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: State the formula for the 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 \(\tan(\beta l)\).Given a line length of \(l = \lambda/8\) and a phase constant of \(\beta = 2\pi/\lambda\):\[ \beta l = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \]\[ \tan(\beta l) = \tan(\pi/4) = 1 \]
Step 3: Substitute the given values into the input impedance formula.With \(Z_0 = 50\Omega\) and \(Z_L = R + j30\):\[ Z_{in} = 50 \frac{(R + j30) + j50(1)}{50 + j(R + j30)(1)} = 50 \frac{R + j80}{50 + jR - 30} = 50 \frac{R + j80}{20 + jR} \]
Step 4: Determine the value of \(R\) that makes \(Z_{in}\) real.To eliminate the imaginary component of \(Z_{in}\), multiply the numerator and denominator by the complex conjugate of the denominator:\[ Z_{in} = 50 \frac{(R + j80)(20 - jR)}{(20 + jR)(20 - jR)} = 50 \frac{(20R + 80R) + j(1600 - R^2)}{400 + R^2} \]For the input impedance to be real, the imaginary part must be zero.\[ 1600 - R^2 = 0 \]\[ R^2 = 1600 \implies R = 40 \, \Omega \]