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900 9011) I ' I ‘ aim] 6:13 " 2.54: " 45's;
FigureSZ.7 : 2.8. Using the convolution integral, :1:(t) * h(t) = f "N 00 2:(1')h(t — T)dT = / h(T):1:(t — T)dT. W 00 Given that h(t) = 6 (t + 2) + 26(t + 1), the above integral reduces to
:1:(t) 1: y(t) = z(t + 2) + 2:1:(t + 1) The signals :r:(t + 2) and 2z(t + 1) are plotted in Figure 82.8. ’ 2.
:‘ eta+1) mom)
a. I
Figure 82.8
Using these plots, we can easily show that
t+ 3, —2 < t s —1 (t)_ t+4, —1<t$0 y ‘ 2 — 2t, 0 < t g 1 0, otherwise 2.9. Using the given deﬁnition for the signal h(t), we may write ' 6‘27, T > 5
h(1) = e2Tu(—'r + 4) + e'27u('r — 5) = e”, T < 4
0, 4 < T < 5 Therefore,
e”, T < —5
h(—r) = e'2’, 1' > 4
0, —5 < 'r < —4
If we now shift the signal h(—'r) by t to the right, then the resultant signal h(t . T) will be €_2(t—T), T < t  5
h(t — 'r) = 62“"), 1' > t  4,
0, (t—5)<1<(t—4) 34 Therefore,
A = t — 5, B = t  4. 2.10. Horn the given information, we may sketch :1:(t) and Mt) as shown in Figure $2.10. (a) With the aid of the plots in Figure 82.10, we can show that y(t) = z(t) a: h(t) is as
shown in Figure 82.10. 1 1—
10:) Ht)
0 l ’l? o N J'
0 at. I Hot t
Figure $2.10
Therefore,
t, 0 S t S a
(t) __ a, a S t S 1
y ' 1+a—t, 15tg(1+a)
0, otherwise (b) From the plot of y(t), it is clear that 1&9 has discontinuities at 0, a, 1, and 1 + oz. If
we want 9352 to have only three discontinuities, then we need to ensure that a = 1. 2.11. (a) From the given information, we see that h(t) is non zero only for 0 S t g 00. Therefore, 00 $(t) at Mt) = / h(‘r)z(t — 1)d7' —00 W)
= [Doe—37w“  ‘T — 3)  u(t — T — 5))dr
0 We can easily show that (u(t — 'r  3) — u(t  'r — 5)) is non zero only in the range
(t — 5) < 'r < (t  3). Therefore, for t _<_ 3, the above integral evaluates to zero. For
3 < t S 5, the above integral is t3
y(t) = / 6—31d1’ =
. o For t > 5, the integral is 1 __ e—3(t3)
3 35 Therefore, the result of this convolution may be expressed as 0, —00 < t S 3
y(t) = 1.133;”, 3 < t s 5
L—L—H—G arms), 5 < t 5 00
(b) By differentiating :r(t) with respect to time we get
‘12?) = 6(t — 3) — 6(t — 5) Therefore, y(t) = ‘13?) * h(t) = e'3“'3’u(t — 3) — e"3““"’u(t  5) (c) From the result of part (a), we may compute the derivative of y(t) to be 133%) = rim3), . 3 < t s 5
(e‘6 —1)e‘3(“5), 5 < t 5 00 This is exactly equal to y(t). Therefore, y(t) = d it . 2.12. The signal y(t) may be written as
y(t) = ~~+e'(‘+6)u(t+6) +e<‘+3)u(t+3)+e‘u(t) +e<‘3>u(t—3) +e<‘6)u(t — 6) + . .
In the range 0 S t < 3, we may write y(t) as
y(t) = + e—('+6)u(t + 6) + e"(‘+3)u(t + 3) + e" (t)
= 6! + e—(z+3) + e—(t+6) + _ _, = e"(l+e‘3+e'6+)
1 1
l—e‘3 = 8 Therefore, A = ‘71::
2.13. (a) We require that @3471]— A (3M) uln — 1]: élnl Putting n = 1 and solving for A gives A = g.
