This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: We see that G'(jw) is periodic with a period of 8. Using the multiplication property,
we know that we = 2—17, [fr {4} * Gem] 7rt
If we denote FT {$793} by A(jw), then Xlﬂu) = U/2w)pﬂjw)*8n E: 6(w~—8k)
k=—oo
= 4 i AW — 8k)
k=—oo X ( jw) may thus be viewed as a replication of 4A(jw)
periodic. Using Table 4.2, we obtain A(jw):{1, IWISI every 8 rad / see. This is obviously 0, otherwise Therefore, we may specify X ( jw) over one period as . _ 4, iwlSl
X(’”)*{0, l<lw$4 4.17. (a) From Table 4.1, we know that a real and odd signal and odd Fourier transform X (jw). Let us now consider the purely imaginary and
odd signal jx(t). Using linearity, we obtain the Fourier transform of this signal to be jX(jw). The function jX(jw) will clearly be real and odd. Therefore, the given
statement is false. signal :1:(t) has a purely imaginary (b) An odd Fourier transform corresponds to an odd signal, while an even Fourier transform corresponds to an even signal. The convolution of an even Fourier transform with an
Odd Fourier may be viewed in the time domain as a multiplication of
odd signal. Such a multiplication will always result in an odd time signal.
transform of this odd signal will always be odd. Therefore, the given an even and The Fourier
statement is true. 8 Using Table 4.2, we see that the rect
”transform X1010)
~~We may Write angular pulse 321 (t) shown in Figure 84.18 has a Fourier
= sin(3w)/w. Using the convolution property of the Fourier transform, . 2
mm = m1(t) * me) 313 Xzew) = X1(jw)X1<jw) = ( 81115.3 a”) 8"" signal $2(t) is shown in Figure 84.18. Using the shifting property, we also note that . 2
émit + 1) (—F—Ta éej‘“ (sm(3w)) LI.) 139 and ~ 2
gm _ 1) a; law (SNM) . Adding the two above equations, we obtain h(t) = w  2
gm“ + 1) + ~12m2(t—— 1) <55) cos(w) (Sln(3w)) ‘
The signal h(t) is as shown in Figure 84.18. We note that h(t) has the given Fourier
transform H ( jw). x (0 ma
1 5/;
l/
—3 O 3 t ‘6 0 6 t
Mt )
5/4
—; —s 0 5 a t Figure $4.18 Mathematically h(t) may be expressed as g, t < 1
.151 + Q, 1 S it] S 5
W) = {a 3
0, otherwise
4.19. We know that Y(' )
. .7“)
H w = . 
(J ) X(Jw) Since it is given that y(t) := e’3tu(t) — e‘4tu(t), we can compute Y(jw) to be Since, H(jw) = 1/(3 +jw), we have X(J'w) = 28:; =1/(4 +y‘w) Taking the inverse Fourier transform of X ( jw), we have m(t) = e‘4tu(t). 4.20. From the answer to Problem 3.20, we know that the frequency response of the circuit is 1
~w2+jw+1' Breaking this up into partial fractions, we may write H (.7141) = H(jw)~—1[ ‘1 + “1 ]
N5 %—‘/7§j+jw é+§j+jw USing the Fourier transform pairs provided in Table 4.2, we obtain the Fourier transform
of H ( jw) to be 1 _1 ﬁ~ _r__\/.'§'
hi =—~—— —e( 2+2J)‘++e( 2 23)” ut
() NJ ] ()
Simplifying,
Mi) 2 —2—3e“§‘sin(—~t)u(t). (a) The given signal is 1  l 
6“”t cos(w0t)u(t) = ﬁew‘tewot’uﬂ) + 56"“te_3“’°l11.(t). Therefore, 1 1
X ' = — ——————————————.
‘5’”) 2m — jwo + jw) 2(a — jwo + M
(b) The given signal is x(t) = e”3" sin(2t)u(t) + e3tsin(2t)u(—t).
We have _ . , . l/2j 1/23"
:, t = 3‘. 21 2‘. ﬁx — — . .
M) e “I“ M) 10“) 3—j2+jw 3+32+jw
Also 3 , . , . 1 2' 1 2‘
12(1) = ersmewuet) = —x1(—t) «53> now) = —X1<—Jw) = :ﬁE—mﬁ
Therefore, ‘ . . 3' 3.7'
XW) = X1(]w + X2(Jw) = 3173,2175)? ” W' 141 (m) We have
__= —t FT _.
h (t) 2t Mt) (#4 Hg(yw) _../'(1 +310)?
