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Unformatted text preview: EE 3:20 CH7 SCLUTIOMS 7.14. (a) T T , ' 1 _ e(J'N+a)T  ..' _ _ + ): _ ,
X(w)=/ol e “‘e Jw'dt—j; e 0”“ dt— jw+a
(b)
T T . 1 _ e—(jw—a)T
z — 'wt _ —(Jw—a) =
X((l))=/;J e"'eJ (it—A e dt jw’a
7.1—5. (a)
' 1 . 2 , . _ jw 2 J'2w
x<w>= / 4e‘1”‘dt+f it: a 1 J“
(b)
° t  T t . 2 .
X(w)= ——e'"‘"dt+/ — "Jw‘dt='—§[coswr+w‘rsmwr— 1]
_, 1' o 1' Tw ‘
This result could also be derived by observing that :c(t) is an‘even function. ‘ . ’
Therefore from the result—in Prob. 7 .1‘1 ‘
2 T . 2 . ' ‘
X(w)= tcoswtdt=—2[cosw+wrs1nwr—l] . s T a T0) 7.21. Figure S7.2—1 shows the plotsrof'various functions. The function in part (a) is a
gate function centereiattheoriginandof width 2. Theﬁrnction in part (b) can be expressed as A This is a triangle pulse centered at the origin and of width 100/3. The function in part (c) is a gate function rect(§)_ delayed by 10. In other
words it is a gate pulse centered at 't =10 and of width 8. The function in part (d) is
a sinc pulse centered at the origin andthe ﬁrst zero occurring at 1%”— = 11', that is at
w = 5. The function in part (8) is a sinc pulse sinc(‘‘5’) delayed by 101r. For the sinc
pulse sinc(§), the ﬁrst zero occurs at 531 = at, that is at w = 51r. Therefore the function
is a sine pulse centered at w = 101r and its zeros spaced at intervals of 51r as shown
in the ﬁgure S'Z.21e. The function in part (f) is a product of a gate pulse (centered
at the origin) of width 101r and a sinc pulse (also centered at the origin) with zeros
spaced at intervals of 57r. . This results in the sinc pulse truncated beyond the interval
i511 (t Z 51) as shown in Fig. f. ~ 7.24. (3) 1 “’° . . 1 we I
I“) = / e"“"°e""""du=_/ eJW(lto)dw
. 21' wo 21f _uo
— _;L_ 'W(¢'o) “o "=' Sinwo(tto) = 32 . t_t
" (we—tad We “raw ,smclwo( 0.)]
(b)
" 1 , M a . M
170) = ﬂ JeJ dw+ —]e7 dw
wo . o
= iejwt'o _ _1_ejut w = lcoswot
21rt ﬂ,“ 21rt o 1.1
7.33. (3) \ T 2 X (to) = Tsmc [eJWTI2 6"”772]
. . wT ' , wT
= 21Tsmc sm 7 :7} sin? w 2 (b) From Figure S7.33b we verify that
z(t) = sintu(t) + sin(t — 1r)u(t — 1r) Note that sin(t—1r)u(t—1r) is sin tu(t) delayed by 1r. Now, sintu(t) <=> 2%.[609
1) — 6(w+‘i)] + 111;; and ’ sing  1r)u(t—1r)<=> {%[6(w  1) — 6(a) + 1)] + 1}e—jwu Therefore X0”) = {zljww  1) — 6(a) + 1)] + (1 + e—jmu) l—w2 Recall that g(z)6(z — :50) = g(zo)6(z — :50). Therefore 6(a) :l: 1)(1 + e'j’m) = 0,
and
1 l—w2 X(w) = (1 +e'jm") 331 ) (C) From Figure S7.3—3c»we verify that
. W A W
I“) = 00“ [11“) ‘ u (t * = costu(t) — costu (t — But sin(t — g = 4— cost. Therefore 20) = costu(t) +sin (t — u(t_ X60) = gm» — 1) +60» + 1)] + j“ 2 + {211.90.} —1)— 6(4) + 1)] + 1 3w2}e'j*”72 1 — 01
A150 because 9(I)6(I  10) = 9&0)“; — $0),
I 6(a) :t 1)e'j*”/2 = 6(u :t 1)e*1'*/ 2 = :tj6(w :t 1) Therefore I I2
