Sine Wave Instantaneous Values

The relationship of the sinusoidal waveform to the trigonometric sine function is useful in determining the instantaneous value of a sinusoidal voltage or current waveform at any electrical degree point. The relationship of instantaneous voltage values to the sine function is expressed mathematically by equation 4–1,

(4–1)

where e_iis the instantaneous voltage or current value at the electrical degree point theta; E_pkis the peak value (maximum amplitude) of the sinusoidal waveform being considered; and sine of theta is the value of the trigonometric sine function of the electrical degree point theta where an instantaneous value is desired.

For example, suppose you wanted to know the instantaneous amplitude of a 40-volt peak sinusoidal voltage such as the one shown in Figure 4.2 30 degrees into the cycle. This can be calculated using equation 4–1:

The instantaneous voltage at 30 degrees is 20 volts.

This calculation was comparatively simple since theta was between 0 and 90 degrees, and therefore, the sine of theta could easily be determined from almost any trigonometric table.

Angles Greater Than 90 Degrees

Suppose, however, the angle theta at which you want to determine the instantaneous voltage is an angle greater than 90 degrees. To find the sine of theta then, you must first determine the first quadrant equivalent angle as discussed in Chapter 2. The chart for determining equivalent angles from Chapter 2 is shown again in Figure 4.3.

For example, suppose you wanted to know the instantaneous voltage amplitude of a 50-volt peak waveform at 200 degrees into the cycle as illustrated in Figure 4.4. It is calculated:

200 degrees is in the third quadrant. It is in the first-half of the negative half-cycle of the waveform. From Figure 4.3, to find the first quadrant equivalent (or first-half of the positive half-cycle equivalent) angle of 200 degrees, 180 degrees must be subtracted from the angle 200 degrees. This yields a result of 20 degrees as follows:

This 20 degrees is the angle of which you must find the sine.

But this value of voltage lies in the negative half-cycle, and therefore, it is a negative value of voltage. Thus, the instantaneous voltage of the 50-volt peak sinusoidal waveform 200 degrees into the cycle is −17.1 volts. Remember, instantaneous values for angles from 180 to 360 degrees will be negative as shown in Figure 4.5.

Plotting a Sine Wave — Electrical Degrees

Since equation 4–1 provides instantaneous values of a sinusoidal waveform, the entire waveform can be plotted using this equation. For example, suppose you wanted to plot a sinusoidal waveform of 20-volts peak amplitude versus time. Using the equation 4–1,

the instantaneous value of voltage at any value of theta into the cycle can be determined.

To help you understand this concept, the instantaneous values for a 20-volt peak waveform will be calculated at 30-degree waveform increments throughout the cycle beginning at 0 degrees. This will be very similar to the way the sine function was plotted in Chapter 2. When the electrical degrees of the cycle progress beyond 90 degrees then equivalent quadrant I angles must be calculated as shown in the following calculations. Either a 0 to 90 degrees trigonometric table or a calculator with trigonometric functions can be used to determine sine theta when performing the calculations.

*First quadrant equivalent angle.

*First quadrant equivalent angle.

All of these calculated instantaneous voltage values can be plotted in graph-form as shown in Figure 4.6. The electrical degrees are plotted on the X axis. If these points are connected with a smooth curve, the result is a sinusoidal waveform with a peak amplitude of 20 volts. Obviously, this is the same waveform as was plotted for the sine function in Chapter 2 modified, of course, by the 20 volts amplitude.

Instantaneous Current

The current that flows as a result of this applied sinusoidal voltage will also be a sinusoidal waveform. The peak value of the current can be calculated. The peak value of current in a resistive circuit is simply the peak value of voltage divided by the resistance in the circuit:

(4–2)

Using equation 4–2, if the 20-volt peak waveform is applied to a resistance of 10-ohms as shown in Figure 4.7a, a peak current of 2 amperes will be produced. The calculation is:

The waveform of this current is shown graphically in Figure 4.7b.

The instantaneous voltage equation 4–1 can be used in Ohm’s law to obtain an equation for instantaneous current in the circuit.

Substituting equation 4–1,

Since

is the peak current, then

(4–3)

Note this equation is similar to the equation used to calculate the value of voltage. Therefore, the instantaneous value of current at any electrical degree point of the current waveform can be calculated in the same way an instantaneous voltage value is calculated.

