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Unformatted text preview: Problem 1 0% «A R ‘9‘ \/‘5 V1 V1 V0  K
11M» 21%! J5! 4 ' mag 51%; ERLV
R Vo _ “,1 '0
“nu evavgor 7. mm 214.922
Vﬂ. z — J5: (Va'tv'h. a)
v" WM Suwpositim
W. ean {of 3 Q
g % {in =33
V11. 3 z $102+ V“— 2)
v3 V‘h‘al
11W. erdv {or '1‘
h Rn. = R
1K
2K 7" ji (V'y'? .3)
V; Vim. Vout 2V“... = Yé+\£3—+\—/—‘+\é_ 2 21 Z3 2"
z :1“; (13V3+11V1+£V1+1°Vo) L) WmG=ngouk <Frow. 941% a) HWl_l Page 1 E: a) A ,Y 4 2. 1 R1 2
7; h; K
Rb Ra, A, :(z 3
5
41: 911'21: (Rad'be
2—3: R1+£3 = Ra“ 31; R‘s‘l' R1 = Rbﬂﬂlad at.) "1—2." — "23" . 2523 = m (RmRb)  M (25+ 2‘)
_‘
"3—1" ‘ ZRQ =  + (Raﬁ'Rc) Mnf RcRb  5 + b +952: 1R1 = Rq+ (5+ RC,
R 1: RC Rh
1 Rod kid‘psc R; and K3, are {wad 1k asundhx
b) mama 1—2 (1—3 slwwk):
Grb4'Gc = G155 (wag ,
km ﬁrmems M series ,ite G1$5G2=
uhhmq 23 (14 slurkd):
em» em = G; SS ( G1+G3)
minding 3" (12 Sl/wv'kll) '.
Gq+ Gt, = G3K(GH+G1)
50M “3 W 0L) bu} only Ydaloduk ,i.e. = m
am am a” G1 0.5.1), CnGz
61+ G2. HWl_l Page 2 <Vout> can be determined by recognizing that the source Vin can be seen as a
superposition of a square wave and DC signal. The square wave ranges from Vm/Z to
—Vm/2. Consequently there is equal charging and discharging of the capacitor: <Vout>
from this is zero. Now the DC signal at Vin is Vm/ 2. With a DC signal the capacitor is fully
charged and passes no current, therefore here <VDIIt>=Vmax/2. Combining the two we get
<Vom>=Vm/2. Using the superposition principle again, we can ignore the DC component in calculating Vpp.
We know from the symmetry of the AC component that during a charging cycle the voltage
must go from some value Vout[t=0]=—V to some value Vout(t=T]=V. le'laX/ vmax/ Based on the charging equations, the functional form of the above curve must be:
v  “£[1 — lie—y“)
2 M with A as an arbitrary constant Taking this form, we substitute in the boundary conditions: VM(t—0)—v?“(l—A]—V lel "'/
v {I=I}=T(l—Ae w)=v
Solving these Linear equations for V we get: V V“ 8V“ —1 m T
V ' —[ ]' T‘aﬂhlﬁl 2 97“ +1
And ﬁnally: vpp =2V = thanh(2;6) In the limit of1>>RC: tanh(x>>1]x1. Therefore, Vppzvm which makes sense because large t corresponds to low frequencies—exactly what a low pass ﬁlter passes. In the limit of
T<<RCE tanh[x<<1]xx. Therefore tipped!max which makes sense because small 1 corresponds to high frequencies. Furthermore, as 'c decreases, Vpp decreases monotonically. HWl_l Page 3 9mb\ou.."{ [ C
W C I R; WWW
—‘\ (6M ___> ’1
l '/ ‘2’“
\'/ V44“
V’tk: Vafjm  5v , Rh = R.IIR1= 5K a) RAkC = 0.’\ \MS b) we WM 0» hf] Pass Wm T’? RwC )
..¢_ 0W UH“. =. 271 t‘lv ‘11..“) mm (6M3 = 5—Z'—=3V
mthwx) =S+1='7V t HWl_l Page4 ...
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 Spring '08
 BOZAROV

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