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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: $\int_{0}^{\infty }{\frac{x\sin ax}{b^{2}+x^{2}} }=\frac{\pi}{2}e^{-ab}$