For example, if the instantaneous current value of this 2-ampere waveform at 30 degrees is desired as shown in Figure 4.8, the known values can be substituted in equation 4–3 to calculate ii:

The instantaneous current value at the 30 degrees point of the current waveform is 1 ampere.

Sinusoidal versus Non-Sinusoidal Waveform

As expressed previously in Chapter 2, a sinusoidal waveform can be defined as any waveform that may be expressed mathematically by using the sine function. The sinusoidal waveform always has the same general appearance shown in Figure 4.9.

There are also, however, non-sinusoidal waveforms. As its name implies, a non-sinusoidal waveform is any waveform that cannot be expressed mathematically by the sine function. For example, the waveforms shown in Figure 4.10 are non-sinusoidal. Each of those waveforms has a distinctly different shape than that of the sine wave.

Modifying the e_i = E_pk sinθ Equation

Recall that the period, T, of a sinusoidal waveform is the time duration of one cycle, and that a cycle is the result of a conductor traveling in a circular path through 360 electrical degrees. Therefore, 360 electrical degrees is equivalent to the period of a cycle as shown in Figure 4.11. Moreover, as illustrated in Figure 4.12, an amount of time, t, less than the period of a cycle, T, can be equated with a number of electrical degrees, 0, less than 360 degrees.

Consider the waveform shown in Figure 4.13 that has a frequency of 1 cycle per second (1 cps or 1 hertz). In previous chapters you recall that the period is obtained from a rearrangement of equation 4–4 as in equation 4–5.

(4–5)

Using equation 4–5, the period is calculated:

The period is 1 second. Therefore, after one second, the waveform has passed through 360 degrees. That is, one second in terms of time is the same as 360 degrees in terms of electrical degrees. After one-fourth second the waveform has passed through 90 degrees as shown in Figure 4.14, and after one-half second it has passed though 180 degrees. One-half second is the same as 180 electrical degrees at this frequency.

Expressing Theta as a Function of f and t

Note that one-half second (an amount of time into the cycle) is the same fraction of 1 second (the period) that 180 degrees (the number of electrical degrees corresponding to the time into the cycle) is of 360 degrees (the number of degrees corresponding to the period). Both ratios equal one-half. Expressed mathematically:

Therefore, the ratio of

is the same as the ratio of

If the frequency of a waveform changes, the electrical degrees of the waveform will correspond to different times (absolute times) into the cycle because the period of the cycle has changed. However if an instantaneous voltage is required at a particular elapsed time, t, into the cycle the ratio of the time elapsed into the period divided by the period will be the same as the electrical degrees at time, t, into the cycle divided by 360 degrees. The relationship of T, t, Θ, and 360 electrical degrees is shown graphically in Figure 4.15.

In terms of a ratio:

(4–6)

If both sides of this equation are multiplied by 360 degrees,

Since frequency equals 1/T, the equation can be rewritten:

(4–7)

(4–8)

Now, when determining the instantaneous voltage of an ac waveform, theta can be replaced in the expression by 360 ft because theta equals 360 times f times t. Thus, equation 4–1 for the instantaneous value of the voltage can be rewritten:

(4–9)

At first glance 360 ft may not appear to yield 0 in degrees. But remember, as long as you express the frequency (f) in hertz, and the time elapsed into the cycle (t) in seconds, that 360 ft will yield 0 in degrees.

Instantaneous Voltages at Specific Times

For example, consider a 50-volt peak, 4-kilohertz waveform such as the one in Figure 4.15. The period is 250 microseconds. Suppose you want to know the instantaneous amplitude of the waveform 100 microseconds into the cycle. Using equation 4–9, the instantaneous voltage can be calculated:

29.39 volts is the instantaneous voltage 100 microseconds into a cycle of an ac voltage of 50 volts peak and a frequency of 4 kilohertz.

