1. One of the complicated parts of solving for the input impedance is the hyperbolic tangent of the product of the propagation constant and distance from the load, because it involves imaginary parts. If the attenuation constant is 0.1167 and the phase shift constant is 0.5532, determine the correct formula to evaluate this complicated part when considering the input impedance 2 m from the load. * tanh(1.1064) +j tan(0.2334) T- tan(1.1064) tanh(0.2334) Lanh(1.1064)+j tan(0.2334) 1+j tanh(1.1064) tan(0.2334) tanh(0.2334)+/ tan(1.1064) 1+j lanh(0.2334) tan(1.1064) tanh(0.2334)+j tan( 1.1064) 1-j lan(0.2334) anh(1.1064) 2. Given the gains below, we can say that. * 6.14 dBd, Apani) 4 dBi O The gain in dBd is greater than or equal to the gain in dBi. O The gain in d8d is equal to the gain in dBi. O The gain in dBd is less than the gain in dBi. The gain in dBd is greater than the gain in dBi.

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Publisher:Robert L. Boylestad
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1. One of the complicated parts of solving for the input impedance is the hyperbolic tangent of the
product of the propagation constant and distance from the load, because it involves imaginary
parts. If the attenuation constant is 0.1167 and the phase shift constant is 0.5532, determine the
correct formula to evaluate this complicated part when considering the input impedance 2 m
from the load. *
tanh(1.1064) +j tan(0.2334)
T- tan(1.1064) tanh(0.2334)
Lanh(1.1064)+j tan(0.2334)
1+j tanh(1.1064) tan(0.2334)
tanh(0.2334)+/ tan(1.1064)
1+j lanh(0.2334) tan(1.1064)
tanh(0.2334)+j tan( 1.1064)
1-j lan(0.2334) anh(1.1064)
2. Given the gains below, we can say that. *
ApdBd)
6.14 dBd, Apani)
4 dBi
O The gain in dBd is greater than or equal to the gain in dBi.
O The gain in d8d is equal to the gain in dBi.
O The gain in dBd is less than the gain in dBi.
The gain in dBd is greater than the gain in dBi.
Transcribed Image Text:1. One of the complicated parts of solving for the input impedance is the hyperbolic tangent of the product of the propagation constant and distance from the load, because it involves imaginary parts. If the attenuation constant is 0.1167 and the phase shift constant is 0.5532, determine the correct formula to evaluate this complicated part when considering the input impedance 2 m from the load. * tanh(1.1064) +j tan(0.2334) T- tan(1.1064) tanh(0.2334) Lanh(1.1064)+j tan(0.2334) 1+j tanh(1.1064) tan(0.2334) tanh(0.2334)+/ tan(1.1064) 1+j lanh(0.2334) tan(1.1064) tanh(0.2334)+j tan( 1.1064) 1-j lan(0.2334) anh(1.1064) 2. Given the gains below, we can say that. * ApdBd) 6.14 dBd, Apani) 4 dBi O The gain in dBd is greater than or equal to the gain in dBi. O The gain in d8d is equal to the gain in dBi. O The gain in dBd is less than the gain in dBi. The gain in dBd is greater than the gain in dBi.
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