《信号与系统——连续时域分析运算》教学资源：Chapter 2 SOLUTIONS MANUAL to accompany

2.15 ∑x2lm=∑(xsm]+xlmn)2 ∑x2 x2an]+2∑ kevIn]xod[n]=∑xvn]+∑x3dl D=-oo as 2ns- Xevin )odiN =0 since Xev In xoaIn is an odd sequence. 2.16xn]=cos(2πkn/N) 0≤n≤N-1. Hence, E、=∑c2nkn/N=∑ n 1 (1+cos(4rkn/N))=3+3>cos(4Ikn/N) Letc=∑cos(4πkn/N,ands=∑sin(4xkn/N n=0 =0 Therefore C+js ∑ 4]kn/N =O. Thus, C=ReC+jS =0 C=Re C+js=0,it E=N 217(a)xn=n]. Energy=∑u2m=∑ Average power=,m、1ns ∑n=kmn32=m1 K 1 K→∞2K+ K2K+12 (b)x2nl=叫 u[n]. Energy=∑n=c(mpm)=∑a=n k-yoo 2K+12n=K(n(n )2=lim1 yk Average power= lim I K→∞2K+1 (e)xin]=Aoejoan. Energy Jn- oeA t Ens_JAo =oo. Average power lim K lim K→∞2K+1 K→∞2K+1 K Aol K→∞2K+1 (d) xn]=Asi M+o=Aoeo+Aye joon where Oo M ?cie and A=A -jo From Part(c), energy = oo and average power= A6+Af+4A2A2-3 218Now叫sJ1,n≥0 1,n<0, Hence,μ-n-1l l0.n≥o. Thus, x[n=山m]+μ-n-1 2.19(a)Consider a sequence defined by x[n]=>8[k]

7 2.15 xn x n x n n ev od n 2 2 [] [] [] =−∞ ∞ =−∞ ∞ ∑ ∑ = + ( ) =++ =+ =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑∑ ∑ ∑∑ x n x n x nx n x n x n ev n od n ev od n ev n od n 22 22 [] [] [] [] [] [] 2 as n= ∞ x nx n ev od ∞ ∑ = – [] [] 0 since x n x n ev od [ ] [ ] is an odd sequence. 2.16 x n kn N [ ] cos( / ), = ≤≤ 2π 0 n N –1. Hence, E kn N x n N = = − ∑cos ( / ) 2 0 1 2π = + ( ) = − ∑ 1 2 0 1 1 4 cos( / ) πkn N n N = + = − ∑ N n N kn N 2 1 2 0 1 cos( / ). 4π Let C kn N n N = = − ∑cos( / ), 4 0 1 π and S kn N n N = = − ∑sin( / ). 4 0 1 π Therefore C jS ej kn N n N + = = − ∑ 4 0 1 π / = − − = e e j k j kN 4 4 1 1 0 π π / . Thus, C C jS = + Re 0 . { } = As C C jS = + Re 0 { } = , it follows that Ex N = 2 . 2.17 (a) xn n 1[ ] [ ]. = µ Energy = µ2 2 [] . n 1 n n =−∞ ∞ =−∞ ∞ ∑ ∑= =∞ Average power = lim ( [ ]) lim lim . K n K K K n K K K n K K →∞ + =− →∞ = →∞ K = + = + ∑ ∑ = 1 2 1 1 2 1 1 2 1 1 2 2 2 0 µ (b) xn nn 2[ ] [ ]. = µ Energy = n n n n n ( µ[] . ) = =∞ =−∞ ∞ =−∞ ∞ ∑ ∑ 2 2 Aveerage power = lim ( [ ]) lim . K n K K K n K K n n K n →∞ =− →∞ = + = + ∑ ∑ = ∞ 1 2 1 1 2 1 2 2 0 µ (c) x n Aeo j no 3[] . = ω Energy = Ae A o j n n o n ωo =−∞ ∞ =−∞ ∞ ∫ = =∞ ∑ 2 2 . Average power = lim lim lim . K o j n n K K K o n K K K o o K A e K A K A K o A →∞ =− →∞ + =− →∞ = + = + ∑ ∑ = 1 2 1 1 2 1 2 2 1 2 2 2 ω 2 (d) xn A n M A e Ae o jn jn o o [ ] sin , = π +     = + 2 − φ 1 ω ω where ωo M= 2π , A A e o j = − 2 φ and A A e j 1 2 = − φ. From Part (c), energy = ∞ and average power = A A AA A o o 2 1 2 2 1 2 2 4 3 4 ++ = . 2.18 Now, µ[ ] , , , . n n n = ≥ <    1 0 0 0 Hence, µ[ ] , , , . −− = < ≥    n n n 1 1 0 0 0 Thus, x[n] = µ µ [ ] [ ]. n n + −− 1 2.19 (a) Consider a sequence defined by x n k k n [ ] [ ]. = =−∞ ∑δ

