{array}{ccc}H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{{\displaystyle \sum _{k=0}^M{b}_k{s}^k}}{{\displaystyle \sum _{k=0}^N{a}_k{s}^k}}& & \left(7.118\right)\end{array}$$

System function, linearity, and initial conditions

Two important observations will be made at this point:

1. The development leading up to the s-domain system function in Eqn. (7.111) has relied heavily on the convolution operation and the convolution property of the Laplace transform. We know from Chapter 2 that the convolution operation is only applicable to problems involving linear and time-invariant systems. Therefore it follows that 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 2.4 of Chapter 2 that a constant-coefficient differential equation corresponds to a linear and time-invariant system only if all initial conditions are set equal to zero.

If we need to use Laplace transform-based techniques to solve a differential equation subject to non-zero initial conditions, that can be done through the use of the unilateral Laplace transform, but not through the use of the system function. The unilateral Laplace transform and its use for solving differential equations will be discussed in Section 7.7.

Example 7.27: Finding the system function from the differential equation

A CTLTI system is defined by means of the differential equation

$$\frac{d^{3}y\left(t\right)}{d{t}^3}+5\frac{d^{2}y\left(t\right)}{d{t}^2}+17\frac{dy\left(t\right)}{dt}+13y\left(t\right)=\frac{d^{2}x\left(t\right)}{d{t}^2}+x\left(t\right)$$

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

Solution: We will assume that all initial conditions are equal to zero, and take the Laplace transform of each side of the differential equation to obtain

$$\begin{array}{ccc}{s}^{3}Y\left(s\right)+5s^{2}Y\left(s\right)+17sY\left(s\right)+13Y\left(s\right)=s^{2}X\left(s\right)+X\left(s\right)& & \left(7.119\right)\end{array}$$

The system function can be obtained from Eqn. (7.119) as

$$H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{s^2+1}{s^3+5s^2+17s+13}$$

Characteristic polynomial vs. the denominator of the system function

Another important observation will be made based on the result obtained in Example 7.27: The characteristic equation for the system considered in Example 7.27 is

$$s^3+5s^2+17s+13=0$$

and the solutions of the characteristic equation are the modes of the system as defined in Section 2.5.3 of Chapter 2. When we find the system function H (s) from the differential 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 s-domain, and consequently, they are identical to the modes of the differential equation of the system.

Recall that in Section 2.5.3 we have reached some conclusions about the relationship between the modes of the differential 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 of the system are real-valued and distinct, then the transient response of the system is in the form

$$y\left[n\right]={\displaystyle \sum _{k=1}^N{c}_k{e}^{pkt}}$$

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

$$\begin{array}{ccc}{p}_{1a}={\sigma }_1+j{\omega }_1,& \text{and}& p_{1b}={\sigma }_1-j{\omega }_1\end{array}$$

yields a response of the type

$$y_1\left(t\right)=d_1{e}^{{\sigma }_{1}t}\cos \left({\omega }_{1}t+{\theta }_1\right)$$

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

$$y_1\left(t\right)=c_{11}{e}^{p_{1}t}+c_{12}t{e}^{p_{1}t}+...+c_{1m}{t}^{m-1}{e}^{p_{1}t}+\text{other}\text{terms}$$

regardless of whether p1 is real or complex-valued. Justifications for these relationships were given in Section 2.5.3 of Chapter 2 through the use of the modes of the differential equation, and will not be repeated here.

Sometimes we need to reverse the problem represented in Example 7.27 and find the differential equation from the knowledge of the system function. The next three examples will demonstrate this.

Example 7.28: Finding the differential equation from the system function

A causal CTLTI system is defined by the system function

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

Find a differential equation for this system.

Solution: The system function is the ratio of the output transform to the input transform, that is,

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

Therefore we can write

$$\begin{array}{ccc}\left(s^2+5s+6\right)Y\left(s\right)=\left(2s+5\right)X\left(s\right)& & \left(7.120\right)\end{array}$$

The differential equation follows from Eqn. (7.120) as

$$\frac{d^{2}y\left(t\right)}{d{t}^2}+5\frac{dy\left(t\right)}{dt}+6y\left(t\right)=2\frac{dx\left(t\right)}{dt}+5x\left(t\right)$$

Example 7.29: Finding the differential equation from input and output signals

The unit-step response of a CTLTI system is

$$y\left(t\right)=\text{Sys}\left\{u\left(t\right)\right\}=\left(2-4e^{-t}+2e^{-2t}\right)u\left(t\right)$$

Find a differential equation for this system.

Solution: Since the input signal is a unit-step function, that is, x (t) = u (t), its Laplace transform is

$$X\left(s\right)=\frac{1}{s}$$

The Laplace transform of the output signal is found as

$$Y\left(s\right)=\frac{2}{s}-\frac{4}{s+1}+\frac{2}{s+2}=\frac{4}{s\left(s+1\right)\left(s+2\right)}$$

The system function can be obtained as the ratio of the output transform to the input transform:

$$H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{4}{\left(s+1\right)\left(s+2\right)}=\frac{4}{s^2+3s+2}$$

The differential equation follows from the system function as

$$\frac{d^{2}y\left(t\right)}{d{t}^2}+3\frac{dy\left(t\right)}{dt}+2y\left(t\right)=4x\left(t\right)$$

Software resources:

ex_7_29.m

Example 7.30: Finding the impulse response from input and output signals

Determine the impulse response of the CTLTI system system the unit-step response of which was given in Example 7.29.

Solution: The system function was found in Example 7.29 to be

$$H\left(s\right)=\frac{4}{s^2+3s+2}$$

which can be expanded into partial fractions to yield

$$H\left(s\right)=\frac{4}{s+1}-\frac{4}{s+2}$$

The impulse response is

$$h\left(t\right)=\left(4e^{-t}-4e^{-2t}\right)u\left(t\right)$$

It can be easily verified that the unit-step response specified in Example 7.29 for this system is indeed the convolution of h (t) found above with the unit-step function.

Software resources:

ex_7_30.m

7.5.2 Response of a CTLTI system to a complex exponential signal

In this section we will consider the response of a CTLTI system to a complex exponential input signal. This will help us gain further insight into the system function concept.

Let a CTLTI system with impulse response h (t) be driven by a complex exponential input signal in the form

$$x\left(t\right)=e^{s_{0}t}$$

where s0 represents a point in the s-plane within the ROC of the system function. The output signal can be determined through the use of the convolution integral

$$\begin{array}{ccc}y\left(t\right)=h\left(t\right)*x\left(t\right)={\displaystyle {\int }_{-\infty }^{\infty }h\left(\lambda \right)x\left(t-\lambda \right)d\lambda }& & \left(7.121\right)\end{array}$$

Substituting $x\left(t-\lambda \right)=e^{s_0\left(t-\lambda \right)}$ into Eqn. (7.121) and simplifying the resulting integral we obtain

$$\begin{array}{lllll}y\left(t\right)\hfill & =\hfill & {\displaystyle {\int }_{-\infty }^{\infty }h\left(\lambda \right)e^{s_0\left(t-\lambda \right)}d\lambda }\hfill & \hfill & \hfill \\ \hfill & =\hfill & e^{s_{0}t}\underset{H\left(s_0\right)}{\underbrace{{\displaystyle {\int }_{-\infty }^{\infty }h\left(\lambda \right)e^{-s_{0}d\lambda }}}}\hfill & \hfill & \left(7.122\right)\hfill \end{array}$$

The integral in Eqn. (7.122) should be recognized as the system function H (s) evaluated at the point s = s0 in the s-plane. Therefore, we reach the following important conclusion:

In Eqn. (7.123) the complex exponential input signal is assumed to have been in existence forever; it is not turned on at a specific time instant. Therefore, the response found is the steady-state response of the system.

Example 7.31: Response to a complex exponential signal

A CTLTI system with the system function

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

is driven by the complex exponential input signal

$$x\left(t\right)=e^{\left(-0.4+j4\right)t}$$

Determine the steady-state response of the system.

Solution: The input signal is complex-valued, and can be written in Cartesian form using Euler’s formula:

$$\begin{array}{ccc}x\left(t\right)=e^{-0.4t}\left[\cos \left(4t\right)+j\sin \left(4t\right)\right]& & \left(7.124\right)\end{array}$$

Real and imaginary parts of x (t) are shown in Fig. 7.34.

The value of the system function at s = s0 = −0.4 + j4 is

$$\begin{array}{lll}H\left(-0.4+j4\right)\hfill & =\hfill & \frac{\left(-0.4+j4\right)+6}{{\left(-0.4+j4\right)}^2+5\left(-0.4+j4\right)+6}\hfill \\ \hfill & =\hfill & 0.0021-j0.3348\hfill \\ \hfill & =\hfill & 0.3348e^{-j1.5645}\hfill \end{array}$$

The output signal is found using Eqn. (7.123) as

$$\begin{array}{lll}y\left(t\right)\hfill & =\hfill & e^{\left(-0.4+j4\right)t}\left(0.3348e^{-j1.5645}\right)\hfill \\ \hfill & =\hfill & 0.3348e^{-0.4t}{e}^{j\left(4t-1.5645\right)}\hfill \end{array}$$

or, in Cartesian form as

$$\begin{array}{ccc}y\left(t\right)=0.3348e^{-0.4t}\cos \left(4t-1.5645\right)+j0.3348e^{-0.4t}\sin \left(4t-1.5645\right)& & \left(7.125\right)\end{array}$$

Real and imaginary parts of the output signal y (t) are shown in Fig. 7.35.

Software resources:

ex_7_31.m

7.5.3 Response of a CTLTI system to an exponentially damped sinusoid

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

$$\begin{array}{ccc}x\left(t\right)=e^{{\sigma }_{0}t}\cos \left({\omega }_{0}t\right)& & \left(7.126\right)\end{array}$$

The derivation outlined in Eqns. (7.127) through (7.137) can be summarized as shown in Fig. 7.36.

Comparison of the input signal in Eqn. (7.126) and the output signal Eqn. (7.137) reveals the following:

Example 7.32: Response to an exponentially damped sinusoid

Consider again the system function used in Example 7.31. Determine the response of the system to the input signal

$$x\left(t\right)=e^{-0.3t}\cos \left(2t\right)$$

Solution: Let s0 be defined as

$$s_0=-0.3+j2$$

The system function evaluated at s = s0 is

$$H\left(-0.3+j2\right)=\frac{\left(-0.3+j2\right)+6}{{\left(-0.3+j2\right)}^2+5\left(-0.3+j2\right)+6}=0.2965-j0.6297$$

or in polar form

$$H\left(-0.3+j2\right)=0.6849e^{-j1.1664}$$

The output signal is found using Eqn. (7.137):

$$y\left(t\right)=0.6849e^{-0.3t}\cos \left(2t-1.1664\right)$$

The input and the output signals are shown in Fig. 7.37.

Software resources:

ex_7_32.m

7.5.4 Pole-zero plot for a system function

A rational system function H (s) can be expressed in pole-zero form as

$$\begin{array}{ccc}H\left(s\right)=K\frac{\left(s-z_1\right)\left(s-z_2\right)...\left(s-z_M\right)}{\left(s-p_1\right)\left(s-p_2\right)...\left(s-p_N\right)}& & \left(7.138\right)\end{array}$$

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 system. The roots z1,..., zM of the numerator polynomial are referred to as the zeros of the system function. In contrast, the roots of the denominator polynomial are the poles of the system function. A pole-zero plot for a system function is obtained by marking the poles and the zeros of the system function on the s-plane. It is customary to use “x” and “o” to mark each pole and each zero respectively.

Example 7.33: Pole-zero plot for system function

Construct a pole-zero plot for a CTLTI system with system function

$$H\left(s\right)=\frac{s^2+1}{s^3+5s^2+17s+13}$$

Solution: The zeros of the system function are the roots of the numerator polynomial, and are found by solving the equation

$$s^2+1=0$$

which yields z1 = j and z2 = −j. The poles of the system function are the roots of the denominator polynomial or, equivalently, the solutions of the equation

$$s^3+5s^2+17s+13=0$$

The three poles are at p1 = −1, p2 = −2 + j3 and p3 = −2 − j3. The pole-zero diagram for the system function is shown in Fig. 7.38.

Software resources:

ex_7_33.m

Example 7.34: ROC and the pole-zero plot

Assume that the system function used in Example 7.33 represents a causal system. Indicate the ROC on the pole-zero plot.

Solution: Following is the knowledge we possess:

1. For a causal system, the ROC of the system function must be the area to the right of a vertical line.

2. The ROC must be bound by one or more poles.

3. There may be no poles within the ROC.

Consequently, the ROC of the system function must be

$$\begin{array}{cc}\text{ROC:}& \mathrm{Re}\left\{s\right\}>-1\end{array}$$

which is shown in Fig. 7.39.

7.5.5 Graphical interpretation of the pole-zero plot

In this section we will explore the significance of the geometric placement of poles and zeros of the system function. The pole-zero plot, introduced in Section 7.5.4, can be used for understanding the behavior characteristics of a CTLTI system, especially in terms of the magnitude and the phase of the system function. The relationship between the locations of the poles and the zeros of the system function and the frequency-domain behavior of the system is significant.