Instantaneous Voltage at Radian Rotation

The instantaneous voltage equation 4–9,

(4–9)

can be rewritten substituting 2π (the number of radians in 360 degrees) for 360. The resulting expression is:

(4–14)

Expressed in this 2πft form of the angle theta has the dimensions of radians of angular rotation. The 2πf portion of this equation is given the name ω (Greek letter omega). It is not the capital omega used for resistance measurements (), but the lower-case omega that looks like a small w. This 2πf form will be used in some equations in the following chapters of this book.

In-Phase and Out-of-Phase

If both generators start at the same time, one wave runs simultaneously with the other as shown in Figure 4.26. This is a zero-degree phase difference. In a case such as this, the waves are said to be in phase.

If generator B waits until generator A has gone through one-half cycle before it starts, a 180-degree phase difference will exist as shown in Figure 4.27. In this case, whenever wave A swings positive, wave B swings negative, and vice versa. Since both waves are always exactly the opposite, they are said to be inverted from each other, or 180 degrees out of phase.

Now, up to this point, we’ve only specified that one wave leads another. This situation is much like two runners in a race as shown in Figure 4.28. You can say runner A is in the lead, but you could just as well say that runner B is lagging behind. Both mean the same thing; both are correct. Similarly, when describing the phase relationship of the two waveforms shown in Figure 4.22 you can say A leads B by 90 degrees or you can just as correctly say, B lags A by 90 degrees. Both statements mean the same thing. The 90-degree phase difference is a common one, and it will be studied in more detail in later chapters. Therefore, it is important that you are able to recognize phase differences.

Determining Phase Difference for Partial Waveforms

In oscilloscope patterns, you may often have two waveforms shown and not really see the beginning of either wave. One of the simplest methods you can use to determine the phase difference is to first choose a point at which both waveforms are of the same instantaneous value of voltage or current. A convenient level to choose is the zero reference level. In Figure 4.29, waveform A and waveform B are the same value at all points where they cross the zero reference: zero volts. But, you must choose two points which are side by side and where waveform A is moving in the same direction as waveform B. That is, both must be either increasing or decreasing in amplitude. Points X and Y are side by side, and both waveforms are changing from their designated positive polarity to negative polarity, or decreasing. Between these two points, then, you can measure the phase angle. In this case that is 90 degrees.

When observing waveforms on a scope, remember that times to the left on the time axis are earlier than times to the right. With this in mind, waveform B crosses the zero reference level sooner in time than does waveform A. Therefore, waveform B leads waveform A. Again, you could just as correctly state that waveform A lags waveform B by 90 degrees.

You could also look at the peaks as the reference level for comparing the phase relationships of two waveforms. In Figure 4.29, for example, you see that waveform B reaches its positive peak to the left of waveform A. Waveform A reaches its positive peak at the point where waveform B reaches a zero point later on. Waveform B leads waveform A by 90 degrees. Waveform B peaks and 90 degrees later waveform A peaks where B reaches a zero point.

Summarizing, then, look for any two points on the waveform that have the same value and are moving in the same direction. Examine where the second waveform is with respect to the first.

Frequency of Waveforms Compared

An important fact that you must remember when comparing the phase differences of two or more waveforms is that all of the waveforms must be of the same frequency. The two waveforms of Figure 4.30 do not have the same frequency. Therefore, there is not a constant phase relationship between the two waveforms. Thus, any attempt to compare phase relationships of the two waveforms is futile.

Note that in this chapter in all discussions concerning the comparison of waveforms A and B, it was stated that generators A and B are identical. Therefore, the frequency of each waveform is identical.

Conventional Phase Difference Specification

There is another interesting point about the phase of waveforms. In Figure 4.31, waveform B leads wave A by 90 degrees. Waveform A crosses the zero reference axis at point W. Note that if you look to the right, waveform B reaches a similar point (point Z) 270 degrees later on. Focusing on this relationship, it can be stated that waveform A leads waveform B by 270 degrees. Originally it was stated that waveform B leads A by 90 degrees. By comparing the phase at points X and Y this is verified.

These different phase relationships illustrate a key point — there are many ways of correctly expressing the phase between two waveforms. By convention, however, usually an angle of less than 180 degrees is used to specify the phase. In other words, if A leads B by 225 degrees, it is more common to say that B lags A by 135 degrees. Other equivalent specifications can be found because of the repetitive nature of the sine function, and these are just as correct.

