Question

Vout 1) For the circuit shown to the right a. Use your knowledge of reactances to determine the high, and low frequency behavior. What kind of filter is this? b. Derive a function for the gain as a function of frequency. Show that it limits correctly at high and low frequencies. c. Find an expression for the characteristic frequency for this circuit

Answer 1

Ans. The given arcuit is :- numRz ovo At very low frequencies, the capacitor acts as an oper circuit, so the equivalent circuit would be i

Since both the inputs of an opamp have the same voltage, Vt no

Since the left side terminal do Ri is open the cument I, is o Since no curent flows through the inputs of an op-amp: - Vo - V+ = I, R2 : Vo-o= 0 R2 Vo=ov So output voltage is o at extremely low frequencies / At very high frequencies, the capacitor acts as a closed switch, so the equivalent circuit would be : - umz This can be further simpliked as:

As proved before, V+=ov Since Vt is directly tied with ve, Vo ou be output voltage is o at extremely high frequencies At intermediate frequencies, the capacitor will have a finite non-zero capacitive inductance Xc. So, intuitively, the output voltage in this case would be non-zero this will be found in part (b)). So output voltage is non-zen at intermediate frequencies So, since the gain is a non-zen Frite number only for L intermediate frequencies the filter is a band-pass filter

C R The Ac equivalent of the arcait above is: - I Zz VowC2 Vjwa R I 1 R2 Vin – – – – – – series The capacitance C, and resistance R, are in parallet. So their equivalent impedance is given by: Zi=L+RI jwa Z, =lt jwR, CL - G) انیان The capacitance ls and resistance Rs are in So their equivalent impedance parallel

Z2 - Ralll jWl₂ = R2. I jwC₂ Rat / jwC₂ i. Z2= R2 - C) Hj w Rala So, the equivalent circuit would be c- Iz Za o Vout TV As proved before, V+=0 Vt-Vin = Izi - Vin = I1 - C11) Z

Yout - V+ = I. Za . Iz - Vaut - (N) Zz Since no current flows through the inputs of an op-amp, II = I2 - Cv) from Cn), v) and Cv) ; 꽑 ... Vout = - Z = - R2 jwel It WR2C2 CltjwRIC) yout=Ar - jWR 2 Cj in C-W²R, C. R₂C2) + jw CR, C +R2(2) I When woo lim Au= lim -j WRC who ci-w ?R, CR₂ (2) tjwCR C + R2C2) = -)(0) R₂ C 1-0²R, G R₂ C2 + j(o) CR. G +R₂C2)

Im Au=o wto When wo lim An = lim -jw Rza wa Ci-wPRI GR2C2)+jW CRG +R>Ca) L = lim (-; R₂ C/w wes C/W²- R, G, R₂ C2) +; CR, C +R2 C2/0 : lim O-RI GR₂C2 tjo) Av=o wooo So, the arcuitse limits corectly at high and low Frequencies. a) The characteristic frequency da bardpaco file always Satishes: - dlAulo dw which also leads to

LAv=o or 180° 29-jwR2 C color Iso Ci-waria RzCa) t;w CRIG tR,C2) For this, it is necessary that the real part of the denominator be o since the numenabor is a purely imagmay number :) - WiRiRaQ Cso i We²=1 Rie RzC2 Wo= VRCR₂C2 Elfo = 1 ZAVR, G RC2