bookstylehw7solution - 5.29. (a) Let the output of the...

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Unformatted text preview: 5.29. (a) Let the output of the system be y[n]. We know that. Y(ej“’) = X(e~7“’)H(e-7“’). ' In this part of the problem HM“) = m- (i) We have - 1 LI) _ X(e’ )— 1— %e_]w Therefore, new) _ 31 . ,1 1~ 4-6—37“) 1 - 56—3“) «2 3 H Taking the inverse Fourier transform, we obtain y['n] = 3 (gnaw — 2 <%)n'u[n]. (ii) We have Therefore, _ 4 _ 2 _ 3 _ l—ée—jw l—le‘j“J (1——%e‘3‘“)2 Taking the inverse Fourier transform, we obtain y[n] = 4 (%)nu[n] — 2 (41)"1414— 3(n +1) <i>nu[n]. X(ej“’) = 27r i 6(w — (2k + 1)7r). (iii) We have k=—oo Therefore, w __ 0° __ 1 my) _ [2:20am (2k+1)w)][1_%e~Jw] = 4.31;: 6(w—(2k+1)7r) Taking the inverse Fourier transform, we obtain :1:[n] = §(—1)". (b) Given we obtain - 1/2 1/2 H 63‘” — ( ) 1 — %ej"/2e-iw + 1 — le-J'W/l’e-J'w' (i) We have new) = 11 1 ~ 56-77“) Therefore, 1 __ EeJW/Ze—Jw 1 __ ée—Jfl/2e—Jw 1 _ ée-Jw __i___ + B + C 1—(1/2)ej7r/2e-2‘w 1—(1/2)e~jw 1—(1/2)e-j7r/2e—Jw’ where A = ~j/[2(1 —j)], B = 1/2, and C = 1/[2(1 +j)]. Therefore, Mn] = 2(1—3 j) (%)nu[n] + 20:3,) (—%>nu[n] + % (%)nu[n]. (ii) In this case, Mn] = [4 —(1)"] um]. 2 (c) Here, We”) = X(ej“’)H(ej“’) = —3e_2j“’ _ ej‘” + 1 —— 2e-J‘2w +6€_jw + 28—3-2“) — 26—13” + 48—75“ +385“ + e74“ — e13“ + Zej‘” Therefore, y[n] = 36[n + 5] + 6[n + 4] — 6[n + 3] — 36[n + 2] +5[n +1]+ 6[n] + 66[n — 1] — 26[n — 3] + 46[n — 5]. 5.50. (a) From the given information, W“) 1— W“ 3‘8”) = X(ej“’) = Taking the inverse Fourier transform, we obtain h[n] = 3 (%)nu[n] — 269391]. (ii) From part (a), we know that new) _ 1 - er” X(ea'w) * (1 —§e-jw)(1—§e-J'w)' Cross-multiplying and taking the inverse Fourier transform yln1- £21m ~ 11+ 115mm — 21: min] — gun — 11. (b) From the given information, Y(ej“’) _ (1 — %e‘j“")2 X(ej“’) — 2(1—fie‘j”)? We now want to find X(ej“’) when Y(ej“’) = (l/2)e"j“’/(l + ée‘j‘”). From the above equation we obtain HM“) = e—a‘wu _ ice—1"“)? (1 — ée—jw)2(1 + ée'j‘”) Taking the inverse Fourier transform of the partial fraction expansion of the above expression, we obtain $[n] = g Ham — 11+ 3 (%)n~lu[n — 11+?» (%)n—1u[n — 1]. 5.51. (a) Taking the Fourier transform of Mn] we obtain new) = 3 1 —jw . _ . a _ 5 * 55 H(e"") — Y(e"”)/X(ej ) - Cross-multiplying and taking the inverse Fourier transform we obtain y[n] — gym —11+§y[n — 2]: 3min] — gm — 1]- (b‘) (i) Let us name the intemediate output w[n] (See Figure 85.51). Figure $5.51 We may then write the following difference equations: 1 l yin] + Elli“ — 11 = Zwlnl + win — 1] and 1 1 —— §w[n — 1] = -— 5:1;[71 — 1]. Taking the Fourier transform of both these equations and eliminating W(ej“’), we obtain , 1 7 ‘ 1 2‘ me“) = We”) 2 _.______z + W“ ' W" J” X(ej“’) 1 — £341“ Cross-multiplying and taking the inverse Fourier transform we obtain 1 1 7 1 — -— — = — — -— — — —~ 2 . y[n] 4mm 2] 4min1+ 8min 1] 2min ] (ii) From (i) I 7 1 2'Lu How) = W“) : z+ 86 w - -e’ X(eJ“’) 1 — 416—21“) (iii) Taking the inverse Fourier transform of the partial fraction expansion of H (6”), we obtain hln] : 26M _ €11; (__:_)nu[n] + % Uln]. ...
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This note was uploaded on 02/13/2012 for the course ECE 301 taught by Professor V."ragu"balakrishnan during the Spring '06 term at Purdue University-West Lafayette.

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bookstylehw7solution - 5.29. (a) Let the output of the...

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