If n <0 then k=0 is not included in the sum and hence xn=0, whereas for n>0, k=0 is included in the sum hence x[n]=1 V n20. Thus x[n]= >8Ik]= 1,n≥0 l0.n<0 u[n] n≥0. (b) Since u[n] it follows that u[n-l n< n≤0 Hene,-p-={0.m≠0=an 2.20 Now x[n]=Asin(@on+o) (a)Given x[n]=0 -12-2-V2 0 v2 2 v2. The fundamental period is N=4 hence o=2兀/8=元/4. Next from x0=Asin(φ)=0 we get o=0, and solving x[l]=Asin(+o)=Asin(/4)=-v2 we get A=-2 (b) Given xn The fundamental period is n= 4, hence ①0=2T4=π/2. Next from x[O]=Asin()=√2 and x[1]=Asin(π/2+9)=Acos()=√2it can be seen that A=2 and o= T/4 is one solution satisfying the above equations (c)x[n]=3 -3. Here the fundamental period is N= 2, hence Oo=T. Next from x[O] Asin(o)=3 and x[l]= Asin(o+I)=-Asin(o)=-3 observe that A=3 and o=t/2 that A=3 and o= T/2 is one solution satisfying these two equations (d)Given x[n]=0 1.5 0-1.5, it follows that the fundamental period of x[n] isN ence @0=2T/4=T/2. Solving x[0]=Asin(o)=0 we get 9=0, and solving x[l=Asin (T/2)=1.5, we get A= 1.5 2.21(a)xI[n]=e Ju 4n. Here, 0=0.4T. From Eq (2.44a), we thus get N- 2rr 2Ir =5r=5 forr=l 0.4兀 (b)x2[n]=sin(0.6n+0.6T). Here, Oo=0.6T. From Eq(2.44a), we thus get 2兀r2πr10 =-r=10 forr= 3 0.6兀3 (c)x3[n]=2cos(1.Intn-05)+2sin(O.7Tn). Here, @1=l.lT and 2=0.7T. From Ec (2. 44a), we thus get Ni ar In 20 2πr2兀 r?. To be periodic l兀11 rI and n. 020.7兀

8 If n < 0 then k = 0 is not included in the sum and hence x[n] = 0, whereas for n ≥ 0 , k = 0 is included in the sum hence x[n] = 1 ∀ ≥ n0. Thus x n k n n n k n [] [] , , , , = = [ ]. ≥ <    = =−∞ ∑δ µ 1 0 0 0 (b) Since µ[ ] , , , , n n n = ≥ <    1 0 0 0 it follows that µ[ –] , , , . n n n 1 1 1 0 0 = ≥ ≤    Hence, µµ δ [n] – [ ] , , , , n [ ]. n n − = n = { ≠ 1 = 1 0 0 0 2.20 Now xn A n [ ] sin = + (ω φ 0 ). (a) Given x n[ ] = − −− { } 0 2 2 2 0 2 2 2 . The fundamental period is N = 4, hence ωo = π =π 28 4 / / . Next from x A [ ] sin( ) 0 0 = = φ we get φ = 0, and solving xA A [ ] sin( ) sin( / ) 1 4 = 4 2 π + = π =− φ we get A = –2. (b) Given x n[ ] = −− { } 22 2 2 . The fundamental period is N = 4, hence ω0 = 2π/4 = π/2. Next from x A [ ] sin( ) 0 2 = = φ and xA A [ ] sin( / ) cos( ) 12 2 = += = πφ φ it can be seen that A = 2 and = /4 is one solution satisfying the above equations. φ π (c) x n[ ] = − { } 3 3 . Here the fundamental period is N = 2, hence ω π 0 = . Next from x[0] = Asin( ) φ = 3 and x A A [ ] sin( ) sin( ) 1 3 = + =− =− φπ φ observe that A = 3 and = /2 φ π that A = 3 and φ = π/2 is one solution satisfying these two equations. (d) Given x n[] . . = − { } 0 15 0 15 , it follows that the fundamental period of x[n] is N = 4. Hence ωπ π 0 = = 24 2 / / . Solving x A [ ] sin( ) 0 0 = = φ we get φ = 0, and solving x A [ ] sin( / ) . 