Question

who can solve this problem? thanks

Problem 3: Chain rule and circuit design Consider the circuit in the below figure. You can as me zero initial conditions. t WW-어 2R L stage 1 stage 3 stage 4 stage 2 Part I: Redraw the above circuit in the s-domain and compute the transfer functions Ti(s), T(s). Tfs), Tds) of each one stages Part II: Somebody with a rusty recollection of linear circuits analyzed the above circuit and concluded that the transfer function T(s) from Vs) to .S) is equal to the product of the transfer functions of the 2RCLS T(s) = T, (s) xT (s) x T, (s) x T4 (s) = RCL$2 + (R2 + L}s +R of the 4 stages identified in the plot. Identify two problems that invalidate this conclusion. Part IlI: Modify the above circuit, keeping all 4 stages, so that the resulting circuit does have trans fer function T(s) by adding at most 2 OpAmps. [Justify your answer] Part IV: Use stage 1, stage 3 and a noninverting OpAmp to design yet another circuit with transfer function T(s). [Provide reasons that justify how you arrived at your design] Part V: What design would you recommend to realize the transfer function T(s), the answer in Part II or the answer in Part IV? Why? Hint: you can provide at least 2 different reasons Hu wWH

Answer 1

circuit in s domaim R - Vou t IS) 2R Vez1s)R Vo31) R Vol's) R Vias SL RIIR Tils)= Volis RIIR+ CS RCS R. TILSJ= R2 2+RCS T21s)- Vo2(S) 2R-2 Vot (s) R T3 (S) Vo3(S = SLR. SLR+ (SLHR) R Vo2ls) T3 (S) S LR 2SLR tR2 T4 IS) Vout (S R :-F VD3 (s)

Vo(S) Vezts) R. R V35) R Vo S 2R O . b a845 stogt 4 is an inver-ting opamp: Hence Vo 3(S) = -Vo S ) Apply kcL at Vo3 s) Vo3(s Vo3 IS) R: Vo2 (S)- Vo3s) 75. Vo2s) R SLR 2SLR R2 Vo3ls) - Vo(S) Vo 2 (5) 2SLR1R2 Vo2s) (2SLR+ R) VOIS) No, stage 2 is on inverting opemp, Hence Vo2(S) - 2 Vois) :19

Vo2IS) =-2 Voi (S) (2 5LR+R) Vo lsy Voi (S) Vo2(S) 2 2 S78 Apply KCL at Vol (S) V2 LS)-Vol (S) Vol(S) + Voi (S) R. R 1 2RES RCS Vol(s) Vi (s) 2+RCS Voi IS) RCS 2+RCS.(2SLR+R) Vols (sle RCS 25LR Vols Vils) 775 (2+Rcs) (2 SLRIR 2 S RCL TIS) Vols) (2+Res) (2SL+ RJ 2s RCL TIs) 2 RCLS2S(4L+ R2)+2 R.

NoW RCS x-2x SLRx- 2 +Rcs TILS) Tr(S) T3s) . Tals) 2 SERC (2+Res) (25L+R) 2 52RC 4 RCLS s(4L+Rc) + 2 R. TIs)7, (s. Ts. T3(s). T4 (S)Vo (S Hener, virs)