Question 11.3: A certain transmission line 2 m long operating at ω = 10^6 r...

(a) Since 1 Np=8.686 dB

\alpha =\frac{8}{8.686}=0.921 Np/m

\gamma =\alpha+j\beta=0.921+j1 /m

\gamma\ell=2(0.921+j1)=1.84+j2

Using the formula for \tanh(x+jy) in Appendix A.3, we obtain

\tanh\gamma\ell=1.033-j0.03929

Z_{in}=Z_{o}\left(\frac{Z_{L}+Z_{o}\tanh\gamma\ell}{Z_{o}+Z_{L}\tanh\gamma\ell} \right)

=(60+j40)\left[\frac{20+j50+(60+j40)(1.033-j0.03929)}{60+j40+(20+j50)(1.033-j0.03929)} \right]

Z_{in}=60.25+j38.79\Omega

(b) The sending-end current is I(z = 0) = I_{o}. From eq. (11.28):

V_{o}=\frac{Z_{in}}{Z_{in}+Z_{g}}V_{g},       I_{o}=\frac{V_{g}}{Z_{in}+Z_{g}}

I(z=0)=\frac{V_{g}}{Z_{in}+Z_{g}}=\frac{10}{60.25+j38.79+40}=93.03\angle-21.15^{\circ} mA

(c) To find the current at any point, we need V_{o}^{+} and V_{o}^{-}. But

I_{o}=I(z=0)=93.03\angle-21.15^{\circ} mA

V_{o}=Z_{in}I_{o}=(71.66\angle32.77^{\circ})(0.09303\angle-21.15^{\circ})=6.667\angle11.62^{\circ} V

From eq. (11.27):

V_{o}^{+}=\frac{1}{2}(V_{o}+Z_{o}I_{o}),       V_{o}^{-}=\frac{1}{2}(V_{o}-Z_{o}I_{o})

V_{o}^{+}=\frac{1}{2}(V_{o}+Z_{o}I_{o})=\frac{1}{2}[6.667\angle11.62^{\circ}+(60+j40)(0.09303\angle-21.15^{\circ})]=6.687\angle12.08^{\circ}

V_{o}^{-}=\frac{1}{2}(V_{o}-Z_{o}I_{o})=0.0518\angle260^{\circ}

At the middle of the line, z=\ell/2,   \gamma z=0.921+j1. Hence, the current at this point is

I_{s}(z=\ell/2)=\frac{V_{o}^{+}}{Z_{o}}e^{-\gamma z}-\frac{V_{o}^{-}}{Z_{o}}e^{\gamma z}

=\frac{(6.687e^{j12.08^{\circ}})e^{-0.921-j1}}{60+j40}-\frac{(0.0518e^{j260^{\circ}})e^{0.921+j1}}{60+j40}

Note that j1 is in radians and is equivalent to j57.3^{\circ}. Thus,

I_{s}(z=\ell/2)=\frac{6.687e^{j12.08^{\circ}}e^{-0.921}e^{-j57.3^{\circ}}}{72.1e^{j33.69^{\circ}}}-\frac{0.0518e^{j260^{\circ}}e^{0.921}e^{j57.3^{\circ}}}{72.1e^{33.69^{\circ}}}

=0.0369e^{-j78.91^{\circ}}-0.001805e^{j283.61^{\circ}}=6.673-j34.456 mA=35.10\angle281^{\circ} mA