R1=47K ohm R3=1k ohm c1=0.015 mu F j= R2=62k ohm R4=30k ohm omega(f)=2pi.f R2=62
ID: 1799222 • Letter: R
Question
R1=47K ohm R3=1k ohm c1=0.015 mu F j= R2=62k ohm R4=30k ohm omega(f)=2pi.f R2=62k ohm R4=30k ohm L1=3.mH sc(f)=j omega(f) Zc(f)=1/sc(f).c1+1/R4 Zc1(f)=R2(R3+Zc(F))/R2+(R3+ZC(f)) Tc(f)=Zc1(f)/Zc1(f)+Zc(f) Plot the Bode Plot for the impedance network equal to R.1 + (R.2//(R.3 + (R.4//(sC)^-1)). Input voltage is across total impedance. Output voltage is across (R.3 + (R.4//(sC)^-1)). Plot the frequency response one decade below the lowest critical frequency and one decade above.Explanation / Answer
total impedance = r1 + r2|| (r3 + r4/r4sc+1) = r1 + r2 || (r3 + r4 + sr3r4c)/sr4c +1 = (r1r2 + r1r3 + r1r4 +r2r3 +r2r4 +sr2r3r4c ) / (r2+r3+r4+ sr3r4c + sr4r2c) now transfer function = {(r3 + r4 + sr3r4c)/sr4c +1} / {(r1r2 + r1r3 + r1r4 +r2r3 +r2r4 +sr2r3r4c ) / (r2+r3+r4+ sr3r4c + sr4r2c)} ie; H(s) = (r3 + r4 + sr3r4c) (r2+r3+r4+ sr3r4c + sr4r2c)/ (sr4c +1)(r1r2 + r1r3 + r1r4 +r2r3 +r2r4 +sr2r3r4c ) H(jw) = (r3 + r4 + jwr3r4c) {r2+r3+r4+ jw(r3r4c + r4r2c)}/ (jwr4c +1)(r1r2 + r1r3 + r1r4 +r2r3 +r2r4 +jwr2r3r4c ) you get zeroes by equating numerator to zero => zeroes at w = j(r3 + r4)/ r3r4c and at w = j(r2 +r3 +r4)/r4c(r2+r3) poles by equating denominator with zero => at w= j/r4c and at w= j(r1r2 + r1r3 + r1r4 +r2r3 +r2r4)/r2r3r4c
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