Let us begin by considering the pole-zero form of the system function given by Eqn. (7.138). Assuming the system is stable, the Fourier transform-based system function H (ω) exists, and can be found by evaluating H (s) for s = jω:

$$\begin{array}{ccc}{H\left(\omega \right)=H\left(s\right)|}_{s=j\omega }=K\frac{\left(j\omega -z_1\right)\left(j\omega -z_2\right)...\left(j\omega -z_M\right)}{\left(j\omega -p_1\right)\left(j\omega -p_2\right)...\left(j\omega -p_N\right)}& & \left(7.139\right)\end{array}$$

The numerator of Eqn. (7.139) has M factors each in the form

$$\begin{array}{cccc}{B}_i\left(\omega \right)=\left(j\omega -z_i\right),& i=1,...,M& & \left(7.140\right)\end{array}$$

Similarly, the denominator of Eqn. (7.139) consists of factors in the form

$$\begin{array}{cccc}{A}_i\left(\omega \right)=\left(j\omega -p_i\right),& i=1,...,N& & \left(7.141\right)\end{array}$$

In a sense, the function Bi (ω) describes the contribution of the zero at s = zi to the frequency response of the system. Similarly, the function Ai (ω) describes the contribution of the pole at s = pi. Using the definitions in Eqns. (7.140) and (7.141), H (ω) becomes

$$\begin{array}{ccc}H\left(\omega \right)=K\frac{B_1\left(\omega \right)B_2\left(\omega \right)...B_M\left(\omega \right)}{A_1\left(\omega \right)A_2\left(\omega \right)...A_N\left(\omega \right)}& & \left(7.142\right)\end{array}$$

If we wanted to compute the frequency response of the system at a specific frequency ω = ω0 we would get

$$\begin{array}{ccc}H\left({\omega }_0\right)=K\frac{B_1\left({\omega }_0\right)B_2\left({\omega }_0\right)...B_M\left({\omega }_0\right)}{A_1\left({\omega }_0\right)A_2\left({\omega }_0\right)...A_N\left({\omega }_0\right)}& & \left(7.143\right)\end{array}$$

The magnitude of the frequency response at ω = ω0 is found by computing the magnitude of each complex function Bi (ω0) and Ai (ω0), and forming the ratio

$$\begin{array}{ccc}\left|H\left({\omega }_0\right)\right|=K\frac{\left|B_1\left({\omega }_0\right)\right|\cdot \left|B_2\left({\omega }_0\right)\right|...\left|B_M\left({\omega }_0\right)\right|}{\left|A_1\left({\omega }_0\right)\right|\cdot \left|A_2\left({\omega }_0\right)\right|...\left|A_N\left({\omega }_0\right)\right|}& & \left(7.144\right)\end{array}$$

The phase of the frequency response is computed as the algebraic sum of the phases of numerator and denominator factors:

$$\begin{array}{ccc}\measuredangle H\left({\omega }_0\right)=\measuredangle B_1\left({\omega }_0\right)+\measuredangle B_2\left({\omega }_0\right)+...+\measuredangle B_M\left({\omega }_0\right)-\measuredangle A_1\left({\omega }_0\right)-\measuredangle A_2\left({\omega }_0\right)-...-\measuredangle A_N\left({\omega }_0\right)& & \left(7.145\right)\end{array}$$

We would like to gain a graphical understanding of the computations in Eqns. (7.144) and (7.145). Let us focus on one of the numerator terms, Bi (ω0):

$$\begin{array}{ccc}{B}_i\left({\omega }_0\right)=j{\omega }_0-z_i& & \left(7.146\right)\end{array}$$

Clearly, B (ω0) is a complex quantity. Its two terms zi and jω0 are shown in Fig. 7.40(a) as points in the s-plane. It is also possible to represent the terms in Eqn. (7.146) with vectors. Treating each term in Eqn. (7.146) as a vector, we obtain the corresponding vector expression

$$\begin{array}{ccc}\overrightarrow{B_i\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{z_i}& & \left(7.147\right)\end{array}$$

which can be written in alternative form

$$\begin{array}{ccc}\overrightarrow{z_i}+\overrightarrow{B_i\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}& & \left(7.148\right)\end{array}$$

We will draw the vector $\overrightarrow{z_i}$ starting at the origin and ending at the point s = zi. Similarly the vector $\overrightarrow{j{\omega }_0}$ will be drawn starting at the origin and ending at the point s = jω0. These two vectors are shown in Fig. 7.40(b). The relationship expressed by Eqn. (7.148) is illustrated in Fig. 7.40(c). The following important conclusions can be drawn from Fig. 7.40(c):

If Fig. 7.40(c) is drawn to scale, the norm and the angle of the vector $\overrightarrow{B_i\left({\omega }_0\right)}$ could simply be measured to determine the contributions of the zero at s = zi to the magnitude and the phase expressions for $\left|H\left({\omega }_0\right)\right|$ and $\measuredangle H\left({\omega }_0\right)$ in Eqns. (7.144) and (7.145).

What if we need to determine the contributions of the zero at s = zi to the magnitude $\left|H\left({\omega }_0\right)\right|$ and phase $\measuredangle H\left({\omega }_0\right)$ not just for a specific frequency ω0 but for all frequencies ω? Imagine the vector $\overrightarrow{B_i\left({\omega }_0\right)}$ to be a piece of rubber band, the tail end of which is permanently attached to the point s = zi. Assume that, as the frequency ω varies, the tip of the rubber band vector moves on the jω-axis. The length of the rubber band and its angle with the positive real axis vary as the tip is moved. The contributions of the zero at s = zi to the magnitude and the phase of the frequency response H (ω) can be graphed by tracking these variations as shown in Fig. 7.41.

Similar analysis can be carried out for determining the contribution of a pole pi to magnitude and phase characteristics of the system. Fig. 7.42 illustrates the effect of a pole on the frequency response.

For a system with M zeros and N poles, magnitude and phase of the system function at ω = ω0 are found by taking into account the contributions of each zero and pole. The next example will illustrate this.

Example 7.35: Frequency response from pole-zero plot

A CTLTI system is described by the system function

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

Construct a pole-zero plot and use it to determine the magnitude and the phase of the frequency response of the system at the frequency ω0 = 1.5 rad/s.

Solution: The system has two zeros and two poles. The zeros of the system function are at

$$\begin{array}{cc}{z}_1=1,& z_2=-2\end{array}$$

and the poles are at

$$\begin{array}{cc}{p}_1=-1+j2,& p_2=-1-j2\end{array}$$

Vector forms of the contribution of each zero and pole to the system function at ω = ω0

can be written using Eqns. (7.140) and (7.141) as

$$\begin{array}{c}\overrightarrow{B_1\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{z_1}=\overrightarrow{\left(-1+j1.5\right)}\\ \overrightarrow{B_2\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{z_2}=\overrightarrow{\left(2+j1.5\right)}\\ \overrightarrow{A_1\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{p_1}=\overrightarrow{\left(1-j0.5\right)}\\ \overrightarrow{A_2\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{p_2}=\overrightarrow{\left(1+j3.5\right)}\end{array}$$

The vectors $\overrightarrow{B_1\left({\omega }_0\right)}$, $\overrightarrow{B_2\left({\omega }_0\right)}$, $\overrightarrow{A_1\left({\omega }_0\right)}$, and $\overrightarrow{A_2\left({\omega }_0\right)}$ are shown in Fig.7.43.

The norm and the phase of each vector are

$$\begin{array}{ll}\left|\overrightarrow{B_1\left({\omega }_0\right)}\right|=1.8028,\hfill & \measuredangle \overrightarrow{B_1\left({\omega }_0\right)}=2.1588\text{rad}\hfill \\ \left|\overrightarrow{B_2\left({\omega }_0\right)}\right|=2.5000,\hfill & \measuredangle \overrightarrow{B_2\left({\omega }_0\right)}=0.6435\text{rad}\hfill \\ \left|\overrightarrow{A_1\left({\omega }_0\right)}\right|=1.1180,\hfill & \measuredangle \overrightarrow{A_1\left({\omega }_0\right)}=-0.4636\text{rad}\hfill \\ \left|\overrightarrow{A_2\left({\omega }_0\right)}\right|=3.6401,\hfill & \measuredangle \overrightarrow{A_2\left({\omega }_0\right)}=1.2925\text{rad}\hfill \end{array}$$

Based on the values measured, the magnitude of the frequency response at ω0 is computed as

$$\begin{array}{ccc}\left|H\left({\omega }_0\right)\right|=\frac{\left|B_1\left({\omega }_0\right)\right|\cdot \left|B_2\left({\omega }_0\right)\right|}{\left|A_1\left({\omega }_0\right)\right|\cdot \left|A_2\left({\omega }_0\right)\right|}=\frac{\left(1.8028\right)\left(2.5000\right)}{\left(1.1180\right)\left(3.6401\right)}=1.1074& & \left(7.149\right)\end{array}$$

The phase of the frequency response is computed as

$$\begin{array}{lll}\measuredangle H\left({\omega }_0\right)\hfill & =\hfill & \measuredangle B_1\left({\omega }_0\right)+\measuredangle B_2\left({\omega }_0\right)-\measuredangle A_1\left({\omega }_0\right)-\measuredangle A_2\left({\omega }_0\right)\hfill \\ \hfill & =\hfill & 2.1588+0.6435-\left(-0.4636\right)-1.2925\hfill \\ \hfill & =\hfill & 1.9735\text{rad}\hfill \end{array}$$

Complete magnitude and phase characteristics for the system can be obtained by repeating this process for all values of ω that are of interest. Fig. 7.44 shows the magnitude and the phase of the system. The values at ω0 = 1.5 rad/s are marked on magnitude and phase plots.

Software resources:

ex_7_35.m

7.5.6 System function and causality

Causality in linear and time-invariant systems was discussed in Section 2.8 of Chapter 2. For a CTLTI system to be causal, its impulse response h (t) needs to be equal to zero for t < 0. Thus, by changing the lower limit of the integral to t = 0 in the definition of the Laplace transform, the s-domain system function for a causal CTLTI system can be written as

$$\begin{array}{ccc}H\left(s\right)={\displaystyle {\int }_{-\infty }^{\infty }h\left(t\right)e^{-st}dt={\displaystyle {\int }_0^{\infty }h\left(t\right)e^{-st}dt}}& & \left(7.150\right)\end{array}$$

As we have discussed in Section 7.2 of this chapter, the ROC for the system function of a causal system is to the right of a vertical line in the s-plane. As a consequence, the system function must also converge at Re {s} → ∞. Consider a system function in the form

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

For the system described by H (s) to be causal we need

$$\begin{array}{ccc}\underset{s\to \infty }{\lim }\left[H\left(s\right)\right]=\underset{s\to \infty }{\lim }\left[\frac{b_M}{a_N}{s}^{M-N}\right]<\infty & & \left(7.151\right)\end{array}$$

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.

7.5.7 System function and stability

In Section 2.9 of Chapter 2 we have concluded that for a CTLTI system to be stable its impulse response must be absolute integrable, that is,

$${\displaystyle {\int }_{-\infty }^{\infty }\left|h\left(t\right)\right|<\infty }$$

Furthermore, we have established in Section 4.3.2 of Chapter 4 that the Fourier transform of a signal exists if the signal is absolute integrable. But the Fourier transform of the impulse response is equal to the s-domain system function evaluated on the jω-axis of the s-plane, that is,

$${H\left(\omega \right)=H\left(s\right)|}_{s=j\omega }$$

provided that the jω-axis of the s-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 to the right of a vertical line in the s-plane, and is expressed in the form

   $$\mathrm{Re}\left\{s\right\}>{\sigma }_1$$

   For the ROC to include the jω-axis we need σ1 < 0. Since the ROC cannot have any poles in it, all the poles of the system function must be on or to the left of the vertical line σ = σ1.

   For a causal system to be stable, the system function must not have any poles on the jω-axis or in the right half s-plane.

2. Anti-causal system:

   If the system is anti-causal, its impulse response is equal to zero for t ≥ 0. The ROC for the system function is to the left of a vertical line in the s-plane, and is expressed in the form

   $$\mathrm{Re}\left\{s\right\}>{\sigma }_2$$

   For the ROC to include the jω-axis we need σ2 > 0. All the poles of the system function must reside on or to the right of the vertical line σ = σ2.

   For an anti-causal system to be stable, the system function must not have any poles on the jω-axis or in the left half s-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 vertical lines at σ = σ1 and σ = σ2, and is expressed in the form

   $${\sigma }_1<\mathrm{Re}\left\{s\right\}<{\sigma }_2$$

   For stability we need σ1 < 0 and σ2 > 0. The poles of the system function may be either

   1. On or to the left of the vertical line σ = σ1
   2. On or to the right of the vertical line σ = σ2

Example 7.36: Impulse response of a stable system

A stable system is characterized by the system function

$$H\left(s\right)=\frac{15s\left(s+1\right)}{\left(s+3\right)\left(s-1\right)\left(s-2\right)}$$

Determine the ROC of the system function. Afterwards find 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 three poles of the system function are at s = −3, 1, 2. Since the system is known to be stable, its ROC must include the jω-axis of the s-plane. The only possible choice is

$$-3<\mathrm{Re}\left\{s\right\}<1$$

as shown in Fig. 7.46(a). Partial fraction expansion of H (s) is

$$H\left(s\right)=\frac{4.5}{s+3}-\frac{7.5}{s-1}+\frac{18}{s-2}$$

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

$$\begin{array}{ccc}h\left(t\right)=4.5e^{-3t}u\left(t\right)+7.5e^{t}u\left(-t\right)-18e^{2t}u\left(-t\right)& & \left(7.152\right)\end{array}$$

which is shown in Fig. 7.46(b). We observe that h (t) tends to zero as t is increased in both directions, consistent with the fact that h (t) must be absolute summable for a stable system.

Software resources:

ex_7_36a.m

ex_7_36b.m

Example 7.37: Stability of a system described by a differential equation

A causal CTLTI system is characterized by the differential equation

$$\frac{d^{2}y\left(t\right)}{d{t}^2}+2\frac{dy\left(t\right)}{dt}+2y\left(t\right)=\frac{dx\left(t\right)}{dt}+x\left(t\right)$$

Determine if the system is stable.

