HW5_Solutions

HW5_Solutions - 4.30. {a} We know that. wfl) =cost 53+...

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Unformatted text preview: 4.30. {a} We know that. wfl) =cost 53+ WUw] = 1r[¢5{w — i) +¢§(w + i) and gum = mm: 53 Cum = 5%{1Uw} - WIML Therefore, Guw) : %X(j(w — 1]) + gxuzhm 13). Since Glju] is 15 shown in Figure 54.30. it is clear from the above equation that Xfiw} is asshown in the Figure 54.30. we) :8qu Figure £4.30 The-reform 2 sin a I“) = wt {h} X1 Um) is as shown in Figure 54.30. 4.33. (9.] Taking the Fourier transform of both sides of the given difl'erential equation, we obtain YUw] _ 2 XUw) '— -w!+2jw +8. HUw) = Using partial fraction expansion. we obtain 1 jw + 2 jw + 4' HUD) = Taking the inverse Fourier trunnion-m. Mr} = c‘z'ufi} —— r“u[r]. {b} For the given signal :[t], we have ‘ 1 m“) = Therefore. 1 _ v ' 2 Yum) = quih'trw) = m (2 Hen)? Using partial fraction expansion, we obtain Taking the inverse Fourier transform, 1 gm = grim) — gig-“um + $26—2‘uU} - Zerilum. (1:) Taking the Fourier transform of both sides of the given difi'ereritial equation, we obtain 1" ‘ — 2 ~ 1 HUM = (w) _ 2( w ) XU‘I’) “ —w2 + flier +11 Using partial fraction expansion, we obtain —J§— efij + —Ji+2fij 3-“ _ —J§+z'v’i jw _ 4J5—3'Ji‘ Taking the inverse Fourier transform, hm = 253) — «in + 2;}e-UH'5‘mum — flu — 2jie“"”"'fiu(i)- .3ij =2+ 4.34. (a) We have YUM jw + 4 Xfiw] 6 — [.02 + ij' Chose-multiplying and taking the inverse Fourier transform, we obtain day“) dyili __ dim. d3: + 5-15“- + 611(1) -- m— + 43”} (b) We have 1 H (in!) = "- 2 + fine 3 + ju- ‘I‘aking the inverse Fourier transform we obtain. Mr) = osmium —-e"uu(t]. {c} We have 1 1 4+jw _ (4+jw)2' XUw): Therefore, 1 [4 + jw}{2 +jw)‘ Finding the partial fraction expansion of YUw) and taking the inverse Fourier tram» Iorm, ] 1 __ __ -2¢ __ _ -u yfl) — 23 11(1) 2: ufl‘). YUM : XiijHUw] = 4.35. (a) P113111 the given information, . V E + Hi {How}! = —“— =1- v'ai +r.i.t5 A150, w dHUw] = —l:ri.:1'1~"'~-J — tram"1 E = -2lan‘1—. B a 0. Also, HUw} = —1 + 2“. =9 hm = —.5(:} + ace—“um. a + 3;.»- (b) If a :1, we have I IHUwil = 1, dHUw)=-2tan‘ w. Therefore. 2 Hit} = magi—i - g1~ mu — g} + mush/5t — 3:}, 4.36. (a) The frequent-2y response is Yfiw} _ 3(3 + jm) H0“) = mu} " (4 +J'w}{2 +31»! (In) Finding the partial fraction expansion of answer in part [a] and taking its inverse Fourier transform. we obtain h(t) = g [6“ + 6—2‘] u[£]. (c) We haw YUw) _ (9+3jw) X(ju:} — 5+6jw —w?' Crwmultiplying and taking the inverse Fourier transform. we obtain (Faro!) dyfl) drill (it? + 6-!“— + 8y(t} = 3H3?- + 91(3). ...
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This note was uploaded on 09/29/2008 for the course EEE 203 taught by Professor Chakrabarti during the Spring '07 term at ASU.

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HW5_Solutions - 4.30. {a} We know that. wfl) =cost 53+...

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