 GATE EC 2010 Electromagnetics Solution Q 5 - Grad Plus

# GATE EC 2010 Electromagnetics Solution Q 5

In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ratio (VSWR) on the 60Ω line is______.

(a) 1.00

(b) 1.64

(c) 2.50

(d) 3.00

Ans :- (b)

Explanation

The given diagram is equivalent to

Where, Z1 is the input impedance seen at λ/8 short line and Z2 is the input impedance seen at λ/4 line. We also know that the input impedance seen for the line of length l and characteristic impedance Z0 is,

Z_{in}=Z_0\left(\frac{Z_L+jZ_0\;\tan\;\beta l}{Z_0+jZ_L\;\tan\;\beta l}\right)

Z_1=30\left[\frac{0+j30\tan\left({\displaystyle\frac{2\pi}\lambda}\right)\left({\displaystyle\frac\lambda8}\right)}{30+0}\right] Z_1=j30

Z_2=30\sqrt2\left[\frac{30+j30\sqrt2\;\tan\left({\displaystyle\frac{2\pi}\lambda}\right)\left({\displaystyle\frac\lambda4}\right)}{30\sqrt2+j30\;\tan\left({\displaystyle\frac{2\pi}\lambda}\right)\left({\displaystyle\frac\lambda4}\right)}\right]

=30\sqrt2\left[\frac{{\displaystyle\frac{30}{\tan\;\pi/2}}+j30\sqrt2}{{\displaystyle\frac{30\sqrt2}{\tan\;\pi/2}}+j30}\right] = 60 \Omega

[ Z2 can also be calculated by formula for λ/4 line]

\therefore Z_L=Z_1+Z_2=j30+60

Magnitude of reflection coefficient is given by,

\left|\Gamma\right|=\left|\frac{Z_L-Z_0}{Z_L+Z_0}\right|=\left|\frac{60+j30-60}{60+j30+60}\right|

=\left|\frac{j30}{120+j30}\right|=\left|\frac{j1}{4+j}\right| =\frac1{\sqrt{16+1}}=\frac1{\sqrt{17}}

\therefore VSWR=\frac{1+\left|\Gamma\right|}{1-\left|\Gamma\right|}=\frac{1+{\displaystyle\frac1{\sqrt{17}}}}{1-{\displaystyle\frac1{\sqrt{17}}}}=1.64
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