Question

a) Insert a resistor R3 in series with the feedback capacitor C in the low pass filter of class. Draw the modified circuit. b) Derive the resulting transfer function. Simplify to the point that the pole and zero coefficients are apparent. c) Plot the transfer function with a DC gain of 0 dB. What is the value of the magnitude (in dB) at the frequency of the pole? At the frequency of the zero? d) If the pole frequency is 1 kHz, the zero frequency is 10 kHz, the DC gain is 0 dB, and C=1 nF, what are the values of R1, R2 and R3? e) This is known as a pole-zero circuit, and has applications in controls. Why does it have this name? AL W

Answer 1

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③ + CSR, I-RCH BCS) + RRCS - HBCS) B(HBCS) Wis R, +RBCS+ RRES = R +CR7 (3+R)S Rt BBCS Blit Rocs) w - H(S= Rz (1+Rmolcs) Po [1+ cb.tn.) 5] let BG= tz &c (23 +222 = Ep , R2 =k (usa depan hot ) TH(S)= K. t ęs taps H (JW) - k (ltico) = k ( lt ju bar) (1+30pcs) (it over) T (W) = K (HiO) Loc=1 / 2 wocze 220 pole Eoefticcent - Ep csak n Zenot coefficient = 22 CamScanner

© DC gains = odB IHCVA) |- k Viet? dc gain at s=0 9600 J Hope 1 Hcpio)] = k & Do gain to do a B . gain in dB + 20log k zo / K=1 o logic to y = B = 70 R2 =12 R = 100-1 J H(S)= It ZS S , HOJW) = 11526 1+ ZpS HjZpw Zero at freaking at atite Wooah H(Joc) = Hjzi wa = Hitzat alti . Thitp.coco tiep. I Hj Za | Hewa | jitlaga = JZ so it DK Eps CCR3+) ſ Magosinte de im disa 2010g Ja Z2= P₂C L Jit (it (2) en Coop M2 - 4x +) = C(Mt 2 - c13 Scanned with CamScanner

of foreniny a pote (2- H (1j4tz). 1+1 Trelocz Thy Thor H (Netz) - Iti tritace = to 22 Huapitan Ho | H&pare) = 11+ @ - Jot feston) 177 1H(32) de = 2010g 11:113} TB ty JE € pue toequonly = 1kHz kate 2000 foequenej = 10Hz WF 2Rf = 228103 &S=Wcz ,(0=224 = 27X107 dls zwey NuwfH634) -trekkergaiterdetičky Jives C= 144095 fco Odz gain, Bari, en frequency at 7000 - Wc, = 1 = L = 2K Klot to RB C = 1 - Z Bo - Rz= 1 = 105 - 50 274100 x 10-9 Rz = 15.913 41037] foebeng.co kule, wczon e(RAR) = 274109 CamScanner X3

27x03 CB+12) = 20 B3+1 = 1 10 : 1000 HOB - 500 NO3 TAB x 10 9 = 2 BytR2= 159.15 x103 15 J15403 R2= 2+159.5X1B 2D 159.15 x103 - B2 = 143.294103 27 R2=8 = /12 = 183.24x103] 8 © co bole- zero circuit brow lead lag Iglead compensator lag lead compersitco Pole nego to coizing lead. lag compenssico 2 200 meas to digin 1 Scanned with CamScanner
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