Chapter 11, Problem 57P

To determine

Find the value of $y_{{\text{in}}_1}$, $y_{{\text{in}}_2}$, and $y_{{\text{in}}_3}$ for the given circuit.

Answer to Problem 57P

The values of $y_{{\text{in}}_1}$, $y_{{\text{in}}_2}$, and $y_{{\text{in}}_3}$ for a shorted section are $20+j15\text{mS}$, $-j10\text{mS}$, and $6.408+j5.189\text{mS}$, and for an open section are $20+j15\text{mS}$, $j10\text{mS}$, and $2.461+j5.691\text{mS}$.

Explanation of Solution

Calculation:

Refer to Figure given in the textbook.

Consider the general expression for input impedance.

$Z_{\text{in}}=\frac{Z_{\text{o}}^2}{Z_L}$        (1)

Consider the general expression for input impedance.

$Z_{\text{in}}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \beta \mathcal{l}}{Z_{\text{o}}+j{Z}_L\tan \beta \mathcal{l}}\right]$        (2)

Consider for ${\mathcal{l}}_1=\frac{\lambda }{4}$.

Consider the general expression for input admittance of line 1.

$y_{{\text{in}}_1}=\frac{Z_L}{Z_{\text{o}}^2}$

Substitute 100 for $Z_{\text{o}}$ and $200+j150$ for $Z_L$ in above equation.

$\begin{array}{c}{y}_{{\text{in}}_1}=\frac{200+j150}{{\left(100\right)}^2}\\ =20+j15\text{mS}\end{array}$

Consider for ${\mathcal{l}}_2=\frac{\lambda }{8}$.

Find the value of electrical length.

$\begin{array}{c}\beta \mathcal{l}=\left(\frac{2\pi }{\lambda }\times \frac{\lambda }{8}\right)\\ =\frac{\pi }{4}\end{array}$

Substitute $\frac{\pi }{4}$ for $\beta \mathcal{l}$ in Equation (2).

$Z_{{\text{in}}_2}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \left(\frac{\pi }{4}\right)}{Z_{\text{o}}+j{Z}_L\tan \left(\frac{\pi }{4}\right)}\right]$

For a shorted line, $Z_L=0$. Therefore, the above equation becomes,

$\begin{array}{c}{Z}_{\text{in}}=Z_{\text{o}}\left(\frac{j{Z}_{\text{o}}}{Z_{\text{o}}}\right)\\ =j{Z}_{\text{o}}\end{array}$

Find the value of input admittance of line 2.

$y_{{\text{in}}_2}=\frac{1}{j{Z}_{\text{o}}}$

Substitute 100 for $Z_{\text{o}}$ in above equation.

$\begin{array}{c}{y}_{{\text{in}}_2}=\frac{1}{j100}\\ =-j10\text{mS}\end{array}$

Consider for ${\mathcal{l}}_3=\frac{7\lambda }{8}$.

Find the value of electrical length.

$\begin{array}{c}\beta \mathcal{l}=\left(\frac{2\pi }{\lambda }\times \frac{7\lambda }{8}\right)\\ =\frac{7\pi }{4}\end{array}$

The input impedance for line 3 is,

$\begin{array}{c}{Z}_{{\text{in}}_3}=Z_{\text{o}}\left[\frac{Z_{\text{in}}+j{Z}_{\text{o}}\tan \left(\frac{7\pi }{4}\right)}{Z_{\text{o}}+j{Z}_{\text{in}}\tan \left(\frac{7\pi }{4}\right)}\right]\\ =\frac{Z_{\text{o}}\left(Z_{\text{in}}-j{Z}_{\text{o}}\right)}{\left(Z_{\text{o}}-j{Z}_{\text{in}}\right)}\end{array}$

The total admittance of line 1 and line 2 is,

$y_{\text{in}}=y_{{\text{in}}_1}+y_{{\text{in}}_2}$

Substitute $20+j15$ for $y_{{\text{in}}_1}$ and $-j10$ for $y_{{\text{in}}_2}$ in above equation.

$\begin{array}{c}{y}_{\text{in}}=\left(20+j15\right)+\left(-j10\right)\\ =20+j5\text{mS}\end{array}$

The input impedance is,

$Z_{\text{in}}=\frac{1}{y_{\text{in}}}$

Substitute $\left(20+j5\right)\times {10}^{-3}$ for $y_{\text{in}}$ in above equation.

$\begin{array}{c}{Z}_{\text{in}}=\frac{1}{\left(20+j5\right)\times {10}^{-3}}\\ =47.06-j11.76\Omega \end{array}$

The input admittance for line 3 is,

$y_{{\text{in}}_3}=\frac{Z_{\text{o}}-j{Z}_{\text{in}}}{Z_{\text{o}}\left(Z_{\text{in}}-j{Z}_{\text{o}}\right)}$

Substitute 100 for $Z_{\text{o}}$ and $47.06-j11.76$ for $Z_{\text{in}}$ in above equation.

