Question

2.63 A 50\Omega lossless line 0.6\lambda long is terminated in a load with Z_(L)=(50 j25)\Omega .At 0.3\lambda from the load, a resistor with resistance R=30\Omega is connected as shown in Fig. P2.63D. Do not Use the Smith chart to find Z_(in ). Figure P2.63 Circuit for Problem 2.63. Do Not Use The Smith Chart to solve this

Name: 263 a 50omega lossless line 06lambda long is terminated in a load with zl50 j25omega at 03lambda from the load a resistor with resistance r30omega is connected as shown in fig p263d do not 86378
Uploaded: 2023-08-01T17:56:31-08:00
Duration: 2 min 30 s
Channel: Sri K
Description: 263 a 50omega lossless line 06lambda long is terminated in a load with zl50 j25omega at 03lambda from the load a resistor with resistance r30omega is connected as shown in fig p263d do not 86378

          2.63 A
50\Omega  lossless line
0.6\lambda  long is terminated in a load with
Z_(L)=(50 j25)\Omega .At
0.3\lambda  from the load, a resistor with resistance
R=30\Omega  is connected as shown in Fig. P2.63D. Do not Use the Smith chart to find Z_(in ).
Figure P2.63 Circuit for Problem 2.63. Do Not Use The Smith Chart to solve this

263 a 50omega lossless line 06lambda long is terminated in a load with zl50 j25omega at 03lambda from the load a resistor with resistance r30omega is connected as shown in fig p263d do not 86378

Added by Adrian C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Sri K

08/01/2023

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63 A 50 $\Omega$ lossless line 0.6$\lambda$ long is terminated in a load with $Z_L = (50 + j25) \Omega$. At 0.3$\lambda$ from the load, a resistor with resistance $R = 30 \Omega$ is connected as shown in Fig. P2.63. Use the Smith chart to find $Z_{in}$. Figure Show more…

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2.63 A 50\Omega lossless line 0.6\lambda long is terminated in a load with Z_(L)=(50 j25)\Omega .At 0.3\lambda from the load, a resistor with resistance R=30\Omega is connected as shown in Fig. P2.63D. Do not Use the Smith chart to find Z_(in ). Figure P2.63 Circuit for Problem 2.63. Do Not Use The Smith Chart to solve this