Solution: The ROC for the system function is not directly stated. On the other hand, we are told that the system is causal. This bit of information should allow us to determine the ROC. Taking the Laplace transform of both sides of the differential equation and using the time differentiation property we have

$$\begin{array}{ccc}{s}^{2}Y\left(s\right)+2sY\left(s\right)+2Y\left(s\right)=sX\left(s\right)+X\left(s\right)& & \left(7.153\right)\end{array}$$

The system function is found by forming the ratio of the output and the input transforms.

$$H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{s+1}{s^2+2s+2}$$

The system has a zero at s = −1 and a pair of complex conjugate poles at s = −1 ± j1. Since it is causal, the ROC must be to the right of a vertical line going through σ = −1, that is

$$\begin{array}{ccc}\text{ROC}:\mathrm{Re}\left\{s\right\}>-1& & \left(7.154\right)\end{array}$$

Since the ROC includes the jω-axis of the s-plane, the system is stable. The impulse response can be determined easily by writing the H (s) in the form

$${H\left(s\right)=\frac{s+1}{{\left(s+1\right)}^2+1}=\frac{s}{s^2+1}|}_{s\to s+1}$$

Using s-domain shifting property of the Laplace transform it follows that

$$h\left(s\right)=e^{-t}\cos \left(t\right)u\left(t\right)$$

Software resources:

ex_7_37a.m

ex_7_37b.m

7.5.8 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

$$\begin{array}{cccc}\left|H\left(\omega \right)\right|=C\left(\text{constant}\right)& \text{for}\text{all}\text{ω}& & \left(7.155\right)\end{array}$$

Consider a first-order CTLTI system with a pole at p1 = −σi + j0 and a zero at z1 = σ1 + j0 so that the system function is

$$\begin{array}{ccc}H\left(s\right)=\frac{s-z_1}{s-p_1}=\frac{s-{\sigma }_1}{s+{\sigma }_1}& & \left(7.156\right)\end{array}$$

For the system to be causal and stable the parameter σ1 needs to be positive so that the resulting pole is in the left half s-plane. The magnitude and the phase of the system function can be expressed using the conventions established in Section 7.5.5. The frequency response H (ω) is found by substituting s = jω:

$$\begin{array}{ccc}H\left(\omega \right)=\frac{B_1\left(\omega \right)}{A_1\left(\omega \right)}=\frac{j\omega -{\sigma }_1}{j\omega +{\sigma }_1}& & \left(7.157\right)\end{array}$$

The magnitude of the frequency response is

$$\begin{array}{ccc}H\left(\omega \right)=\frac{\left|B_1\left(\omega \right)\right|}{\left|A_1\left(\omega \right)\right|}=\frac{\sqrt{{\omega }^2+{\sigma }_1^2}}{\sqrt{{\omega }^2+{\sigma }_1^2}}=1& & \left(7.158\right)\end{array}$$

The phases of numerator and denominator terms are

$$\begin{array}{cccc}\measuredangle B_1\left(\omega \right)=\text{π}-{\tan }^{-1}\left(\frac{\omega }{{\sigma }_1}\right),& \text{and}& \measuredangle A_1\left(\omega \right)=-{\tan }^{-1}\left(\frac{\omega }{{\sigma }_1}\right)& \left(7.159\right)\end{array}$$

and the phase of the frequency response is

$$\begin{array}{ccc}\measuredangle H\left(\omega \right)=\measuredangle B_1\left(\omega \right)-\measuredangle A_1\left(\omega \right)=\text{π}-2{\tan }^{-1}\left(\frac{\omega }{{\sigma }_1}\right)& & \left(7.160\right)\end{array}$$

The pole-zero diagram and the vector representation of the system function are shown in Fig. 7.48(a) along with the vectors

$$\begin{array}{ccccc}\overrightarrow{B_1\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{{\sigma }_1}& and& \overrightarrow{A_1\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}+\overrightarrow{{\sigma }_1}& & \left(7.161\right)\end{array}$$

Fig. 7.48(b) shows the phase characteristics for σ1 = 1, σ1 = 2 and σ1 = 3.

Since the shape of the phase response can be controlled by the choice of parameter σ1 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. Consider a first-order system with a pole at p1 = −α1 + jω1 and a zero at z1 = σ1 + jω1. Again we choose σ1 > 0 to obtain a system that is both causal and stable. The corresponding system function is

$$\begin{array}{ccc}H\left(s\right)=\frac{s-z_1}{s-p_1}=\frac{s-{\sigma }_1-j{\omega }_1}{s+{\sigma }_1-j{\omega }_1}& & \left(7.162\right)\end{array}$$

The frequency response of the system, found by substituting s = jω into the system function, is

$$\begin{array}{ccc}H\left(\omega \right)=\frac{j\omega -z_1}{j\omega -p_1}=\frac{j\omega -{\sigma }_1-j{\omega }_1}{j\omega +{\sigma }_1-j{\omega }_1}& & \left(7.163\right)\end{array}$$

It is a trivial matter to show that the magnitude of H (ω) is still equal to unity. Its phase

$$\begin{array}{ccc}\measuredangle H\left(\omega \right)=\measuredangle \left(j\omega -{\sigma }_1-j{\omega }_1\right)-\measuredangle \left(j\omega +{\sigma }_1-j{\omega }_1\right)=\text{π}-2{\tan }^{-1}\left(\frac{\omega -{\omega }_1}{{\sigma }_1}\right)& & \left(7.164\right)\end{array}$$

The pole-zero diagram and the vector representation of the system function are shown in Fig. 7.49(a) along with the vectors

$$\begin{array}{ccccc}\overrightarrow{B_1\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}-\overrightarrow{z_1}& \text{and}& \overrightarrow{A_1\left({\omega }_0\right)}=\overrightarrow{j{\omega }_0}+\overrightarrow{p_1}& & \left(7.165\right)\end{array}$$

The phase response is shown in Fig. 7.49(b) for ω1 = 1.5 rad/s and for three different values of σ1, namely σ1 = 1, σ1 = 2 and σ1 = 3.

Naturally, the system function in Eqn. (7.162) has complex coefficients. If an allpass system with real coefficients is desired, complex poles and zeros must occur in conjugate pairs. A second-order system with zeros at z1, 2 = σ1 ± jω1 and poles at p1, 2 = −σ1 ± jω1 has allpass characteristics. As before, σ1 > 0 for a causal and stable system. The system function is

$$H\left(s\right)=\frac{\left(s-{\sigma }_1-j{\omega }_1\right)\left(s-{\sigma }_1+j{\omega }_1\right)}{\left(s+{\sigma }_1-j{\omega }_1\right)\left(s+{\sigma }_1+j{\omega }_1\right)}=\frac{{\left(s-{\sigma }_1\right)}^2+{\omega }_1^2}{{\left(s+{\sigma }_1\right)}^2+{\omega }_1^2}$$

The pole-zero diagram for a second-order allpass system is shown in Fig. 7.50(a). The phase response is shown in Fig. 7.50(b) for ω1 = 1.5 rad/s and for three different values of σ1.

7.5.9 Inverse systems

The inverse of a system is another system which, when connected in cascade with the original system, forms an identity system. This relationship is depicted in Fig. 7.51.

The output signal of the original system is

$$\begin{array}{ccc}y\left(t\right)=\text{Sys}\left\{x\left(t\right)\right\}& & \left(7.166\right)\end{array}$$

We require the inverse system to recover the original input signal x (t) from the output signal y (t), therefore

$$\begin{array}{ccc}x\left(t\right)={\text{Sys}}_i\left\{y\left(t\right)\right\}& & \left(7.167\right)\end{array}$$

Combining Eqns. (7.166) and (7.166) yields

$${\text{Sys}}_i\left\{\text{Sys}\left\{x\left(t\right)\right\}\right\}=\begin{array}{ccc}x\left(t\right)& & \left(7.168\right)\end{array}$$

Let the original system and its inverse be both CTLTI systems with impulse responses h (t) and hi (t) respectively as shown in Fig. 7.52. For the output signal of the inverse system to be identical to the input signal of the original system, the impulse response of the cascade combination must be equal to δ (t), that is,

$$\begin{array}{ccc}{h}_{eq}\left(t\right)=h\left(t\right)*h_i\left(t\right)=\delta \left(t\right)& & \left(7.169\right)\end{array}$$

or, using the convolution integral

$$\begin{array}{ccc}{h}_{eq}\left(t\right)={\displaystyle {\int }_{-\infty }^{\infty }h\left(\lambda \right)h_i\left(t-\lambda \right)d\lambda =\delta \left(t\right)}& & \left(7.170\right)\end{array}$$

The corresponding relationship between the system functions of the original system and the inverse system is found by taking the Laplace transform of Eqn. (7.169):

$$\begin{array}{ccc}{H}_{eq}\left(s\right)=H\left(s\right)H_i\left(s\right)=1& & \left(7.171\right)\end{array}$$

Consequently, the system function of the inverse system is

$$\begin{array}{ccc}{H}_i\left(s\right)=\frac{1}{H\left(s\right)}& & \left(7.172\right)\end{array}$$

Two important characteristics of the inverse system are causality and stability. We will first focus on causality. Consider the system H (s) function in the form

$$\begin{array}{ccc}H\left(s\right)=\frac{B\left(s\right)}{A\left(s\right)}=\frac{b_M{s}^M+b_{M-1}{s}^{M-1}+...+b_{1}s+b_0}{a_N{s}^N+a_{N-1}{s}^{N-1}+...+a_{1}s+a_0}& & \left(7.173\right)\end{array}$$

The system function for the inverse system is

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

If the original system with system function H (s) is causal then M ≤ N as we have established in Section 7.5.6. By the same token, causality of the inverse system with system function Hi (s) 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 (s) in pole-zero form. Using M = N we have

$$\begin{array}{ccc}H\left(s\right)=\frac{b_N\left(s-z_1\right)\left(s-z_2\right)...\left(s-z_N\right)}{a_N\left(s-p_1\right)\left(s-p_2\right)...\left(s-p_N\right)}& & \left(7.174\right)\end{array}$$

If the original system is both causal and stable, all its poles must be in the left half s-plane (see Section 7.5.7), therefore

$$\begin{array}{cccc}\mathrm{Re}\left\{p_k\right\}<0,& k=1,...,N& & \left(7.175\right)\end{array}$$

The system function of the inverse system, written in pole-zero form, is

$$\begin{array}{ccc}{H}_i\left(s\right)=\frac{a_N\left(s-p_1\right)\left(s-p_2\right)...\left(s-p_N\right)}{b_N\left(s-z_1\right)\left(s-z_2\right)...\left(s-z_N\right)}& & \left(7.176\right)\end{array}$$

For the inverse system to be stable, its poles must also lie in the left half s-plane. 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 in the left half s-plane. In addition to Eqn. (7.175) we also need

$$\begin{array}{cccc}\mathrm{Re}\left\{z_k\right\}<0,& k=1,...,N& & \left(7.177\right)\end{array}$$

Example 7.38: Inverse of a system described by a differential equation

A causal CTLTI system is described by a differential equation

$$\frac{dy\left(t\right)}{dt}+2y\left(t\right)=\frac{dx\left(t\right)}{dt}+x\left(t\right)$$

Determine if a causal and stable inverse can be found for this system. If yes, find a differential equation for the inverse system.

Solution: Taking the Laplace transform of both sides of the differential equation we get

$$\left(s+2\right)Y\left(s\right)=\left(s+1\right)X\left(s\right)$$

The system function for the original system is

$$H\left(s\right)=\frac{s+1}{s+2}$$

The system function for the inverse system is found as the reciprocal of H (s):

$$H_i\left(s\right)=\frac{1}{H\left(s\right)}=\frac{s+2}{s+1}$$

The inverse system is also causal and stable. It leads to the differential equation

$$\frac{dy\left(t\right)}{dt}+y\left(t\right)=\frac{dx\left(t\right)}{dt}+2x\left(t\right)$$

7.5.10 Bode plots

Bode plots of the frequency response are used in the analysis and design of feedback control systems. A Bode plot consists of the dB magnitude $20{\log }_{10}\left|H\left(\omega \right)\right|$ and the phase $\measuredangle H\left(\omega \right)$, each graphed as a function of ${\log }_{10}\left(\omega \right)$. Because of the use of the logarithm, individual contributions of the zeros and the poles of H (s) to the magnitude of the frequency response are additive rather than multiplicative. Also, since ${\log }_{10}\left(\omega \right)$ is used for the horizontal axis, only positive values of ω are of interest.

Consider again the pole-zero form of the system function H (s) given by Eqn. (7.138). A slightly modified form of it will be more convenient for use in deriving the Bode plot. Let us scale each factor in the numerator and the denominator, and write Eqn. (7.138) as

$$\begin{array}{ccc}H\left(s\right)=K_1\frac{\left(1-s/z_1\right)\left(1-s/z_2\right)...\left(1-s/z_M\right)}{\left(1-s/p_1\right)\left(1-s/p_2\right)...\left(1-s/p_N\right)}& & \left(7.178\right)\end{array}$$

The new gain factor K1 is chosen to compensate for all the scale factors used in individual terms. The dB magnitude of H (ω) is obtained as

$$\begin{array}{ccc}20{\log }_{10}\left|H\left(\omega \right)\right|=20{\log }_{10}\left|K_1\right|+20{\log }_{10}\left|1-j\omega /z_1\right|+20{\log }_{10}\left|1-j\omega /z_2\right|+...+20{\log }_{10}\left|1-j\omega /z_M\right|& & \left(7.179\right)\end{array}$$

and the phase is

$$\begin{array}{ccc}\measuredangle H\left(\omega \right)=\measuredangle K_1+\measuredangle \left(1-j\omega /z_1\right)+\measuredangle \left(1-j\omega /z_2\right)+...+\measuredangle \left(1-j\omega /z_M\right)-\measuredangle \left(1-j\omega /p_1\right)-\measuredangle \left(1-j\omega /p_2\right)-...-\measuredangle \left(1-j\omega /p_N\right)& & \left(7.180\right)\end{array}$$

To facilitate the analysis of individual contributions from zeros and poles of the system function, let us write H (s) as a cascade combination of M + N subsystems such that

$$\begin{array}{ccc}H\left(s\right)=K_1{H}_1\left(s\right)H_2\left(s\right)...H_M\left(s\right)H_{M+1}\left(s\right)H_{M+2}\left(s\right)...H_{M+N}\left(s\right)& & \left(7.181\right)\end{array}$$

with

$$\begin{array}{cccc}{H}_i\left(s\right)=1-s/z_i,& i=1,...,M& & \left(7.182\right)\end{array}$$

and

$$\begin{array}{cccc}{H}_{M+i}\left(s\right)=\frac{1}{1-s/p_i},& i=1,...,N& & \left(7.183\right)\end{array}$$

The Bode plot for magnitude and phase can be constructed by computing the contribution of each term in Eqns. (7.182) and (7.183). We will consider four different cases.

Zero at the origin

Let Hk (s) = s. Corresponding dB magnitude and phase are computed as

$$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|=20{\log }_{10}\left|j\omega \right|=20{\log }_{10}\left(\omega \right)& & \left(7.184\right)\end{array}$$

and

$$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)=\measuredangle j\omega ={90}^{\circ }& & \left(7.185\right)\end{array}$$

The magnitude characteristic, when graphed as a function of ${\log }_{10}\left(\omega \right)$, is a straight line that goes through 0 dB at ω = 1. Its slope is 20 dB per decade.2 This is illustrated in Fig. 7.53(a).

