Consider the transmission line with resistive load of 50Ω. The characteristic impedance of the line is 100Ω. The line is very long, then the minimum and maximum impedance on line is
A
75Ω,300Ω
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B
100Ω,200Ω
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C
100Ω,100Ω
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D
50Ω,200Ω
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Solution
The correct option is D50Ω,200Ω ΓL=ZL−Z0ZL+Z0=50−10050+100=−13=13e−jx SWR(ρ)=1+|ΓL|1−|ΓL|=3+13−1=2 Zmin=Z0ρ=1002=50Ω Zmax=Z0ρ=100×2=200Ω