EECE 290 Homework, Spring 2008

16.26 The full-wave rectified sine-wave voltage shown in Fig. P16.26(a) is applied to the circuit shown in Fig. P16.26(b). Figure PiG.26 (— ; >) v4 (V) 340 L t (s) 0 1/120 1/60 1/40 (a) Gr 16H L. 12.5 pF RS i, 3 1kO Ug( -O- (b) XY wy, a) Find the first four nonzero terms in the Fourier series representation of v,. b), Does your solution for v, make sense? Ex- plain. P 16.26 [a] Note — find i,(t) v-V, 16s Yo _ 1000 1 Vg 1 J, | —— a —6 —_| = — Vo | +125 x 10 5+ T5099 ls + Vo(12.5 x 1078s) + 0 \,,(1000 + 0.2s° + 16s) = 1000V, —_ 50001, "52-4 80s + 5000 7, 10 5V, "1000 s?2 + 80s + 5000 I _ H(s)= = ——— Vy, 82+ 80s + 5000 5 H ; (V0) = 5 = nae) & SOc oF oo) 9 <0 = Fe = Hor: wa = 57.6007: SOxvo = 19.2007 5 F( jrey) — 2, T ; (5000 = 57.60072n2) + j19.200Rn HU) = 1074 H( jay) = 8.82 x 107°/— 173.89° A ( j22y) = 2.20 x 107°/— 176.96° H(j3xy) = 9.78 x 107' /— 177.97° A (jtuy) = =5.5 x 107 7 — 178.48° 680 1360 1 1 1 ty = | 3 + coswint + 75 Quyt + 35 COS Swot + 33 cos duit +. 680 1360 ip = x 1077 — (8.82 x 107°) cos(wyt — 173.89°) Tv oT 1360 — Ep (2:20 x 107%) cos(2uot — 176.96") Jil 1360 ~ 3e (9.78 x 107") cos(3wipt — 177.97°) O07 _ 1360 637 = 216.45 x 107? — 1.27 x 107? cos(wot — 173.89°) (5.5 x 1077) cos(4uigt — 178.48°) — — 6.35 x 107° cos(2upt — 176.96°) — 1.21 x 107° cos(3wpt — 177.97°) — 3.8 x 107° cos(4wot — 178.48) — , ig = 216.45 — 1.27 cos(wot — 173.89°) mA Note that the sinusoidal component is very small compared to the de component, so ig = 216.45mA (adc current) [b] Yes, the solution makes sense. The circuit is a low-pass filter which nearly eliminates all but the dc component. 16.37 a) b) Use the first four nonzero terms in the Fourier series approximation of the peri- odic voltage shown in Fig. P16.37 to esti- mate its rms value. Calculate the true rms value of the voltage. Calculate the percentage of error in the es- timated value. Figure P1G.37 la v(V) 107 4 2.597 T/4 T/2 37/4 P 16.37 [a] v has half-wave symmetry, quarter-wave symmetry, and is odd dy = 0, a, = Oall hk. b, = 0 k-even 8 Ts. == / f(t)sinkuytdt, k-odd Jo 8 TV “m == Ul 1 sin hut dt + po J, Sin utah ~ T/8 T/A T/ ] — 8Vin cos kwt 4T htuto T/8 _ 8Vin | COS kwot ' 0 kw — 8VYn i-e kr 8Vin co hi 0 ~ thal | OT | Fh, [O° TF SVin 1 1 kr ki = ——— cos + COS hwoT [4 4 I Vin 1 a ker = Tk {5 +0.75c08 7 | [10 + 30 cos kir/-4)] — Il 0+ 30cos(a/4) = 31.21 3[10 + 30 cos(3n 7/4)] = —3.74 = ae II | bs = z {10 + 30 cos(5m/4)| = —2.24 by = =[10 + 30.cos(77/4)] = 4.46 31.212 + 3.742 + 2.242 + 4.467 Vv (rms) = Vj, / + + + = 22.51 5) 2 LT 2 {[b] Area under v- = 2 [2 (2. 5a)? (5 ) +10 “ = §3.1250°T 1 V(ms) = \ 7 (53. 12577)T = V53.1257 22.51 22.90 [c] % Error = ( - 1) (100) = —1.7% 46.43 The periodic voltage source in the circuit shown in Fig. P16.43(a) has the waveform shown in Fig. P16.43(b). a) Derive the expression for Cy. b) Find the values of the complex coefficients Co; C1, C1, C_2, C2, C_3, C3, C24; and C4 for the input voltage v, if V, = 54 V and T = 107 ps. Figure P16.43 . 62.5 kO AW- @ | : + 7@ 25 wH ipF 22509 v, | ¥ il -e (a) ‘ Ug A Vin | | -t -T —T/2 T/2 T b L (b) J