Question

(2)Find the transfer function Hw). Also obtain the magnitude and phase angle of H(w), and the cutoff frequency. Plot Hw) and the phase angle. What kind of filter is it? 0 + Oro +6 Q1

Answer 1

. + + 9 > 8! 4 hi Replace capacitor with s-domain impedance' t's and Vin with Vics), Vo with Vols) + IR 1C VOCs) Vics) ettes a mi Apply virtual short concept ie V+ = Vo. Apply Kel at node Vt IR = Ic. IR = Ic+I Vie) was) - Yote Vics) -VO(S)= RCS Vors) (1 +RCs) Vols) = Vics) Vo(s) : 1 1+RCS Transfer function. Vics)

Video - TARS 7= RC → Time constant - = HCS) Vois) vics) HTS | Replace şejw. Hljw)= How) 1 + RC (16) H(W) = Hew) = tjwRc Magnitude = /HC) = () - [tañ(WRC) phase angle = ZH6w) = Zoo [tant (WRC) HIW) = 1+jWRC it juwezo cutoff frequency L = fc =- - 27TRC WC- - RC

1 H(W) plot :- Huo) - viturg Vitam) How) 3 alHew) 0 Code) 0.995 Š W=0.1 Wc - 0.98 (-3dB) 0.707-------- W=o.zuc W=auc . 3 - 0.89 Tool 0.2 0.5 0.7 1 2 3 W=0.5Wc 0.819 3 W=0.7 wc 0.74 0.9 wc 3 3dB frequency cut off frequency >w=wc = the - 2TR 0.707 3 W=Wc 0.447 3 w=2uc 0.316 3 W=3WL From magnitude plot it's clearly a Low pass filter. It allows Low 0.196 3 w=5Wc frequencies with low attenuation and high frequencies with high attenuation.

phose plot Chow) :- <H(we-tar(WRC) = -tan'wc) w LH (W) 3 n <Hew) 0 out 0.30.50.71_23_ų awc On WC - 5.70 -16.70 o 30c -45 --- ------ -450 angoo -26.560 -350 0.7 We -41,90 o 0.9 wc -45 Wc 63.43 quc -71.50 3wo -75.960 yWC -78.699 5 we