EE4150_2003_hw3sol

EE4150_2003_hw3sol - (a) (3) (4) (5) (5) I (7) ‘2 4-K H V...

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Unformatted text preview: (a) (3) (4) (5) (5) I (7) ‘2 4-K H V ll 2 h(k)z(fi — k) k _ I I Zyfit) = ZZh(k)z(n-k)=2h(k) Z x(n—k) I: k n' fl=-OO (m) (w) I 9(a) = h(n) a: 201) =-'{1,3,7,7,7,6,4} Z) 9(a) = 35. 2 has) = 5, 23m = 7 n a k I you) = {1,4,2,—4, 1} n I, k you) = {avg-rig, 2,o,—-:-,—2} 2 9(3) = -5, 231(11): 2 5, 230;) = -2 9(a) = {1,2,3,4,5} 23:01) = 15, 2 Mn) = 1, - 2201) = 15 9(a) = {010: 1s ""1: 2: 2: 1:3} Z 90') = 8' E M“) = 4. 22:01) = 2 9(a) = {o,o,1,—1,2,2,'1,3} _ :90“) :8? Zh(")=2. Z:(n)=4 1 n 90“) = {01114: _4: _5, “'1, 21 (8) (9) (m) (11) 2 .1 7 (a) 3(a) Mn) yo!) ‘ 31(0) 5(1) 9(2) 9(3) 3(4) 3(5) 31(6) K7) 9(8) yo!) y(n) Ii 1! II II I! ll II II II II y(n) = u(n) + u(n — 1) + 21:01 - 2) 290‘) = 00: zhfiz) = 00, 230:) = 4 II 90") = {1! _1) “512:3:—5i 1: 29(3) = 0, 2M1!) = 0, I 23(11): 4 ‘1 a 2:01): {1.4.4.4,10,4,4,4,1} 2 yo!) = 36’ z M“) 2 6: z 3201) = 6 1‘! you) = {2030‘ -(fi)“1u(n) 29(11):; ':h(n)=§, 23(n)=2 )111?1} I5l4l312? { ,{fl 2 z(k)h(n — 1:) no z(0)h(0) = e, 3(0):;(1) + 3(1)}:(0) == 11 =(0)h(2) + a:(1)"I(1)+ =(2)h(0) = 15 - 3(0)}:(3) + z(1)h(2) + z(2)h(1) + 3(3)}:(0) = 18 z(0)h(4) + =(1)h(3) + x(2)h(2) + x(3)h(1) + z(4)h(0) =14 z(0)h(5) + :c(1)h(4) + 3(2):;(3) + =‘(3)h(2) + 2(4):;(1) + 2(5)::(0) = 10 - z(1)h(5) + z(2)h(4) + 2(3)::(2) = s 3(2)};(5) + 3(3)}:(4) = 3 ' £(3)h(5) = 1 0,1: 2 9 Ha an— {g, 11, 15,18,14,10, 6, 3, 1} 22 . TI? _ y(n) = {6,11,15,1r8,14,10,6,3,1} r) By following the same procedure as in (a), we obtain 310'!) = 2,2,1} I (d) By following the Same procedure as in-(a), we obtain y(n) = {%,2,2,2,1} 2.24 (3) 60¢) = 50'! - k) = 2(3) = 201) = 2(a) = Thus, C; = (b) (c) 703) - MO: —— I) and, 7(73 - JE) — 670: — x.» .. 1). Then, 2 a:(k)6(n - 3:) *=-oo him =(k)[7(n — k) .. “(a _ k __1)] It; 3(k)7(n'._ k) _ a: 2mm * k _~ 1) gm sum ~ k) _ a: :0» - mo: — k) i [am - azfl: —1)}7(n _'L.) azfl' -— 1) Mn) = T[z(n)] = 77 Z “flu-'le k=—og oo 2 ctT[7(n e— H] kB-um 00 2 mo: -. k) k=-m I! ll Mn) 79502)] 1171'?!) - 070'; —— 1)] flm~ww—n II II II 2.30 (a) 9(a) -- 0.6y(n -- 1) + 0.08y(n -- 2) = 2(a). I The characteristic equation is A3 — 0.6A + 0.08 = o. A = 0.2, 0.4 Hence, 1” 2” 9M“) — 61-5- 4-62-5- - With 3(a) = 5(n), the initial conditions are y(0) = 1, I y(1) - {169(0) = 0 =:r y(1) = 0.6. Hence, c1 + c; = 1 and 1 2 gc1+-5- = 0.6:c1=—1,c2=3. H 5' 9”: .O a” A 3 v II [4?" + 2(3)“] um) The step response is 7] s01) = 2310: -—k),n 2 o 1': )3 [mgr-i ~ (§)"-*] i=0 {€13[(gn-thjfifig[(%n+1_1]}u(n) y(n) — 0.7y(n - 1) + 0.ly(n — 2) = 230:) - 3(11 - 2). The characteristic equation is H (b) A2 — 0.7A + 0.1 = o. A 2 %,§- Hence, . I“ 1" 3M”) = 61-2- + £23 - With 3(a) = 6(a), we have 33(0) _= 2r . y(1) —— 0.7y(0) = 0 => ya) = 1.4. Hence,c1 + c; z 2am] I 1 7 '2—61 + *5- -— 1.4 -— E 2 14 2} C; + 3-03 - These equations yield 6 _ £9 6 -es 1 — 3 ) 2 —' I _ 10 1 n _ :1- }. n nun-UT- k=0 10 “ 1 4 “ 1 = — (*)““*--Z(“)“"‘ 3 kg) 2 3 i=0 5 10 1 " k 4 1 " k = —(-)"):2 ——<—)"):5 3 2 k=0 3 5 i=0 10 1" - 1 1" = 515 (W - 1)u<n)— 5(3 (5W — mm 2.47 901) = 0.9y(n -- 1) + 3(a) + 2::(1: — 1) + 32:0: - 2) (a)For :03) = 6(a), we haye 9(0) = 1: 3(1) = 2.9, y(2) = 5.61, y(3) .—. 5.049, y(4) = 4.544, y(5) = 4.090,... (b) 8(0) = y(0)=l. 8(1) = y(0)+y(1)=3.91 8(2) = 1:60 + W) + 9(2) = 9-51 3(3) = 3(0) + 1(1) + y(2) + 9(3) = 14-56 - 4 5(4) = Zy(n)=19.10 5 3(5) = 2909:2319 0 (c) Mn) 8(a) + 2.960: — 1) + 5.61(o.9)"‘1u(n — 2) (U.9)"u(n) + 2(0.9)"“1u(n — 1) + 3(o.9)""u(n — 2) 2.52 Obviously, the length of His) is 2, Le. Mn) = {hmhl} ho —— 3% +h1 : ’10 = 1, ’31 r: 1 I! rill—- 2.55 From problem 2.54, Mn) = [1312" + cgn2“} u(n) With y(0) = 1, y(1) = 3, we have c1 = 1 261 + 203 3 - 1 l => ca — '2' . ‘ Thus Mn) '- [2" + $1124 110:) ______ _________ ...
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This note was uploaded on 02/26/2012 for the course EE 4150 taught by Professor Wu during the Fall '10 term at LSU.

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EE4150_2003_hw3sol - (a) (3) (4) (5) (5) I (7) ‘2 4-K H V...

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