1 2 15 = = π , we get A = 1.5. 2.21 (a) ˜ [] . . xn e j n 1 0 4 = − π Here, ωo = π 0 4. . From Eq. (2.44a), we thus get N r r r o = π = π π = = 2 2 0 4 5 5 ω . for r =1. (b) xn n ˜ [ ] sin( . . ). 2 = π+ π 06 06 Here, ωo = π 0 6. . From Eq. (2.44a), we thus get N r r r o = π = π π = = 2 2 0 6 10 3 10 ω . for r = 3. (c) xn n n ˜ [ ] cos( . . ) sin( . ). 3 = π − π+ π 2 11 05 2 07 Here, ω1 = π 1 1. and ω2 = π 0 7. . From Eq. (2.44a), we thus get N r r r 1 1 1 1 1 2 2 1 1 20 11 = π = π π = ω . and N r r r 2 2 2 2 2 2 2 0 7 20 7 = π = π π = ω . . To be periodic

we must have Ni=n2. This implies, 111=2. This equality holds for r=ll and r,=7 and hence n= ni= n2=20 2πn120 2 20 (d)N1 1.3丌13 n and N 0.3I 32. It follows from the results of Part(c), N=20 with n=13 and r2=3 (e)N,、2 1.2π3 0.8I22 and N3=N2. Therefore, N=N=N2=N2=5 for 3 and (fx6n]=n modulo 6. Since x6[n+6]=(n+6) modulo 6= n modulo 6=xgn Hence n= 6 is the fundamental period of the sequence xIn 2.22(a)Oo=0.14T. Therefore, N 2r2πr100 r=100 for r=7 0.14 (b)(o=0.24T. Therefore, N 一r=2iorr 0。0.24兀3 (c)0o=0.34T. Therefore, N 2Tr2Tr =100 forr=17 0。0.34兀17 (d)oo=0.75兀. Therefore,Ns2rr2兀r8r=8for=3 ①0.75兀3 2.23 x[n]=xa(nT)=cos(SonT) is a periodic sequence for all values of T satisfying SoT.N=2TI for r and n taking integer values. Since, Bot=2r /N and r/N is a rational number, BoT must also be rational number. For Qo=18 and T=T/6, we get N= 2r/3. Hence, the smallest value of n=3 forr= 3 224(a)xn]=3n+3]-28n+2]+n]+4n-1+5n-2]+2m-3] (b)yn]=78n+2]+8n+1-38n]+48n-1]+98n-2]-28n-3] (c)wn]=-58n+2]+4δ[n+2]+36[n+1]+6δn]-5δn-1+n-3] 2.25 (a) For an input xi[n], i= l, 2, the output is iln]=axi[n]+axin-1]+aixi [ n-2]axI[n-4], for i= 1, 2. Then, for an input x[n]=AX1[n]+Bx2[n], the output is yln]=a,(Ax1[n]+Bx2[n])+a2 (AX1[n-1]+BX2(n-1) +a3(Ax1{n-2]+Bx2n-2)+a4(Axn-3]+Bx2n-3) A(aix[n]+a2X1[n-1]+a3x1[n-2]+a4X1[n-4]) +B(a1x2m]+a2x2n-11+x3x2n-2]+a4x2{n-4])=Ay1n+By2n Hence, the system is linear

9 we must have N N 1 2 = . This implies, 20 11 20 7 1 2 r r = . This equality holds for r1 = 11 and r2 = 7, and hence N = N N 1 2 = = 20. (d) N r r 1 1 1 2 1 3 20 13 = π π = . and N r r 2 2 2 2 0 3 20 3 = π π = . . It follows from the results of Part (c), N = 20 with r1 = 13 and r2 = 3. (e) N r r 1 1 1 2 1 2 5 3 = π π = . , N r r 2 2 2 2 0 8 5 2 = π π = . and N N 3 2 = . Therefore, N N N N == = = 122 5 for r 1 = 3 and r2 = 2. (f) x n ˜ [ ] 6 =n modulo 6. Since x n ˜ [ ] 6 + = 6 (n+6) modulo 6 = n modulo 6 = x n ˜ [ ] 6 . Hence N = 6 is the fundamental period of the sequence x n ˜ [ ] 6 . 