Bridge voltages with open and short lines

What are the bridge voltages $V_z$ and $V_a$ if the reflectometer is connected to a lossles open or short circuit transmission line?

Working in normalised units ($Z_0=1.0+j0.0$),

$$V_a=\frac{1}{\vert Z+1\vert }$$

$$V_z=\frac{\vert Z\vert }{\vert Z+1\vert }$$

For a short-circuited transmission line, the input impedance is

$$Z=j\tan(\beta l)$$

So, $\vert Z \vert = \tan(\beta l)$, and

$$\vert Z+1\vert = \vert 1 + j\tan(\beta l) \vert $$

As $\tan$ is $\frac{\sin}{\cos}$:

$$\vert Z+1\vert = \vert 1 + j\frac{\sin(\beta l)}{\cos(\beta l)} \vert = \frac{1}{\cos(\beta l)} \vert \cos(\beta l) + j\sin(\beta l) \vert $$

and we know that

$$\vert \cos(\beta l) + j\sin(\beta l) \vert = 1 $$

So

$$\vert Z+1\vert = \frac{1}{\cos(\beta l)} $$

Hence,

$$V_a = \frac{1}{\vert Z+1\vert } = \cos(\beta l)$$

and

$$V_z = \frac{\vert Z\vert }{\vert Z+1\vert } = \frac{\tan(\beta l)}{\cos(\beta l)} = \sin(\beta l)$$

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