3015_Homework_3_Solutions

3015_Homework_3_Solutions - 2.32 A discrete—time LTI...

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Unformatted text preview: 2.32. A discrete—time LTI system has the impulse response Mn] depicted in Fig. P232 (a). Use linear— ity and time invariance to determine the system output y[n] if the input is Use the fact that: §[n — k] Si Mn] [s31 + h£2[11]] Si Mn] Mn — in] ﬁlm] at Mn] + hmgfn] an M11] [a] = 36]“:1] — 25[n — l] 3Mn] — 2M1: — l] +1]+ T’Mn] — ?§[n — 2] + 545m — 3] — 25]}: — 4] sin] [c] I[1|‘1] as given in Fig. P232 [1)] lie] yln] 25h: — 3] + 2§[n] — 5].": + 2] BMW}. — 3] + ﬂMn] — Mn + 2] —<5[n + 3] — 3£[n+2] +7§[n] + 35hr, — 1] +85[n — 3] +4.5[n — =1]— 25hr, — 5] + 25[n— 6] 2.34. Consider the discrete-time signals depicted in Fig. 132.34. Evaluate the convolution sums indi- cated below. (a) mln] = Ilnl * zln] forn+5<tﬂ Edi—5 m[n]=ﬂ forn+5<f4 —55n{—1 n+5 m[n]=Zl=n+l3 k=U forn—l-el —15n<:2 3 n+5 m[n]=zl+2zl=2n+s e=u k==l forn+5<t9 23n<4 3 n+5 [d] m[ﬂ] = Ilnl * Elm] fern—Iii forn—lnig farm—1:39 m[n] forn+5-c:—8 forn—l-at—T forn+5<t4 forn—l-ﬁ—l [urn—1:31 forn—l-dll fern—1311 4311-55 3 E m[n]— Z 1+221=15—n k=n—1 k: 5:11:10 k=:|1-—1 1'1ng Inf-'1]: El 11-5—5 n+6 —5£n{—l 2n+8 —1£n-¢t2 9+1: an-iri 15—n din-:5 213—271 Sin-1:11] D 11310 “rid—13 m[n]= —14-::n-::—ﬁ n+5 m[n]= Zl=n+14 k=—3 —ﬁ-::n<:—l —2 m[n]= 2 12—11 k=ﬂs—1 —1-::n<:U —ﬁ n+5 m[n]= Z l+Zl=—2 k=fb—'I k=4 Dina-:5 n+5 m[n]=Zl=n+2 =‘-‘I Eﬁncilﬂ 1n m[n]= Z 1:12—1r1 k=n—1 113-12 m[n: = [II E] n «t: —13 n+l4 —133nc:—{5 —n —{5 :2 n «r: —l m[?1] = —2 —l :2 n {{1 ?1+9 {liar-«:5 lﬂ—n 5-571»; 12 E] r1312 Ith mm] = ma] * ﬁn] Intervals n-a: —§: —§: Inf—4 —-1 Urch (13nd? 2t n-zEr b -=_~Z n an: ID v12 1K] mII'I] = 11"[I'|]"|I'|I|1r'l] Figure P234. Figures of Mn] and f :n — k] P2_33[h3 rn[n] = ﬁnl'ﬁnl M | I mhin‘ aleItude D | —9 —e I O i B Figurt‘ P234. m[n: = y[?r.] -= ﬂu] Lj} m[rr: = u=[rr: -= in: Intvn-‘als, —{3 i: H. —3 —3 -:_2 H. —l —l g n. Ii] {)5 rr 3 3 1: rr 5 rr 7 rr 9 9 rr 11 LL 1: n. 1-3 |.-"'-. 5 l' |.:'"-. |a"'a r1215 n1[n] = W[|'|].9["'] Fjgm’ﬂ PEAS-L. Flgurw ul' rt'iu' and (Jill - J-'| 52mm rn[n] = winil'mn] rn_:."1:- MEI-110513 Fjglu'r- 11-2454. min' whrl silerl 2.39. Evaluate the continuous—time convolution integrals given below. Eh}! yt't) = 42““ mix) *2 u[r — 3) Figure P239. {th Graph of 1'[T] and h[t — T] fort+3crﬂ 34—3 yftl' = U for f :_=- —3 M3} — { _ Era-n+3] 133;: _3 {G} ME} = cosmrﬁuﬁt + 1] — at? — 1)) *- um XV] h[t — I] I T T —1 1 Figure P239. {(3} Graph of and h} — T] fort<—1 yﬂ=ﬂ for i <1 I- y t} =f cosﬁfrth'r —1 t}— lsinlh‘ri‘} y _ J'T fort:>1 1 y t} 2/- coshrtm'r —1 yﬂ=ﬂ 1‘ a —1€t<1 ym = 1rsmﬂilr _ [II otherwise [nyu}=zﬁmu+JJ—uu—1n*2mr+m fort+2<—l t<:—3 MH= fart+2<il —35t<:—1 {+2 4 mn=2]- kﬁr=g[ﬁ+m3+ﬂ —1 fort+23—l rib—1 1 8 yﬂt} = 2/ ﬁrﬂdr = — -1 3 ﬂ tc—E wit] = §[{:+2}3+1] 494—1 g {3—1 ﬁ]ﬂﬂ={ﬂﬂ+¢}+JH—5H*Mt—H fort—lett—l till] y££l=ﬂ fart—1:35 Ugtéﬁ By the sifting pmperty. y{rj=f_12al:c+1)¢r = 2 —I.‘I'LI U t-ctiﬂ ...
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3015_Homework_3_Solutions - 2.32 A discrete—time LTI...

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