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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 haifpawer '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 halfpawer 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 opencircuit 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 Transferfunctinn 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 .
ﬂﬂill I
EDIE 0.01 fffa 136.33 (a) Applying the vel'l'egedivisien 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 firsTer'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:
QuaIL 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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 Spring '08
 TedK.Higman
 Electrical Engineering

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