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20062ee113_1_Spring%202006%20EE%20113%20Problem%20HW%202%20solution_051406

# 20062ee113_1_Spring%202006%20EE%20113%20Problem%20HW%202%20solution_051406

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Unformatted text preview: BB 113 Problem #2 Solution (Due: 4/28/06) Problem 1. a) y(n) = cos[(x(n)+1)l It is static and causal. It is not linear. 005(0) is nonlinear function. It is time-invariant. T (x(n — k» = cos(x(n — k)+ 1) = y(n — k) = cos(x(n — k)+1). It is BIBO stable. |cos(0)| S 1, if lx(n)| S M < oo 2 |y(n)l S1 < 00. b) y(n) = x(n) It is static and causal. It is linear. T (ax1 (n) + bx2 (11)) = ax1(n)+ bx2 (n) = ay1(n)+ by2 (n) . It is time—invariant. T (x(n — k» = x(n — k) = y(n —— k). It is also BIBO stable. ]y(n)' = |x(n)| S M < 00. 0) W) = |x(n)| It is static and causal. It is not linear. H is non—linear function. It is time-invariant. T (x(n — k» = |x(n — k] : y(n — k). It is BIBO stable. if Mn] S M < oo 2 |y(n)[ S M < 00. d) y(n) = (n ~1)x(n —— l) + n It is not static. y(n) depends on x(n — 1). It is non—linear. T(ax1(n)+ 5x2 (11)) = (n — l)(axl (n ~1)+ bx2 (n -l))+ 11 \$01 ~l)a.x1(n —l)+ (n ——1)bx2(n —l)+ 2n It is not time-invariant. T(x(n-—k))=(n~1)x(n—-k—1)+n¢(n—~k—l)x(n——k—l)+n—k It is causal. It is not BIBO stable. although Mn) S M < oo , but as n ——) oo 2 |y(n)| ——> 00 Problem 2. a) y(n)= fame—k) k=—m = ;(1/4)*(1/2)"-k =(1/2)"§(2/4V =(1/2)"11‘_(1é12/)2")” :[(1/2).-._(1/2)..]_u(n) b) y(n) = kZX(k)h(n — k) : kick/5y u(k)~— (1/4)k u(k __ 5)] (1/2)n—k u(n __ k) = gunk (1/2)” gem (1/2)“ = (1/2)“ 2 (2/5)]: +(1/2)";(2/4)k = [(5/3) (1/ 2)" -(2/3)-(1/5)”J~u(n)~4"5 [20/2715 -(1/ 2)2"”1°J-u(n —5) Problem 3. a) y(n)— :- y(n — 2) 2 x(n —— 1), and assume zero initial condition. i.e. y(n) = 0, Vn < 0 First, ﬁnd the impulse response h(n) Let x(n) = 6(11), then the system becomes y(n)~ %y(n — 2) = 6(n ~ 1) which is unforced for n > 1. i.e. y(n)— %y(n —- 2) = 0, Vn >1 To ﬁnd y(n), for n 2 0 , we need yh (n), y(0) and y(l) The characteristic equation for y(n)— % y(n —- 2): 0 is 22,2 —1 = 0 which has 2 distinct roots Al = \/1/12 ,/12 = ”M . Therefore yh (n) 2 01 (My. + c2 (— x/T/E)‘ . From y(n)~%y(n—2)= 6(11—1), we can derive y(0)= 0 and y(1)=l y(0)=c +c2=0 _> _ __ {y(l)=(\/l/—2)-(cl—~cz)=l "‘ Q-Mﬂz— M So, h(n)= [c1 (JD—2y +c2 (— M), ]u(n)= “My” +(—\/1/‘2)ﬂ+1]-u(n). Next, ﬁnd step response S(n) The step function u(n) can be represented u(n) = 5(n)+ 6(n ~1)+ 6(n — 2) + ........ Apply linearity and time—invariant property, step response s(n) can be derived S(”)= ihGI—k): gﬁﬁrk” +(~/\/1—§Y~k+1] (M 2m ( n 2( m l/ﬂ— 1/ MH b) y(n) =2 \$3201 — 1)— ﬁﬂn — 2)+ x(n) , with zero initial condition. i.e. y(n) = 0, Vn < 0 First, ﬁnd the impulse response h(n) Let x(n) = 6(n), then the system becomes y(n) = % y(n — l)— 1—16—y(n ~e 2)+ 6(11) which is unforced for n > O. i.e. y(n)—%y(n —l)+%y(n ~2)= O,Vn > 0 To ﬁnd y(n), Where n 2 0 , we need yh (n), y(0) and y(— 1) The characteristic equation for y(n) — % y(n — 1) + 116— y(n — 2) = 0 is (4/1 — l)2 = 0 which has 2 identical roots 1,1 = \$12 = 211—. Therefore yh (n) = c1 (1/ 4)" + 02 -n - (1/4)". From y(n) = % y(n — 1)— %y(n — 2)+ 6(n) , we can derive y(0) = l and y(— 1) = 0 { no): c1 =1 y(—1)=(1/4)-(c1-c)=o So, h(n)= 01(1/4)" +c2 -n.