Question

Which one of the following is not true of a lossless line?

• A. \( Z_{in} = -jZ_0 \) for a shorted line with \( l = \frac{\lambda}{8} \)
• B. \( Z_{in} = j\infty \) for a shorted line with \( l = \frac{\lambda}{4} \)
• C. \( Z_{in} = Z_0 \) for a matched line
• D. At a half-wavelength from load \( Z_{in} = Z_L \) and repeats for every half wavelength there after

Hint:

Input impedance of a lossless line: \(Z_{in} = Z_0 \frac{Z_L + jZ_0 \tan(\beta l)}{Z_0 + jZ_L \tan(\beta l)}\).
Shorted line (\(Z_L=0\)): \(Z_{in} = jZ_0 \tan(\beta l)\).
Open-circuited line (\(Z_L=\infty\)): \(Z_{in} = -jZ_0 \cot(\beta l)\).
Matched line (\(Z_L=Z_0\)): \(Z_{in} = Z_0\).
For \(l=\lambda/8\), \(\beta l = \pi/4\), \(\tan(\pi/4)=1\). For shorted line, \(Z_{in} = jZ_0\).
For \(l=\lambda/4\), \(\beta l = \pi/2\), \(\tan(\pi/2)=\infty\). For shorted line, \(Z_{in} = j\infty\) (open circuit).
For \(l=\lambda/2\), \(\beta l = \pi\), \(\tan(\pi)=0\). For shorted line, \(Z_{in} = 0\) (short circuit). Input impedance repeats every \(\lambda/2\).

Answer 1

The Correct Option is A

For a lossless transmission line of characteristic impedance \(Z_0\) and length \(l\), terminated by a load \(Z_L\), the input impedance \(Z_{in}\) is given by: 

\( Z_{in} = Z_0 \frac{Z_L + jZ_0 \tan(\beta l)}{Z_0 + jZ_L \tan(\beta l)} \) where \(\beta = \frac{2\pi}{\lambda}\).

Let's analyze each statement for a lossless line:

• (a) Shorted line (\(Z_L = 0\)), \(l = \lambda/8\):

\(\beta l = \left(\frac{2\pi}{\lambda}\right) \left(\frac{\lambda}{8}\right) = \frac{\pi}{4}\). So \(\tan(\beta l) = \tan(\pi/4) = 1\).

\(Z_{in} = Z_0 \frac{0 + jZ_0(1)}{Z_0 + j(0)(1)} = Z_0 \frac{jZ_0}{Z_0} = jZ_0\).

The statement says \(Z_{in} = -jZ_0\). This is FALSE. \(Z_{in}\) should be \(+jZ_0\).

• (b) Shorted line (\(Z_L = 0\)), \(l = \lambda/4\):

\(\beta l = \left(\frac{2\pi}{\lambda}\right) \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}\). So \(\tan(\beta l) = \tan(\pi/2) = \infty\).

For this case, it's easier to use \(Z_{in} = jZ_0 \tan(\beta l)\) for a shorted line.

\(Z_{in} = jZ_0 \tan(\pi/2) = jZ_0 (\infty) = j\infty\). This represents an open circuit.

The statement says \(Z_{in} = j\infty\). This is TRUE.

• (c) Matched line (\(Z_L = Z_0\)): 

\(Z_{in} = Z_0 \frac{Z_0 + jZ_0 \tan(\beta l)}{Z_0 + jZ_0 \tan(\beta l)} = Z_0 \frac{Z_0(1 + j\tan(\beta l))}{Z_0(1 + j\tan(\beta l))} = Z_0\).

The statement says \(Z_{in} = Z_0\). This is TRUE.

• (d) At a half-wavelength from load (\(l = \lambda/2\)), \(Z_{in} = Z_L\), and repeats for every half wavelength thereafter:

If \(l = n\lambda/2\) (where n is an integer), then \(\beta l = \left(\frac{2\pi}{\lambda}\right) (n\lambda/2) = n\pi\).

\(\tan(\beta l) = \tan(n\pi) = 0\).

Then \(Z_{in} = Z_0 \frac{Z_L + jZ_0(0)}{Z_0 + jZ_L(0)} = Z_0 \frac{Z_L}{Z_0} = Z_L\).

This is TRUE. The input impedance repeats every half-wavelength.

The question asks: "Which function does NOT satisfy the wave equation?"

All options (a), (b), (c), and (d) correctly follow the expected formulas for a lossless transmission line with the given parameters.

Therefore, the statement that is "not true" is:

\( \boxed{Z_{in} = -jZ_0 \text{ for a shorted line with } l = \frac{\lambda}{8}} \)