Pole at the origin

Let Hk (s) = 1/s. Corresponding dB magnitude and phase are computed as

$$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|=20{\log }_{10}\left|\frac{1}{j\omega }\right|=-20{\log }_{10}\left(\omega \right)& & \left(7.186\right)\end{array}$$

and

$$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)=\measuredangle \left(\frac{1}{j\omega }\right)=-{90}^{\circ }& & \left(7.187\right)\end{array}$$

The magnitude characteristic, when graphed as a function of ${\log }_{10}\left(\omega \right)$, is a straight line that goes through 0 dB at ω = 1. Its slope is −20 dB per decade. This is illustrated in Fig. 7.53(b).

Single real zero

Let zk be a real-valued zero of the system. Substituting s = jω into Eqn. (7.182) we obtain

$$\begin{array}{ccc}{H_k\left(\omega \right)=H_k\left(s\right)|}_{s=j\omega }=1-j\omega /z_k& & \left(7.188\right)\end{array}$$

Corresponding dB magnitude is

$$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|=20{\log }_{10}\sqrt{1+{\left(\frac{\omega }{z_k}\right)}^2}=10{\log }_{10}\left[1+{\left(\frac{\omega }{z_k}\right)}^2\right]& & \left(7.189\right)\end{array}$$

and the corresponding phase characteristic is

$$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)=\measuredangle \left(1-\frac{j\omega }{z_k}\right)={\tan }^{-1}\left(-\frac{\omega }{z_k}\right)=-{\tan }^{-1}\left(\frac{\omega }{z_k}\right)& & \left(7.190\right)\end{array}$$

Keep in mind that we need to graph $20{\log }_{10}\left|H_k\left(\omega \right)\right|$ and $\measuredangle H_k\left(\omega \right)$ as functions of ${\log }_{10}\left(\omega \right)$, and we are only interested in positive values of ω.

1. If $\omega \ll \left|z_k\right|$ we have

   $$\begin{array}{ccc}{\left(\frac{\omega }{z_k}\right)}^2\ll 1& & \left(7.191\right)\end{array}$$

   The magnitude in Eqn. (7.189) can be approximated as

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|\approx 10{\log }_{10}\left(1\right)=0& & \left(7.192\right)\end{array}$$

   The phase angle can be approximated as

   $$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)\approx -{\tan }^{-1}\left(0\right)& & \left(7.193\right)\end{array}$$

2. At the opposite extreme, for $\omega \gg \left|z_k\right|$ we have

   $$\begin{array}{ccc}{\left(\frac{\omega }{z_k}\right)}^2\gg 1& & \left(7.194\right)\end{array}$$

   The magnitude in Eqn. (7.189) can be approximated as

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|\approx 10{\log }_{10}{\left(\frac{\omega }{z_k}\right)}^2=20{\log }_{10}\left(\frac{\omega }{\left|z_k\right|}\right)& & \left(7.195\right)\end{array}$$

   which can also be written in the form

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|\approx 20{\log }_{10}\left(\omega \right)-20{\log }_{10}\left|z_k\right|& & \left(7.196\right)\end{array}$$

   In this case the phase angle depends on the sign of zk:

   $$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)\approx \{\begin{array}{cc}{90}^{\circ },& z_k<0\\ -{90}^{\circ },& z_k>0\end{array}& & \left(7.197\right)\end{array}$$

3. At the frequency $\omega =\left|z_k\right|$ the magnitude is

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|=10{\log }_{10}\left(2\right)\approx 3\text{dB}& & \left(7.198\right)\end{array}$$

   The phase angle depends on the sign of zk:

   $$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)\approx \{\begin{array}{cc}{45}^{\circ },& z_k<0\\ -{45}^{\circ },& z_k>0\end{array}& & \left(7.199\right)\end{array}$$

The conclusions obtained above can be summarized as follows:

This is illustrated in Figs. 7.54 and 7.55.

Single real pole

Derivation of the Bode plot for a single real pole is similar. Let pk be a real-valued pole of the system. Substituting s = jω into Eqn. (7.183) yields

$$\begin{array}{ccc}{H_k\left(\omega \right)=\frac{1}{1-s/p_k}|}_{s=j\omega }=\frac{1}{1-j\omega /p_k}& & \left(7.200\right)\end{array}$$

Corresponding dB magnitude is

$$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|=20{\log }_{10}\frac{1}{\sqrt{1+{\left(\frac{\omega }{p_k}\right)}^2}}=-10{\log }_{10}\left[1+{\left(\frac{\omega }{p_k}\right)}^2\right]& & \left(7.201\right)\end{array}$$

and the corresponding phase characteristic is

$$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)=\measuredangle \left[\frac{1}{1-j\omega /p_k}\right]=-{\tan }^{-1}\left(-\frac{\omega }{p_k}\right)={\tan }^{-1}\left(\frac{\omega }{p_k}\right)& & \left(7.202\right)\end{array}$$

1. If $\omega \ll \left|p_k\right|$ we have

   $$\begin{array}{ccc}{\left(\frac{\omega }{p_k}\right)}^2\ll 1& & \left(7.203\right)\end{array}$$

   The magnitude in Eqn. (7.201) can be approximated as

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|\approx -10{\log }_{10}\left(1\right)=0& & \left(7.204\right)\end{array}$$

   The phase angle can be approximated as

   $$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)\approx {\tan }^{-1}\left(0\right)=0& & \left(7.205\right)\end{array}$$

2. At the opposite extreme, for ${}^{\omega \gg \left|p_k\right|}$ we have

   $$\begin{array}{ccc}{\left(\frac{\omega }{p_k}\right)}^2\gg 1& & \left(7.206\right)\end{array}$$

   The magnitude in Eqn. (7.201) can be approximated as

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|\approx -10{\log }_{10}{\left(\frac{\omega }{p_k}\right)}^2=-20{\log }_{10}\left(\frac{\omega }{\left|p_k\right|}\right)& & \left(7.207\right)\end{array}$$

   which can also be written in the form

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|\approx -20{\log }_{10}\left(\omega \right)+20{\log }_{10}\left|p_k\right|& & \left(7.208\right)\end{array}$$

   In this case the phase angle depends on the sign of pk:

   $$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)\approx \{\begin{array}{cc}-{90}^{\circ },& p_k<0\\ {90}^{\circ },& p_k>0\end{array}& & \left(7.209\right)\end{array}$$

3. At the frequency $\omega =\left|p_k\right|$ the magnitude is

   $$\begin{array}{ccc}20{\log }_{10}\left|H_k\left(\omega \right)\right|=-10{\log }_{10}\left(2\right)\approx -3\text{dB}& & \left(7.210\right)\end{array}$$

   The phase angle depends on the sign of pk:

   $$\begin{array}{ccc}\measuredangle H_k\left(\omega \right)\approx \{\begin{array}{cc}-{45}^{\circ },& p_k<0\\ {45}^{\circ },& p_k>0\end{array}& & \left(7.211\right)\end{array}$$

The conclusions obtained above can be summarized as follows:

This is illustrated in Figs. 7.56 and 7.57.

Example 7.39: Constructing a Bode plot

A CTLTI system is described by the system function

$$H\left(s\right)=\frac{s\left(1+s/300\right)}{\left(1+s/5\right)\left(1+s/40\right)}$$

Plot the Bode magnitude and the phase characteristics of the system.

Solution: Let us express the system function in the form of four subsystems connected in cascade:

$$H\left(s\right)=H_1\left(s\right)H_2\left(s\right)H_3\left(s\right)H_4\left(s\right)$$

The subsystems are

$$\begin{array}{cccc}{H}_1\left(s\right)=s,& H_2\left(s\right)=1+s/300,& H_3\left(s\right)=\frac{1}{1+s/5},& H_4\left(s\right)=\frac{1}{1+s/40}\end{array}$$

Magnitude and phase contributions of the four subsystems are shown in Fig. 7.58.

Bode plot for the magnitude of the system function is obtained by adding the magnitude contributions in Fig. 7.58(a), (c), (e) and (g). Asymptote lines of the magnitude characteristics can also be added together. Four sections of asymptotes are shown in Fig. 7.59(a) numbered 1 through 4.

1. Section 1 covers frequencies up to ω = 5 rad/s, and has a slope of 20 dB per decade which is the sum of slopes of all four contributing asymptote lines up to that point. At ω = 5 rad/s, its value is 13.98 dB (see Problem 7.43 at the end of this chapter).

2. Section 2 covers frequencies from ω = 5 rad/s to ω = 40 rad/s, and has a slope of zero (20 dB per decade from Hi and −20 dB per decade from H3.)

3. Section 3 covers frequencies from ω = 40 rad/s to ω = 300 rad/s, and has a slope of −20 dB per decade (20 dB per decade from Hi; −20 dB per decade each from H3 and H4). At the endpoint ω = 300 rad/s, its value is −3.52 dB (see Problem 7.43 at the end of this chapter).

4. Section 4 covers frequencies greater than ω = 300 rad/s, and has a slope of zero (20 dB per decade each from Hi and H2; −20 dB per decade each from H3 and H4).

The actual Bode magnitude plot is also shown in Fig. 7.59. Notice how it passes approximately 3 dB above or below each corner point depending on whether it belongs to a zero or a pole.

Bode plot for the phase of the system function is obtained by adding the phase contributions in Fig. 7.58(b), (d), (f) and (h), and is shown in Fig. 7.59(b).

Software resources:

ex_7_39a.m

ex_7_39b.m

Conjugate pair of poles

Consider a causal and stable second-order system with a pair of complex conjugate poles, that is, $p_2=p_1^{*}$. The system function is in the form

$$\begin{array}{ccc}H\left(s\right)=\frac{1}{\left(1-s/p_1\right)\left(1-s/p_1^{*}\right)}& & \left(7.212\right)\end{array}$$

H (s) can be written in a slightly different form by multiplying both the numerator and the denominator of Eqn. (7.212) with the product ($p_1{p}_1^{*}$):

$$\begin{array}{ccc}H\left(s\right)=\frac{{\left|p_1\right|}^2}{\left(s-p_1\right)\left(s-p_1^{*}\right)}& & \left(7.213\right)\end{array}$$

Let us put H (s) into the standard form

$$\begin{array}{ccc}H\left(s\right)=\frac{{\omega }_0^2}{s^2+2\zeta {\omega }_{0}s+{\omega }_0^2}& & \left(7.214\right)\end{array}$$

with

$$\begin{array}{ccc}{\omega }_0^2={\left|p_1\right|}^2,\text{and}\text{2}\zeta {\omega }_0=-2\mathrm{Re}\left\{p_1\right\}& & \left(7.215\right)\end{array}$$

It follows from Eqn. (7.215) that

$$\begin{array}{ccc}\zeta =-\frac{\mathrm{Re}\left\{p_1\right\}}{\left|p_1\right|}& & \left(7.216\right)\end{array}$$

Since the system is causal and stable, Re {p1} < 0. Consequently, when the poles of the system form a complex conjugate pair, we have 0 < ζ < 1.

Analysis of the second-order system

The form of the second-order system function given by Eqn. (7.214) can be used for a system with either two real poles or a complex conjugate pair. Let us write H (s) using two poles p1 and p2:

$$\begin{array}{ccc}H\left(s\right)=\frac{1}{\left(1-s/p_1\right)\left(1-s/p_2\right)}=\frac{p_1{p}_2}{\left(s-p_1\right)\left(s-p_2\right)}& & \left(7.217\right)\end{array}$$

Equating H (s) in Eqn. (7.217) with the form given in Eqn. (7.214) we obtain the relationships

$$\begin{array}{ccc}{\omega }_0^2=p_1{p}_2,\text{and}\text{2}\zeta {\omega }_0=-\mathrm{Re}\left\{p_1\right\}-\mathrm{Re}\left\{p_2\right\}& & \left(7.218\right)\end{array}$$

The parameter ${\omega }_0=\sqrt{p_1{p}_2}$ is called the natural undamped frequency of the system. The parameter ζ is called the damping ratio, and is computed as

$$\begin{array}{ccc}\zeta =\frac{-\mathrm{Re}\left\{p_1\right\}-\mathrm{Re}\left\{p_2\right\}}{2\sqrt{p_1{p}_2}}& & \left(7.219\right)\end{array}$$

The poles of the system function can be related to the parameters ω0 and ζ as

$$\begin{array}{ccc}{p}_{1,2}=-\zeta {\omega }_0\pm {\omega }_0\sqrt{{\zeta }^2-1}& & \left(7.220\right)\end{array}$$

Since the system is causal and stable, both poles are in the left half of the s-plane. As a result the numerator of Eqn. (7.219) is positive, and therefore the damping ratio ζ must be positive. Locations of the poles depend on the value of ζ. We will observe three distinct possibilities:

Fig. 7.60 illustrates the placement of the two poles based on the values of the parameters ζ and ω0.

It is also interesting to look at the movement of the two poles as the parameter ζ is changed. Fig. 7.61 illustrates the trajectories of pi and p2.

