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Prelab: Operational Amplifier Integrator and Active Filter

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Lab #7 Prelab: Operational Amplifier Integrator and Active Filter Section 507 Jeanine Saygi Consider the circuit in Figure 7.1 to be an op amp integrator. Using the ideal op amp model and KCL, sho ... w that: Vout= -1/RC * the integral of Vin(t) dt B. If Vin is a square wave, what will the shape of Vout be? -Integration of a square wave is a triangular wave b/c Vout=t/RC C. If Vin is a triangle wave, what will the shape of Vout be? -integration of a triangle wave would result in a t^2 which would be a parabolic wave D. If Vin is a sine wave, what will the shape of Vout be? -Integration of a sine wave would result in a cosine wave E. Now consider the circuit in Figure 7.1 to be an active low-pass filter. Show that the amplitude response and phase response are: -Vin(t)/ R1 – Vout(t)/ R2 -L dVout(t)/dt =0 --- laplace domain -Vin(s)/ R1 – Vout(s)/R2 -CsVout(s) =0 Vout(s)(1/R2 + Cs) = -Vin(s)/ R1 Vout(s)/Vin(s)= (-R2/R1)/(1+sCR2) s=jw  Vout(jw)/Vin(jw)= (-R2/R1)/(1+jwCR2)  Amplitude |Vout(w)/Vin(w)| = (R2/R1)/sqrt((1+wC R2)^2 ) = A(w) Phase: φ(ω) = -arctan(ωCR2) = arctan(-ωC R2) [Show More]

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