Therefore, 7r
YUw) ’ Xiawwmw) = 3M» 4 1) , 5(a) ~ 1)]
we obtain 2. Note that Mt) = In“ .. 1), where 4.3 is the impulse re Therefore, h(t) i d is in the range
' ing the shift property, . (a) We have .
X100): “6’ 12
It is clear that This result is equivaie (b) We have
Xzijw) = 71. Xﬂ‘aﬂaw ~
‘7 k=0 Therefore,
' ' Tuummw This implies that 1
ygm == 5 sin(3t — 1).
iy the sinusoid with frequen at on (c) We have
6]“), {col < 4
0, otherwise X300) : {
new) = X3(jw)H(jw) = awe—1"“
This implies that sin(4t)
7rt ' 313(15): 153(t'"1)= We may have obtained the same result by noting that X3(jw) lies entirely in the
passband of H(jw). (d) X4(jw) is as shown in Figure 84.32. “.le ~¢+ ° + t»
Figure $4.32 Therefore, 4
HOW = X4(jw)H(jw) = X4(jw)6”’w. This implies that ‘ __ 2 We may have obtained the same result by noting that X4(jw) lies entirely in the
passband of H(jw). 33 (a) Taking the Fourier transform of both sides of the given differential equation, we obtain Y(jw) 2 X(jw) ~w2+2jw+8. H (N) = Using partial fraction expansion, we obtain 1 1 jw+2 jw+4' HOW) = 151 e inverse Fourier transform, Taking th
Mt) = e‘QtuU) w e'4tu(t).
(b) For the given signal $(t), we have
X( m)  #3.,
3 (2 +21»)?
Therefore,
. . 2 1
/. ff 2. Y(jw) = X(jw)H(le = (W2 + 2,,» + 8) (2 + Jw) pansion, we obtain Using partial fraction ex
. 1/4 1/2 1 1/4
Y = __.._——— — / ._.—————~ _. .
(9w) jw+2 (jw+2)2 + (jw+2)3 jw+4 Taking the inverse Fourier transform,
1 1.
”Wm — atea—”mm + {la—mum — Ze"4‘u(t). 1
1N) = 718 ansform of both sides of the given differential equation, we obtain (c) Taking the Fourier tr Using partial fraction expansion, we obtain
—— 2 — 2+ 2 2"
H(jw)=2+_._‘—/—'——f\/‘Jf+jw ﬂ «3/3.
3L0 __ — 2+l —' 2;2\/§ Taking the inverse Fourier transform, h(t) = 25m « f (1 + 2 j‘)e—< WWW) — ﬂu — 2j)e’(1"j)"/‘/§u(t). 4.34. (a) We have
Y(jw _ jw + 4
X(jw) 6~w2 +5jw‘ Fourier transform, we obtain d taking the inverse d2 t d t d,t
3:2)“; sighting: 294—41170). Crossmultiplying an (b) We have
. 2 “ 1 Taking the inverse Fourier t (c) We have 1 1
X ' = — M.
(W) 4 +jw (4 +jw)2 Therefore,
1
(4 +jw)(2 + jw)’ Finding the partial fraction expansion of Y(jw)
form, YUW) = X(jw)H(jw) = and taking the inverse Fourier trans . 1 _ 1 _
y(t) 2 ie 2tu(t) — 5e 4tu(t). 4.35. (a) From the given information, «m
mew): = ——~——— = 1.
ﬁﬁ$§
Also,
LU
<1H(jw) = ~tan‘1 E — tan“1 2 = —2 tan’1 —.
a a (1.
Also, 2
H(jw) = —1 + a +“jw => h(t) = ~6(t) + aleaims. (b) Ifa = 1, we haveH(jw)[ = 1, <IH(jw) = —2tan”1 w.
Therefore, y(t) = cos(\% — g) — cos(t ~ 32:) + cosh/gt ~ 233:) 4.36. (a) The frequency response is . _me_ m+m>
”W * qu)  m (b) Finding the partial fraction expansion of answer in part (a) and t aking its inverse
Fourier transform, we obtain h(t) = [0(0) [6—4t + 8—2t] u(t). (c) We have
Y(jw) _ (9 + 3jw)
qu) * 8 +6jw —w2'
Crossmultiplying and taking the inverse Fourier transform, we obtain 2 x '
gagawigaaymﬂddi” +9241» 153 ...
View
Full Document
 Fall '06
 V."Ragu"Balakrishnan

Click to edit the document details