Question 9.4: A 2[m]-transmission line is terminated in a load impedance Z......
A 2[m]-transmission line is terminated in a load impedance Z_{L} = 10 + j40[Ω] , and connected to a generator with v_{g} (t) = 20 \cos (10^{7} t)[V] and Z_{g} = 50[Ω] . On the line γ = 0.15 + j3.5[m^{-1}] and Z_{o} = 110 + j80[Ω] . Find
(a) input impedance,
(b) input current, and
(c) input voltage.
(a) From the values given in the problem, we write
\gamma \ell = 0.3 +j 7.0 \\ \tanh (\gamma \ell) = \frac{e^{\gamma \ell} – e^{ – \gamma \ell}}{e^{\gamma \ell} + e^{ – \gamma \ell }} = \frac{1 – e^{ – 0.6 – j14 }}{1 + e^{- 0.6 – j14}} = 0.482 +j 0.749From Eq. (9-38), we obtain
\boxed{Z_{in} = Z_{o} \frac{Z_{L} + Z_{o} \tanh \gamma \ell}{Z_{o} + Z_{L} \tanh \gamma \ell}} [Ω] (lossy line) (9-38)
Z_{in} = Z_{o} \frac{Z_{L} + Z_{o} \tanh \gamma \ell}{Z_{o} + Z_{L} \tanh \gamma \ell} \\ \quad = (110+j 80) \frac{(10 +j 40) + (110 +j 80) ( 0.482 + j 0.749)}{(110 +j 80) + (10 +j 40) ( 0.482 + j 0.749)} \\ \quad = 45.86 +j 153.9 (9-40a)
(b) From the equivalent circuit, the input current is obtained as
I_{in} = \frac{V_{g}}{Z_{g} + Z_{in}} = \frac{20}{50 + 45.86 + j153.9} \\\quad =0.0583 – j 0.0936 [A] (9-40b)
(c) Input voltage is thus
V_{in} = Z_{in} I_{in} = (45.86 + j153.9) (0.0583 – j 0.0936) \\ \quad = 17.08 + j 4.68 [V] (9-40c)