Transmission lines

I'm not going to type out the derivation of the transmission line equations. I'm okay with just skimming through that part.

But lets summarize the good parts:

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

Where, $\gamma$ is the propagation constant. $\alpha$ is the attenuation constant and $\beta$ is the phase constant.

The transmission line equation is a solution of linear homogeneous differential equations:

$$V_s(z) = V_o^+ e^{-\gamma_z} + V_o^- e^{\gamma_z} \\ I_s(z) = I_o^+ e^{-\gamma_z} + I_o^- e^{\gamma_z}$$

Assuming an infinite line, we realize tha the $e^{\gamma z}$ terms must vanish. We come up with a cool new formula:

$$Z_o = \sqrt{\frac{R+j\omega L}{G+j\omega C}}$$

We can come up with 2 special cases:

1. Lossless Line ($R = 0 = G$)

$$\alpha = 0 \\ \beta = \omega \sqrt{L C} \\ \text{and}\\ R_o = \sqrt{\frac{L}{C}}\\ X_o = 0$$

2. Distortionless Line ($\frac{R}{L} = \frac{G}{C}$)

Signal consists of multiple frequencies. The different frequency components get attenuated differently when travelling through the medium.

In a distortionless line, teh attenuation constant $\alpha$ is frequency independent of the phase constant $\beta$.

For such a line

$$\frac{R}{L} = \frac{G}{C} \\ \text{thus} \\ \alpha = \sqrt{R G} \\ \beta = \omega \sqrt{L C} \\ \text{and} \\ R_o = \sqrt{\frac{R}{G}} = \sqrt{\frac{L}{C}} \\ X_o = 0 \\$$