Question

Problem #5 - (20%) A circuit has the transfer function: H(S) = S. (s + 5623 (s + 31.62) · (s + 17778) (a) Use asymptotic analysys to compute (HS) at infinite frequency by inspection of the circuit (not by computation). Express your answer in dB. (4%) (b) Determine the phase of the transfer function at infinite frequency. (4%) (c) Rewrite the transfer function in the form used for creating a Bode plot. (4%)
Please answer all parts

Answer 1

45) Tol) Given HB) = 10.5 (5+5673) (5431162) (5+17778) 1051 ost1) x5613 10 x 31.62412778 (het) ( +1) X 3163719425 21.62 17948 standard time constant form OoIS (1.7788104 S+1) HEIZ (0.0316 5+1) 15062571855+) Bode plot Cradlieee) corner freferency diolace) slope of the term (dBldec) effective slipot the plot term ons 20 20 31.62 0.0165ti 20 90 5623 (1.77880745+1) - 20 14,778 (5. 625 416 55+1) starting freavenly w=0.) Cradlec).

1H (jw) I wrol - 2010g 6.1xw) The lym (d) - 90logo.Ix0.1). 40dB) - -- zolde) 36 o doldec) 10dB) o Ta Bldee) o la 20681dee) 20 Goldee) -40 t 17,778 5623 31.62 T e 104 plo logowy o Gradisec ) io 100 103 01 ona wa 3libr , slope = 20 (dBldec) 20 -40-a logo: 1-1ogskrár - 49099 _-40-9=> a=10 (dB) 56234 wx 17,178 , slope = 20 de dec) 10- b 40 - logE622) -109 (71773) -9.998 - 10-b =) b= 19.998 by 2002) at infinity (H) 1 = 20 (dB) b) (Heftet) =

b) HGW)2 gortan, Hlu)2 0.190 (1.7784164 Jw+1) (30.0316 9+1 ) (1516Xboʻ10+1) <Hc8w)2 nottany 1.7784109 2010) - tam 10.031640) – tan (5.6254185w) por tanton – tanlas) - tant loos <H (800) = = 40+90 -90-46 LH lion z oo Ts) ars (47784 109 (+1) $00316S+) (567581s+1)