(1/4)” = (n+1)-(1/4)"u(n) => c1=1,c2 :1 Next, ﬁnd step response s(n) Let x(n) = u(n), then the system becomes y(n) = % y(n — 1) — % y(n —- 2)+ u(n) The complete solution has the form y(n) = yp (n)+ yh (n) y p (n) = ku(n), Plug x(n), y p (n) into difference equation, ekiln):é.k.u(n_1)_f.6u.k.u(n_2)+u(n) To solve k, n should be picked such that n 2 2. ::> k = 199. Therefore, the particular solution y p (n) = l9?— - u(n) Because y k (n) 2 01 (1/4)" + 02 -n - (1/ 4)" was previously derived, therefore y(n) 2 c1 (1/ 4)" + 02 'n (1/ 4)" + %u(n). Because n 2 2 was used to calculate k , we need y(0) and y(1) to solve cpc2 y(0), y(1) can be derived from y(n) = % y(n — 1)— % y(n —- 2)+ u(n) with zero initial condition, i.e. y(—~ l) = 0 , y(— 2) = O y(o>=—;—y<—1)-fgy(—2)+u(o)=1 y(1)=§y(o)-%y(—1)+um=§_ ~7 ~1 :> cl = ?,c2 _. ? s(n)= ((7—9?) (1/4)" +6) . ”(1/4)" +199).u(n) Problem 4. a) The characteristic equation is 10/12 — 7 /1 +1 = 0 The roots are £1 =1/5,/12 = 1/2 b) The homogeneous solution yh (n) = (cl (1/ 5)” + c2 (1/ 2)" ) u(n) c) x(n) = (1/3)" u(n),.'. y p (n) = k - (1/3)" u(n) . Plug x(n), y p (11) into difference eq, :3 k- (1/3)"u(n)~——1-76-k .(1/3)"-1u(n ~1)+ % . k . (1/3)"~2u(n _ 2) = (1/3)"u(n) To solve k , 71 should be picked such that n 2 2 . :> k = —~5. Therefore, the particular solution y P (n) 2 —5 - (1/ 3)" u(n) d) The complete solution has the form y(n) = y h (n)+ y P (11). Based on b) and c) y(n)= c1(1/5)" +cz(1/2)" 411/3)" -u(n) n 2 2 has been used to calculate k , we need y(0), y(1) to solve c1 and 02 32(0), y(1) can be derived from y(n)— 126 y(n — 1)+ 1—16 y(n —' 2) = (1/ 3)" u(n) with the initial conditions y(— 1) = 1 , y(— 2) = 0 7 1 17 y(0)=16y(‘1)“16y(-2)+1=T6=c1+02—5 7 1 1 427 c c 5 y() 10y(0)10y( )+3 300 5+2 3 01:13/15,cz = 35/6 . . 13 n 35 n It So the complete solutlon IS y(n) = (B (1/5) +?(1/ 2) ~ 5 - (1/3) J u(n) Problem 5. a) X(Z) 2501 1)Z" :2-1 ROC: [z|¢o b) X(Z)= =:30; u(n)+/3" u(—n— 1))2 =Z(aZ 7+ 32” =Z(aZ“T+n§:(ﬁ“ZY n=0 _ 1 ,6”1Z a Z + Z 1—02"1 1342 2—05 ,B—Z’ ROC: lal < 12} < [A c) x(zi((n+1)(_1) u(n))Z =z(zz)~1)+zn(~1) 2— ~00 : Z4 °° 1 Z 1 1+ZZ“ =ZO(Z Y Z 32(sz )WJ=1+Z‘I+(1+Z‘1)2:(1+Z")2 ROC: [Z[>1 Problem 6. a) x1(n)= (1/4)" u(n) X(Z)=Z(1/4)"u(n)Z'"241/42)": Z Z1/4 ROC. {2| >1/4 b) x2(n)=x1(n—3) _ X2(Z)=Z_3 ‘X1(Z)= ZZ_1/4 ROC: ]Z| >1/4 c) x(n)= (1/2)"u(n)+ (1/2)"‘u(— n) X(Z)=;(1/2)"Z‘"+”;O(1/2)_"Z‘” Z—(l/ZZ)" +Z(Z/2)" =27—22T2— E ROC:%<‘Z‘<2 Problem 7. A a) A = 5.3284 forN = 4 A b) A =5.3284 forN= 8 A c) A = 5.3284 for N = 4000 Unlike HW#1, the errors introduced by rouding repeat every period of the signal. And since we are selecting the same samples of each period, the errors did not average to zero in this case. Problem 8. ( Sayed 3.8 ) For sin(wn) to be periodic, if and only if sin(wn)= sin(w(n + kN» Vk,N : integer ie. w-N-k = 27r-l where all k,N,l: integer 2 N - k . . . 2) l = — Whlch is a ratlonal number. W Problem 9. ( Sayed 4.9) Solution #1: If lain-kl SM <oo,then ; y(")1 = game, _ mi g 214%” - k] g 2110.14.41. _ k)‘ s M . gym.) 11(11): (1/2)” 2) I y(n)1 S 1—1? = 2M < co, .'