Slu to Ch7

Slu to Ch7 - EE- 3:20 CH7 SCLUTIOMS 7.1-4. (a) T T , ' 1 _...

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Unformatted text preview: EE- 3:20 CH7 SCLUTIOMS 7.1-4. (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) 7.2-1. 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 and-the 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.2-1e. 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. ~ 7.2-4. (3) 1 “’° . . 1 we I I“) = -/ e"“"°e""""du=_/ eJW(l-to)dw . 21'- -wo 21f _uo —- _;L_ 'W(¢-'o) “o "=' Sinwo(t-to) = 32 . t_t "- (we—tad We “raw ,smclwo( 0.)] (b) " 1 , M a -. M 170) = fl JeJ dw+ —]e-7 dw -wo . o = iejwt'o _ _1_ejut w = l-coswot 21rt fl,“ 21rt o 1.1 7.3-3. (3) \ T 2 X (to) = Tsmc [eJWTI2 6"”772] . . wT ' , wT = 21Tsmc sm 7 :7} sin? w 2 (b) From Figure S7.3-3b 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 e-aT _ .UT 1 _(¢+ .un, X0») = eJ = [l—e J 1 Jw+a_jw+a jw+a z'ih ’ )‘\‘a, (by shut-1T3 cut--57) ' a 9M? tut-3. Figure 37.33 7.3-6. (a) Ehe signal r‘(t) in this case is a triangle pulse A(%) (Figure S7.3-6) multiplied y cos 10t. t :r(t) — A cos 10t Also from Table 4.1 (pair 19) A(%) <=> 1rsinc2(%) Iii-om 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 S7.3-6a.' (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 S7.3-6b. the amplitude spectrum in this case as shown in Figure S7.3-6b. 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 the-signal 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)]}e-j2mn 7.3—7. (a) . X(w) = rect (‘0 g 4) + rect (w :- 4) Also ' 1 l . w ;smc(t) <=> rect Therefore :c(t) = fr-sincfl) cos 4t 334 (b) 7.3-8. (3.) Figure S7.3-6 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'w-A) =A(_“_+_“:)+A(E_:E-\ 4. 4/ ;r1—sincz(t) <=> A 1:(t) = gsincza) cos 4t 1 and u(t) <=> 1r6(w) + 1 jw-A (my-3) __l l [—A + j“: A] I because g(a:)6(1:) = g(0)6(1:) -(«6<w>+j%)] 335 ...
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This note was uploaded on 02/08/2012 for the course EE 3220 taught by Professor Audiffred during the Fall '06 term at LSU.

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Slu to Ch7 - EE- 3:20 CH7 SCLUTIOMS 7.1-4. (a) T T , ' 1 _...

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