# 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 ${3\lambda \over 8}$:
$$Z_L = j95$$