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## Q. 4.9

As an example of using this method, let us examine a simple circuit shown in Fig. 4.23(a). (This circuit may be considered as a first moment simplified equivalent circuit of a power transformer, which is a capacitor, C, connected to a cable transmission line, represented by its characteristic resistance, R (see further on in Chapter 7).) Assume that a pulse voltage of rectangular form, 4.23(b), is applied to this circuit and find the voltage across the cable. ## Verified Solution

We first should find the real part of the transmission coefficient for $\upsilon _{2}(t)$

$K_R(\omega )=Re\left[\frac{R}{R+1/j\omega C} \right] =Re\left[\frac{j\omega CR}{1+j\omega CR} \right] =\frac{(\omega CR)^2}{1+(\omega CR)^2}=\frac{(\omega \tau )^2}{1+(\omega \tau )^2},$

where τ = RC (time constant). By treating the voltage pulse as two constant voltages shifted by time interval t, we will have for the first voltage applied at t = 0

$\upsilon^{\prime}_2(t)=\frac{2 V_0}{\pi}\int_{0}^{\infty }{K_R(\omega )}\frac{\sin \omega t}{\omega }d\omega =\frac{2V_0}{\pi}\int_{0}^{\infty }{\frac{(\omega \tau )^2}{1+(\omega \tau )^2} }\frac{\sin\omega t}{\omega } d\omega .$

By assigning $x = \omega \tau we have \omega = x/\tau , d\omega = (1/\tau) dx$ and

$\upsilon^{\prime}_2 (t)=\frac{2V_0}{\pi}\int_{0}^{\infty }{\frac{x\sin(x/\tau )t}{1+x^2} }dx=\frac{2V_0}{\pi}\frac{\pi}{2} e^{-\frac{t}{\tau } }=V_0e^{-\frac{t}{\tau } }.^{(*)}$

The second part of the voltage differs from the first one by the sign and the shifting factor $e^{−j \omega d}$. Therefore,

$\upsilon ^{\prime \prime}_2 (t)=V_0e^{-\frac{t-d}{\tau } }u(t-d).$

Finally, we have

$\upsilon _{2}(t)=\upsilon ^{\prime }_2 +\upsilon ^{\prime \prime}_2(t)=V_0\left[e^{-\frac{t}{\tau } }u(t)-e^{-\frac{t-d}{\tau } }u(t-d)\right].$

A plot of this waveform is shown in Fig.4.23(c).

$^{(*)}$ The integral in this expression is tabulated integral: $\int_{0}^{\infty }{\frac{x\sin ax}{b^{2}+x^{2}} }=\frac{\pi}{2}e^{-ab}$ 