Let us determine the magnitude and phase responses of the system. Substituting s = jω into Eqn. (7.214) yields

$$\begin{array}{ccc}H\left(\omega \right)=\frac{{\omega }_0^2}{{\left(j\omega \right)}^2+2\zeta {\omega }_{0}j\omega +{\omega }_0^2}=\frac{1}{1-\left(\frac{\omega }{{\omega }_0}\right)+j2\zeta \left(\frac{\omega }{{\omega }_0}\right)}& & \left(7.221\right)\end{array}$$

The dB magnitude of the system function is

$$\begin{array}{ccc}20{\log }_{10}\left|H\left(\omega \right)\right|=-10{\log }_{10}\left\{{\left[1-{\left(\frac{\omega }{{\omega }_0}\right)}^2\right]}^2+{\left[2\zeta \left(\frac{\omega }{{\omega }_0}\right)\right]}^2\right\}& & \left(7.222\right)\end{array}$$

and the phase characteristic is

$$\begin{array}{ccc}\measuredangle H\left(\omega \right)=-{\tan }^{-1}\left[\frac{2\zeta \left(\frac{\omega }{{\omega }_0}\right)}{1-{\left(\frac{\omega }{{\omega }_0}\right)}^2}\right]& & \left(7.223\right)\end{array}$$

If ω ≪ ω0 then we have

$$\begin{array}{ccc}20{\log }_{10}\left|H\left(\omega \right)\right|\approx 0,& \text{and}& \measuredangle H\left(\omega \right)\approx 0\end{array}$$

If ω ≫ ω0, the dB magnitude becomes

$$20{\log }_{10}\left|H\left(\omega \right)\right|\approx -10{\log }_{10}{\left(\frac{\omega }{{\omega }_0}\right)}^4=-40{\log }_{10}\left(\omega \right)+40{\log }_{10}\left(\omega \right)$$

and the phase becomes

$$\measuredangle H\left(\omega \right)\approx -{\tan }^{-1}\left[2\zeta \left(\frac{{\omega }_0}{\omega }\right)\right]=-{180}^{\circ }$$

It will also be interesting to check the magnitude and the phase at the corner frequency. Substituting ω = ω0 into Eqn. (7.221) we obtain

$$H\left({\omega }_0\right)=\frac{1}{j2\zeta }$$

In this case the phase angle is $\measuredangle H\left(\omega \right)=-90\text{degrees}$. Before we compute the dB magnitude at the corner frequency we will define the quality factor as

$$\begin{array}{ccc}Q=\frac{1}{2\zeta }& & \left(7.224\right)\end{array}$$

so that H (ω0) = −jQ. The dB magnitude at ω = ω0 is

$$20{\log }_{10}\left|H\left({\omega }_0\right)\right|=20{\log }_{10}Q$$

The conclusions obtained above can be summarized as follows:

The asymptotic behavior of the dB magnitude characteristic is illustrated in Fig. 7.62 for ω0 = 3 rad/s and for two different values of ζ, namely ζ = 0.1 and ζ = 1.2.

The definitions of overdamped, critically damped and underdamped systems can be related to the new parameter Q as follows:

$$\begin{array}{rrrr}\hfill \text{Overdamped}& \hfill \zeta >1& \hfill \Rightarrow & \hfill Q<0.5\\ \hfill \text{Critically}\text{damped}& \hfill \zeta =1& \hfill \Rightarrow & \hfill Q=0.5\\ \hfill Underdamped& \hfill \zeta <1& \hfill \Rightarrow & \hfill Q>0.5\end{array}$$

Bode plots for dB magnitude and phase of the second-order system are shown in Fig. 7.63 for several values of Q.

The responses of the second-order system to unit-impulse and unit-step input signals are also of interest. Let us use partial fraction expansion to find the impulse response h (t). For the case of two distinct poles (real or complex) the system function H (s) in partial fraction form is

$$\begin{array}{ccc}H\left(s\right)=\frac{k_1}{s-p_1}+\frac{k2}{s-p_2}& & \left(7.225\right)\end{array}$$

The residues in Eqn. (7.225) are found as

$$\begin{array}{ccc}{k}_1=\frac{p_1{p}_2}{p_1-p_2}=\frac{{\omega }_0}{2\sqrt{{\zeta }^2-1}},& \text{and}& k_2=\frac{p_1{p}_2}{p_2-p_1}=-\frac{{\omega }_0}{2\sqrt{{\zeta }^2-1}}\end{array}$$

and the impulse response is

$$\begin{array}{ccccc}h\left(t\right)& =& k_1{e}^{p_{1}t}+k_2{e}^{p_{1}t}& & \\ & =& \frac{{\omega }_0}{2\sqrt{{\zeta }^2-1}}{e}^{-\zeta {\omega }_{0}t}\left[e^{\left({\omega }_0\sqrt{{\zeta }^2-1}\right)t}-e^{-\left({\omega }_0\sqrt{{\zeta }^2-1}\right)t}\right]u\left(t\right)& & \left(7.226\right)\end{array}$$

If ζ < 1, the result in Eqn. (7.226) becomes

$$\begin{array}{ccc}h\left(t\right)=\frac{{\omega }_0}{\sqrt{1-{\zeta }^2}}{e}^{-\zeta {\omega }_{0}t}\sin \left({\omega }_0\sqrt{1-{\zeta }^{2}t}\right)u\left(t\right)=\frac{{\omega }_0}{\sqrt{1-{\zeta }^2}}\sin \left({\omega }_{0}t\right)u\left(t\right)& & \left(7.227\right)\end{array}$$

If ζ = 1 then the expression in Eqn. (7.226) or the one in Eqn. (7.227) will not work since p1 = p2, and the partial fraction expansion in Eqn. (7.225) is not valid in this case. For ζ = 1 we have

$$H\left(s\right)=\frac{{\omega }_0}{{\left(s+{\omega }_0\right)}^2}$$

Using the s-domain differentiation property of the Laplace transform it can be shown that

$$\begin{array}{ccc}h\left(t\right)={\omega }_0^{2}t{e}^{-{\omega }_{0}t}u\left(t\right)& & \left(7.228\right)\end{array}$$

The unit-step response can be found by convolving the impulse response found with the unit-step input signal:

$$\begin{array}{cccc}\text{Sys}\left\{u\left(t\right)\right\}=h\left(t\right)*u\left(t\right)={\displaystyle {\int }_0^{t}h\left(\lambda \right)d\lambda ,}& t\ge 0& & \left(7.229\right)\end{array}$$

For ζ ≠ 1 the use of Eqn. (7.226) in Eqn. (7.229) yields

$$\begin{array}{llll}\text{Sys}\left\{u\left(t\right)\right\}\hfill & =\hfill & 1+\frac{1}{p_1-p_2}\left[p_2{e}^{p_{1}t}-p_1{e}^{p_{2}t}\right]u\left(t\right)\hfill & \hfill \\ \hfill & =\hfill & 1+\frac{e^{-{\omega }_0\zeta t}}{2{\omega }_0\sqrt{1-{\zeta }^2}}\left[\left(-{\omega }_0\zeta -{\omega }_0\sqrt{1-{\zeta }^2}\right)e^{\left({\omega }_0\sqrt{1-{\zeta }^2}\right)t}+\left({\omega }_0\zeta -{\omega }_0\sqrt{1-{\zeta }^2}\right)e^{\left(-{\omega }_0\sqrt{1-{\zeta }^2}\right)t}\right]u\left(t\right)\hfill & \left(7.230\right)\hfill \end{array}$$

If ζ = 1, the unit-step response is found by integrating the result Eqn. (7.228):

$$\begin{array}{ccc}\text{Sys}\left\{u\left(t\right)\right\}=\left[1-e^{-{\omega }_{0}t}-{\omega }_{0}t{e}^{-{\omega }_{0}t}\right]u\left(t\right)& & \left(7.231\right)\end{array}$$

Impulse response of the system is shown in Fig. 7.64(a) for several values of the damping ratio ζ. Fig. 7.64(b) shows the unit-step response of the system for the same set of values for ζ.

7.6. Simulation Structures for CTLTI Systems

Development of simulation structures for continuous-time systems was discussed briefly in Section 2.6 of Chapter 2 in the context of obtaining a block diagram from a differential equation. Our discussion in this section will parallel that of Section 2.6, and will utilize the s-domain system function as the starting point.

7.6.1 Direct-form implementation

The method of obtaining a block diagram from an s-domain system function will be derived using a third-order system, but its generalization to higher-order system functions is quite straightforward. Consider a CTLTI system described by a system function H (s).

$$\begin{array}{ccc}H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}=\frac{b_2{s}^2+b_{1}s+b_0}{s^3+a_2{s}^2+a_{1}s+a_0}& & \left(7.232\right)\end{array}$$

X (s) and Y (s) are the Laplace transforms of the input and the output signals respectively. Let us use an intermediate function ω (s) and express the system function as

$$\begin{array}{ccc}H\left(s\right)=\frac{Y\left(s\right)}{X\left(s\right)}\frac{W\left(s\right)}{X\left(s\right)}=\frac{b_2{s}^{-1}+b_1{s}^{-2}+b_0{s}^{-3}}{1+a_2{s}^{-1}+a_1{s}^{-2}+a_0{s}^{-3}}& & \left(7.233\right)\end{array}$$

where we have also multiplied both the numerator and the denominator of the system function with s−3 to ensure that no positive powers of s appear. The relationships described by Eqns. (7.232) and (7.233) are illustrated in Fig. 7.65.

If we choose to associate W (s) /X (s) with the denominator of Eqn. (7.233), that is,

$$\begin{array}{ccc}{H}_1\left(s\right)=\frac{W\left(s\right)}{X\left(s\right)}=\frac{1}{1+a_2{s}^{-1}+a_1{s}^{-2}+a_0{s}^{-3}}& & \left(7.234\right)\end{array}$$

then we have

$$\begin{array}{ccc}{H}_2\left(s\right)=\frac{Y\left(s\right)}{W\left(s\right)}=b_2{s}^{-1}+b_1{s}^{-2}+b_0{s}^{-3}& & \left(7.235\right)\end{array}$$

to satisfy Eqn. (7.233). Rearranging the terms in Eqn. (7.234) yields

$$\begin{array}{ccc}W\left(s\right)=X\left(s\right)-a_2{s}^{-1}W\left(s\right)-a_1{s}^{-2}W\left(s\right)-a_0{s}^{-3}W\left(s\right)& & \left(7.236\right)\end{array}$$

The relationship in Eqn. (7.236) can easily be translated to a simulation diagram as shown in Fig. 7.66.

Solving Eqn. (7.235) for the output transform y (s) yields

$$\begin{array}{ccc}Y\left(s\right)=b_2{s}^{-1}W\left(s\right)+b_1{s}^{-2}W\left(s\right)+b_0{s}^{-3}W\left(s\right)& & \left(7.237\right)\end{array}$$

The terms s−1W (s), s−2W (s) and s−3W (s) are already available in the simulation diagram of Fig. 7.66. Utilizing them to implement the relationship described by Eqn. (7.237), the diagram can be completed as shown in Fig. 7.67.

Using the integration property of the Laplace transform stated by Eqn. (7.86), multiplication by s−1 in the transform domain corresponds to integration of the signal in the time domain. Consequently, the blocks with system function 1/s in the diagram of Fig. 7.67 represent integrators.

It should be noted that the diagram in Fig. 7.67 could easily have been obtained directly from the system function in Eqn. (7.233) by inspection, using the following set of rules:

Example 7.40: Obtaining a block diagram from system function

A CTLTI system is described through the system function

$$H\left(s\right)=\frac{2s^3-26s+24}{s^4+7s^3+21s^2+37s+30}$$

Draw a block diagram for simulating this system.

Solution: Using the technique outlined above, the block diagram shown in Fig. 7.69 is obtained.

7.6.2 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 or parallel combination smaller diagrams. Consider a system function of order M that can be expressed in the form

$$\begin{array}{lllll}H\left(s\right)\hfill & =\hfill & H_1\left(s\right)H_2\left(s\right)...H_M\left(s\right)\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{W_1\left(s\right)}{X\left(s\right)}\frac{W_2\left(s\right)}{W_1\left(s\right)}...\frac{Y\left(s\right)}{W_{M-1}\left(s\right)}\hfill & \hfill & \left(7.238\right)\hfill \end{array}$$

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. 7.70.

An easy method of sectioning a system function in the style of Eqn. (7.238) would be to determine the poles and the zeros of the system function, and to 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 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 7.41: Cascade form block diagram

Develop a cascade form block diagram for simulating the system used in Example 7.40.

Solution: The system function specified in Example 7.40 can be factored into the form

$$H\left(s\right)=\frac{2\left(s+4\right)\left(s-3\right)\left(s-1\right)}{\left(s+1-j2\right)\left(s+1+j2\right)\left(s+3\right)\left(s+2\right)}$$

The roots can be found easily using MATLAB (see MATLAB Exercise 7.3). Let us write H (s) as

$$H\left(s\right)=H_1\left(s\right)H_2\left(s\right)H_3\left(s\right)$$

by choosing

$$\begin{array}{ccc}{H}_1\left(s\right)=\frac{2\left(s+4\right)}{\left(s+1-j2\right)\left(s+1+j2\right)}=\frac{2s+8}{s^2+2s+5},& H_2\left(s\right)=\frac{s-3}{s+3},& H_3\left(s\right)=\frac{s-1}{s+2}\end{array}$$

The cascade form simulation diagram is shown in Fig. 7.71.

It is also possible to consolidate the neighboring adders although this would cause the intermediate signals W1 (s) and W2 (s) to be lost. The resulting diagram is shown in Fig. 7.72.

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:

$$\begin{array}{lllll}H\left(s\right)\hfill & =\hfill & {\overline{H}}_1\left(s\right)+{\overline{H}}_2\left(s\right)+...+{\overline{H}}_M\left(s\right)\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{{\overline{W}}_1\left(s\right)}{X\left(s\right)}+\frac{{\overline{W}}_2\left(s\right)}{X\left(s\right)}+...\frac{{\overline{W}}_M\left(s\right)}{X\left(s\right)}\hfill & \hfill & \left(7.239\right)\hfill \end{array}$$

A simulation diagram can be constructed by implementing each term in Eqn. (7.239) using the direct-form approach, and then connecting the resulting subsystems in a parallel configuration as shown in Fig. 7.73.

A rational system function H (s) can be sectioned in the form of Eqn. (7.239) using partial fraction expansion. If some 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 7.42: Parallel form block diagram

Develop a parallel form block diagram for simulating the system used in Example 7.40.

Solution: The system function specified in Example 7.40 can be expanded into partial fractions as

$$H\left(s\right)=\frac{-2+j3}{s+1-j2}+\frac{-2-j3}{s+1+j2}+\frac{12}{s+2}+\frac{-6}{s+3}$$

Since the first two terms have complex poles and complex conjugate residues, we will combine them back into a second-order section to avoid the need for complex gain factors in the diagram. This results in

$$H\left(s\right)={\overline{H}}_1\left(s\right)+{\overline{H}}_2\left(s\right)+{\overline{H}}_3\left(s\right)$$

with

$$\begin{array}{ccc}{\overline{H}}_1\left(s\right)=\frac{-2+j3}{s+1-j2}+\frac{-2-j3}{s+1+j2}=\frac{-4s-16}{s^2+2s+5},& {\overline{H}}_2\left(s\right)=\frac{12}{s+2},& {\overline{H}}_3\left(s\right)=\frac{-6}{s+3}\end{array}$$

The parallel form simulation diagram is shown in Fig. 7.74.

