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Tema6- Respuesta en frecuencia - Introduccin Tema 6...

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Electrónica I 2007-08 Tema 6: Respuesta en Frecuencia Electrónica I Por Daniel Izquierdo Gil Electrónica I 2007-08 Introducción • Concepto: plano s=j ω donde ω =2 π f • Capacidad ( ω ) Zc= 1/j ω C • Función de transferencia Av=V O /V I • (dB)=20Log|Av| frente ω • Módulo y Fase de f.d.t Electrónica I 2007-08 Filtro Paso Bajo http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/lowpass/lpf.html Electrónica I 2007-08 Filtro Paso Bajo (Modulo) ω c = ω ω c >> ω ω c << ω =1/ 2 =1 = ω c / ω (Bajas Frecuencias) (Altas Frecuencias) (Frecuencia Corte)
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Electrónica I 2007-08 Filtro Paso Bajo (Modulo) =1/ 2 =1 = ω c / ω 20log(1/ 2)=-3dB 20log(1)=0dB 20log( ω c / ω ) (Bajas Frecuencias) (Altas Frecuencias) (Frecuencia Corte) Electrónica I 2007-08 Filtro Paso bajo (Fase) -Arctg(1)=-45º Arctg(0)=0º ψ =-Arctg( ω / ω c ) -Arctg( )=-90º (Bajas Frecuencias) (Altas Frecuencias) (Frecuencia Corte) ω c = ω ω c >> ω ω c << ω Electrónica I 2007-08 Filtro Paso Bajo (Bode) Electrónica I 2007-08 Filtro Paso Bajo (Respuesta) (Bajas Frecuencias ω c >> ω )
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Electrónica I 2007-08 Filtro Paso Bajo (Respuesta)
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