Amplitude versus Phase

Another important point about phase relationships is that the amplitude of a sine wave has no bearing on its phase. The phase relationship of two sine waves of different amplitude is determined in the same way as two sine waves of the same amplitude. For example in Figure 4.32 waveform B leads A by 90 degrees. B has a greater amplitude than A, but the frequency of both is the same. Therefore, it is possible to determine accurate phase relationships between the two waveforms.

Phasor Notation

Thus far in this discussion of phase, sine wave plots of both waveforms have been used to determine and describe the phase. You probably have realized that if you wanted to decipher the phase relationships of three or more waveforms using these diagrams, it could become very tedious if not virtually impossible.

For example, in Figure 4.33 are three waveforms, A, B, and C. C leads A by 90 degrees, and is out of phase with B by 100 degrees. Although the phase relationships are simple, these diagrams are very complicated and there must be an easier way. There is. A method you can use to simplify the examination of phase relationships is called phasor notation. The word phasor is a word meaning phase vector. What are phase vectors, and how can they help you keep track of phase relationships? First, you have to know what a vector is.

Vectors

A vector is defined as a line that represents a direction and whose length represents a magnitude. In Chapter 2, trigonometry was discussed and you were told how to find the value of a trigonometric function by using a 90-degree trigonometric table. Also, a circle was divided into four quadrants and a line, A, rotated through various degree points as shown in Figure 4.34. This line will now be referred to as a vector. The vector extends from point 0 to some point A out on the circumference. By rotating the vector, 0–A, any degree point from 0 to 360 degrees can be indicated. By lengthening or shortening the vector line, any magnitude can be indicated. Vectors are used in this fashion to represent voltage and current phase relationships, and when they do they are called phasors.

Using Phasor Notation

Phasor notation can be used to represent the phase relationship between two sinusoidal waveforms such as those shown in Figure 4.35.

When using phasor notation, first one waveform must be chosen as the reference. In this example, the reference will be waveform A. The reference waveform phasor, E_A, is then positioned along the X axis, as shown in Figure 4.36, at the zero-degree rotational reference. This phasor is a vector representing the voltage of an ac generator as its conductors are rotated through a magnetic field. By convention, this rotation will always be rotated in a counterclockwise (ccw) direction.

If you look at the waveform diagram of Figure 4.35, you see that waveform B leads waveform A by 90 degrees. Waveform B, then, can be represented by another phasor, E_B, placed so that it leads phasor E_Aby 90 degrees. Since E_Ais at the 0-degree point, and the phasors are to be rotated counter-clockwise, this means that phasor E_B should be placed at the 90-degree point as shown in Figure 4.37.

Since the two phasors, E_A and E_B, represent voltages generated in conductors, adding E_B is like adding another conductor to the ac generator. Therefore, when the two conductors are rotated in the magnetic field, of the ac generator, the voltage from conductor E_Bwill lead conductor E_Aby 90 degrees throughout the cycle as shown in Figure 4.38, and the waveform E_B will lead the waveform E_A by 90 degrees throughout the entire cycle. Thus, since the phasors represent voltages generated by conductors, the phasor representation of the generator action graphically shows the phase relationship of two or more waveforms throughout an entire cycle.

Remember that phasors that are positioned counter-clockwise from E_A will lead E_A, and phasors positioned clockwise from E_A will lag E_A.

Phasor Magnitude

Recall that previously it was stated that a phasor is a vector that is a line whose position on a set of axis represents a direction and whose length represents a magnitude. As a result, the length of a phasor is proportional to the quantity it represents. For example, phasor E_A could be 8 units long to represent an 8-volt peak voltage of waveform A, and phasor E_B could be 5 units long to represent a 5-volt peak voltage of waveform B. This is shown in Figure 4.39. Any waveform amplitude specification (peak, peak-to-peak, rms) can be used when specifying phasor length as long as you are consistent. If one phasor represents the peak amplitude, the other phasor must also represent peak amplitude.

In summary, the phasor diagram can show graphically both phase relationships and amplitude relationships of two or more sinusoidal waveforms. The phasor diagram can be interpreted more easily than a waveform diagram.

