Chapter 10, Problem 10.14P

A lossless transmission line having characteristic impedance $Z_0=50\Omega$ is driven by a source at the input end that consists of the series combination of a 10-V sinusoidal generator and a $50-\Omega$ resistor. The line is one-quarter wavelength long. At the other end of the line, a load impedance $Z_L=50-j50\Omega$ is attached. (a) Evaluate the input impedance to the line seen by the voltage source-resistor combination. (b) Evaluate the power that is dissipated by the load. (c) Evaluate the voltage amplitude that appears across the load.

To determine

(a)

The input impedance to the line seen by the combination of voltage-source and resistor.

Answer to Problem 10.14P

   $z_{in}=25+j25\Omega$

Explanation of Solution

Given:

The characteristic impedance of lossless line is $z_0=50\Omega$ . The supply voltage by which the load is driven is 10 V and it is connected in series with 50 $\Omega$ resistor. Length of line is quarter wavelength and load impedance is $z_L=50-j50\Omega$.

Concept Used:

For quarter wavelength input impedance is

   $z_{in}=\frac{z_0{}^2}{z_L}$

Calculation:

Although

   $z_{in}=\frac{z_0{}^2}{z_L}$

Plugging value of load impedance and character impedance in above formula

   $\begin{array}{l}{z}_{in}=\frac{{\left(50\right)}^2}{50-j50}\\ z_{in}=25+j25\Omega \end{array}$

Conclusion:

Hence, the input impedance to the line seen by the voltage-source resistor combination is $\left(25+j25\right)\Omega$.

To determine

(b)

The power dissipated by the load.

Answer to Problem 10.14P

   $P_{in}=1.25W$

Explanation of Solution

Given:

   $z_g=50\Omega$

   $z_{in}=25+j25\Omega$

   $v_{so}=10V$

Concept Used:

Power dissipated by the load is given by

   $P_{in}=\frac{1}{2}\mathrm{Re}\left[\frac{v_{in}{v}^{\ast }{}_{in}}{z^{\ast }{}_{in}}\right]$

Calculation:

Voltage across input impedance is given by

   $v_{in}=v_{so}\frac{z_{in}}{z_g+z_{in}}$

Plugging value of input $z_g$ , $z_{in}$ and $v_{so}$ in the formula shown above

   $\begin{array}{l}{v}_{in}=10\left[\frac{25+j25}{50+50+j50}\right]\\ v_{in}=10+5jV\end{array}$

Hence power dissipated by the load is

   $\begin{array}{l}{P}_{in}=\frac{1}{2}\mathrm{Re}\left[\frac{v_{in}{v}^{\ast }{}_{in}}{z^{\ast }{}_{in}}\right]\\ P_{in}=\frac{1}{2}\mathrm{Re}\left[\frac{\left(10+j5\right)\left(10-j5\right)}{\left(25-j25\right)}\right]\\ P_{in}=1.25W\end{array}$

Conclusion:

Hence,power dissipated by the load is 1.25W.

To determine

(c)

The voltage amplitude that appears across the load.

Answer to Problem 10.14P

   $v_{in}=\left(-5+j15\right)\text{V}$

Explanation of Solution

Given:

   $z_L=50-j50\Omega$

   $z_0=50\Omega$

Length = $\frac{\lambda }{4}$

Concept Used:

   $v_s\left(z\right)=v_0{}^{+}{e}^{-j\beta z}+v_0{}^{-}{e}^{+j\beta z}$

   $\begin{array}{l}{v}_0{}^{-}=\Gamma l{v}_0{}^{+}\\ \Gamma l=\frac{z_L-z_0}{z_L+z_0}\end{array}$

Calculation:

In the line phasor voltage at any point is given by sum of forward and backward wave

   $v_s\left(z\right)=v_0{}^{+}{e}^{-j\beta z}+v_0{}^{-}{e}^{+j\beta z}$

Although

   $\begin{array}{l}{v}_0{}^{-}=\Gamma l{v}_0{}^{+}\\ \text{Where}\\ \Gamma l=\frac{z_L-z_0}{z_L+z_0}\\ \text{Plugging}\text{value}\text{of}{z}_L\text{and}{z}_0\text{in}\text{above}\text{formula}\\ \Gamma l=\frac{50-j50-50}{50-j50+50}\\ \Gamma l=0.2-j0.4\end{array}$

Assume load is located at z = 0 therefore $v_{in}$ is therefore given by the above expressionevaluated at z = - l

Hence l = $-\frac{\lambda }{4}$

Thus $\beta l=\frac{\pi }{2}$

So

   $\begin{array}{l}{v}_{in}\left(z\right)=v_0{}^{+}{e}^{j\beta z}+v_0{}^{-}{e}^{-j\beta z}\\ v_{in}\left(z\right)=v_0{}^{+}\left[e^{j\beta l}+\Gamma l{e}^{-j\beta l}\right]\\ v_{in}\left(z\right)=v_0{}^{+}\left[j+\left(0.2-j0.4\right)\left(-j\right)\right]\\ v_{in}\left(z\right)=v_0{}^{+}\left[-0.4+j0.8\right]\end{array}$

Using $v_{in}$ from part b,

   $v_0{}^{+}=\frac{\left(10+j5\right)}{\left(-0.4+j0.8\right)}$

Therefore, voltage at load is

   $\begin{array}{l}{v}_L=v_0{}^{+}\left(1+\Gamma l\right)\\ v_L=\frac{\left(10+j5\right)}{\left(-0.4+j0.8\right)}\left(1+0.2-J0.4\right)\\ v_L=-5+j15V\end{array}$

Conclusion:

Hence voltage amplitude that appears across the load is $-5+j15V$.