2.22 (a) ωo = π 0 14 . . Therefore, N r r r o = π = π π = = 2 2 0 14 100 7 100 ω . for r = 7. (b) ωo = π 0 24 . . Therefore, N r r r o = π = π π = = 2 2 0 24 25 3 25 ω . for r = 3. (c) ωo = π 0 34 . . Therefore, N r r r o = π = π π = = 2 2 0 34 100 17 100 ω . for r = 17. (d) ωo = π 0 75 . . Therefore, N r r r o = π = π π = = 2 2 0 75 8 3 8 ω . for r = 3. 2.23 x n x nT nT a o [ ] ( ) cos( ) = =Ω is a periodic sequence for all values of T satisfying ΩoTN r ⋅ =π2 for r and N taking integer values. Since, ΩoT rN = π2 / and r/N is a rational number, ΩoT must also be rational number. For Ωo = 18 and T = π/6, we get N = 2r/3. Hence, the smallest value of N = 3 for r = 3. 2.24 (a) xn n n n n n n [] [ ] [ ] [] [ ] [ ] [ ] = +− + + + −+ − + − 3 32 2 4 15 22 3 δ δ δδ δ δ (b) yn n n n n n n [] [ ] [ ] [] [ ] [ ] [ ] = + + +− + −+ − − − 7 2 13 4 19 22 3 δ δ δδ δ δ (c) wn n n n n n n [] [ ] [ ] [ ] [] [ ] [ ] =− + + + + + + − − + − 5 24 23 16 5 1 3 δ δ δ δδ δ 2.25 (a) For an input x n i[ ], i = 1, 2, the output is yn xn xn xn xn ii i i i [ ] [ ] [ ] [ ] [ ], = + −+ − + − αα α α 12 3 4 1 2 4 for i = 1, 2. Then, for an input x n Ax n Bx n [ ] [ ] [ ], = + 1 2 the output is y n A x n Bx n A x n Bx n [] [] [] [ ] [ ] = + α α 11 2 21 2 ( ) + −+ − ( 1 1 ) + −+ − α α 31 2 41 2 (Ax n Bx n Ax n Bx n [ ] [ ] [ ] [ ] 22 33 ) + −+ − ( ) = + −+ − + − A xn xn xn xn (αα α α 11 21 31 41 [ ] [ ] [ ] [ ] 12 4 ) + + −+ − + − B xn xn xn xn (αα α α 12 22 32 42 [ ] [ ] [ ] [ ] 12 4 ) = + Ay n By n 1 2 [ ] [ ]. Hence, the system is linear

(b) For an input xi[n ,i=l, 2, the output is yin]=boxi[n]+bixi[n-1]+baXi[n-2]+ayi[n-1]+ayi[n-2], i= 1, 2. Then, for an input x[n]=AXI[n]+ Bx2In], the output is y[n]=A(boxI[n]+biXI[n-1]+b2X1[n-2]+aiyI[n-1]+a2y1[n-2]) h-b(box21nJ+ bX2[n-1+b2x2In-2]+a1y2[n-1]+a2y2In-20)=Ayi[n]+By2In] ce, the system is linear (c)For an input xiIn], i=1, 2, the output is yin *i[n/L],n=0,tL,+2L,K Consider the input x[n]=AXI[n]+ Bx2[n], Then the output yIn] for n=0, tL, +2L,K given by yIn]= x[n/l]=AXIn/L]+ Bx2In/L]=Ayin]+By2ln]. For all other values of n, yIn]=A.0+B-0=0. Hence, the system is linear (d) For an input xi[n],i= l, 2, the output is yi[n]=xi[n/M]. Consider the input xn]=Ax,In]+ Bx2In], Then the output yIn]=AXIn/M]+Bx2ln/M=AyIIn]+ By2ln] Hence, the system is linear (e) For an input xi[n],i= l, 2, the output is yi[n] M2k=0XiLn-k. Then, for an input x(n =Ax[n]+Bx2[], the output is yi[n]=LEM-(Ax[n-k]+Bx2In-kJ) lica m 2k=o xi[n-K+M2k=0 x2In-k1=Ayi[n]+By2[n]. Hence, the system is (f) The first term on the RHs of Eq(2.58)is the output of an up-sampler. The second term on the rhs of Eq(2.58) is simply the output of an unit delay followed by an up- ereas he