Question 11.9: A 75 Ω transmission line of length 60 m is terminated by a 1...

In the previous example, the switching on of a battery created a step function, a pulse of infinite width. In this example, the pulse is of finite width of 5\mu s. We first calculate the voltage reflection coefficients:

\Gamma_{G}=\frac{Z_{g}-Z_{o}}{Z_{g}+Z_{o}}=\frac{25-75}{25+75}=\frac{-1}{2}

\Gamma_{L}=\frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}}=\frac{100-75}{100+75}=\frac{1}{7}

The initial voltage and transit time are given by

V_{o}=\frac{Z_{o}}{Z_{o}+Z_{g}}V_{g}=\frac{75}{100}(4)=3V

t_{1}=\frac{\ell}{u}=\frac{60}{0.1(3\times10^{8})}=2\mu s

The time taken by V_{o} to go forth and back is 2t_{1}=4\mu s, which is less than the pulse duration of 5\mu s. Hence, there will be overlapping.

The current reflection coefficients are

-\Gamma_{L}=-\frac{1}{7}   and  -\Gamma_{G}=\frac{1}{2}

The initial current

I_{o}=\frac{V_{g}}{Z_{g}+Z_{o}}=\frac{4}{100}=40mA

Let i and r denote incident and reflected pulses, respectively. At the generator end:

0\lt t\lt5\mu s,       I_{r}=I_{o}=40mA

4\lt t\lt 9,       I_{i}=-\frac{1}{7}(40)=-5.714

I_{r}=\frac{1}{2}(-5.714)=-2.857

8\lt t\lt 13,       I_{i}=-\frac{1}{7}(-2.857)=0.4082

I_{r}=\frac{1}{2}(0.4082)=0.2041

12\lt t\lt 17,       I_{i}=-\frac{1}{7}(0.2041)=-0.0292

I_{r}=\frac{1}{2}(-0.0292)=-0.0146

and so on. Hence, the plot of I(0,t) versus t is as shown in Figure 11.35(a).

At the load end:

0\lt t\lt 2\mu s,       V=0

2\lt t\lt 7,       V_{i}=3

V_{r}=\frac{1}{7}(3)=0.4296

6\lt t\lt 11,       V_{i}=-\frac{1}{2}(0.4296)=-0.2143

V_{r}=\frac{1}{7}(-0.2143)=-0.0306

10\lt t\lt 14,       V_{i}=-\frac{1}{2}(-0.0306)=0.0154

V_{r}=\frac{1}{7}(0.0154)=0.0022

and so on. From V(\ell,t), we can obtain I(\ell,t) as

I(\ell,t)=\frac{V(\ell,t)}{Z_{L}}=\frac{V(\ell,t)}{100}

The plots of V(\ell,t) and I(\ell,t) are shown in Figure 11.35(b) and (c).