Question

Figure 2 below shows a bode-plot of a Butterworth response filter, with cut-off frequency, fc of 95 kHz and damping factor, a of 1. Define roll-off rate and explain how it effects the frequency response of this filter. Then, modify the frequency response to have a -80 dB/decade roll-off rate by redesigning the filter with appropriate structure and components value. Draw your filter design. Gain (normalized to 1) OdB -3 dB Actual response of a single-pole RC filter – Passband -20 dB Transition region -40 dB -20 dB/decade BW Stopband. region -60 dB 0.01 0.1% 10. 100% 1000% Figure 2 (15 Marks)

Answer 1

magnitude for Ath order butterworth response filter is given by 146f) = (fe cutoff frequency) 204 V +(8) as forfu => 1 then, to THCA)/= (feln N f to fi ren, IHCE) = so log i ja = son log (a) HCf) &s changed by -200 dB, it'll changes by factor tol. as 1st order -nodB /decade Colloff) → for sodb/decade roll off N=4 is needed. pren, H(s) Ito.765 S ( 2 her)") (1+1.846 - 4 4th order t Butterworth LPF Egn. 2 and order filter cascade connection, slunce they 3 R3 + Ving Vout I 요 V4= V = Vout Cideal opamp) and V, = Vout ( 1t5cky Applying KCL at vi V-Vin R, scout + (vi-Vout) 54,70 +

(% ): ving - Vout (sc, 8(₂) Vout (1+25 (sh, t5R1R2(162) = Vin RI quen, (itaso Vin 1+25C, R, +5R,R,C,C) & (RRGG 1 Rici 2RxC Q we Rrly . Taking C,=0,-C-Inf. Then, we= / RI Then, for han stage For 4th stage CURRY CVR, R2 We = 25x95x²= I 0.75 & - JR, Rw = 1675.3 and Ri= 6.8382 By solving, R = 4.38ks and R2 = 641_, calhf. For and stage, a = 1 0.541== Then, Ri=1.548k2, R1 = 1.81k. Hence, Filter is givenby, vin m -4.38k 6412 -0.54) - 1.8648 ove Rr in 누 Vol + -Inf THE I Inf + Yout 1.slika 1.548km 4 order Butterworth LPF"
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