_ in: 12"“ _ 1 r ﬁn,”
X(w)—1_w2+ 1_w2 —1_w2gw+e ]
(d)
2(t) =, e‘°‘[u(t) — u(t — T)! = e'“‘U(t) — e""‘u(t — T)
i:— e""‘u(t) — eﬂTe'°(‘_T)u(t — T)
1 eaT _ .UT 1 _(¢+ .un,
X0») = eJ = [l—e J 1 Jw+a_jw+a jw+a z'ih ’
)‘\‘a,
(by shut1T3 cut57) ' a
9M? tut3. Figure 37.33 7.36. (a) Ehe signal r‘(t) in this case is a triangle pulse A(%) (Figure S7.36) multiplied
y cos 10t. t
:r(t) — A cos 10t Also from Table 4.1 (pair 19) A(%) <=> 1rsinc2(%) Iiiom the modulation prop
erty (4.41), it follows that zit) = A cos 10t <=> g{sinc2 + sinc2["(w;'10)]} The Fourier transform in this case is a real function and we need only the
amplitude spectrum in this case as shown in Figure S7.36a.' (b) The signal :r(t) here is the same as the signal in (a) delayed by From time
shifting property, its Fourier transform is the same as in part (a) multiplied by . , ,..
e'j“(”')‘. Therefore xt) = g {sinc’ [——"‘“’; 10)] +5ch [__._"<’w ; 10>] } The Eonrier transform in this case is the same as that in part (a) multiplied by
e’”"“’. This multiplying factor represents a linear phase spectrum 24rw. Thus
we have an amplitude spectrum [same as in part (a)] as well as a linear phase
spectrum 1X00) = —21rw as shown in Figure S7.36b. the amplitude spectrum
in this case as shown in Figure S7.36b. Note: In the above solution, we ﬁrst multiplied the triangle pulse A(2—‘;) by
cos 10: and then delayed the result by 22r. This means the signal in (b) is
expressed as AU}? ) cos 10(t — 21r) We couid have interchanged the operation in this particular case, that is, the
triangle pulse A(2+‘1r) is ﬁrst delayed by 21r and then the resuit is multiplied
by cos lat. In this alternate procedure, the signal in (b) is expressed as
A(%) cos lot. This interchange of operation is permissible here oniy because the sinusoid
cos 10: executes integral number of cycles in the interval 21r. Because of this
both the expressions are equivalent since cos 1‘06  21r) = cos 10L (c) In this case thesignal is identical to that in (b), except that the basic pulse is
rect( instead of a triangle pulse A(2+‘r—). Now i .
rect <=:> 21r s1nc(1rw) Using the same argument as for part (b), we obtain X (to) = 1r{sinc[1r(w + 10)] + sinc[1r(w _ 10)]}ej2mn 7.3—7. (a) .
X(w) = rect (‘0 g 4) + rect (w : 4)
Also ' 1 l
. w
;smc(t) <=> rect Therefore :c(t) = frsincﬂ) cos 4t 334 (b) 7.38. (3.) Figure S7.36 X (to) Also Therefore e“u(t) <=>
If :z:(t) = e‘“u(t) * u(t), then (Jul—A) 1r6(w)
jw  A ’X(w) + jwo'wA) =A(_“_+_“:)+A(E_:E\ 4. 4/ ;r1—sincz(t) <=> A 1:(t) = gsincza) cos 4t 1
and u(t) <=> 1r6(w) + 1
jwA (my3) __l l
[—A + j“: A] I because g(a:)6(1:) = g(0)6(1:) («6<w>+j%)] 335 ...
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 Fall '06
 Audiffred

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