third term is the output of an unit adavance operator followed by an up-sampler We have shown in Part(c)that the up-sampler is a linear system. Moreover, the delay and the advance operators are linear systems. A cascade of two linear systems is linear and a linear combination of linear systems is also linear. Hence, the factor-of-2 interpolator is a linear system. (g) Following thearguments given in Part (f), it can be shown that the factor-of-3 interpolator is a linear system 226(a)ym]=n2x[n For an input xin] the output is yin]=n[],i=l, 2. Thus, for an input x3[n]= AxIn +Bx2[], the output y3[n] is given by y3[n]=n(AxI[n]+Bx2[n])=AyI[n]+By2[n] Hence the system is linear. Since there is no output before the input hence the system is causal. However, ly[n] being proportional to n, it is possible that a bounded input can result in an unbounded output. Let x[n]=1yn, then yIn]=n2. Hence yIn]→∞asn→o, hence not bibo stable Let yIn] be the output for an input x[n], and let yIIn] be the output for an input x,[n]. If X,In]=xIn-noI then y,n]=nx,[n]=n*xIn-nol. However, yln-nol=(n-no)-xn-nol Since y,[n]+yIn-nol, the system is not time-invariant

10 (b) For an input x n i[ ], i = 1, 2, the output is y n b x n bx n b x n ay n a y n i ii i i i [ ] [ ] [ ] [ ] [ ] [ ], = + −+ − + −+ − 01 2 1 2 1 2 1 2 i = 1, 2. Then, for an input x n A x n Bx n [ ] [ ] [ ], = + 1 2 the output is yn A b x n bx n b x n ay n a y n [] [] [ ] [ ] [ ] [ ] = + −+ − + −+ − ( 01 11 21 11 21 1 2 1 2 ) + + −+ − + −+ − Bb x n bx n b x n ay n a y n ( 02 12 22 12 22 [ ] [ ] [ ] [ ] [ ] 1 21 2 ) = + Ay n By n 1 2 [ ] [ ]. Hence, the system is linear. (c) For an input x n i[ ], i = 1, 2, the output is y n xnL n L L otherwise i i [] , [ / ], , , , , , =  =± ±   0 2 0 K Consider the input x n A x n Bx n [ ] [ ] [ ], = + 1 2 Then the output y[n] for n LL =± ± 0 2 ,, , K is given by y[n] = x n L A x n L Bx n L A y n By n [ / ] [ / ] [ / ] [] [] = + =+ 1 2 12 . For all other values of n, y n A B [] . = ⋅+ ⋅= 0 0 0 Hence, the system is linear. (d) For an input x n i[ ], i = 1, 2, the output is y n x n M i i [] [ / ] = . Consider the input x n Ax n Bx n [ ] [ ] [ ], = + 1 2 Then the output y n A x n M Bx n M A y n By n [ ] [ / ] [ / ] [ ] [ ]. = + =+ 1 2 12 Hence, the system is linear. (e) For an input x n i[ ], i = 1, 2, the output is y n M i i xn k k M [] [ ] = − = − ∑1 0 1 . Then, for an input x n A x n Bx n [ ] [ ] [ ], = + 1 2 the output is y n M i Ax n k Bx n k k M [] [ ] [ ] = −+ − ( ) = − ∑1 1 2 0 1 = −     + −     = + = − = − A ∑ ∑ M xn k B M x n k Ay n By n k M k 1 1 M 1 0 1 2 0 1 1 2 [ ] [ ] [ ] [ ]. Hence, the system is linear. (f) The first