Question

0.552 +0.5 102 s² + +1 S 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,(s) H(S) = I(S) 1012 10 (a) Plot the pole-zero diagram of H(s). (b) What filter is given by H(s)? Why? (c) Determine the resonant frequency 0., the quality factor Q, the cut-off frequencies 01 and 02 and the bandwidth B. i (t) w ift) R Fig. 4.4 jo

Answer 1

V13) kal apply at hode U13) th Yec 1018) vie) - 1968) + V13) Ilt (3 411 CAR 18[** IBC = Tils) Bitte +R =) (8) [ 821c41 8² RLCHILAR له = Lils) =) VLB) [ SPRLC +9L+R +32RLC+R] = 11(3): R (87RLC+BL+R) ☺ =1 Ion z (8) 013) 341 de TR IL TR 118) [1+3460] 3248²RCAR SPLC+ =) from ① & ② If siz Iols) - ViB) [8RLC + 5L+2R] R( 82 RLC & SL4R) ols) [14846] 82 RLC FILAR. urb) [1+8226] RC 92 g 43L4R) S2RLASLAR (28FRIC +BL+2R) y 82+ Yec 8² Yuc 232RLL APLIOR 28²+3 he he 2[$2+sitet H13) = Io(3) If(s) 3) HCP) = (1482LCR are tú H(3) = 8²7 (0.04)(25) (15129 2[sts. I (2)(02x106) (25x102) 0.04*25*102 HOS) = 8² + 1012 2[$+$(108) + 1012]

H(3) rent] Toiser 1017 lol 71 [ i 102tL 107 +1] pole zero diagram +1069 4998149.22 -59000 -998749.225 X 1069 = ) at wzo c =) Hewl اليا at wal. ایسا رستا ( = band resect bitter. resonant frequency (wo) = Jidă z 106 radhee. quality factor (Q) = 1/24 =) 8²7 28 unstun²=0 lor 83 loret 1d2 =0 १६ 2 & un 0.1 Q = You 10 او+ (کم السلام (اسدار - cutout douquencies Thews) == 1-w²+102) =) ( 20102_2w7+83wC103) = so [id2wny =) (20 (1)] +Lexid 22 w22) = 2(1012 wej? =) Law (105)]?. = -221d? wye 4W²X1000 = -2(104 twt_2x12² w²) awt+w2 (4:04X102) + 2x10²4 =0=44= 931788.607/24 Hy z 1072210.134/27 =)