(b) Piom part (a), we know that 5lnl ll h[n] — %h[n — 1]
Mn] =~ (6M — gain  11) = an}
From the deﬁnition of an inverse system, we may argue that g[n] = 6[n] — gap; — 1]. 36 2.22. (a) The desired convolution is Then Figure 82.21 W) W) = (b) The desired convolution is W) This may be written as Therefore, 0, { $(T)h(t — T)d1 
'\.
8 —00 ll um
te‘ﬁ‘uU) Q
II foo :1:(T)h(t — T)d7‘ 00 t
/ e'ate‘w‘TM‘r, t 2 0
o 1;th —— T)dT — /:h(t —— T)dT. '2 5
/ e2(‘"7)d'r / e2(“’)dr,
02 2 5
/ e2(t—r)d.r__/ e2(t—r)d.r,
t—l 2 5
/ ext”(11, t—l (1/2)[e2t _ 282(t—2) + 620—5)],
(1/2)[e2 + e2(“‘5) — 262(“‘2)],
(1/2)[e2“—5> — e21, 0, 43 (c) The desired convolution is N
y(t) = / :z:('r)h(t  T)dT
—oo
2
= / sin(1r'r)h(t — T)d7'.
0
This gives us
0, t < 1 t (2/7T)[1cos{1r(t—1)}], l<t<3
””’ (2/7r)[cos{1r(t—3)}1], 3<t<5 a 5 < t
(d) Let
he) = mm  §6<t  2),
where 4/3 0 < < 1
, ._ t _
him 2 { 0, otherwise '
Now, 1
y(t) = h(t) * $(t) = [h1(t) * $(t)l " §$(t  2)
We have
mm It z(t) = /t :(a’r +b) dT—  [—at2 — 1a( t— l)2 + bt — b(t — 1)].
1—1
Therefore,
y( = ggat'z — $00 — 1)2 + bt — b(t —1)]— —[a(t — 2) + b] = at + b = :r(t) (e) m(t) periodic implies y(t) periodic. determine 1 period only. We have .1 t /2(tT—l)d1‘+/ (1t+7)dr=;—+t—t2, —%<t<%
_ _l y(t): ‘11 t
/2 (1—t+7)d1+£(t—1—1)dr=t2—3t+7/4, %<t<%
t—l 5
The period of y(t) is 2.
2.23. y(t) is sketched in Figure 82.23 for the different values of T.
2.24. (a) We are given that h2[n] = 6[n] + 6[n  1]. Therefore, h2[n] * h2[n] = 6[n.] + 26[n — 1] + 6[n — 2]. 44 (b) Now,
ylﬂl = ialnl * yllnl = ylln] — sill"  1] Therefore,
2{1 — (1/2)"+3} +2{1 —(1/2)"+4} = (1/2)"+3, n 2 —2
yln’l = 11 n = —3
0, otherwise Therefore, y[n] = (1 / 2)"+3u[n + 3]. (c) We have
y2[n] = $291.11.: x3[n] = u[n + 3] — u[n + 2] = 6[n + 3]. (d) From the result of part (c), we get ylnl = yzlnl * wxln] = Ill" + 3] = (1/2)"+3U[n + 3] 2.27. The proof is as follows. 5
t:
ll /_:y(t)dt ‘
1:1:z(1)h(t — T)det
[00 $(T)[—:h(t  11(1th = foo :1:(1')Ahdr oo = AIAh ll 00
2.28. (a) Causal because h[n] = 0 for n < 0. Stable because 2%)" = 5/4 < oo. n=0 00
(b) Not causal because h[n] 5!: 0 for n. < 0. Stable because 2 (0.8)" = 5 < oo. n=—‘2 o
(c) Anticausal because h[n] = 0 for n > 0. Unstable because 2: (1 / 2)" = 00 "2—00 3
(d) Not causal because h[n] 7e 0 for n < 0. Stable because 2 5" = 6—33 < oo n=—oo
(e) Causal because h[n] = 0 for n < 0. Unstable because the second term becomes infimte
as n —> oo. 00
(1‘) Not causal because h[n] # 0 for n < 0. Stable because 2 h[n] = 305/3 < 00
n w 47 2.29. 2.30. 2.31. 00
(g) Causal because h[n] = 0 for n < 0. Stable because 2 h[n] = 1 < oo. n=—oo 00
(a) Causal because h(t) = 0 for t < 0. Stable because / h(t)dt = e‘8/4 < oo. 00
(b) Not causal because h(t) aé 0 for t < 0. Unstable because loo h(t) = 00.
