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## Q. 4.10

As an example of using this method let us assume that, at the time t = 0, an exponential pulse $V_0 e^{−at},$ shown in Fig. 4.26(a), is applied to RL equivalent circuit, Fig. 4.26(b). Our goal is to find the voltage across the inductance, i.e., an output voltage. The Fourier transform of the output voltage may be found as

$\pmb{v}_0(j\omega )=\pmb{V}_{in}(j\omega )\pmb{K}_{oi}(j\omega ).$

## Verified Solution

Here:

$\pmb{V}_{in}(j\omega )=\frac{V_{0}}{a+j\omega }$

(see 7th entry in Table 4.3) and

 Table 4.3 Fourier transform pairs f(t) $\pmb{F}( j \omega)$ 1 $\delta (t)$ 1 2 $\delta (t-t_0)$ $e^{−j\omega t_0}$ 3 1 $2\pi \delta (\omega)$ 4 u(t) $\pi \delta (\omega) +\frac{1}{j\omega }$ 5 sgn(t) $\frac{2}{j\omega }$ 6 $e^{j\omega_0 t}$ $2\pi \delta(\omega −\omega_0)$ 7 $e^{−at}u(t)$ $\frac{1}{a+j \omega }$ 8 $te^{−at}u(t)$ $\frac{1}{(a+j \omega) ^2}$ 9 $\sin \omega_0 t$ $j\pi [\delta(\omega +\omega_0) − \delta(\omega −\omega_0)]$ 10 $\cos \omega_0 t$ $\pi [\delta(\omega +\omega_0) − \delta(\omega −\omega_0)]$ 11 $\sin \omega_0 t u(t)$ $\frac{j \pi}{2}[\delta (\omega +\omega _{0})-\delta (\omega -\omega _0)] +\frac{\omega _0}{\omega ^2_0-\omega ^2}$ 12 $\cos \omega_0 t u(t)$ $\pi [\delta (\omega +\omega _{0})-\delta (\omega -\omega _0)] +\frac{\omega _0}{\omega ^2_0-\omega ^2}$ 13 $e^{−at} \sin \omega_0 t u(t)$ $\frac{\omega _{0}}{(a+j\omega )^2+\omega ^2_0}$ 14 $e^{−at} \cos \omega_0 t u(t)$ $\frac{a+\omega _{0}}{(a+j\omega )^2+\omega ^2_0}$ 15 $u(t+\tau /2)−u(t−\tau/2)$ $\tau\frac{\sin\omega \tau /2}{\omega \tau /2}$

$\pmb{K}_{oi}(j\omega )=\frac{j\omega L}{R+j\omega L}=\frac{j\omega \tau }{1+j\omega \tau }$

where τ =L /R. The real part of $V_o( j\omega)$ then will be

$G(\omega )=V_0 \left|\frac{j\omega \tau }{(a+j\omega )(1+j\omega \tau )} \right|=V_{0}\frac{-\omega ^2\tau (1+a\tau )}{(a-\omega ^2\tau)^2 +(\omega (1+a\tau ))^2}.$

The positive plot of this function, for a = 1 ms and τ = 5 ms, is shown in Fig. 4.26(c). This plot might be divided into 4 trapezoids, as shown in Fig. 4.26(d). Then in accordance with equation 4.104 and the data obtained from Fig. 4.26(d) the time-domain response of the output voltage can be calculated, and the result is shown in Fig. 4.26(c), curve 1. Note that at the first moment the whole voltage applied to the circuit is transferred to the output: $\upsilon_o (0) = \upsilon_in (0)$, since the current, i(0), is equal to zero.
This example is, of course, simple enough to use approximate methods and can be easily solved analytically, for instance with switching formula in equation 4.93. (Th result is $0.25·(5e^{−t}−e^{−0.2t})$, which is also shown in Fig. 4.26(d), curve 2. However, we brought this example to illustrate the above method, which can be used for solving complicated problems using appropriate computer programs.

$\frac{\pmb{F}_{1}(j\omega )}{\pmb{F}_2(j\omega )}\leftrightarrow \sum\limits_{k=1}^{n}{\frac{\pmb{F}_1(j\omega _k)e^{j\omega_kt}}{\left[\frac{d}{d_j\omega }\pmb{F}_2(j\omega ) \right]_{\omega=\omega_k } } }=j\sum\limits_{k=1}^{n}{\frac{\pmb{F}_1(j\omega _{k})e^{j\omega _kt}}{\left[\frac{d}{d\omega }\pmb{F}_2(j\omega ) \right]_{\omega =\omega _k} } },$                                    (4.93a)

$\frac{\pmb{F}_1(j\omega )}{j\omega \pmb{F}_{3}}\leftrightarrow \frac{\pmb{F}_1(0)}{\pmb{F}_2(0)}+\sum\limits_{k=1}^{n-1}{\frac{\pmb{F}_1(j\omega _k)e^{j\omega _kt}}{\omega _k\left[\frac{d}{d\omega }\pmb{F}_3 (j\omega )\right]_{\omega =\omega_k}} } .$                                (4.93b)

$f(t)=\sum{f_i}(t)=\frac{2}{\pi}\sum{g_{0i}\omega _i}Sa(\omega _it)Sa(\delta _it).$                                              (4.104)