HW5_Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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). ...
View Full Document

## This note was uploaded on 09/29/2008 for the course EEE 203 taught by Professor Chakrabarti during the Spring '07 term at ASU.

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online