Chapter 11, Problem 29P

(a)

To determine

Find the value of $Z_{\text{in}}$ and V at the sending end.

Answer to Problem 29P

The value of $Z_{\text{in}}$ and V at the sending end is $j29.375\Omega$ and $5.75\angle 102°\text{V}$ respectively.

Explanation of Solution

Calculation:

Find the value of electrical length of the line.

$\begin{array}{c}\beta \mathcal{l}=\frac{1}{4}\times 100\\ =25\text{rad}\\ =\text{25}\times 57.2958\left\{\because 1\text{rad}=57.2958\mathrm{deg}\right\}\\ =1432.4°\end{array}$

The above equation becomes,

$\beta \mathcal{l}=352.4°\left\{\because 1432.4°-1080°=352.4°,360°\times 3=1080°\right\}$

Consider the general expression for input impedance.

$Z_{\text{in}}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \beta \mathcal{l}}{Z_{\text{o}}+j{Z}_L\tan \beta \mathcal{l}}\right]$        (1)

Substitute 60 for $Z_{\text{o}}$, $j40$ for $Z_L$, and $352.4°$ for $\beta \mathcal{l}$ in Equation (1).

$\begin{array}{c}{Z}_{\text{in}}=\left(60\right)\left[\frac{\left(j40\right)+j\left(60\right)\tan \left(352.4°\right)}{\left(60\right)+j\left(j40\right)\tan \left(352.4°\right)}\right]\\ =j29.375\Omega \end{array}$

Consider the conditions at the input for the initial voltage,

$V_{\text{o}}\left(z=0\right)=V\left(z=0\right)=\frac{Z_{\text{in}}}{Z_{\text{in}}+Z_g}{V}_g$

Substitute $j29.375$ for $Z_{\text{in}}$, $10\angle 0°$ for $V_g$, and $50-j40$ for $Z_g$ in above equation.

$\begin{array}{c}V=\frac{\left(j29.375\right)}{\left(j29.375\right)+\left(50-j40\right)}\left(10\angle 0°\right)\left\{\because V=V_{\text{o}}\right\}\\ =\frac{293.75\angle 90°}{51.116\angle -12°}\\ =5.75\angle 102°\text{V}\end{array}$

Find the value of current $I_{\text{o}}$.

$I_{\text{o}}=\frac{V_{\text{o}}}{Z_{\text{in}}}$

Substitute $5.75\angle 102°$ for $V_{\text{o}}$ and $j29.375$ for $Z_{\text{in}}$.

$\begin{array}{c}{I}_{\text{o}}=\frac{5.75\angle 102°}{j29.375}\\ =0.1957\angle 12°\text{A}\end{array}$

Conclusion:

Thus, the value of $Z_{\text{in}}$ and V at the sending end is $j29.375\Omega$ and $5.75\angle 102°\text{V}$ respectively.

(b)

To determine

Find the value of $Z_{\text{in}}$ and $V_s$ at the receiving end.

Answer to Problem 29P

The value of $Z_{\text{in}}$ and $V_s$ at the receiving end is $j40\Omega$ and $7.75\angle 102°\text{V}$ respectively.

Explanation of Solution

Calculation:

At the receiving end,

$Z_{\text{in}}=Z_L=j40\Omega$

Consider the general expression for voltage wave.

$V_s\left(z\right)=V_{\text{o}}^{+}{e}^{-\gamma z}+V_{\text{o}}^{-}{e}^{\gamma z}$

$V_s\left(z\right)=V_{\text{o}}^{+}{e}^{-j\beta z}+V_{\text{o}}^{-}{e}^{j\beta z}$        (2)

Find the value of $V_{\text{o}}^{+}$.

$V_{\text{o}}^{+}=\frac{1}{2}\left[V_{\text{o}}+Z_{\text{o}}{I}_{\text{o}}\right]$

Substitute $5.75\angle 102°$ for $V_{\text{o}}$, $0.1957\angle 12°$ for $I_{\text{o}}$, and 60 for $Z_{\text{o}}$ in above equation.

$\begin{array}{c}{V}_{\text{o}}^{+}=\frac{1}{2}\left[\left(5.75\angle 102°\right)+\left(60\right)\left(0.1957\angle 12°\right)\right]\\ =5.145+j4.0328\\ =6.537\angle 38.09°\end{array}$

Find the value of $V_{\text{o}}^{-}$.

$V_{\text{o}}^{-}=\frac{1}{2}\left[V_{\text{o}}-Z_{\text{o}}{I}_{\text{o}}\right]$

Substitute $5.75\angle 102°$ for $V_{\text{o}}$, $0.1957\angle 12°$ for $I_{\text{o}}$, and 60 for $Z_{\text{o}}$ in above equation.

$\begin{array}{c}{V}_{\text{o}}^{-}=\frac{1}{2}\left[\left(5.75\angle 102°\right)-\left(60\right)\left(0.1957\angle 12°\right)\right]\\ =-6.34+j1.592\\ =6.537\angle 165.909°\end{array}$

Substitute $6.537\angle 38.09°$ for $V_{\text{o}}^{+}$ and $6.537\angle 165.909°$ for $V_{\text{o}}^{-}$ in Equation (2).

$V_s\left(z\right)=\left(6.537\angle 38.09°\right)e^{-j\beta z}+\left(6.537\angle 165.909°\right)e^{j\beta z}$

Find the value of $V_s\left(z=0\right)$.

$\begin{array}{c}{V}_s\left(z=0\right)=V_{\text{o}}\left(z=0\right)\\ =\left(6.537\angle 38.09°\right)\left(1\right)+\left(6.537\angle 165.909°\right)\left(1\right)\\ =5.75\angle 102°\text{V}\end{array}$

Find the value of $V_s\left(z=\mathcal{l}\right)$.

