Question

1. Please do the hand calculation to find out the transfer function for the circuit in Fig.1. Then, note down the Poles and Zeros if it has. 2. Using Bode-rules to draw the Bode Plot (Magnitude), make some assumption or discussion if needed.

Answer 1

Solution : Transforming the circuit in to s-domain then using KCL analysis we can obtain the transfer function in s-domain.

Now depending on the values of components used in the given circuit the location of poles and zero will be varied which changes the behaviour of the given circuit as shown below-

Solution :- girem circuit into s-domain- l/se node А "Transform the ranking 'o' as ground • Apply get = VALT [ + aiku, to meet chata ] Vines) KOL At Node 'A' eve R2} Youtis) 312 m Vinis) I+1 Ri (189) A VA (5) Vols) (Due to Parallel connection) vinis) [ Itsel RI ) ** Volls [ 66 454) + (detse? * WAT [, 4916) + (1854. c) Page 459 e] >> Voks) R3 (1+546) +R, (1+5c2Rz) Pz ( 1+SC,RI) Vines) Voce) [(e +Rx) + SCQR2 + 562 R2RI] = Ri ( Itsc,Ro) Win es) Vols) (R1+R2) 1 + SRR2, (C,+(2) 1 tske . +62)] = R4 (1159R) Vince) > voltage Transfer ofnican konu) (lites) in sodomain R2 1+ SC, R, HUS) = Vois) vinus) L R1+R2) 1 + S R. R2 (G+C2) R1+R2)

To obraun Leos of His) ! to Zero : equating numerator of trame fer function d+ SCARI =0 S. cz To obrain poles २८ HUS): Equaning denominators of tranfer function → t+ S R, R2 (9+(2) equal to zero. = 0 R1+R2) → S R, R2 C9+(2)= -1 R1+R2) (R, TR2 C,+2) Now ff R = R₂ =R Garg - thew transfer function 1 + $49)(20) H13) 1+ SRC 1 2 HIS) 1+SRC TtSRC Hes) I As sago Hijos 0.5 rejowe dBA 0.5 Magnitude of reju) Thejas! = 0.5 0.1 to (nog scale)

Pill Ra -1 Zeros Sz 와 the New As R, R2 RitR₂ Location of Poles Sp (Cit(a) (R, 11R2) Location -1 CARI Depending on the components values to be choosen , location Pole of Zero may be vasied case (1): If Pole is lying toorigin - Ispl < 18₂1 (08) (6+2) BR2) (R+R) Ar ci nearer ie) < RICA X (C+2) WRZ 11 + Ri R2 <((1+(2) » X+ RL <x+62 R2 7 R, G < Rq (2

branch to e ue, if time constant of impedance of R, in parallel is less than the time constant of impedance branch of Re in parallel. to 12 The nature of response will be obtained as Low pass filter کمال | * o Sp P.2 Plot A sajo 41 His) (R2/R1+R2) (1 +SCRI) [ 1t s R R2 - (Cit(a) (Ritra) rejal Rz/r+ez (1+jwcRo) ['t juo leo (C+02) => Magnitude fjs = ? 1 + (wcR, 22 Pitre hijos) (Rithe) 4 wa projeyle wat ! (58,5't not) 'R, R2 G, Aca))) 2 1+[ WI RitR2 As woo R2 RITRE + W2 19+R2 & xl Rot R2 ) x (CK) x 1 eta2) KR2 (C+(2) I Mejor F C (9+)

It Zeno wereying near to Origin - Iszl<1 spl (or = R₂ C2 < Rico del circuit behaves as 1njw high pass filter 0 * SP Sz (P-2 Plot) sketching the Bode magnitude plot 149) Thomas R2 R, FR, - 20dB decade Citea +2008 → decade CI (C1+C2) R2 Ritra 1 x RR2 → W (Cit(2) RIR2) 1 Rico XRitra (C1+C2) RI R2 (log scale) RIG g Zero ↑ (Togetale) Pole Zero Pole 149 <R,C2) (Approximated Low-Pars filter response) If (R, ex > R2 (2) (Approximated high pass Axtar responeel) where magnitude How! is in Linear scale shows the absolute while W-aris in log scale. Answer'