Topic: Transmission Lines
An RF signal is applied to a $50\Omega$ lossless transmission line which is terminated in a load with impedance, $Z_L=j50\Omega$. The wavelength is 8 cm. Find the position of voltage and current maximum respectively nearest to the load measured from load end?
- A. 1 cm, 3 cm
- B. 3 cm, 1 cm
- C. 3 cm, 5 cm
- D. 5 cm, 3 cm
Answer: A. 1 cm, 3 cm
Explanation: The reflection coefficient is $\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{j50 - 50}{j50 + 50} = \frac{-1+j}{1+j} = j = 1\angle(\pi/2)$. The phase angle of the reflection coefficient is $\theta = \pi/2$. The distance from the load to the first voltage maximum is $d_{vmax} = \frac{\theta \lambda}{4\pi} = \frac{(\pi/2) \times 8}{4\pi} = 1$ cm. The current maximum corresponds to the voltage minimum, which occurs at a distance of $\lambda/4 = 2$ cm from the voltage maximum. Therefore, $d_{imax} = 1 + 2 = 3$ cm.