Question

The input impedance of a \(\frac{\lambda}{8}\) section of a lossless transmission line of the characteristic impedance \(50\Omega\) is found to be real when the other end is terminated by a load \(Z_L = (R + jX)\Omega\). If \(X\) is \(30\Omega\) the value of R (in \(\Omega\)) is

• A. 40\(\Omega\)
• B. 50\(\Omega\)
• C. 80\(\Omega\)
• D. 100\(\Omega\)

Hint:

For an impedance transformation problem where the result must be purely real, after substituting into the main equation, rationalize the complex fraction. Then, simply set the resulting imaginary part of the numerator to zero and solve.

Answer 1

The Correct Option is A

Step 1: Write the formula for input impedance of a transmission line. \[ Z_{in} = Z_0 \frac{Z_L + jZ_0 \tan(\beta l)}{Z_0 + jZ_L \tan(\beta l)} \]
Step 2: Calculate \(\tan(\beta l)\). The length of the line is \(l = \lambda/8\). The phase constant is \(\beta = 2\pi/\lambda\). \[ \beta l = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \] \[ \tan(\beta l) = \tan(\pi/4) = 1 \]
Step 3: Substitute the known values into the impedance formula. \(Z_0 = 50\Omega\), \(Z_L = R + j30\). \[ Z_{in} = 50 \frac{(R + j30) + j50(1)}{50 + j(R + j30)(1)} = 50 \frac{R + j80}{50 + jR - 30} = 50 \frac{R + j80}{20 + jR} \]
Step 4: Set the imaginary part of \(Z_{in}\) to zero. To make \(Z_{in}\) real, we multiply the numerator and denominator by the complex conjugate of the denominator: \[ Z_{in} = 50 \frac{(R + j80)(20 - jR)}{(20 + jR)(20 - jR)} = 50 \frac{(20R + 80R) + j(1600 - R^2)}{400 + R^2} \] The input impedance is real, so its imaginary part must be zero. \[ 1600 - R^2 = 0 \] \[ R^2 = 1600 \implies R = 40 \, \Omega \]