$\begin{array}{c}{V}_s\left(z=\mathcal{l}\right)=\left(6.537\angle 38.09°\right)e^{-j25}+\left(6.537\angle 165.909°\right)e^{j25}\\ =-1.5+j7.0948\\ =7.75\angle 102°\text{V}\end{array}$

Consider the conditions at the load.

$\begin{array}{c}{V}_L=V_s\left(z=\mathcal{l}\right)\\ =7.75\angle 102°\text{V}\end{array}$

Conclusion:

Thus, the value of $Z_{\text{in}}$ and $V_s$ at the receiving end is $j40\Omega$ and $7.75\angle 102°\text{V}$ respectively.

(c)

To determine

Find the value of $Z_{\text{in}}$ and $V_s$ at 4 m from the load.

Answer to Problem 29P

The value of $Z_{\text{in}}$ and $V_s$ at 4 m from the load is $-j3487.1\Omega$ and $13.07\angle 102°\text{V}$ respectively.

Explanation of Solution

Calculation:

Find the value of electrical length of the line.

$\begin{array}{c}\beta \mathcal{l}=\frac{1}{4}\times 4\\ =1\text{rad}\\ =1\times 57.295°\\ =57.295°\end{array}$

Substitute 60 for $Z_{\text{o}}$, $j40$ for $Z_L$, and $57.295°$ for $\beta \mathcal{l}$ in Equation (1).

$\begin{array}{c}{Z}_{\text{in}}=\left(60\right)\left[\frac{\left(j40\right)+j\left(60\right)\tan \left(57.295°\right)}{\left(60\right)+j\left(j40\right)\tan \left(57.295°\right)}\right]\\ =-j3487.1\Omega \end{array}$

From Part (b),

$V_{\text{o}}^{+}=6.537\angle 38.09°$

$V_{\text{o}}^{-}=6.537\angle 165.909°$

Substitute $6.537\angle 38.09°$ for $V_{\text{o}}^{+}$ and $6.537\angle 165.909°$ for $V_{\text{o}}^{-}$ in Equation (2).

$V_s\left(z\right)=\left(6.537\angle 38.09°\right)e^{-j\beta z}+\left(6.537\angle 165.909°\right)e^{j\beta z}$        (3)

4 m from the load is the same as 96 m from the source. That is,

$\begin{array}{c}\mathcal{l}=100-4\\ =96\text{m}\end{array}$

Find the value of $\beta \mathcal{l}$.

$\begin{array}{c}\beta \mathcal{l}=\left(0.25\right)\left(96\right)\\ =24\text{rad}\\ =24\times 57.2958\mathrm{deg}\left\{\because 1\text{rad}=57.2958\mathrm{deg}\right\}\\ \cong 1375°\end{array}$

Find the value of $V_s\left(z=\mathcal{l}\right)$.

$\begin{array}{c}{V}_s\left(z=\mathcal{l}\right)=\left(6.537\angle 38.09°\right)e^{-j24}+\left(6.537\angle 165.909°\right)e^{j24}\\ =13.07\angle 102°\text{V}\end{array}$

Conclusion:

Thus, the value of $Z_{\text{in}}$ and $V_s$ at 4 m from the load is $-j3487.1\Omega$ and $13.07\angle 102°\text{V}$ respectively.

(d)

To determine

Find the value of $Z_{\text{in}}$ and $V_s$ at 3 m from the source.

Answer to Problem 29P

The value of $Z_{\text{in}}$ and $V_s$ at 3 m from the source is $-j18.2178\Omega$ and $3.8\angle -78°\text{V}$ respectively.

Explanation of Solution

Calculation:

$3\text{m}\left(\mathcal{l}\right)$ from the source is the same as 97 m from the load. That is,

$\begin{array}{c}\mathcal{l}=100-3\\ =97\text{m}\end{array}$

ss

Find the value of electrical length of the line.

$\begin{array}{c}\beta \mathcal{l}=\frac{1}{4}\times 97\\ =24.25\text{rad}\\ =309.42°\end{array}$

Substitute 60 for $Z_{\text{o}}$, $j40$ for $Z_L$, and $309.42°$ for $\beta \mathcal{l}$ in Equation (1).

$\begin{array}{c}{Z}_{\text{in}}=\left(60\right)\left[\frac{\left(j40\right)+j\left(60\right)\tan \left(309.42°\right)}{\left(60\right)+j\left(j40\right)\tan \left(309.42°\right)}\right]\\ =-j18.2178\Omega \end{array}$

From Part (b),

$V_{\text{o}}^{+}=6.537\angle 38.09°$

$V_{\text{o}}^{-}=6.537\angle 165.909°$

Substitute $6.537\angle 38.09°$ for $V_{\text{o}}^{+}$ and $6.537\angle 165.909°$ for $V_{\text{o}}^{-}$ in Equation (2).

$V_s\left(z\right)=\left(6.537\angle 38.09°\right)e^{-j\beta z}+\left(6.537\angle 165.909°\right)e^{j\beta z}$        (4)

Find the value of $\beta \mathcal{l}$.

$\begin{array}{c}\beta \mathcal{l}=\left(0.25\right)\left(3\right)\\ =0.75\text{rad}\\ =43°\end{array}$

Find the value of $V_s\left(z=\mathcal{l}\right)$.

$\begin{array}{c}{V}_s\left(z=\mathcal{l}\right)=\left(6.537\angle 38.09°\right)e^{-j0.75}+\left(6.537\angle 165.909°\right)e^{j0.75}\\ =3.8\angle -78°\text{V}\end{array}$

Conclusion:

Thus, the value of $Z_{\text{in}}$ and $V_s$ at 3 m from the source is $-j18.2178\Omega$ and $3.8\angle -78°$ respectively.