The examples are divided into three groups. Most of the problems in first group deals with the rms and the average values of some periodic signals. The problems in second are on decomposing the given waveforms and developing expressions using the singularity functions and operations on them. The problems in third group deal with developing sketches of signals, given their expressions in time domain.
Example 1:
A sinusoidal signal has a period of 10 ms. At t = 0, the signal has a value of 20 V. It reaches its first positive peak 2 ms later. Determine the equation of the signal.
Solution:
Equation (7.1) presents the equation of a sinusoid. It is necessary to determine the values of E and j. The value of w can be obtained from the period of the waveform, as shown by equation (7.2). When the signal reaches its peak, the argument of the cosine function has zero value, as shown by equation (7.3) and the value of j can be obtained, as shown. When the signal lags the reference signal that has zero phase, the value of j is negative. If the value j is less than p, then we can add 2p to j such that - p < j < p radians. Let t = 0 in equation (7.1). Then we can obtain the value of E, as shown by equation (7.4). Equation (7.5) presents the equation of the sinusoid that needs to be determined.
Example 2:
A sinusoidal signal has a period of 10 ms. At t = 0, the signal has a value of 58.779 V. It reaches its first positive peak 8.5 ms later. Determine the equation of the signal.
Solution:
The solution is presented below without any explanation, since it is similar to the first example.
Example 3:
Obtain the rms and the absolute average values of the periodic waveforms in Fig. 36.
Solution:
The waveform of the square-wave shown on the left side of Fig. 36 has already been presented in the previous page. Its absolute average and rms values can be obtained, as shown below by equations (7.11a) and (7.11b).
The waveform of the square-wave shown on the right side of Fig. 36 has already been presented in the previous page. Its absolute average and rms values can be obtained, as shown below by equations (7.12) and (7.13). The duty cycle of this waveform is less than 1. The waveform on the left-side of Fig. 36 has a duty cycle of one. Hence the average value of waveform, with D less than one, can be obtained by multiplying the average value of waveform that has unity duty cycle by the duty cycle. In the same way, the rms value of of waveform, with D less than one, can be obtained by multiplying the the rms value of of waveform that has unity duty cycle by the square root of duty cycle. You can understand what is stated by understanding equations (7.11b) and (7.13).
Example 4:
Express the waveform in Fig. 37 by an equation. Obtain the rms and the average values of the periodic waveform in Fig. 37.
Solution:
The waveform displayed in Fig. 37 is a periodic signal that exists for t < 0 and for t ≥ 0 and hence equation (7.14) that defines this waveform does not contain u(t) as part of it. The average value and the rms value can be obtained as shown by equations (7.15) and (7.16).
Example 5:
Express the waveform in Fig. 38 by an equation. Obtain the rms and the average values of the periodic waveform in Fig. 38.
Solution:
From the average and the rms values of the full-wave rectified waveform, the average and the rms values of the half-wave rectified waveform can be obtained. The duty cycle of the half-wave rectified waveform is 0.5. Hence equations (7.18) and (7.19) can be derived from equations (7.15) and (7.16) respectively.
Example 6:
Express the waveforms in Fig. 39 by suitable equations. Obtain the rms and the average values of the periodic waveforms in Fig. 39.
Solution:
First, we deal with the waveform in Fig. 39a. We can express f1(t) as the sum of two signals, g1(t) and g2(t), as shown in Fig. 40.
We can express f1(t) as shown by equation (7.20). The average value of this waveform can be obtained, as shown by equation (7.21).
The rms value of f1(t) can be obtained as shown below by equations (7.220, (7.23) and (7.24).
As shown in Fig. 40, f1(t) is the sum of two signals, g1(t) and g2(t). Signal g1(t) expresses the average value of f1(t). Signal g2(t) expresses the difference between f1(t) and its average value. Equation (7.25) defines g1(t) and g2(t).
The average value of g2(t) is zero and the value of the product of g1(t) and g2(t) integrated over a cycle period is zero. Then the signals, g1(t) and g2(t), are said to be orthogonal. When two signals are orthogonal, the rms value of the sum of these two signals can be obtained as expressed by equation (7.27). Equation (7.28) expressed the rms value of g1(t) and g2(t). Then the rms value of f1(t) can be obtained, as shown by equation (7.29). It can be verified that the two expressions shown for the rms value of f1(t) are equal to each other. Expand the terms in the first expression and then simplify. Then you get the second expression, obtained earlier in equation (7.24).
We can extend what we have learned so far to obtain the rms value of waveform in Fig. 39b. We know how we can determine the average and the rms value of a waveform with D < 1, given the average and the rms value of the same waveform with D = 1. Secondly, we know how we can decompose a given waveform into the sum of orthogonal signals and then obtain the rms value of the given waveform.
We can decompose the signal in Fig. 39b, as shown in Fig. 41.
We can obtain equations (7.30) and (7.31) from equations (7.21) and (7.29).
We can work out again. Equations (7.32) and (7.33) show how we can obtain the rms value of f2(t).
We can obtain the average and the rms value of f2(t), staring from the definition of average and rms values. Equation (7.34) defines f2(t) and its average value is obtained, as shown by equation (7.35).
The rms value of f2(t) can be obtained, as shown by equations (7.36) and (7.37).
Example 7 :
Express the waveform in Fig. 42 in time domain and s-domain.
Solution:
The solution is explained with the help of a sketch presented in Fig. 43.
The expressions in in time domain and s-domain can be obtained, as shown in equations (7.39). and (7.40).
Example 8:
Express the function shown in Fig. 44 in terms of exponential functions and unit step functions. Express the function in s-domain also.
Solution:
The equations are presented below.
Example 9 :
Express the function shown in Fig. 45 in terms of exponential singularity functions. Express the function in s-domain also.
Solution:
The solution is explained with the help of a sketch presented in Fig. 46.
Example 10 :
Sketch the following functions:
Solution:
Example 11 :
Sketch the following functions.
Solution:
Example 12:
A periodic saw-tooth signal, f(t) with a period of T is defined as follows:
Sketch the waveform. Find its average value and its rms value. What is its form factor ?
Solution:
This page has presented some examples to illustrate the concepts associated with the topic of signals. The topic of Network Functions is taken up for study in the next page.