7.7 Unilateral Laplace Transform

It was mentioned in earlier discussion that the Laplace transform as defined by Eqn. (7.1) is sometimes referred to as the bilateral Laplace transform.

We use the subscript “u” to distinguish the unilateral Laplace transform from its bilateral counterpart. Comparing the definition in Eqn. (7.240) with the definition of the bilateral transform given by Eqn. (7.1) it is clear that the only difference is the lower limit of the integral. In fact, if the signal x (t) is causal, that is, if x (t) = 0 for t < 0, then both definitions of the Laplace transform produce the same result.

In many engineering applications we work with causal signals, and may not need to pay attention to which Laplace transform definition we use. On the other hand, for a non-causal signal, the two definitions produce different results. Consider a signal x (t) that is non-causal. The unilateral Laplace transform of x (t) is

$$\begin{array}{lllll}{ℒ}_u\left\{x\left(t\right)\right\}\hfill & =\hfill & {\displaystyle {\int }_{0^{+}}^{\infty }x\left(t\right)e^{-st}dt}\hfill & \hfill & \hfill \\ \hfill & =\hfill & {\displaystyle {\int }_{-\infty }^{\infty }x\left(t\right)u\left(t\right)e^{-st}dt=ℒ\left\{x\left(t\right)u\left(t\right)\right\}}\hfill & \hfill & \left(7.242\right)\hfill \end{array}$$

Thus, the unilateral Laplace transform of a signal x (t) is the same as the bilateral Laplace transform of the signal [x (t) u (t)].

Because of the way Xu (s) is defined, its region of convergence is always to the right of a vertical line in the s-plane, and does not have to be explicitly stated. The ambiguity that we have observed with the bilateral Laplace transform in Eqns. (7.22) and (7.23) does not exist with the unilateral transform.

Most of the properties discussed in Section 7.3 for the bilateral Laplace transform apply to the unilateral Laplace transform as well. A few of the properties need to be modified, and a few new ones need to be introduced. These will be discussed briefly.

7.7.1 Time shifting

The use of the time shifting property with the unilateral Laplace transform requires special care. Recall the time shifting property derived in Eqn. (7.52) and repeated here:

$$\begin{array}{cc}\text{Eqn}\text{.}\left(7.52\right):& ℒ\end{array}\left\{x\left(t-\tau \right)\right\}=e^{-s\tau }ℒ\left\{x\left(t\right)\right\}$$

Using Eqn. (7.242), the unilateral Laplace transform of x (t − τ) is

$$\begin{array}{ccc}{ℒ}_u\left\{x\left(t-\tau \right)\right\}=ℒ\left\{x\left(t-\tau \right)u\left(t\right)\right\}& & \left(7.243\right)\end{array}$$

Correspondingly, the time shifting property will work for the unilateral Laplace transform only if the shift by τ does not cause any signal components to move from the negative time territory to positive time territory or vice versa. Mathematically we have the following:

Fig. 7.75 depicts a signal x (t) and three of its shifted versions. It can be shown that the time shifting property in Eqn. (7.244) holds for x1 (t) and x2 (t), but not for x3 (t).

In contrast, the time shifting property does not hold for any amount of shift of the signal x (t) in Fig. 7.76 due to signal components that cross from left of the vertical axis to the right or vice versa.

7.7.2 Differentiation in time

This property will be very important in using the unilateral Laplace transform for solving differential equations with initial conditions. The proof is similar to the proof of the corresponding property for the bilateral Laplace transform, and will be given here.

The unilateral Laplace transforms of higher-order derivatives can be found through repeated use of Eqn. (7.245). For example,

$$\begin{array}{lllll}{ℒ}_u\left\{\frac{d{x}^2\left(t\right)}{d{t}^2}\right\}\hfill & =\hfill & {ℒ}_u\left\{\frac{d}{dt}\left(\frac{dx\left(t\right)}{dt}\right)\right\}\hfill & \hfill & \hfill \\ \hfill & =\hfill & {s{ℒ}_u\left\{\frac{dx\left(t\right)}{st}\right\}-\frac{dx\left(t\right)}{dt}|}_{t=0^{+}}\hfill & \hfill & \left(7.249\right)\hfill \end{array}$$

Using Eqn. (7.245) in Eqn. (7.249) yields

$$\begin{array}{lllll}{ℒ}_u\left\{\frac{d{x}^2\left(t\right)}{d{t}^2}\right\}\hfill & =\hfill & {s\left[s{X}_u\left(s\right)-x\left(0^{+}\right)\right]-\frac{dx\left(t\right)}{dt}|}_{t=0^{+}}\hfill & \hfill & \hfill \\ \hfill & =\hfill & {s^2{X}_u\left(s\right)-sx\left(0^{+}\right)-\frac{dx\left(t\right)}{dt}|}_{t=0^{+}}\hfill & \hfill & \left(7.250\right)\hfill \end{array}$$

Similarly it can be shown that

$$\begin{array}{ccc}{{ℒ}_u\left\{\frac{d{x}^3\left(t\right)}{d{t}^3}\right\}=s^3{X}_u\left(s\right)-s^{2}x\left(0^{+}\right)-s\frac{dx\left(t\right)}{dt}|}_{t=0^{+}}{-\frac{d^{2}x\left(t\right)}{d{t}^2}|}_{t=0^{+}}& & \left(7.251\right)\end{array}$$

and for the general case

$$\begin{array}{ccc}{{ℒ}_u\left\{\frac{d{x}^n\left(t\right)}{d{t}^n}\right\}=s^n{X}_u\left(s\right)-s^{n-1}x\left(0^{+}\right)-s^{n-2}\frac{dx\left(t\right)}{dt}|}_{t=0^{+}}{-...-\frac{d^{n-1}x\left(t\right)}{d{t}^{n-1}}|}_{t=0^{+}}& & \left(7.252\right)\end{array}$$

As mentioned before, the primary utility of the unilateral Laplace transform is in solving differential equations with specified initial conditions. The next couple of examples will illustrate this.

Example 7.43: Using Laplace transform to solve a differential equation

Consider the circuit shown in Fig. 7.77(a) driven by the pulse signal shown in Fig. 7.77(b). Determine the output signal y (t) for t > 0 subject to the initial condition y (0+) = −2 V.

Solution: The system can be modeled with the following differential equation (see Example 2.9 in Chapter 2 for derivation):

$$\frac{dy\left(t\right)}{dt}+4y\left(t\right)=4x\left(t\right)$$

Taking the unilateral Laplace transform of each side of the differential equation we obtain

$$\begin{array}{ccc}s{Y}_u\left(s\right)-y\left(0^{+}\right)+4Y_u\left(s\right)=4X_u\left(s\right)& & \left(7.253\right)\end{array}$$

The input signal x (t) can be written as

$$x\left(t\right)=u\left(t\right)-u\left(t-2\right)$$

and its unilateral Laplace transform is

$$\begin{array}{ccc}{X}_u\left(s\right)=\left(1-e^{-2s}\right)\frac{1}{s}& & \left(7.254\right)\end{array}$$

Using Eqn. (7.254) in Eqn. (7.253) and substituting the initial value y (0+) leads to

$$Y_u\left(s\right)=\frac{-2}{s+4}+\left(1-e^{-2s}\right)\frac{4}{s\left(s+4\right)}$$

which can be written in the form

$$Y_u\left(s\right)=\frac{-2s+4}{s\left(s+4\right)}-e^{-2s}\frac{4}{s\left(s+4\right)}$$

Let

$$\begin{array}{ccc}{Y}_{1u}\left(s\right)=\frac{-2s+4}{s\left(s+4\right)},& \text{and}& Y_{2u}\left(s\right)=\frac{4}{s\left(s+4\right)}\end{array}$$

so that

$$Y_u\left(s\right)=Y_{1u}\left(s\right)-e^{-2s}{Y}_{2u}\left(s\right)$$

and consequently

$$y\left(t\right)=y_1\left(t\right)-y_2\left(t-2\right)$$

Using partial fraction expansion, we can write

$$\begin{array}{ccccc}{Y}_{1u}\left(s\right)=\frac{1}{s}-\frac{3}{s+4}& \Rightarrow & y_1\left(t\right)=\left(1-3e^{-4t}\right)u\left(t\right)& & \left(7.255\right)\end{array}$$

$$\begin{array}{ccccc}{Y}_{2u}\left(s\right)=\frac{1}{s}-\frac{1}{s+4}& \Rightarrow & y_2\left(t\right)=\left(1-e^{-4t}\right)u\left(t\right)& & \left(7.256\right)\end{array}$$

The output signal is

$$y\left(t\right)=\left(1-3e^{-4t}\right)u\left(t\right)-\left(1-e^{-4\left(t-2\right)}\right)u\left(t-2\right)$$

and is shown in Fig. 7.78.

Example 7.44: Solving second-order differential equation using Laplace transform

Find the solution of the second-order homogeneous differential equation

$$\frac{d^{2}y\left(t\right)}{d{t}^2}+9y\left(t\right)=0$$

subject to initial conditions

$$\begin{array}{ccc}y\left(0\right)=1,& \text{and}& {\frac{dy\left(t\right)}{dt}|}_{t=0^{+}}=1\end{array}$$

Solution: Taking the unilateral Laplace transform of the differential equation leads to

$${s^2{Y}_u\left(s\right)-sy\left(0^{+}\right)-\frac{dy\left(t\right)}{dt}|}_{t=0^{+}}+9Y_u\left(s\right)=0$$

Substituting initial conditions we obtain

$$s^2{Y}_u\left(s\right)-s-1+9Y_u\left(s\right)=0$$

which can be solved for Yu (s) to yield

$$Y_u\left(s\right)=\frac{s+1}{s^2+9}$$

The transform Yu (s) can be written in partial fraction form as

$$Y_u\left(s\right)=\frac{k_1}{s+j3}+\frac{k_2}{s-j3}$$

with residues

$$\begin{array}{ccc}{k}_1=\frac{1}{2}+j\frac{1}{6},& \text{and}& k_2=\frac{1}{2}-j\frac{1}{6}\end{array}$$

Therefore the solution is

$$\begin{array}{lll}y\left(t\right)\hfill & =\hfill & \left(\frac{1}{2}+j\frac{1}{6}\right)e^{-j3t}+\left(\frac{1}{2}-j\frac{1}{6}\right)e^{j3t}\hfill \\ \hfill & =\hfill & \cos \left(3t\right)+\frac{1}{3}\sin \left(3t\right),t\ge 0\hfill \end{array}$$

The solution y(t) is shown in Fig. 7.79. It can easily be verified that y (t) satisfies the differential equation and the specified initial conditions.

7.7.3 Initial and final value theorems

The initial value theorem is stated as follows:

The initial value theorem does not apply to rational transforms in which the order of the numerator is equal to or greater than that of the denominator, since the limit in Eqn. (7.257) would not exist in that case.

The final value theorem is stated as follows:

The final value theorem does not apply to a transform Xu (t) if the corresponding signal x (t) is unbounded, or if it has undamped oscillations. An unbounded signal indicates the presence of poles in the right half s-plane. Undamped oscillations are associated with poles on the jω axis of the s-plane other than at the origin s = 0. Consequently, the final value theorem should not be used with transforms that have poles on the jω axis or in the right half s-plane. A single pole at the origin s = 0 is permissible.

Example 7.45: Application of initial and final value theorems

Consider the transform

$$X_u\left(s\right)=\frac{s}{s^2+4}$$

Can initial and final values be determined using initial and final value theorems?

Solution: Since the numerator order is lower than the denominator order, the initial value theorem can be used.

$$x\left(0^{+}\right)=\underset{s\to \infty }{\lim }\left[s\left(\frac{s}{s^2+4}\right)\right]=\underset{s\to \infty }{\lim }\left[\frac{s^2}{s^2}\right]=1$$

The final value theorem does not apply since Xu (s) has poles on the jω axis at s = ±j2. We may also recognize that Xu (s) is the Laplace transform of the signal

$$x\left(t\right)=\cos \left(2t\right)u\left(t\right)$$

which does not have a final value due to the undamped oscillating term.

Example 7.46: Further exploring initial and final value theorems

Determine initial and final values of the signal with the Laplace transform

$$X_u\left(s\right)=\frac{2}{s\left(s^2+4s+13\right)}$$

Solution: The numerator order is lower than the denominator order; therefore, the initial value theorem can be used.

$$x\left(0^{+}\right)=\underset{s\to \infty }{\lim }\left[\frac{2}{s^2+4s+13}\right]=0$$

The transform Xu (s) has a pair of complex conjugate poles at s = −2 ± j3 and a single pole at s = 0. Complex conjugate poles are in the left half s-plane. The single pole at the origin does not cause any problems in the use of the final value theorem since it is canceled by the factor s in Eqn. (7.258). The final value of x (t) is

$$\underset{t\to \infty }{\lim }\left[x\left(t\right)\right]=\underset{s\to 0}{\lim }\left[\frac{2}{s^2+4s+13}\right]=\frac{2}{13}$$

7.5 Further Reading

[1] J. Bak and D.J. Newman. Complex Analysis. Undergraduate Texts in Mathematics. Springer, 2010.

[2] R.J. Beerends. Fourier and Laplace Transforms. Cambridge University Press, 2003.

[3] W.R.L. Page. Complex Variables and the Laplace Transform for Engineers. Dover Books on Electrical Engineering Series. Dover Publications, 1980.

[4] J.L. Schiff. The Laplace Transform: Theory and Applications. Springer Undergraduate Texts in Mathematics and Technology. Springer, 1999.

[5] J.L. Taylor. Complex Variables. Pure and Applied Undergraduate Texts. American Mathematical Society, 2011.

MATLAB Exercises

Problems

1. 7.1. Using the Laplace transform definition given by Eqn. (7.1) determine the transform of each signal listed below. For each transform construct a pole-zero diagram and specify the ROC.

