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svp [43]
3 years ago
9

A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative p

ermittivity of the line is 2.56 and the frequency is 3.0 GHz, find the input impedance to the line, the reflection coefficient at the load, the reflection coefficient at the input, and the SWR on the line
Engineering
1 answer:
Cerrena [4.2K]3 years ago
6 0

Answer:

4.26

Explanation:

The wavelength λ is given by:

\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\

Phase constant (β) = 2π/λ

βl = 2π/λ × l

l = 2 cm = 0.02 m

βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°

1 rad = 180/π degrees

Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j

\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\  \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26

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