Phasor Notation Examples — Two Vectors

Suppose you want to represent in phasor notation the relationship of the two waveforms shown in Figure 4.40a. Note that waveform A leads waveform B by 90 degrees, and they both have the same peak amplitude, 5 volts. The phasor representation of their phase relationship would appear as shown in Figure 4.40b. Waveform A has been selected as the reference waveform. Thus, phasor E_A(5 units in length) is placed at the 0-degree point.

Since waveform A leads waveform B, phasor E_B(5 units in length) is placed at 270 degrees, 90 degrees out of phase with E_A. Both phasors have the same length since both represent the same peak amplitude, 5 volts. From the phasor diagram, then, you can interpret that E_Aleads E_B by 90 degrees or that E_Blags E_A by 90 degrees.

As another example, suppose that you are shown the phasor diagram illustrated in Figure 4.41a. What is represented? That diagram shows that E_Aleads E_Bby 90 degrees. It also shows that E_Aand E_Bare the same amplitude because phasor E_Aand E_Bare the same length. What would the waveforms represented by the phasor diagram look like?

The sinusoidal waveform representation of these waveforms is shown in Figure 4.41b. The phasor diagram of Figure 4.41a represents the voltages at the instantaneous degree point of point X showing E_Aat the 90-degree point and E_Bat the 0-degree point in Figure 4.41b.

Figure 4.42a shows a phasor diagram of two voltages which are the same amplitude, but 180 degrees out of phase. The sinusoidal waveforms showing this phase relationship are as shown in Figure 4.42b. Note that the waveform E_Ais passing through zero amplitude at the 0-degree point and is increasing in a positive direction. To correspond to this, the phasor E_Ais positioned at the 0-degree reference on the phasor diagram. The E_Bwaveform also is passing through zero amplitude but is increasing negatively, so the axis crossing point is at 180 electrical degrees. To represent this, the phasor is positioned at 180 degrees on the phasor diagram.

Figure 4.43 shows two voltages of different amplitude that are in phase. Note that E_Ais greater in amplitude than E_B, since it is longer than E_Bon the phasor diagram.

Phasor Notation Examples — Multiple Vectors

So far, phase relationships between only two phasors or waveforms have been considered. Suppose, however, that the phase relationships between three phasors, representing three waveforms must be considered. Figure 4.44 shows a phasor diagram of three voltages out of phase with one another, but of the same amplitude. The phase relationships can be described in several ways. E_Bleads E_Aby 45 degrees. E_C leads E_Aby 90 degrees. And E_Cleads E_Bby 45 degrees. It could also be said that E_Blags E_C by 45 degrees, or that E_Alags E_Bby 45 degrees, or that E_Alags E_Cby 90 degrees. All of these statements about the phasors in the diagram are true.

The sinusoidal waveforms showing these phase relationships would appear as shown in Figure 4.45. It should be obvious that reading the phasor diagram is much easier than deciphering a sine wave diagram.

Figure 4.46 is a phasor diagram of another example of three voltages out of phase. The phasor lines indicate that each voltage is of the same amplitude, but 120 degrees out of phase with each other. This type of phase relationship exists between the voltages in a three-phase power system as shown by the waveforms in Figure 4.47.

It should be obvious that phasor notation greatly simplifies the visual description of the phase relationships of voltages. Phasors can also be used to show current relationships, or voltage-current relationships, or even power relationships. In fact, phasor diagrams showing voltage-current phase relationships are used extensively when working with ac. A typical example is the phasor diagram of Figure 4.48, which shows a voltage leading a current by 90 degrees. In fact, as you will see in later chapters, one of the quantities in an ac circuit that is important is the phase angle — the phase difference in electrical degrees between the voltage applied to a circuit and the total circuit current.

1. If the amplitude of a sine wave is 20 V_pk, determine the instantaneous amplitude of the sine wave 10°, 150°, 230°, and 300° into the cycle.

    Solution:

    The equation which relates the instantaneous amplitude of a sine wave at various degree points into the cycle is

    Since the amplitude is 20 V_pkthe equation can be rewritten as

    Now, all you must do is evaluate the value of sin θ at 10°, 150°, 230°, and 300° into the cycle and use the equation to determine the instantaneous amplitudes at these points:

    For clarity, these values are plotted on the sine wave below:

2. If the amplitude of a sine wave is 45 V_pk, and has a frequency of 10 kHz, determine the instantaneous value of the sine wave at 4 μs and 80 μs into the cycle.