   1. x (t)= e−2t u (t)
   2. x (t)= e−2t u (t − 1)
   3. x (t) = e2t u (−t)
   4. x (t) = e2t u (−t + 1)
   5. $x\left(t\right)=\{\begin{array}{ll}1,\hfill & 0<t<1\hfill \\ -1,\hfill & 1<t<2\hfill \\ 0,\hfill & t<0\text{or}t>2\hfill \end{array}$
   6. $x\left(t\right)=\{\begin{array}{ll}1,\hfill & 0<t<1\hfill \\ -1,\hfill & t>1\hfill \\ 0,\hfill & t<0\hfill \end{array}$

2. 7.2. Using the Laplace transform definition given by Eqn. (7.1) determine the transform of the signal

   $$x\left(t\right)={\displaystyle \sum _{n=0}^{\infty }{e}^{-anT}\delta \left(t-nT\right)}$$

   where a > 0 and t > 0. Put the transform into a closed form using the appropriate formula. Afterwards construct a pole-zero plot and indicate the ROC.

3. 7.3. Pole-zero diagrams for four transforms are shown in Fig. P.7.3. For each, determine the ROC if it is known that the Fourier transform of x (t) exists.

   Figure P. 7.3

   

4. 7.4. Consider again the four transforms with pole-zero diagrams shown in Fig. P.7.3. Determine the ROC for each of the transforms if x (t) is known to be causal in each case.

5. 7.5. Repeat Problem 7.3 if x (t) is known to be anti-causal for each case.

6. 7.6. The Laplace transform of a signal x (t) is given by

   $$\begin{array}{ccc}X\left(s\right)=\frac{s-1}{s^2-s-2},& \text{ROC:}& -1<\mathrm{Re}\left\{s\right\}<2\end{array}$$

   Which of the following Fourier transforms can be obtained from X (s) without actually determining the signal x (t)? In each case, either determine the indicated Fourier transform or explain why it cannot be determined.

   1. ℱ{xt)}

   2. ℱ{xt) et}

   3. ℱ{xt) e−3t}

7. 7.7. The Laplace transform of a signal x (t) is given by

   $$\begin{array}{ccc}X\left(s\right)=\frac{1}{s^2+2s+5},& \text{ROC:}& \mathrm{Re}\left\{s\right\}>-1\end{array}$$

   Which of the following Fourier transforms can be obtained from X (s) without actually determining the signal x (t)? In each case, either determine the indicated Fourier transform or explain why it cannot be determined.

   1. ℱ{xt)}

   2. ℱ{xt) e−t}

   3. ℱ{xt) et}

   4. ℱ{xt) e3t}

8. 7.8. Construct a pole-zero diagram and specify the ROC for each of the transforms given below. Also, determine the Fourier transform X (ω) if it exists.

   1. $\begin{array}{cc}X\left(s\right)=\frac{s-2}{s^2+3s+2},& x\left(t\right)\text{is causal}\end{array}$
   2. $\begin{array}{cc}X\left(s\right)=\frac{s}{s^2-1},& x\left(t\right)\text{is causal}\end{array}$
   3. $\begin{array}{cc}X\left(s\right)=\frac{s+1}{s^2-4s+3},& x\left(t\right)\text{is anti-causal}\end{array}$
   4. $\begin{array}{cc}X\left(s\right)=\frac{s^2-s}{s^2-s-6},& x\left(t\right)\text{is anti-causal}\end{array}$
   5. $\begin{array}{cc}X\left(s\right)=\frac{s^2-s}{s^2+8s+15},& ℱ\left\{x\left(t\right)e^{4t}\right\}\text{exists}\end{array}$

9. 7.9. Using the linearity property of the Laplace transform, determine X (s) for each of the signals listed below. Also indicate the ROC in each case.

   1. x (t) = 3 e−t u (t) − 5 e−3t u (t)

   2. x (t) = 3 e−t u (t) + 2 e3t u (−t)

   3. x (t) = 5 (t) + 2 e−t u (t)

   4. x (t) = (1 − e−t) u (t)

   5. x (t) = cos (2t) u (t) + 2 sin (3t) u (t)

   6. x (t) = e−2t cos (3t) u (t)

10. 7.10. Using time shifting and linearity properties of the Laplace transform as needed, determine X (s) for each of the signals listed below. Also indicate the ROC in each case.

    1. x (t) = e−2(t−1) u (t − 1)

    2. x (t) = e−2t u (t − 1)

    3. x (t) = e2(t+1) u (−t − 1)

    4. x (t) = e2t u (−t − 1)

    5. x (t) = e2t u (−t + 1)

11. 7.11. Determine the Laplace transform of the signal

    $$x\left(t\right)=\cos \left(3t+\text{π/6}\right)u\left(t\right)$$

    using two different methods as specified below. Show that the same result can be obtained through each.

    1. Use a trigonometric identity to express x (t) as the sum of a cosine and a sine term. Afterwards use the entries for cos (3t) and sin (3t) from the Laplace transform table.

    2. Use Euler’s formula to express x (t) as a sum of complex exponentials. Afterwards use the entry for eat from the Laplace transform table.

12. 7.12. Determine the Laplace transform of each signal listed below. Specify the ROC in each case. Sketch each signal and justify the ROC based on the characteristics of each signal.

    1. x (t) = u (t) − u (t − 1)

    2. x (t) = u (t) − 2u (t − 1) + u (t − 2)

    3. x (t) = u (t) − 2u (t − 1)

13. 7.13. Using the linearity and time shifting properties, determine the Laplace transform of each signal shown graphically in Fig. P.7.13. Specify the ROC in each case.

    Figure P. 7.13

    

14. 7.14. Express each of the signals shown in Fig. P.7.14 in terms of the unit-ramp function and its scaled and/or time shifted versions. Afterwards, determine the Laplace transform of each signal using linearity and time shifting properties, Specify the ROC in each case.

    Figure P. 7.14

    

15. 7.15. The signal x (t) is in the form of one cycle of a sinusoidal waveform as shown in Fig. P.7.15.

    Figure P. 7.15

    

    Determine its Laplace transform using two different methods:

    1. Apply Laplace transform definition directly. Hint: Express the sinusoidal signal using Euler’s formula.

    2. Use the transform for x1 (t) = sin (ω0t) u (t) and then apply linearity and time shifting properties.

16. 7.16. The transforms given below correspond to causal signals. Using the table of Laplace transforms, determine the inverse Laplace transform of each.

    Hint: For each case find a similar entry in the table and adjust its parameters to obtain the desired result.

    1. $X\left(s\right)=\frac{1}{{\left(s+2\right)}^2}$
    2. $X\left(s\right)=\frac{s+3}{{\left(s+3\right)}^2+2}$
    3. $X\left(s\right)=\frac{1}{{\left(s+1\right)}^2+6}$

17. 7.17. Determine the inverse Laplace transform of each function listed below using partial fraction expansion.

    1. $$\begin{array}{cc}X\left(s\right)=\frac{1}{s^2+3s+2},& \text{ROC}:\mathrm{Re}\left\{s\right\}>-1\end{array}$$
    2. $$\begin{array}{cc}X\left(s\right)=\frac{1}{s^2+3s+2},& \text{ROC}:\mathrm{Re}\left\{s\right\}<-2\end{array}$$
    3. $$\begin{array}{cc}X\left(s\right)=\frac{1}{s^2+3s+2},& \text{ROC}:-2<\mathrm{Re}\left\{s\right\}<-1\end{array}$$
    4. $$\begin{array}{cc}X\left(s\right)=\frac{\left(s-1\right)\left(s-2\right)}{\left(s+1\right)\left(s+2\right)\left(s+3\right)},& x\left(t\right):\text{causal}\end{array}$$
    5. $$\begin{array}{cc}X\left(s\right)=\frac{\left(s+1\right)\left(s-2\right)}{\left(s-1\right)\left(s+2\right)\left(s+3\right)},& ℱ\left\{x\left(t\right)\right\}\text{exists}\end{array}$$

18. 7.18. Use partial fraction expansion to find the inverse Laplace transform of each function listed below.

    1. $$\begin{array}{cc}X\left(s\right)=\frac{2s+3}{s+2},& \text{ROC}:\mathrm{Re}\left\{s\right\}>-2\end{array}$$
    2. $$\begin{array}{cc}X\left(s\right)=\frac{s\left(s+1\right)}{\left(s+2\right)\left(s+3\right)},& \text{ROC}:\mathrm{Re}\left\{s\right\}<-3\end{array}$$
    3. $\begin{array}{cc}X\left(s\right)=\frac{s+5}{s^2+4},& \text{ROC}:\mathrm{Re}\left\{s\right\}>0\end{array}$
    4. $\begin{array}{cc}X\left(s\right)=\frac{s+6}{{\left(s+1\right)}^2+4},& \text{ROC}:\mathrm{Re}\left\{s\right\}>0\end{array}$
    5. $\begin{array}{cc}X\left(s\right)=\frac{s\left(s-1\right)}{{\left(s+1\right)}^2\left(s+2\right)},& \text{ROC}:\mathrm{Re}\left\{s\right\}>-1\end{array}$

19. 7.19. Given the transform

    $$X\left(s\right)=\frac{s\left(s-3\right)}{\left(s-1\right)\left(s+1\right)\left(s+2\right)}$$

    1. Construct a pole-zero diagram for X (s).

    2. List all possibilities for the ROC.

    3. For each choice of the ROC listed in part (b) determine the inverse transform x (t), and indicate whether it is square integrable or not.

20. 7.20. For each Laplace transform given below, find all possible signals that might have led to that transform.

    1. $X\left(s\right)=\frac{s+1}{\left(s+2\right)\left(s+3\right)}$
    2. $X\left(s\right)=\frac{s+1}{{\left(s+2\right)}^2\left(s+3\right)}$
    3. $X\left(s\right)=\frac{s\left(s-1\right)}{{\left(s+3\right)}^2}$
    4. $X\left(s\right)=\frac{s}{{\left(s+2\right)}^2+9}$

21. 7.21. Use partial fraction expansion and the time shifting property to find the inverse Laplace transforms of functions listed below. Assume each transform corresponds to a causal signal.

    1. $X\left(s\right)=\frac{1-e^{-s}}{s+1}$
    2. $X\left(s\right)=\frac{\left(1-e^{-s}\right)s}{s+1}$
    3. $X\left(s\right)=\frac{1-e^{-s}}{s\left(s+1\right)}$
    4. $X\left(s\right)=\frac{1}{s}\left(1-e^{-s}+e^{-2s}-e^{-3s}\right)$

22. 7.22. Find a system function for each CTLTI system described below by means of a differential equation.

    1. $3\frac{dy\left(t\right)}{dt}+2y\left(t\right)=7x\left(t\right)$
    2. $\frac{d^{2}y\left(t\right)}{d{t}^2}+4\frac{dy\left(t\right)}{dt}+3y\left(t\right)=\frac{dx\left(t\right)}{dt}-x\left(t\right)$
    3. $\frac{d^{2}y\left(t\right)}{d{t}^2}+4y\left(t\right)=\frac{d^{2}x\left(t\right)}{d{t}^2}+\frac{dx\left(t\right)}{dt}+3x\left(t\right)$

23. 7.23. Find a differential equation for each CTLTI system described below by means of a system function.

    1. $H\left(s\right)=\frac{1}{s+4}$
    2. $H\left(s\right)=\frac{s}{s+4}$
    3. $H\left(s\right)=\frac{s+1}{s^2+5s+6}$
    4. $H\left(s\right)=\frac{s\left(s-1\right)}{s^3+5s^2+8s+6}$
    5. $H\left(s\right)=\frac{\left(s-1\right)\left(s+2\right)}{\left(s+1\right)\left(s^2+6s+13\right)}$

24. 7.24. Let yu (t) represent the response of a CTLTI system to a unit-step input signal x (t) = u (t). Determine the system function for each system the unit-step response of which is given below. Afterwards find a differential equation for each system.

    1. yu (t) = e−t u (t)

    2. yu (t) = (1 − e−t) u(t)

    3. yu (t)= (e−t − e−2t) u (t)

    4. yu (t)=(1 − e−t + 2 e−2t) u (t)

    5. yu (t) = [1 − 0.3 e−t cos (2t)] u (t)

25. 7.25. Let yr (t) represent the response of a CTLTI system to a unit-ramp input signal x (t) = tu(t). Determine the system function for each system the unit-step response of which is given below. Afterwards find a differential equation for each system.

    1. yr (t) = (1 − e−t) u (t)

    2. yr (t) = (e−t + 2 e−2t − 4 e−3t) u (t)

    3. yr (t) = [1 − 0.3 e−t cos (2t)] u (t)

26. 7.26. A CTLTI system is described by means of the system function

    $$H\left(s\right)=\frac{s+3}{s^2+3s+2}$$

    Determine the response of the system to the following signals:

    1. x (t) = e−0.5t

    2. x (t) = e(−0.5+j2)t

    3. x(t) = ej3t

    4. x (t) = e−j3t + ej3t

    5. x (t) = ej2t + ej3t

27. 7.27. Show that, for a CTLTI system with a real-valued impulse response, the value of the system function at the point $$s=s_0^{*}$$ is the complex conjugate of its value at the point s = s0. Given

    $$H\left(s_0\right)=H_0{e}^{j{\Theta }_0}$$

    where

    $$H_0=\left|H\left(s_0\right)\right|$$

    and

    $${\Theta }_0=\measuredangle H\left(s_0\right)$$

    prove that

    $$H\left(s_0^{*}\right)={\left[H\left(s_0\right)\right]}^{*}=H_0{e}^{-j{\Theta }_0}$$

    Hint: Use the Laplace transform definition with the impulse response h (t) and evaluate the result at $$s=s_0^{*}$$. Manipulate the resulting expression to obtain the desired proof.

28. 7.28. A CTLTI system is described by means of the system function

    $$H\left(s\right)=\frac{s\left(s+1\right)}{s^2+4s+13}$$

    Determine the response of the system to the following signals:

    1. x (t) = cos (2t)

    2. x (t) = e−0.5t cos (2t)

    3. x (t) = e−t sin (2t)

29. 7.29. A CTLTI system has the impulse response

    $$h\left(t\right)=e^{-t}u\left(t\right)$$

    Using Laplace transform techniques, determine the response of the system to each input signal listed below. Identify transient and steady-state components of the output signal in each case.

    1. x (t) = cos (2t) u (t)

    2. x (t) = sin (3t) u (t)

30. 7.30. Determine the impulse response of each system specified below by means of a system function.

