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Centro de enseñanza técnica industrial





Zp1=20kΩ
Zp2=10kΩ
Zs=60kΩ

Encontrar las impedancias de entrada encorto y en abierto
Y también la impedancia de entrada
Z´a= Zp1//(Zs+Zp2)

Z´a= 20kΩ //(60kΩ +10kΩ) =51.851KΩ

Z´c= Zp1//Zs

Z´c= 20kΩ//60kΩ = 15KΩ

Z´=Z´a*Z´c

Z´=51.851kΩ*15kΩ= 27.88KΩ

Encontrar las impedancias de salida en corto y en abierto e impedancia de salida

Z´´a= Zp2//(Zs+Zp1) ,

Z´´a= 10kΩ//(60kΩ+20kΩ)= 8.888kΩ

Z´´c= Zp2//Zs , Z´´c= 10kΩ//60kΩ= 8.5714kΩ

Z´´=Z´´a*Z´´c , Z´´=8.888kΩ*8.5714kΩ = 8.7287kΩ

Calcular X
X=Z'CZ'a =Z''CZ''a=8.88888KΩ8.5714KΩ=1.01829


Calcular la atenuación
AT=10log1-1.018291+1.01829=-20.4227dBCalcular la impedancia característica
ZO=Z'*Z''=27.88KΩ*8.7287KΩ=15.599KΩ

Red T
Calcular ZP
1dB = 0.115nep
20.4227dB=2.3486nepZP=ZOsinhAT=15.599KΩsinh2.3486nep=3.0069KΩ
ZS1=Z'*cothAT-ZP=27.88KΩ*coth2.3486-3.0069KΩ=25.386kΩ
ZS2=Z''*cothAT-ZP=8.7287KΩ*coth2.3486-3.0069KΩ=5.8824KΩ

Desarrollo practico
Circuito π
R(Ω) | Vi | Vo | At |
40 | 10.607 |0.007038 | -40.832 |
50 | 10.607 | 0.008788 | -38.904 |
60 | 10.607 | 0.010533 | -37.331 |
100 | 10.607 | 0.01747 | -32.934 |
200 | 10.607 | 0.03454 | -27.013 |
300 | 10.607 | 0.05124 |-23.589 |
400 | 10.607 | 0.06755 | -21.188 |
5000 | 10.607 | 0.558 | -2.845 |
60000 | 10.607 | 1.326 | 4.66 |
80000 | 10.607 | 1.369 | 4.944 |
100000 | 10.607 | 1.502 | 5.754 |
| | | |Circuito T
R(Ω) | Vi | Vo | At |
40 | 10.607 | 0.005218 | -43.432 |
50 | 10.607 | 0.006515 | -41.503 |
60 | 10.607 | 0.006515 | -39.93 |
100 | 10.607 | 0.012955 | -35.53 |
200 | 10.607 |0.025614 | -29.612 |
300 | 10.607 | 0.03798 | -26.189 |
400 | 10.607 | 0.050085 | -23.787 |
5000 | 10.607 | 0.41385 | -5.445 |
60000 | 10.607 | 0.46254 | -4.479 |
80000 | 10.607 | 0.54228...