    Solution:

    The equation which relates the time, frequency, and amplitude of a sine wave to an instantaneous value of voltage at that point is

    Since the amplitude and frequency are specified, the equation can be rewritten:

    Now, all that must be done to evaluate the instantaneous voltage at times into the cycle is to use (360 ×10 kHz ×t) to determine the number of degrees that specified times into the cycle (t) represent and solve for the sine of these angles.

    The equation can then be used to determine the instantaneous value of voltage as in example 1.

    For clarity, these values are plotted on the sine wave below:

3. Plot a sine wave having an amplitude of 12 VAC. Tabulate values for each 20° between 0° and 360°, then plot the sine wave. Use e_i=E_pksinθ. Also plot values at 0°, 90°, 180°, and 270°.

    The table of values for angles between 0° and 360° is shown below. Note that a 12 VAC sine wave has a peak value of 17 V_pk(E_pk=1.414 ×E_rms=1.414 ×12 V =17 V). Thus, the equation used is ei =17 V sin θ.

    Solution:

4. For the following angles specified in radians, determine the equivalent angle in degrees: 1 rad, 3.6 rad, 4.7 rad, and 5.9 rad.

    Solution:

    Since there are 2 π radians in 360°, there must be π radians in 180°. Therefore, there aredegrees in each radian. Thus, 57.29578° or 57.3° equals 1 radian. To determine the number of degrees a certain number of radians represents, simply multiply the number of radians by 57.3°.

    1 rad: 1 rad ×57.3°/rad = 57.3°

    3.6 rad: 3.6 rad ×57.3°/rad = 206.3°

    4.7 rad: 4.7 rad ×57.3°/rad = 269.3°

    5.9 rad: 5.9 rad ×57.3°/rad = 338°

5. For the following angles specified in degrees, determine the equivalent angle in radians: 15°, 88°, 150°, 230°, 300°.

    Solution:

    Since there aredegrees in each radian and therefore 57.3° equals 1 radian, simply divide the number of degrees by 57.3° to determine the number of radians corresponding to the specified number of degrees.

6. For the following pairs of waveforms, specify the phase relationships of the waveforms stating lead/lag and phase difference.

7. Sketch the equivalent phasor diagram for the waveforms in question 6. Use waveform A as reference at 0° position.

8. For the two following phasor diagrams, state the phase relationship of the phasors stating lead/lag and phase difference.

9. For the following phasor diagrams shown below, sketch the sinusoidal waveform representation of the pair showing appropriate amplitudes and phase relationships.

1. Determine the instantaneous amplitude of the following sine waves at the specified number of degrees into the cycle:

2. Determine the instantaneous amplitude of the following sine waves at the specified time elapsed into the cycle:

3. Plot a sine wave having an amplitude of 48 V_ppusing 18° increments, from 0° to 360°.

4. For the following angles expressed in radians, determine the equivalent angle in degrees:

5. For the following angles expressed in degrees, determine the equivalent angle in radians:

6. For the following pairs of waveforms, specify the phase relationships of the waveforms stating lead/lag and phase difference:

7. Sketch the equivalent phasor diagram for the waveforms in question 6. Use waveform A as reference at the 0° position.

8. For the following phasor diagrams, state the phase relationship of the phasors stating lead/lag and phase difference:

9. For the following phasor diagrams, sketch the sinusoidal waveform representation of the pair showing appropriate amplitudes and phase relationships:

1. Determine the instantaneous amplitude of the following sine waves at the specified number of degrees into the cycle:

2. Determine the instantaneous amplitude of the following sine waves at the specified time elapsed into the cycle:

3. For the following angles expressed in radians, determine the equivalent angle in degrees:

4. For the following angles expressed in degrees, determine the equivalent angle in radians:

5. For the following pairs of waveforms, specify the phase relationships of the waveforms stating lead/lag and phase difference, then sketch the equivalent phasor diagram using waveform A as reference at the 0° position.

6. For the following phasor diagrams, sketch the sinusoidal waveform representation of the pair showing appropriate amplitudes and phase relationships.