Rowan-Classes/5th-Semester-Fall-2023/EEMAGS/Homework/Homework-08.md

Homework 8 - Aidan Sharpe

1

A transmission line is terminated with a matched 50$\Omega$ load. The transmitter puts out 100W of power, and the transmission line is 100ft long. What for what value of \alpha will the power loss be 10W over the length of the line?

\alpha = {\ln\left({P_\text{in} \over P_\text{out}}\right) \over l} = {\ln\left({100 \over 90}\right) \over 30.48[\text{m}]} = 0.00345671\left[{\text{np} \over \text{m}}\right]

2

Evaluate the phase velocity and attenuation constant for a distorionless line, and compare it to a lossless line.

\gamma = \alpha + j\beta = \sqrt{(R + j\omega L)(G + j\omega C)}

\gamma ^2 = (R + j\omega L)(G + j\omega C)

\gamma^2 = \omega^2 LC(j + {R \over \omega L})(j + {G \over \omega C})

For a distorionless line:

{R \over L} = {G \over C}

(j + {R \over \omega L}) = (j + {G \over \omega C})

So \gamma^2 becomes:

\gamma^2 = \omega^2 L C (j + {R \over \omega L})^2

Solving for \gamma

\gamma = \omega\sqrt{L C}(j + {R \over \omega L})

Separating the real and imaginary components:

\alpha = R \sqrt{C \over L}

\beta = j\omega\sqrt{L C}

For a lossless line, there is no attenuation (by definition). Therefore:

\alpha = 0

To actually be able to build this lossless line, R=0 and G=0.

\gamma^2 = (R + j\omega L)(G + j\omega C)

Since R and G are both 0:

\gamma^2 = (j\omega L)(j \omega C)

\gamma^2 = -\omega^2 LC

\gamma = \sqrt{-\omega^2 LC} = j\omega\sqrt{LC}

Separating out real and imaginary:

\alpha = 0

\beta = j\omega\sqrt{LC}

For lossless and distortionless lines, the attenuation constant differs, but the phase constant does not. Since the phase velocity only depends on \omega and \beta, v_p is the same for both lossless and distortionless lines.

v_p = {\omega \over \beta} = {-j \over \sqrt{LC}}

3

\hat{Z}_L = 75 + j150 \Omega

f = 2[\text{MHz}]

\omega = 2\pi f = 4\pi \times 10^6

r = 150\left[{\Omega \over \text{km}}\right]

l = 1.4\left[{\text{mH} \over \text{km}}\right]

c = 88\left[{\text{nF} \over \text{km}}\right]

g = 0.8\left[{\mu\text{s} \over \text{km}}\right]

\hat{V}_G = 100 e^{j0^\circ}

z = 100[\text{m}]

R = 15[\Omega]

L = 140[\mu\text{H}]

C = 8.8[\text{nF}]

G = 80[n\Omega]

a)

Find \hat{Z}_0:

Z_0 = \sqrt{R + j\omega L \over G + j\omega C} = \sqrt{15 + j(4\pi\times10^6)(140\times10^{-6}) \over 80\times10^{-9} + j(4\pi \times 10^6)(8.8\times10^{-9})}

Z_0 = 126.1324 - j0.5377

\hat{\gamma} = \alpha + j\beta = \sqrt{(R + j\omega L)(G + j\omega C)}

\hat{\gamma} = 0.0595 + j13.9482

b)

Find the input imedance

Z_\text{in} = Z_0{Z_L + Z_0 \tanh(\gamma z) \over Z_0 + Z_L \tanh(\gamma z)}

Z_\text{in} = -126.1324 + j0.5377

c)

Find the average power delivered

\alpha = {\ln\left({P_\text{in} \over P_\text{out}}\right) \over z}

\alpha = 0.0595

P_\text{in} = {V_G^2 \over Z_\text{in}} = 79.2802 + j0.3373

P_\text{out} = P_\text{in} e^{-\alpha z} = 0.2073 + j0.0009

4

\Gamma = {Z_L - Z_0 \over Z_L + Z_0}

a)

Z_L = 3 Z_0

\Gamma = 0.5

b)

Z_L = (2 - j2)Z_0

\Gamma = 0.5385 - j0.3077

c)

Z_L = -j2Z_0

\Gamma = 0.6 - j0.8

d)

Z_L = 0

\Gamma = -1

5

a)

\Gamma = 0.06+j0.24

b)

\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|} = 1.657

c)

Z_\text{in} = Z_0 {Z_L + Z_0\tanh(\gamma l) \over Z_0 + Z_L\tanh(\gamma l)} 30.50 - j1.09

d)

Y_\text{in} = {1 \over Z_in} = 0.03274 + j0.0012

e)

0.106\lambda

f)

z = -0.106\lambda

6

L = {3\lambda \over 8}

Z_\text{in} = -j2.5

Z_L = {-j2.5 \over 100} = -j0.025

At {3\lambda \over 8}:

Z_L = j95