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A transmitter is connected to an antenna with a coaxial cable with length \$l=1.5 \:\text{m}\$. The characteristic impedance of the cable is \$Z_0 = 75 \: \Omega\$ and the scattering parameters are \$\textbf{S} = \Bigg[ \begin{matrix} 0 & e^{-j27} \\ e^{-j27} & 0 \end{matrix} \Bigg]\$. The distance from antenna to the first maximum voltage on the transmission line is \$|z_{max}| = 10 \: \text{cm}\$.

Plot the magnitude of the voltage along the TM-line normalized to the magnitude of the forward voltage wave along the TM-line when VSWR = 1.5

Comparing the scattering parameters with the general expression of \$\text{S}\$ for a transmission line \$\textbf{S} = \Bigg[ \begin{matrix} 0 & e^{-j\beta l} \\ e^{-j \beta l} & 0 \end{matrix} \Bigg] \$ we see that \$\beta = \frac{27}{1.5}=18 \$.

From the definition of VSWR we can find the reflection coefficient:

$$VSWR=\frac{1+|\Gamma_L|}{1-|\Gamma_L|} \Rightarrow \Gamma_L = \pm 0.2 \: \: \: \: \text{for} \: \: \: \: VSWR=1.5 $$

We also know the relation $$|V(z)| = |V_0^+| \cdot |1+\Gamma_Le^{2j\beta z}| $$ which can be manipulated into what we want $$\frac{|V(z)|}{|V_0^+|} = |1+\Gamma_Le^{2j\beta z}|$$

Plotting this function yields: -

enter image description here

I have reason to believe that my method is incorrect. I don't use the information about \$Z_0\$ or \$z_{max}\$. I also think my determination of \$\Gamma_L\$ is incorrect since there is some information lost when we look at \$|\Gamma_L|\$ instead of just \$\Gamma_L\$.

My question is am I on the right track? Are my determinations valid or am I misunderstanding the transmission line theory?

Carl
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1 Answers1

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$$VSWR=\frac{1+|\Gamma_L|}{1-|\Gamma_L|} \Rightarrow \Gamma_L = \pm 0.2 \: \: \: \: \text{for} \: \: \: \: VSWR=1.5 $$

This isn't correct. You know \$|\Gamma_L|=0.2\$, but \$\Gamma_L\$ is not necessarily real. It could be a complex number with any phase. It could be any number with the form \$0.2e^{j\theta}\$.

You need to use the information about the distance to the first maximum in the standing wave pattern to work out what \$\theta\$ is.

The Photon
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