HW1soln - Problem 1 0% «A R ‘9‘ \/‘5 V1 V1 V0 - K...

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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 4-1: 911'21: (Rad'be 2—3: R1+£3 = Ra“ 3-1; R‘s‘l' R1 = Rbflfllad at.) "1—2." — "2-3" -. 25-23 = m (Rm-Rb) - 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 firm-ems M series ,ite- G1$5G2= uhhmq 2-3 (14 slurkd): em» em = G; SS ( G1+G3) minding 3" (1-2 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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HW1soln - Problem 1 0% «A R ‘9‘ \/‘5 V1 V1 V0 - K...

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