Chapter 11, Problem 4RQ

To determine

Choose the correct options from the given choices to mention which of the given conditions are not true of a lossless line.

Answer to Problem 4RQ

The correct options from the given choices are option (a) and (c).

Explanation of Solution

Calculation:

Consider the general expression of input impedance for a lossless line.

$Z_{\text{in}}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \beta \mathcal{l}}{Z_{\text{o}}+j{Z}_L\tan \beta \mathcal{l}}\right]$

For a shorted line with $\mathcal{l}=\frac{\lambda }{4}$:

Consider the general expression for electrical length.

$\begin{array}{c}\beta \mathcal{l}=\left(\frac{2\pi }{\lambda }\right)\left(\frac{\lambda }{4}\right)\\ =\frac{\pi }{2}\end{array}$

Therefore, the input impedance is,

$\begin{array}{c}{Z}_{\text{in}}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \beta \mathcal{l}}{Z_{\text{o}}+j{Z}_L\tan \beta \mathcal{l}}\right]\\ =Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\left(\frac{\sin \beta \mathcal{l}}{\cos \beta \mathcal{l}}\right)}{Z_{\text{o}}+j{Z}_L\left(\frac{\sin \beta \mathcal{l}}{\cos \beta \mathcal{l}}\right)}\right]\\ =Z_{\text{o}}\left[\frac{Z_L\cos \beta \mathcal{l}+j{Z}_{\text{o}}\sin \beta \mathcal{l}}{Z_{\text{o}}\cos \beta \mathcal{l}+j{Z}_L\sin \beta \mathcal{l}}\right]\\ =Z_{\text{o}}\left[\frac{\left(Z_L\right)\left(0\right)+j\left(Z_{\text{o}}\right)\left(1\right)}{\left(Z_{\text{o}}\right)\left(0\right)+j\left(Z_L\right)\left(1\right)}\right]\left\{\begin{array}{l}\because \cos \beta \mathcal{l}=\cos \left(\frac{\pi }{2}\right)=0,\\ \sin \beta \mathcal{l}=\sin \left(\frac{\pi }{2}\right)=1\end{array}\right\}\end{array}$

The above equation becomes,

$Z_{\text{in}}=\frac{Z_{\text{o}}^2}{Z_L}$        (1)

Therefore, a transmission line with length $\mathcal{l}=\frac{\lambda }{4}$, the input impedance is inversely proportional to the load impedance.

For a short circuit, $Z_L=0$:

Substitute 0 for $Z_L$ in Equation (1).

$\begin{array}{c}{Z}_{\text{in}}=\frac{Z_{\text{o}}^2}{0}\\ =\infty \end{array}$

Hence, a transmission line with one-quarter wavelength long transforms a short circuit into an open-circuit and vice versa.

For an open line with $\mathcal{l}=\frac{\lambda }{2}$:

The electrical length is,

$\begin{array}{c}\beta \mathcal{l}=\left(\frac{2\pi }{\lambda }\right)\left(\frac{\lambda }{2}\right)\\ =\pi \end{array}$

The input impedance is,

$\begin{array}{c}{Z}_{\text{in}}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\left(\frac{\sin \beta \mathcal{l}}{\cos \beta \mathcal{l}}\right)}{Z_{\text{o}}+j{Z}_L\left(\frac{\sin \beta \mathcal{l}}{\cos \beta \mathcal{l}}\right)}\right]\\ =Z_{\text{o}}\left[\frac{Z_L\cos \beta \mathcal{l}+j{Z}_{\text{o}}\sin \beta \mathcal{l}}{Z_{\text{o}}\cos \beta \mathcal{l}+j{Z}_L\sin \beta \mathcal{l}}\right]\\ =Z_{\text{o}}\left[\frac{Z_L\left(-1\right)+j{Z}_{\text{o}}\left(0\right)}{Z_{\text{o}}\left(-1\right)+j{Z}_L\left(0\right)}\right]\left\{\begin{array}{l}\because \cos \beta \mathcal{l}=\cos \pi =-1,\\ \sin \beta \mathcal{l}=\sin \pi =0\end{array}\right\}\\ =Z_L\end{array}$

The above equation becomes,

$Z_{\text{in}}=Z_{\text{o}}\left\{\because Z_L=Z_{\text{o}}\right\}$

For a shorted line with $\mathcal{l}=\frac{\lambda }{8}$,

The electrical length is,

$\begin{array}{c}\beta \mathcal{l}=\left(\frac{2\pi }{\lambda }\right)\left(\frac{\lambda }{8}\right)\\ =\frac{\pi }{4}\end{array}$

The input impedance is,

$\begin{array}{c}{Z}_{\text{in}}=Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}\tan \beta \mathcal{l}}{Z_{\text{o}}+j{Z}_L\tan \beta \mathcal{l}}\right]\\ =Z_{\text{o}}\left[\frac{Z_L+j{Z}_{\text{o}}}{Z_{\text{o}}+j{Z}_L}\right]\left\{\because \tan \beta \mathcal{l}=\tan \left(\frac{\pi }{4}\right)=1\right\}\\ =Z_{\text{o}}\left[\frac{Z_{\text{o}}+j{Z}_{\text{o}}}{Z_{\text{o}}+j{Z}_{\text{o}}}\right]\left\{\because Z_L=Z_{\text{o}}\right\}\\ =Z_{\text{o}}\end{array}$

For a matched line, $Z_L=Z_{\text{o}}$. Therefore, the equation for input impedance for a lossless transmission line reduces to $Z_{\text{in}}=Z_{\text{o}}$. And at a half-wavelength from a load, $Z_{\text{in}}=Z_L$ and repeats for every half-wavelength.

Therefore, from the above explanation, option (a) $Z_{\text{in}}=-j{Z}_{\text{o}}$ for a shorted line with $\mathcal{l}=\frac{\lambda }{8}$ and option (c) $Z_{\text{in}}=j{Z}_{\text{o}}$ for an open line with $\mathcal{l}=\frac{\lambda }{2}$ is not true for a lossless line.

Conclusion:

Thus, the correct options from the given choices are option (a) and (c).