Question

0.55 +0.5 102 S Problem 4.4 In Fig. 4.4, R=0.2 M2, C=25 pF and L=0.04 H. Show that the transfer function H(s) is: 1 (5) H(S)= (5) + +1 L102 107 (a) Plot the pole-zero diagram of H(s). (b) What filter is given by H(s)? Why? (e) Determine the resonant frequency 0o, the quality factor Q, the cut-off frequencies 01 and 02 and the bandwidth B. i (0) it) R Fig. 4.4

Answer 1

mu And cierto LLO R R Fiolt) 410) @ R TL es ER -SLXLS DO y cost ER { sults su szc+1 fiolt) 1,Lt) ER 7 Lolt) FR SELCH til) क) R titet int RLCS2+LSER SZLETH M R

9 Loot iwe Now applying current division rule in branch get Colt) R xict) RtRLCS2+LS +R LCS2+1 Lolt H(S) - lilt) R (Les²+1) S²RL C+R+ RLCS² + SL+R R(Les ²1) a R Les + S2+OR Colt) = H (6) Cilt) 2 KULCS2+1) X(ale s ²+ 25 +2 R Now, Put the exact values which are given R = 0.2x106, L=0.04 H, C = 25x152F. H(s) = lolt) $10.04 x 25x10 +1 ilt) S²X004 Kg 5X151242 +0.04 s + 2 0.9X106 -12 H(s) s?+ 1 1012 252 +0.25 +2 1012 106 4(5) ors 11012 2 10 S2 + S To Sy1 015732 +0.5 H(S) = 1012

a) In ordu to plot pole zero diagram we must simplify the HCS HC) 0.5s 2 to.5 2012 2 S 2 + s +1 7 Jo 10 -> 0.55 +0.5x1012 tole $2+ sx10.5 +1x1012 lotz +(s) = 0.5(8:27102) s2410s s' tiora =0 Now, for zero's Numrator must be equal to zero.ie 0.565271012) or si zji o G or 1 j 10x108 for Poles of the system, Denominator must be equal to zuo suice it is a quadratic will discriment ie S - bt b²-4ac ga 105 t 1012 4 xide وطرق we of a 105 105 1-400 q

4 Or 10 st 109 2 1-399 1091109x7 19:27 2 Pig Ps 0 5xlo I 1999 x105 Julxtos) +2 y Z 40.5+19.99)x+0 px jioxtos zi to 8 2 15 6.5 +(x105) I -2 -4 -8 Zg -j 10x105 40.5=39.99cto? Pa Moto -12 dajo

5 for the ditumination of filter Conside the simplified egn of H(s) from H15) o.s (521012) s2105s +1012 egno i Now, at S = 0 H(s) 0.581012 0.5 lotz and at SE 00 H(S) it S300 0.5 $! ( 1+1013 Sex (1110s +10! 12 S 4(3) 0.5 (1+ o) I to HD from the value of above H(S) at s=of s=0 get the filter as below TH03) W which is BAND STOP FILTER >s

Ango → Resonant frequency wo= tic 6 Wo - SO Jo.04 X2 5x1512 we=100 a parallel RLC circut Quality factor of guen ky q: R JE = 0.28106 [25x1.512 0.04 OY TO 5 Parallel Ric cicuity Bandwidth of guen by A10 RC - AW 2x10 o 2X100X95X1512 - Cut of frequency w = wo- AW 9 6 -10 - 4x2x105 Twi= 9x105T but of frequency wg +42X* X 100 wotlaw 6. 10 W2 = 1x10 5