Question

Consider a lossless transmission line terminated with a short circuit as shown in the figure below. As one moves towards the generator from the load, the normalized impedances \(z_{inA}\), \(z_{inB}\), \(z_{inC}\), and \(z_{inD}\) (indicated in the figure) are: \begin{center} \includegraphics[width=8cm]{15.png} \end{center}

• A. \(z_{inA} = +j \, \Omega, \, z_{inB} = \infty, \, z_{inC} = -j \, \Omega, \, z_{inD} = 0\)
• B. \(z_{inA} = \infty, \, z_{inB} = +0.4j \, \Omega, \, z_{inC} = 0, \, z_{inD} = +0.4j \, \Omega\)
• C. \(z_{inA} = -j \, \Omega, \, z_{inB} = 0, \, z_{inC} = +j \, \Omega, \, z_{inD} = \infty\)
• D. \(z_{inA} = +0.4j \, \Omega, \, z_{inB} = \infty, \, z_{inC} = -0.4j \, \Omega, \, z_{inD} = 0\)

Hint:

To analyze lossless transmission lines, use the tangent function for normalized impedance calculations, and carefully consider the phase shifts introduced by the distance from the load.

Answer 1

The Correct Option is A

Step 1: Understand the properties of the lossless transmission line.
For a lossless transmission line terminated with a short circuit, the normalized input impedance at a distance \(z = -l\) from the load is given by: \[ z_{in} = j \tan\left(\frac{2\pi l}{\lambda}\right), \] where \(\lambda\) is the wavelength of the signal. Step 2: Calculate the normalized impedances at the specified points.
- At \(z = -\lambda/8\) (\(z_{inD}\)): \[ z_{inD} = j \tan\left(-\frac{\pi}{4}\right) = j(0) = 0. \] - At \(z = -\lambda/4\) (\(z_{inC}\)): \[ z_{inC} = j \tan\left(-\frac{\pi}{2}\right) = -j \infty. \] - At \(z = -3\lambda/8\) (\(z_{inB}\)): \[ z_{inB} = j \tan\left(-\frac{3\pi}{4}\right) = \infty. \] - At \(z = -\lambda/2\) (\(z_{inA}\)): \[ z_{inA} = j \tan(-\pi) = +j. \] Step 3: Summarize the normalized input impedances.
The calculated values are: \[ z_{inA} = +j \, \Omega, \, z_{inB} = \infty, \, z_{inC} = -j \, \Omega, \, z_{inD} = 0. \] Final Answer: \[\boxed{{(1) } z_{inA} = +j \, \Omega, \, z_{inB} = \infty, \, z_{inC} = -j \, \Omega, \, z_{inD} = 0.}\]