Question 9.8: A 75[Ω] -lossless line is terminated in an unknown impedance......
A 75[Ω] -lossless line is terminated in an unknown impedance Z_{L}. Measurements show that S = 3 on the line, and that the first and second voltage maxima occur at distances 4[cm] and 14[cm] from the load, respectively. Find (a) Γ , and (b) Z_{L}.
(a) The distance between two adjacent maxima is a half-wavelength. Thus, the wavelength and the phase constant are
λ = 2 × (14 – 4)= 20 [cm]
β = 2π / λ = 10 π [rad/m]
Substituting \ell^{\prime }_{max} =0.04 [m] , m = 0 , and β = 10π into Eq. (9-47a) we get
\boxed{\ell^{\prime }_{max} = \frac{1}{2\beta } (\phi + 2 m\pi )} (m = 0,1,2..) (9-47a)
Φ = 0.8π [rad] (9-64a)
Substituting S = 3 into Eq. (9-50) we get
\boxed{\left|\Gamma \right| = \frac{S – 1}{S + 1} } (9-50)
\left|\Gamma \right| = \frac{3 – 1}{3 + 1} =0.5 (9-64)
From Eqs. (9-64a) and (9-64b) we get
\Gamma = 0.5 e^{j0.8 \pi } (9-65)
(b) Inserting Eq. (9-65) into Eq. (9-42) we get
\boxed{\Gamma = \frac{Z_{L} – Z_{o}}{Z_{L} + Z_{o}} \equiv \left|\Gamma \right| e^{j\phi }} (9-42)
\frac{Z_{L} – 75}{Z_{L} + 75} =0.5 e^{j0.8\pi } \\ Thus \\ Z_{L}= 46 + j 88 [\Omega ]