Question 17.13: If vs = 12sin(200πt)u(t) V in the circuit of Fig. 17.38, f......

1. Define. Although the problem appears to be clearly stated, it might be advisable to check with the individual who assigned the problem to make sure he or she wants the transient response rather than the steady-state response; in the latter case the problem becomes trivial.
2. Present. We are to determine the response i(t) given the input  v_{s} (t) , using PSpice and Fourier analysis.
3. Alternative. We will use DFT to perform the initial analysis. We will then check using the FFT approach.
4. Attempt. The schematic is shown in Fig. 17.39. We may use the DFT approach to obtain the Fourier coefficents of i(t). Since the period of the input waveform is  T = 1 / 100 = 10 ms , in the Transient dialog box we select Print Step: 0.1 ms, Final Time: 100 ms, Center Frequency: 100 Hz, Number of harmonics: 4, and Output Vars: I(L1). When the circuit is simulated, the output file includes the following:(table 1)

With the Fourier coefficients, the Fourier series describing the current i(t) can be obtained using Eq. (17.73); that is,

v_{o} (t) = a_{o} + \sum\limits^{9}_{n = 1}{A_{n} \sin (nω_{o}t  +  Ψ_{n})}                            (17.73)

i(t) =  8.5833  +   8.73  sin(2π .  100t – 89.84°)

 +  0.1017 sin(2π .  200t –  83.06°)

 + 0.068 sin(2π .  300t –  82.35°) + ··· mA

5. Evaluate. We can also use the FFT approach to cross-check our result. The current marker is inserted at pin 1 of the inductor as shown in Fig. 17.39. Running PSpice will automatically produce the plot of I(L1) in the PSpice A/D window, as shown in Fig. 17.40(a). By double clicking the FFT icon and setting the range of the X-axis from 0 to 200 Hz, we generate the FFT of I(L1) shown in Fig. 17.40(b). It is clear from the FFT-generated plot that only the dc component and the first harmonic are visible. Higher harmonics are negligibly small. One final observation, does the answer make sense? Let us look at the actual transient response, i(t) = (9.549e^{-0.5t} – 9.549) \cos (200πt)u(t) mA .    The period of the cosine wave is 10 ms while the time constant of the exponential is 2000 ms (2 seconds). So, the answer we obtained by Fourier techniques does agree.
6. Satisfactory? Clearly, we have solved the problem satisfactorily using the specified approach. We can now present our results as a solution to the problem.