Question 9.5: With reference to the transmission line given in Example 9-4......
With reference to the transmission line given in Example 9-4, find
(a) reflection coefficient,
(b) expression for the total voltage on the line,
(c) power dissipated at the load, and
(d) total power dissipated in the line.
(a) The reflection coefficient is, from Eq. (9-42),
\boxed{ \Gamma = \frac{Z_{L} – Z_{o} }{Z_{L} + Z_{o} } ≡\left|\Gamma \right| e^{j \phi }} (9-42)
\Gamma = \frac{Z_{L} – Z_{o} }{Z_{L} + Z_{o} } = \frac{(10 + j 40) – (110 + j 80 )}{(10 + j 40) + (110 + j 80 )} = 0.635 e^{j 2.74}(b) The input voltage is obtained from Eq. (9-43a) by substituting z = 0 and \ell ^{\prime } = 2[m]:
V(z) = V^{+}_{o} e^{\gamma z}[1 + \Gamma e^ { – 2\gamma \ell^{\prime } }] [V] (9-43a)
V_{in}(z = 0 , \ell^{\prime } = 2 ) = V^{+}_{o} [1 + \Gamma e^ { – 4\gamma }] = V^{+}_{o} [1 +0.635 e^{j 2.74} e^{ – 0.6 – j14 }] \\ \quad \quad \quad \quad \quad \quad \quad = V^{+}_{o} ( 1.0911 + j 0.336) [V]Equating this with Eq. (9-40c), the amplitude of the forward wave is
V_{in} = Z_{in} I_{in} = (45.86 + j 153.9)(0.0583 – j 0.0936) \\ \quad \quad = 17.08 +j 4.68 [V] (9-40c)
V^{+}_{o} = 15.51 e^{ – j0.031 } [V]Total voltage on the line is therefore
V(z) = V^{+}_{o} e^{ – \gamma z}[1 + \Gamma e^{ – 2\gamma \ell^{\prime } }] \\ \quad \quad = 15.51 e^{ – j 0.031 } e^{ – (0.15 + j 3.5)z} [1 + 0.635 e^{j 2.74} e^{ – (0.3 + j7.0) \ell^{\prime }}] [V] (9- 53)
(c) By substituting z = 2[m] and \ell ^{\prime } =0 into Eq. (9-53), the load voltage is
V_{L} = 15.51 e^{ – j 0.031 } e^{ – (0.3 + j 7.0)} [1 + 0.635 e^{j 2.74}] = 5.56 e^{- j 0.210} [V]Time-average power dissipated at the load is therefore
\left\langle\mathscr{P}_{L}\right\rangle = \frac{1}{2} Re\left[ V_{L}\left\lgroup\frac{V_{L}}{Z_{L}} \right\rgroup^{*}\right] = \frac{1}{2} Re \left[\frac{(5.56)^{2}}{10 – j 40} \right] = 90.9 [ mW](d) By using the input voltage and current given in Eqs. (9-40b) and (9-40c), we compute the total power dissipated in the line and the load:
I_{in} = \frac{V_{g}}{Z_{g} + Z_{in}} = \frac{20}{50 + 45.86 + j 153.9} \\ \quad = 0.0583 – j 0.0936 [A] (9-40b)
\left\langle\mathscr{P}_{in}\right\rangle = \frac{1}{2} Re [V_{in}I^{*}_{in}] = \frac{1}{2} Re [ (17.08 + j 4.68) (0.0583 + j 0.0936)] \\ \quad \quad = 279 [ mW]Thus the power dissipated in the line is
279 – 90.9 = 188.1[mW]