—00
(c) Not causal because h(t) 94 0 for t < 0. a Stable because foo h(t)dt = e1°°/2 < oo.
—00
(d) Not causal because h(t) # 0 for t < 0. Stable because [00 h(t)ldt = e‘2/2 < oo.
00
(e) Not causal because h(t) aé 0 for t < 0. Stable because [00 h(t)dt = 1/3 < oo.
oo 00
(f) Causal because h(t) = 0 for t < 0. Stable because / h(t)dt = 1 < 00. ‘M (X)
(g) Causal because h(t) = O for t < 0. Unstable because / h(t)dt = 00.
00 We need to ﬁnd the output of the system when the input is a:[n] = 6[n]. Since we are asked
to assume initial rest, we may conclude that y[n] = 0 for n < 0. Now, ylnl = min] — 2yln — 11.
Therefore,
W] = 110]  2yl1] = 1, ylll = lel  2yl01 = ~2, yl2l = zl2] + 2yl2] = 4 and so on. In closed form,
ylnl = (2)"u{nl This is the impulse response of the system.
Initial rest implies that y[n] = O for n < —2. Now y[n] = :c[n] + 21171  2]  2y[n — 1].
Therefore, yl—2] = 1, yl—l] = 0, gm] = 5.
gm = 56,y[5] = —110, y[n] = 110(—2)"5 fom 2 5. 2.32. (a) If yh[n] = A(1/2)", then we need to verify 1" 1 1"1
AG) 74(5) ‘0' Clearly this is true. 48 ylll 2.37. 2.38.
2.39.
2.40. Let us consider two inputs
271(t) = 0, for all t and
z2(t) = et[u(t)  u(t — 1)]. Since the system is linear, the response y1(t) = 0 for all t. Now let us ﬁnd the output y2(t) when the input is 32(t). The particular solution is of the form
yp(t) = Yet, 0 < t < 1. Substituting in eq. (P2.331), we get
3Y = 1. ‘2‘, we get the Now, including the homogeneous solution which is of the form yh(t) = Ae
overall solution: y2(t) = Ae‘22 + get, 0 < t < 1. Assuming ﬁnal rest, we have y(1) = 0. Using this we get A = ~63/3. Therefore,
y2(t) = —%e‘2t+3 + ée‘, 0 < t < 1. (82.37—1) For t < 0, we note that $2(t) = 0. Thus the particular solution is zero in this range and
y2(t) = Be_2t, t < 0. (5237.2) Since the two pieces of the solution for y2(t) in eqs. (32371) and (82.372) must match at
t = 0, we can determine B from the equation 113_ _...e __ 3 3 which yields
1 1 3 = —— '2‘ 0.
Wm (3 3e)e , t< Now note that since 31(t) = 220) for t < 0, it must be true that for a causal system
y1(t) = y2(t) for t < 0. However, the results of obtained above show that this is not true. Therefore, the system is not causal.
The block diagrams are as shown in Figure 82.38.
The block diagrams are as shown in Figure 82.39. (a) Note that t
y(t) = / e“(“7)$(r — 2)d1 = / e_("2'7’)z(r')d‘r'.
