Question 6.11: A load impedance ZL = 50 + j50 Ω terminates a transmission l...
A load impedance ZL = 50 + j50 Ω terminates a transmission line that is 5 meters long and has a characteristic impedance of Z_c = 25 Ω. Using the Smith chart, find the impedance at the signal generator if the frequency of oscillation f = 1 x 105 Hz. The phase velocity for this transmission line is v = 2 x 106 m/s.
The wavelength λ is
\lambda=\frac{v}{f}=\frac{2 \times 10^{6}}{1 \times 10^{5}}=20 m
The distance between the load and the generator is \lambda / 4 . The normalized load impedance is
Z_{L}=\frac{50+j 50}{25}=2+j 2
The normalized load impedance is first located on the Smith chart.
Using a compass located at the center of the Smith chart, rotate this point a distance \lambda / 4 toward the generator. The normalized load impedance at the generator Z_{\text {in }} as read from the Smith chart is Z_{\text {in }}=0.25-j 0.25 . Therefore, the input impedance that is connected to the signal generator is
Z_{\text {in }}=Z_{c} Z_{\text {in }}=25(0.25-j 0.25)=6.25-j 6.25 \Omega