$\begin{array}{c}{y}_{{\text{in}}_3}=\frac{100-j\left(47.06-j11.76\right)}{100\left(47.06-j11.76-j100\right)}\\ =\frac{100-j47.06-11.76}{100\left(47.06-j11.76-j100\right)}\\ =6.408+j5.189\text{mS}\end{array}$

Consider, if the shorted section were open,

Consider for ${\mathcal{l}}_1=\frac{\lambda }{4}$.

$y_{{\text{in}}_1}=20+j15\text{mS}$

Consider for ${\mathcal{l}}_2=\frac{\lambda }{8}$.

For a shorted line, $Z_L=\infty$.

Find the value of input admittance of line 2.

$y_{{\text{in}}_2}=\frac{1}{Z_{{\text{in}}_2}}$        (3)

The input impedance of line 2 is,

$\begin{array}{c}{Z}_{{\text{in}}_2}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \left(\frac{\pi }{4}\right)}{Z_{\text{o}}+j{Z}_L\tan \left(\frac{\pi }{4}\right)}\right]\\ =Z_{\text{o}}\left[\frac{1+j\left(\frac{Z_{\text{o}}}{Z_L}\right)\tan \left(\frac{\pi }{4}\right)}{\left(\frac{Z_{\text{o}}}{Z_L}\right)+j\tan \left(\frac{\pi }{4}\right)}\right]\\ =Z_{\text{o}}\left[\frac{1}{j\tan \left(\frac{\pi }{4}\right)}\right]\left\{\because Z_L=\infty ,\frac{1}{\infty }=0\right\}\end{array}$

Substitute $Z_{\text{o}}\left[\frac{1}{j\tan \left(\frac{\pi }{4}\right)}\right]$ for $Z_{{\text{in}}_2}$ in Equation (3).

$\begin{array}{c}{y}_{{\text{in}}_2}=\frac{1}{Z_{\text{o}}\left[\frac{1}{j\tan \left(\frac{\pi }{4}\right)}\right]}\\ =\frac{j\tan \left(\frac{\pi }{4}\right)}{Z_{\text{o}}}\end{array}$

Substitute 100 for $Z_{\text{o}}$ in above equation.

$\begin{array}{c}{y}_{{\text{in}}_2}=\frac{j\tan \left(\frac{\pi }{4}\right)}{100}\\ =j10\text{mS}\end{array}$

Consider for ${\mathcal{l}}_3=\frac{7\lambda }{8}$.

The input impedance for line 3 is,

$\begin{array}{c}{Z}_{{\text{in}}_3}=Z_{\text{o}}\left[\frac{Z_{\text{in}}+j{Z}_{\text{o}}\tan \left(\frac{7\pi }{4}\right)}{Z_{\text{o}}+j{Z}_{\text{in}}\tan \left(\frac{7\pi }{4}\right)}\right]\\ =\frac{Z_{\text{o}}\left(Z_{\text{in}}-j{Z}_{\text{o}}\right)}{\left(Z_{\text{o}}-j{Z}_{\text{in}}\right)}\end{array}$

The total admittance of line 1 and line 2 is,

$y_{\text{in}}=y_{{\text{in}}_1}+y_{{\text{in}}_2}$

Substitute $20+j15$ for $y_{{\text{in}}_1}$ and $j10$ for $y_{{\text{in}}_2}$ in above equation.

$\begin{array}{c}{y}_{\text{in}}=\left(20+j15\right)+\left(j10\right)\\ =20+j25\text{mS}\end{array}$

The input impedance is,

$Z_{\text{in}}=\frac{1}{y_{\text{in}}}$

Substitute $\left(20+j25\right)\times {10}^{-3}$ for $y_{\text{in}}$ in above equation.

$\begin{array}{c}{Z}_{\text{in}}=\frac{1}{\left(20+j25\right)\times {10}^{-3}}\\ =19.51-j24.39\Omega \end{array}$

The input admittance for line 3 is,

$y_{{\text{in}}_3}=\frac{Z_{\text{o}}-j{Z}_{\text{in}}}{Z_{\text{o}}\left(Z_{\text{in}}-j{Z}_{\text{o}}\right)}$

Substitute 100 for $Z_{\text{o}}$ and $19.51-j24.39$ for $Z_{\text{in}}$ in above equation.

$\begin{array}{c}{y}_{{\text{in}}_3}=\frac{100-j\left(19.51-j24.39\right)}{100\left(19.51-j24.39-j100\right)}\\ =\frac{100-j19.51-24.39}{100\left(19.51-j24.39-j100\right)}\\ =\frac{75.61-j19.51}{100\left(19.51-j124.39\right)}\\ =2.461+j5.691\text{mS}\end{array}$

Conclusion:

Thus, the values of $y_{{\text{in}}_1}$, $y_{{\text{in}}_2}$, and $y_{{\text{in}}_3}$ for a shorted section are $20+j15\text{mS}$, $-j10\text{mS}$, and $6.408+j5.189\text{mS}$, and for an open section are $20+j15\text{mS}$, $j10\text{mS}$, and $2.461+j5.691\text{mS}$.