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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). 2.47. 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) = ew) * e2*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) 2.49. '
= e'z‘u(t) * u(t)  2te'2‘u(t) a: u(t) = [(2%)<;;e2*—tem>]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 82.412b 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, $2.4 , I e.
‘2.4.17. For 0 s t 521 (see Figure 32.4.1711)
' t
z(t)*g(t)=/ s‘inrd1‘=‘1—cost ogtszw
. 0
For 21r S t S 41r (Figure S2.4—17b)
2w
a:(t)*g(t)=/ sinrdr=cost1 21r$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 32.417 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) ‘ '  «ea4:23) By inspection, we ﬁnd Figure 82.419 shows a:(t) t W(t) Figure 82.419 * 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 zeroinput response has form yo(t)= ale—”(30. Using the IC, 170(0) = 2 =
61. Thus, yo(t) = 284/0“”. The zerostate response is 2:(t) 4: h(t), where h(t) = bo6(t) + [P(D)§o(t)]u(_t).
For this ﬁrstorder 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 zerostate response is (f; ﬁEr/(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 thezeroinput response and the zero
state response, ' y(t) = (1 + e*/<RC>) u(t). (b) Rom 2.4:25a, we know the zeroinput response is yo(t) = yo(0)e“/(RC). Since
the system is timeinvariant, the unit step response from 2.4—25a is shifted by one to provide the response to :r(t) = u.(t —1) Thus, the zerostate 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%" 2.433. Ea) KCL at the negative terminal of the op—amp yields LUR—M + Cy(t) = 0. Thus, y(t) = 313:0). (b) The zerostate 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; mil01d?) u(t) = —Rt—Cu(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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 Fall '06
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