Chapter 10, Problem 10.15P

Figure 10.29 See Problem 10.15.

For the transmission line represented in Figure 10.29, find V_s,out if f = (a) 60 Hz; (b) 500 kHz.

To determine

(a)

Value of $V_{s,out}$ at the given frequency.

Answer to Problem 10.15P

The $V_{s,out}$ for $f=60\text{ Hz}$ is $104\text{V}$.

Explanation of Solution

Given:

The transmission line is shown below:

The given frequency is 60 Hz.

Calculation:

The value is,

$\begin{array}{l}\beta =\frac{\omega }{v_p}\\ =\frac{2\pi f}{v_p}\end{array}$

Substitute ${\text{2c/3 for v}}_p$ ,

$\beta =\frac{2\pi f}{2c/3}\mathrm{.........}(1)$

Consider the speed of light, $c\text{=}\text{3}\times {\text{10}}^8\text{m}/\text{s}$

Substitute 60 Hz for f in equation (1).

$\begin{array}{l}\beta \text{=}\frac{2\pi \times 60}{(2/3)\times (3\times {10}^8)}\\ \text{ =1}\text{.9}\times {\text{10}}^{-6}\text{ rad/m}\end{array}$

To find $\beta l$ ,

   $\begin{array}{l}\beta l=(1.9\times {10}^{-6})(80)\\ \text{ =1}\text{.52}\times {\text{10}}^{-4}\text{ rad}\end{array}$

As $\beta l<<1$ , the line is lumped circuit.

Therefore,

   $\begin{array}{l}{\text{Z}}_{in}=Z_L\\ Z_L=80\Omega \end{array}$

Find the expression for ${\text{V}}_{s,out}$.

   $\begin{array}{l}{V}_{sout}=V_S\left[\frac{Z_L}{Z_8+Z_L}\right]\\ \text{Here},\\ V_s\text{ is the source voltage,}\\ {\text{Z}}_s\text{ is the source impedance,}\\ {\text{Z}}_L\text{ is the output impedance}\text{.}\end{array}$

Substitute 120 for $V_S$ , 80 for ${\text{Z}}_L$ and 12 for ${\text{Z}}_S$

   $\begin{array}{l}{V}_{sout}=120\left[\frac{80}{12+80}\right]\\ \text{ =104V}\end{array}$

Conclusion:

Thus, V_sout for f = 60Hz is $104\text{V}$.

To determine

(b)

Value of $V_{s,out}$ at the given frequency.

Answer to Problem 10.15P

The $V_{s,\text{out}}$ for f = 500 kHz is $\left(52.6-j123\right)\text{V}$.

Explanation of Solution

Given:

The transmission line represented is given below:

The given frequency is 500kHz.

Calculation:

Substitute $5\times {10}^5$ for f in equation (1).

$\begin{array}{l}\beta =\frac{2\pi \times (5\times {10}^5)}{(2/3)\times (3\times {10}^8)}\\ =15.7\times {10}^{-3}\text{ rad/m}\end{array}$

To find $\beta l$ ,

   $\begin{array}{l}\beta l=(15.7\times {10}^{-3})(80)\\ =1.26\text{ rad}\end{array}$

Find the expression of input impedance,

   $Z_{in}=Z_0\left[\frac{Z_L\cos (\beta \mathcal{l})+j{Z}_0\sin (\beta \mathcal{l})}{Z_0\cos (\beta \mathcal{l})+j{Z}_L\sin (\beta \mathcal{l})}\right]$

Substitute 50 for ${\text{Z}}_0$ , 80 for ${\text{Z}}_L$ and 1.26 for $\beta l$.

   $\begin{array}{l}{Z}_{in}=50\left[\frac{80\cos (1.26)+j50\sin (1.26)}{50\cos (1.26)+j80\sin (1.26)}\right]\\ =50\left[\frac{24.46+j47.6}{15.3+j76.17}\right]\\ =33.13-j9.4\\ =34.44\angle -0.28\end{array}$

Write the expression for voltage across ${\text{Z}}_{in}$.

   $V_{in}=V_s\left[\frac{Z_{in}}{Z_s+Z_{in}}\right]$

Substitute 120 for $V_S$ , $33.13-j9.4$ for ${\text{Z}}_{in}$ and 12 for ${\text{Z}}_s$

   $\begin{array}{l}{V}_{in}=120\left[\frac{33.13-j9.4}{12+(33.13-j9.4)}\right]\\ =89.41-j6.36\\ =89.64\angle -0.071\end{array}$

   $V_{in}=V_0^{+}{e}^{-j\beta z}+V_0^{+}{e}^{j\beta z}\mathrm{......}(2)$

   $\begin{array}{l}{V}_0^{-}={\Gamma }_L{V}_0^{+}\\ =\frac{Z_L-Z_0}{Z_L+Z_0}{V}_0^{+}\\ =\frac{80-50}{80+50}{V}_0^{+}\\ =\frac{30}{130}{V}_0^{+}\\ V_0^{-}=\frac{3}{13}{V}_0^{+}\end{array}$

Therefore,

   $V_{s,out}=V_0^{+}(1+{\Gamma }_L)$

   $\begin{array}{l}{\text{V}}_0^{+}=\text{42}\text{.88 - j100}\\ {\Gamma }_L=3/13\end{array}$

Put above values in the equation

   $\begin{array}{l}{V}_{s,out}=(42.88-j100)(1+3/13)\\ =52.6-j123\end{array}$

Conclusion:

Hence, $V_{s,out}$ for f = 500 kHz is $\left(52.6-j123\right)\text{V}$.