    1. $H\left(s\right)=\frac{s+1}{s+3}$
    2. $H\left(s\right)=\frac{s^2+1}{s^2+3s+2}$
    3. $H\left(s\right)=\frac{s^2-1}{\left(s+2\right)\left(s^2+2s+2\right)}$
    4. $H\left(s\right)=\frac{s-3}{s^2+2s+1}$

31. 7.31. Consider again the system used in Problem 7.29. Using the Laplace transform, determine the response of the system to the signal

    $$x\left(t\right)=\cos \left(2t\right)$$

    which is similar to the input signal in part (a) of Problem 7.29 except it exists for all t instead of being turned on at t = 0. Compare the output signal to that obtained in Problem 7.29(a) and comment on the result.

32. 7.32. A CTLTI system has the impulse response

    $$h\left(t\right)=e^{-t}\left[u\left(t\right)-u\left(t-2\right)\right]$$

    Using Laplace transform techniques, determine the response of the system to the input signal x (t) = e−0.5t cos (2t) u (t).

33. 7.33. Construct a pole-zero diagram for each of the system functions given below.

    1. $H\left(s\right)=\frac{s+1}{\left(s+2\right)\left(s+3\right)}$
    2. $H\left(s\right)=\frac{s^2+1}{s^2+5s+6}$
    3. $H\left(s\right)=\frac{s^2-1}{\left(s+2\right)\left(s^2+4s+13\right)}$

34. 7.34. Pole-zero diagrams for four system functions are shown in Fig. P.7.34. For each, determine the the system function H (s). Set |H (0)| = 1 for each system.

35. 7.35. For each system function given below, sketch the magnitude and phase characteristics using the graphical method outlined in Section 7.5.5.

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

       Figure P. 7.34

       

36. 7.36. The system functions given below correspond to causal systems. For each system, construct a pole-zero plot and determine whether the system is stable or not.

    1. $H\left(s\right)=\frac{s+1}{s^3+2s^2+16s+12}$
    2. $H\left(s\right)=\frac{s^2-1}{s^3+s-6}$
    3. $H\left(s\right)=\frac{s^2+1}{s^2+2s+10}$
    4. $H\left(s\right)=\frac{s^2+1}{s^2-2s+10}$

37. 7.37. Determine which of the system functions given below could represent a system that is both causal and stable.

    1. $H\left(s\right)=\frac{s-1}{s^2+3s+2}$
    2. $H\left(s\right)=\frac{s\left(s+1\right)}{s^2+2s+5}$
    3. $H\left(s\right)=\frac{s^2+1}{s\left(s^2+5s+6\right)}$
    4. $H\left(s\right)=\frac{s+3}{s^2-6s+10}$

38. 7.38. A causal system is described by the differential equation

    $$\frac{d^{2}y\left(t\right)}{d{t}^2}=-2\frac{dy\left(t\right)}{dt}-ay\left(t\right)+x\left(t\right)$$

    1. Determine the system function H (s).

    2. Determine which values of a lead to a stable system.

39. 7.39. Consider a first-order allpass system with system function

    $$H\left(s\right)=\frac{s-2}{s+2}$$

    1. Compute and sketch the phase response of this system.

    2. Determine the response y (t) of the system to the input signal x (t) = e−t u (t).

    3. Find analytical expressions for the magnitude responses X (ω) and Y (ω) of the input and the output signals.

    4. Sketch each magnitude spectrum.

40. 7.40. Repeat Problem 7.39 using the second-order allpass system with the system function

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

41. 7.41. A CTLTI system has the system function

    $$H\left(s\right)=\frac{\left(s+1\right)\left(s-2\right)}{\left(s+3\right)\left(s+4\right)}$$

    1. Does the system have an inverse that is both causal and stable?

    2. Express the system function in the form of the product of two system functions

       $$H\left(s\right)=H_2\left(s\right)H_2\left(s\right)$$

       such that H1 (s) corresponds to a minimum-phase system and H2 (s) corresponds to a first-order allpass system.

    3. Find the inverse system $H_1^{-1}\left(s\right)$. Is it causal? Is it stable?

    4. Find the overall system function of the cascade combination of H (s) and $H_1^{-1}\left(s\right)$.

42. 7.42. For each CTLTI system defined by means of a differential equation below, determine whether a causal and stable inverse exists. If the answer is yes, find a differential equation for the inverse system.

    1. $\frac{dy\left(t\right)}{dt}+5y\left(t\right)=2\frac{dx\left(t\right)}{dt}+3x\left(t\right)$
    2. $\frac{d^{2}y\left(t\right)}{d{t}^2}+7\frac{dy\left(t\right)}{dt}+12y\left(t\right)=\frac{d^{2}x\left(t\right)}{d{t}^2}+\frac{dx\left(t\right)}{dt}$
    3. $\frac{d^{2}y\left(t\right)}{d{t}^2}+3y\left(t\right)=\frac{dx\left(t\right)}{dt}+5x\left(t\right)$
    4. $\frac{dy\left(t\right)}{dt}-3y\left(t\right)=\frac{d^{2}x\left(t\right)}{d{t}^2}+2\frac{dx\left(t\right)}{dt}+x\left(t\right)$

43. 7.43. Refer to the Bode magnitude plot in Example 7.39.

    1. Show that the value of the asymptote for the Bode magnitude plot is 13.98 dB at ω = 5 rad/s.

    2. Also show that, at ω = 300 rad/s, the corner of the asymptote is at −3.52 dB.

44. 7.44. Develop a direct-form block diagram for each system specified below by means of a system function. Assume that each system is causal and initially relaxed.

    1. $H\left(s\right)=\frac{s+1}{s+2}$
    2. $H\left(s\right)=\frac{s^2+1}{s^3+9s^2+26s+24}$
    3. $H\left(s\right)=\frac{s+5}{2s^3+4s+7}$

45. 7.45. 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. $H\left(s\right)=\frac{s^2+3s+2}{s^3+4s^2+8s+6}$
    2. $H\left(s\right)=\frac{s^2+1}{s^3+9s^2+26s+24}$
    3. $H\left(s\right)=\frac{s^2-3s+2}{s^2+3s+2}$

46. 7.46. Develop a parallel-form block diagram for each system specified in Problem 7.45. Assume that each system is causal and initially relaxed. Use first- and second-order parallel sections and ensure that all coefficients are real.

47. 7.47. Let Xu (s) be the unilateral Laplace transform of the signal x (t), that is

    $$x\left(t\right)\stackrel{{ℒ}_u}{\leftrightarrow }{X}_u\left(s\right)$$

    For each of the signals listed below, write the unilateral Laplace transform Gu (s) in terms of Xu (s). In each case, indicate the condition for obtaining the transform directly from Xu (s) without having to find x (t) first.

    1. g (t) = x (t − 1)

    2. g (t) = x (t + 2)

    3. g (t) = x (2t)

    4. g (t) = e−2t x (t)

    5. g (t) = tx (t)

48. 7.48. For the RLC circuit shown in Fig. P.7.48, the initial values are

    $$\begin{array}{ccc}y\left(0^{+}\right)=2\text{V,}& \text{and}& i\left(0^{+}\right)=0.5\text{A}\end{array}$$

    Figure P. 7.48

    

    1. Find a differential equation between the input x (t) and the output y (t).

    2. Obtain the system function H (s) from the differential equation found in part (a).

    3. Using unilateral Laplace transform determine the output signal if x (t) = u (t).

    4. Determine the output signal if x (t) = e−2t u (t).

49. 7.49. For the circuit shown in Fig. P7.49, the initial values are

    $$\begin{array}{ccc}y\left(0^{+}\right)=2\text{V,}& \text{and}& v_0\left(0^{+}\right)=-2\text{V}\end{array}$$

    Figure P. 7.49

    

    1. Find a differential equation between the input x (t) and the output y (t).

    2. Obtain the system function H (s) from the differential equation found in part (a).

    3. Using unilateral Laplace transform determine the output signal if x (t) = u (t).

    4. Determine the output signal if x (t) = e−t u (t).

MATLAB Problems

1. 7.50. Consider the transform

   $$\begin{array}{ccc}X\left(s\right)=\frac{s\left(s+2\right)}{{\left(s+1\right)}^2+9}& & \left(7.259\right)\end{array}$$

   1. Determine the poles and zeros of X (s). Manually construct a pole-zero diagram.

   2. Write a MATLAB script to evaluate the magnitude of X (s) at a grid of complex points in the s-plane. Use the function meshgrid(..) to generate the grid of complex points within the ranges −5 < σ < 5 and −5 < ω < 5 with increments of 0.1 in each direction.

   3. Use the function mesh(..) to produce a three dimensional mesh plot of |X (s)|.

   4. Evaluate the Laplace transform for s = jω and use the function piot3(..) to plot it over the three-dimensional mesh plot.

2. 7.51. Refer to Problem 7.8. Using MATLAB, construct a pole-zero plot for each transform listed.

3. 7.52. Refer to Problem 7.8. For each signal that has a Fourier transform, write a script to compute X (ω) from the Laplace transform X (s) given, and to graph it in the frequency range −10 ≤ ω ≤ 10 rad/s.

4. 7.53. Each transform given below corresponds to a causal signal x (t). Use symbolic mathematics capabilities of MATLAB to find the inverse Laplace transform of each. Refer to MATLAB Exercise 7.5 for an example of the solution technique.

   1. $$X\left(s\right)=\frac{1}{s^2+3s+2}$$
   2. $$X\left(s\right)=\frac{\left(s-1\right)\left(s-2\right)}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}$$
   3. $$X\left(s\right)=\frac{s\left(s-1\right)}{{\left(s+1\right)}^2\left(s+2\right)}$$

5. 7.54. A CTLTI system is described by means of the system function

   $$H\left(s\right)=\frac{s\left(s+1\right)}{s^2+4s+13}$$

   Compute and graph the response of the system to the signal

   $$x\left(t\right)=e^{-0.2t}u\left(t\right)$$

   using the following steps:

   1. Define a system object “sys” for H (s).

   2. Create a vector with samples of the input signal.

   3. Use the function isim(..) to compute the output signal.

6. 7.55. Using the function residue(..) obtain a partial fraction expansion for the transform X (s) given by

   $$X\left(s\right)=\frac{s^4-9s^3+30s^2-42s+20}{s^5+12s^4+59s^3+152s^2+200s+96}$$

   Use the results from MATLAB to write X(s) in partial fraction format.

7. 7.56. Using system objects and the function bode(..), construct Bode magnitude and phase plots for the system functions given below:

   1. $H\left(s\right)=\frac{s+1}{s^2+5s+6}$
   2. $H\left(s\right)=\frac{s\left(s+1\right)}{s^3+5s^2+8s+6}$
   3. $H\left(s\right)=\frac{\left(s-1\right)\left(s+2\right)}{\left(s+1\right)\left(s^2+6s+13\right)}$

8. 7.57. Consider the second-order system function

   $$H\left(s\right)=\frac{{\omega }_0^2}{s^2+2\zeta {\omega }_{0}s+{\omega }_0^2}$$

   Let ω0 = 5 rad/s. Using system objects and the function bode(..), construct Bode magnitude and phase plots for the system for ζ = 0.01, 0.1, 0.5, 1, 2.

9. 7.58. Refer to Problem 7.41.

   1. Using system objects of MATLAB, construct objects to represent H (s), H1 (s) and $H_1^{-1}\left(s\right)$.

   2. Compute and graph the frequency responses (magnitude and phase) for H (s), H1 (s), $H_1^{-1}\left(s\right)$ and $H\left(s\right)H_1^{-1}\left(s\right)$.

   3. Compute and graph the unit-step responses of H (s) and $H\left(s\right)H_1^{-1}\left(s\right)$. Comment on the results.

10. 7.59. Refer to the RLC circuit in Problem 7.48. Using symbolic mathematics capabilities of MATLAB, write a script to compute and graph the response of the circuit for the input signal and the initial conditions specified in the problem statement. Use MATLAB Exercise 7.10 for an example of the solution technique.

11. 7.60. Refer to the RLC circuit in Problem 7.49. Using symbolic mathematics capabilities of MATLAB, write a script to compute and graph the response of the circuit for the input signal and the initial conditions specified in the problem statement. Use MATLAB Exercise 7.10 for an example of the solution technique.

MATLAB Projects

1. 7.61. Consider the second-order system analyzed in Section 7.5.10. Its system function is

   $$H\left(s\right)=\frac{{\omega }_0^2}{s^2+2\zeta {\omega }_{0}s+{\omega }_0^2}$$

   Let y (t) be the unit-step response of this system.

   1. Determine the Laplace transform of the unit-step response y (t). Afterwards use the final value theorem to verify that the final value of the unit-step response is unity.

   2. Develop a MATLAB script to compute the unit-step response of this system. Construct a system object for specified values of parameters ω0 and ζ. Use the function step(..) to compute the step response for 0 ≤ t ≤ 30 s with increments of T = 0.01 s.

   3. One performance measure of the responsiveness of a system is the rise time which is defined as the time it takes for the unit-step response to progress from 10 percent to 90 percent of its final value. Investigate the dependency of the rise time (tr) on parameter ζ. Set ω0 = 1 rad/s, and use the script from part (b) repeatedly with varying values of ζ. Measure the rise time from the graph and/or the data vector for each ζ value. Roughly sketch the relationship between tr and ζ.

2. 7.62. Refer to Problem 7.61. Another performance measure for a system is the settling time which is defined as the time it takes for the unit-step response to settle within a certain percentage of its final value. Let ts represent the 5 percent settling time, that is, the time it takes for the response to settle in the interval 0.95 < y (t) < 1.05 and stay. Using the MATLAB script developed in Problem 7.61 investigate the dependency of the settling time on parameter ζ. Use ω0 = 1 rad/s. Roughly sketch the relationship between ts and ζ.

1 Care must be taken in certain circumstances. If the time shift makes a causal signal non-causal then the points s = ∞ + jω would need to be excluded from the ROC. Similarly, if an anti-causal signal loses its anti-causal property as the result of a shift, then the points s = −∞ + jω need to be excluded.

2 One decade corresponds to a tenfold change in ω. If ω2 = 10ω1, then ${\log }_{10}\left({\omega }_2\right)={\log }_{10}\left({\omega }_1\right)+1$.