Slu to Ch2

# Slu to Ch2 - QE 3820 Hm CHADTBR Z §ocumw£ 2.4.2 2(a g(at...

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Unformatted text preview: QE 3820 Hm CHADTBR Z §ocumw£ 2.4.2. _ 2(a) ,. g(at) = f” 2(a1j)g[a(t — ¢)] (11' = l f z(w)g(at — w) dw a 1 = — > C(at) a 0 When a < 0, the limits of integration become from 00 to ~oo,—which is equivalent to the limits from —oo to co witha negative sign. Hence, a:(at) a: g(a.t) =~ lildat). 2.4-7. In this problem, we use Table 2.1 to ﬁnd the desired convolution. (5) ya) = W) * w) = e-‘um * nu) = (1 — e-‘)u(t) (b) y(t) = h(t) * 1:(t) = e“u(t) =0: e“u(t) = te“u(t) (c) ya) = e-w) * e-2*u(t) = (8-: — e"“‘)u£t) (d) y(t) = sin 3tu(t) a: e“u(t) ' Here we use pair 12 (Table 2.1) with a =«0-, ﬂ = 3, i9 = -90° and /\ = —-1. This yields 4; = tau‘1[:-:ii-] = —1os.4° and ‘ os18.4° “ — cos 3t +.18..4° sin 3t u(t) * e“u(t) = MTG—(#110) 0.9486e“‘ -— cos(3t + 18.4°) —T/T_o—_—u(t) 2.4-9. ' = e'z‘u(t) * u(t) - 2te'2‘u(t) a: u(t) = [(2%)-<;-;e-2*—te-m>]u<~ = te‘z‘uﬂ) ya) = (1 - 2i)'e~"u(t) * nu) 2.4.12. (1:) 31(1‘) [750) + 28"“(01 * €‘u(-t) _ —6(t) * e‘u(—t) + 2e“u(t) a: e‘u(—t) —e‘u(—t) + Ie“u(t) + e‘u(~t)l e”‘u(t) (b) Refer to Figure S2.4—12b. t 9 Figure 82.4-12b For t 2 21r, the area of one cycle is zero and 2.4—16. For t < 21r (see Figure 32.4—16) ‘ ) .' t V c(t)=z(t)*g(t)=/ sinrdr=1-—cost 05t\$27r . o . z(t)*g(t)=0 t227r and t<0 "" Z 1 1T ZTF t—D Pl <5, \$2.4 , I e. ‘2.4.17. For 0 s t 521 (see Figure 32.4.1711) ' t z(t)*g(t)=/ s‘inrd1‘=‘1—cost ogtszw . 0 For 21r S t S 41r (Figure S2.4—17b) 2w a:(t)*g(t)=/ sinrdr=cost-1 21r\$t\$41r ‘ ‘ t—21r For t > 41r (also for t < 0), z(t) =0: g(t) = 0. Figure S2.4—17c shows C(t). « (b 3 Figure 32.4-17 2.4—19. :i:(t) = 6(t) — 6(t , 2) t - / W(T)dr = A (5—1) - 0 2 Therefore “‘ mm mm = [60) — an -2); *A (L91) ‘ ' - «ea-4:23) By inspection, we ﬁnd Figure 82.4-19 shows a:(t) t W(t) Figure 82.4-19 * 2.4—25. (a) Using KVL, :r(t) = RCy(t) +y(t) or y(t) + 72151;“) = 31—01:“). The, characteristic 5 root is A = 7%. The zero-input response has form yo(t)-= ale—”(30. Using the IC, 170(0) = 2 = 61. Thus, yo(t) = 28-4/0“”. The zero-state response is 2:(t) 4: h(t), where h(t) = bo6(t) + [P(D)§o(t)]u(_t). For this ﬁrst-order system, 370(t) = 516-” (RC) and ﬂow) = 1 = 51. Using 1700‘.) = e'mRC), b0 = 0, and P(D) = %, the impulse response is h(t) = jlzlde—e/(Rcmu). Thus, the zero-state response is (f; ﬁE-r/(RCMT) u(t) = l '_ 9 - 1 (-6 ’/‘RC’|r=o) up) = (1 .. e é/<RC)) u(t). For t 2 0, the total response is the sum of the-zero-input response and the zero state response, ' y(t) = (1 + e-*/<RC>) u(t). (b) Rom 2.4:25a, we know the zero-input response is yo(t) = yo(0)e“/(RC). Since the system is time-invariant, the unit step response from 2.4—25a is shifted by one to provide the response to :r(t) = u.(t —1) Thus, the zero-state response is (1 — e‘("1)/(RC)) u'(t - 1). Summing the two parts together and evaluating at... . ﬁ ‘22 ﬂlﬂiés 3(1): yz : 3° (0) e—z/Rc‘ré—e’l/RC)' Seamus Fol 3°03) Vie—Lbs fjvéo): erc __ 0'5 6%" 2.4-33. Ea) KCL at the negative terminal of the op—amp yields LUR—M + Cy(t) = 0. Thus, y(t) = 313:0). (b) The zero-state response is y(t) = \$(t)*h(t), where h(t) = bo6(t)+[P(D)370(-t)]u(t). This is a ﬁrst order system with A = 0, thus 370(t) = the” = 51. Since yo(0) = 1 = 51, b0 = 0, and P(D) = —F15, the impulse response is h(t) = —%u(t). Thus , y(t) = (f; mil-01d?) u(t) = —-Rt—C-u(t). ' ' i ‘ ‘ tuitively this makes d nﬁm as tune increases. In , ‘ 5 town. I tZhould output an unbounded ramp function. Notice, [y(t)] ramp _ sense; a DC input to an integrator ...
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