Z_{in1} \lambda/4 short Z_0 circuit \lambda/2 2Z_0 Z_{in} If Z_0 = 50 \Omega, calculate: (a) Z_{in1} (b) Z_{in2} (c) Z_{in} Z_0 Z_{in2} 2\lambda/3 300 \Omega
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Given that Zin = 2/4 and Za = 2/2, we can substitute these values into the formula: 2/4 = 2/2 + Zin1 Simplifying the equation, we have: 1/2 = 1 + Zin1 Subtracting 1 from both sides, we get: Zin1 = 1/2 - 1 Zin1 = -1/2 Show more…
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