umn EE 3005 (Imbertson) HW 4 (5th edition), Spring 2011

umn EE 3005 (Imbertson) HW 4 (5th edition), Spring 2011 -...

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Unformatted text preview: 136.14“ Given emit) = Fm eesiEsfr] T vmt'f] = [Fm eesiZn HM? = Vfisinflsfi‘} a 231:!” The pheser's are 231 H}, = Vmifln V = fl 5— DD V” 23:1" 9 The Transfer Funei‘ien is V 1 — j H 1" =fl=—x_’— D" =— { ) V- 2m” 9 23:3” HT Piefs e‘F 1he magnitude and phase sf this Wensfer funetien are: 'l' I—I'LI' I—m: M Plnfs of The magnitude and phase. nf This Transfer functinn are: mm] 43"" P621 The airauif diagram is: VIII C v” The haif-pawer 'Fren::.Iuer'm:'3.nr is: 1 f = 3 are: Plafe af The magnitude and phase af The Transfer Funafian are shawn in Figure 6.3 in “the Text ”LEE" The half-pawer frequency at the filter is 1 f = = sac H 3 can: I The transfer functian is given by Equatian 6.9 in the text: 1 ”l” = —1 + we) The given input signal is rm {1"} = 5 (205(5Dfli'1'f) + 5casflflflflat] + 5 cas[EDDD:rt) which has campanents with frequencies af 250, 500, and lflflfl H2. Evaluating the transfer functian far these Frequencies yields: 1 H 250 = — = a3 44; — 23.5?“ { ) 1+ IEEU/EDD) 9 H{caa) = 0.71331; — 45“ Hucaa) = [3.44723 — 3343* Applying the apprapriate value cf the transfer functian ta each campanent cf the input signal yields the autput: vmfi] = 4.4?2 cas[5i]fl:rf — 23.57" )+ 3.535 cashmfln r — 45°] +3333 caslzflflflat 453.43“) P627 Rearmnging Equnfiun 6.3 in the Text yieids: 1 1 5' = = BEIGE 313103 flflfl = 31.33 nF 136.32 (a) First we find The Thévenin equiveien‘l' fer The source and resisi'nnees. '1’; 9 E’ 'I" l. 1’3" The open-circuit veii'nge is given by it? Ff{f] = Paeif) = “grim In Terms of pheser's This becomes: R ”= ”5m {11* Zeroing The seur'ee, we find The Thevenin r'esis'l'unee: ins—E: 1 R’ = we; +1;(e + e) Thus, 'l'he er'iginei circuit has the equivalent 12: "' Val“ C Vane ‘- The ‘l'r'ensfer' Funefien fer This circuit is: Vi _ 1 vi 1+ ans) 1 2.111524? {3} where f3: ___- 1. _ Using Equufian (1} in subsi'itui'a For Vfin Equui'iun {2] and rearranging, we have: 2 arm a 1 Hif) V — {3} 5 _Efimg Haw/1;) {in} Evaluu‘fing for “the. circuii campunenfs given, we. have: a = 930:: r; = 512.»: H: 0.02 1+ JWIE) A MATLAB pragram 11: pin? the Transfer-functinn magnifude is: fuverfb=fl:fl.01:3; Hmag=ubs(fl.02.f {1 + i*fnverfb)}; pluflfuverfmeug] axisflfl 3 U [102]} Hfi‘) The: resulting pla'l' is: lHtfll one m . flfli-ll- I EDIE 0.01 fffa 136.33 (a) Applying the vel'l'ege-divisien principle, we have: H{f}=vfl_ 9: _ flax/(191+le 92/{RI+RZ} vm _ A; +92 Hem. Ewart/(g +92) _ 1+ “EV/1;) where if; = [R1 + fill/(23¢) {b} Evaluating fer 1'he eempenen'l' values given, we have: a“; =1.061MH: 0.5 = 1+ fife) A skefeh ef ‘l'he 'l'rensfer funefien magnitude is: HE} I {F} 55 H ‘ 136.54 This is e fir-sT-er'der' lewpuss Rf filter. The break frequency is: 1 f=—=5.B 5MH 3 2m: 9 I The Bede plats are like Figures 6.15 and 6.16 in the fex'l'. 136.5? First, we find the Thevenin equivalent far the sauree and the resistances. The Thevenin resistance is 1 Zngflm 1% and the Thevenin vaitage is = ”2 =01 vi 191+}?sz .vm Thus, an equivalent far the ariginai circuit is: Qua-IL V: C “I" This is a iavvpass Fiiter having a transfer funetian given by Eguatian 6.8 {with changes in natatian}: h 2 4 V? 1+ JTf/fis) where r; =1/{exe:]=17s.e H1. Using The fact that '9} = 0.1V,” , we. have HF) = The Bade plants are: ...
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