.the system is BIBO stable _ 2 Solution #2: 1102): 1 1—0524’ , inside unit circle. Therefore the system is BIBO stable. so this system only has single pole at Z = 0.5 which is Problem 10. ( Sayed 5.1 ) x(n) = 5(2n ~ 4): 5(n - 2):: 12(11): h(n ~ 2). Problem 11. ( Sayed 7.1 ) a) Modes of the system = roots of the characteristic equation 1223—822 —2+1=0:>2=%9y,~% b) Initially relaxed and causal means y(n) = O,Vn < 0 For x(n) = 501), the system y(n)~ gye _1)_ 132%, _ 2)+ éye _ 3) = (y(n) is unforced Vn > 0 . T o solve y(n) Vn 2 O , we need we need Mn), y(0)= y(-1) and y(— 2) From a), y(n): yh(n)= 01(é)" +c2 41(1)” +641)” 3 2 3 y(— 2): 0 = 4c1 ~8c2 +903 y(—l)= 0 = 2c1 «2c2 ~3c3 y(O):1:cl +03 21 3 4 3C1='2—5,L’2 =‘5‘,C3 =§§ TheimpuiseW h(n)=[21 [1)" 4341(1)" +1.(:1]"].u(n). 2325 2253 2 1 1 ... _____ _ .___ *2 _ —- : y(n) 3w 1) 12w )+12y(n 3) An) 1 .-.H(Z)=W 1—3‘Z*1—iz‘2 +—1~Z"3 3 12 12 Because the poles for this system are Z = %’ y f 13 , which are all inside the unit circle, so it is stable. 0) Zero input response is the homogeneous solution with the constant determined by initial conditions. Now, given y(— 3) = y(—— 2) = 0, y(~— l) = l and y(n):yh(n): 018:] +02 Hn[%j +c3[%1) , we can solve 01,02,63 y(— l): l = 201 — 202 - 303 y(— 2) = o = 401—802 +9c3 y(_. 3) = 0 = 801 - 24c2 ~— 2703 l8 3 —~4 m1 WE We =75 The zero input response is 18 1 " 3 1 " 4 —1 " ‘ = . + .n- - . . ”'01) [25 [2) 10 (2) 75 (3) i ”(n) Problem 12. ( Sayed 8.2 ) a) The complete solution for the system is: y1(n)= y p (n)+ yh (11) Form the characteristic equation, 1223’ — 8/12 - 2, +1 2 0 :> .2. = % y, "% yh(n)= (g) WE] (3;) For x(n)= u(n), the particular solution y p (n) has the form y p (n) = k ~ u(n) Plug y p (n) and x(n) into the original difference equation, we get k-u(n)—§k-u(n—l)——il—2—k‘u(n~2)+il—2—ku(n—3)=u(n) To solve k, n should be picked such that n23 => k= 3 1 " 1 " —1 ” y1(n)=cl(§) +cz~n-(—2—) +c3[—3—] +3u(n) forn>2 To ﬁnd 51,62,639 we need y1(”)>y1(2)>.}’1(1)9y1(0) The zero state response yzs (n) is the complete solution yl (n) of an initially relaxed system, i.e. y1 (~ 3): y1(- 2) = y;(—- 1): 0 - -.~y(n)7§y(n*1)—1—12—y(n~2)+§y(n~3)=utn)and yl(~ 3): y,(—- 2): y1 (~ 1) =10 are given y1(2),y1_(l),y1(0) can be deriged y<o)= §y(~1)+§2—y<—2)—~115y<—3)+u(o)=1 y<1>=§y<o>+\$ye1>~1gyewe):g 79 y<2>=§y<1>+igy<o>~lgy<4>+u<zp3_6 Now we can calculated cpcz,c3 From y(0)=1=cl+c3+3 5 c c c 1=_=_1_ .2"..§_+3 3’0 3 2+2 3 79 c c c 2 =—=—‘+—2—+~3—+3 y() 36 4 2 9 —51 __—_3: 1 => H3; 5 “3‘5; yarn—55.11;)"raw-W =[B3g3—JGJ"+(-I—§J-n‘(—;-J"+(;—:J£%J"W") b) y(n)=yzs(n)+yzi(n) Where yzi(n) is ﬁom prob. 11. Transient response y,, (n) = [[H 33X 1 j +[__ 3 :31) +3].u(n) 25 2 10 Steady State response y” (n) == 3 d) If x(n) = u(n)— u(n — k), apply LTI property J -n-(:J"+(;;J(:ﬂ y<n>=[£:§J[;J"+£a—§)[email protected]"+(—;%J(:;J"+3]u<n>+ U%[email protected]""‘+(a—3—J-<n~k>(::-J”reef”+3Ju<n~k> ...
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20062ee113_1_Spring%202006%20EE%20113%20Problem%20HW%202%20solution_051406

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