A 50-loss-less line 0.6long is terminated in a load with ZL = (50 + j25) W. At

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A 50-loss-less line 0.6long is terminated in a load with ZL = (50 + j25) W. At 0.3from the load, a resistor with resistance R =30 is connected as shown in Fig. 2-43 (P2.45 (a). Use the Smith chart to find Zin.

Zin Zo = 50 2 Zo = 50 2 30 2 ZL 0.32 +/4 0.3%- ZL = (50 +j25) N Figure P2.45: (a) Circuit for Problem 2.45. 0.300 A t Al

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Elements of Electromagnetics

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Authors: Matthew

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Question Posted: Mar 26, 2010 11:26 PM

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A 50-loss-less line 0.6long is terminated in a load with ZL = (50 + j25) W. At

Question:

Zin Zo = 50 2 Zo = 50 2 30 2 ZL 0.32 +/4 0.3%- ZL = (50 +j25) N Figure P2.45: (a) Circuit for Problem 2.45. 0.300 A t Ala Y-IN Z-LOAD KSETANE CIMPONETR OR CONDUCTANzCOMPON / Y-LOAD Z-IN L0300 2 Figure P2.45: (b) Solution of Problem 2.45. amaRorao

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Refer to Fig P245b Since the 302 resistor is in parallel with the input impedance at that point it ...View the full answer

Elements of Electromagnetics

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