09 —00 Therefore,
h(t) = e<‘2)u(t — 2). 55 2.41. Figure 82.38 m
8* x0e) Figure 82.39 (b) We have y(t) /°° h(r)z(t  T)dT = / for”[11“ — T +1)  u(t — 'r — 2)
2 h(7‘) and x(t — 'r) are as shown in the ﬁgure below.
Using this ﬁgure, we may write 0, t l (r—2) _ —(t—l)
e dr 1 e ,
y“)  f2 t+1
/ e"("'2)d‘r = e'("4)[l  6‘3],
t—2 (a) We may write gln] xIn] — axln — 1] a"u[n] —~ a"u[n — 1]
6[n]. 56 t<1 1<t<4 t>4 HT} 1:13—11 t—L a I:+I T a...» a 1 1':
Figure 31.10
[b] ”at: that 51h] = SIN»! 1: {1![11] — 115:“ — 1]} Therefarm [mm part. [a]. we km?! that
::[11] 11 {H11} — nrﬁln.  11} n H11]. Using thiﬂ 'Hll may write
2111: {din r 1]— ail": — 2n  aln  11.
=11  111: + 11 — natal} = 11: + 1].
II'ﬂ] 1: {H11 +2] — nib1+ 1]} = 15in + 2].
New nut: that
:1:[11] 1 11111} = 4111: +1] +111: + 1] + ﬁn] + $11: 1].
Theda1:,
Ila]  hln] = 1:12] 1 {ﬁln + 11  can + 11}
+ 211111: {ﬂ[11 + 1] — milnl}
+ zin] 1 {H111  1:151:11 — 1]}
+ {1111:5111  {15111  1] — aﬁlﬂ — 1]}
This may be written as
:[11] = hfn} = :[11] : {Ilﬂn + ‘3] — Ilmiln +1]+ HEn + 1}
 3113[11]+ H11] — Iraﬂu _, 1]
41 (112m:  1] — [11:11:11 ﬂ 2]
i If Thererun. '
Mn} = “In + 2} + I!  damn + 11+{1  ia‘jﬁln]
+ I:1.I"1"— “”1““ 1] ‘ [1315111 '3]
2.13. We 11:11 yﬂ] I: 3111:] I Hi]: ' ﬁﬂthﬂdr. 5'? Therefore, 0.5 . 2
11(0) = / e”“’°'dT = —— sin(wo/2).
—o.5 wo (a) If we = 21r, then y(0) = 0.
(b) Clearly, our answer to part (a) is not unique. Any wo = 2k1r, k E I and k aé 0 will sufﬁce.
2.43. (a) We ﬁrst have
[1:(t) a: h(t)] * g(t) = /_ : /_ :z(1)h(a' — T)g(t — a')drdo’
= [:[:x(1)h(a)g(t — a — 1’)d'rda
Also, 8 :c(t) * [h(t) :0: g(t)] x(t  a’)h(1’)g(a' — T)da'dr 8 II \'\. 83 8
\\8\ :L'(cr)h(1)g(t — T — a)d1do ‘W —00
00 (X) z(7)h(a)g(t — a — T)deO’ ll
\. ~00 00
The equality is proved.
(b) (i) We ﬁrst have 1 k=0 CAIN [1 — (—éwl] u[n]. Now,
yln] = wlnl * hzln] = (n + 1)u[n]~ (ii) We ﬁrst have 7 I; 11—1
gln] = 111an . h2[n] = 2: (—g) + $2}? = uln]
k=0 k=0
Now,
yin] = u[71] * gin] = 14”] * u["1 = (n + 1)Ulnl
The same result was obtained in both parts (i) and (ii). (c) Note that
xlnl * (h2lﬂl * hllnl) = (xlnl * hzlnl) * hllnl Also note that
:i:[n.] It h2[n] = a"u[n] — a"u[n  1] = 6[n]. Therefore,
1:[n] * h, [n] in MM = 6[n] 1! sin 8n = sin 811. 58 ...
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 Fall '10
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