Question

Measurements from a 99Ω transmission line were as follows: .Standing wave ratio 23 Distance from the load to the first voltage minimum 2.3m Distance from the load to the second voltage minimum 19.0m What is the load impedance? Select one: 0 a. 0.79+)(6.91)2 b. 5.22+j(45.62)2 b. 5.22+j(45.62)92 Oc. 48.54+j(-212.07)Q d. 8.99+j(-102.95)92

Answer 1

Given, z_0 = 99 Ohm s = 23. As corresponding voltage minimum are lambda/2 a part. i. e . Each minimum with next corresponding minimum are differ in lambda/2 distance (same is valid for maximum also). rightwardsdoublearrow lambda/2 = 19 - 23 = 16.7 m. In case of minimum, phi - 2 beta l)min = -(2n + 1) pi. & n = 0 for first minimum. rightwardsdoublearrow phi - 2 times 2 pi/lambda times 2.3 = -pi rightwardsdoublearrow phi - 4 pi/2 times 16.7 times 2.3 = -pi rightwardsdoublearrow phi = -0.72 pi. Let for case in calculation, phi almostequalto -0.5 pi (-pi/2) \r\n Also, vert n vert = r - 1/r + 1 = 23 - 1/23 + 1 = 0.916 rightwardsdoublearrow n = 0.916 e^-n = j pi/2 = -j 0.916. finally, Z_L = Z_0 [1 + n/1 + n] = -99[L - j 0.916/L + j 0.916] rightwardsdoublearrow Z_L = 99[L + 0.916^2 -0 2 times 0.916/1^2 + 0.916^2] [Rationalize] rightwardsdoublearrow Z_L = (8.61 - 98.43 j) Ohm Hence (d) is most approximate answer.