term on the RHS of Eq. (2.58) is the output of an up-sampler. The second term on the RHS of Eq. (2.58) is simply the output of an unit delay followed by an up￾sampler, whereas, the third term is the output of an unit adavance operator followed by an up-sampler We have shown in Part (c) that the up-sampler is a linear system. Moreover, the delay and the advance operators are linear systems. A cascade of two linear systems is linear and a linear combination of linear systems is also linear. Hence, the factor-of-2 interpolator is a linear system. (g) Following thearguments given in Part (f), it can be shown that the factor-of-3 interpolator is a linear system. 2.26 (a) y[n] = n2x[n]. For an input xi[n] the output is yi[n] = n2xi[n], i = 1, 2. Thus, for an input x3[n] = Ax1[n] + Bx2[n], the output y3[n] is given by y n n A x n Bx n A y n By n 3 2 12 12 [] [] [] [] [] = + ( ) = + . Hence the system is linear. Since there is no output before the input hence the system is causal. However, y n[ ] being proportional to n, it is possible that a bounded input can result in an unbounded output. Let x[n] = 1 ∀ n , then y[n] = n2. Hence y n[ ] →∞ →∞ as n , hence not BIBO stable. Let y[n] be the output for an input x[n], and let y1[n] be the output for an input x1[n]. If x n xn n 1 0 [] [ ] = − then y n n x n n x n n 1 2 1 2 0 [] [] [ ] = =− . However, y n n n n x n n [ ]( )[ ] − =− − 0 0 2 0 . Since y n y n n 1 0 [] [ ] ≠ − , the system is not time-invariant

(b) yIn]=x"[n For an input xi[n] the output is yin]=xiIn],i=l, 2. Thus, for an input XiN]= Ax[n]+ Bx2In], the output y3In] is given by y3[n]=(AxI[n]+ Bx2In*A[n]+AxiN Hence the system is not linear Since there is no output before the input hence the system is causal Here, a bounded input produces bounded output hence the system is biBo stable too Let yIn] be the output for an input x[n], and let 1 [n] be the output for an input xi[n]. If X,n]=xn-noI then yin]=Xi[n]=x[n-no]=yln-noJ. Hence, the system is time- (c)yn]=+∑xn-1 1=0 For an input xi[n] the output is yi[n]=B+>[], i=1, 2. Thus, for an input x3[n] Ax,In]+ Bx2In], the output y3In] is given by 3 y叫]=B+∑(Ax1n-1+Bx2n-1=β+∑Axn-1+∑Bx2n-1 ≠Ay1n]+By2[n] Since阝≠0 hence the system is not linear Since there is no output before the input hence the system is causal Here, a bounded input produces bounded output hence the system is BIBo stable too Also following an analysis similar to that in part(a) it is easy to show that the system is time- Invariant (d)yn=B+∑xn-1 For an input xi[n] the output is yi[n]=B+>xi[n-1],i= 1,2. Thus, for an input x3[n] 1=-3 AxI[n]+ Bx2ln, the output y3ln] is given by y]=+∑(Ax1m-1+Bx2n-1)=B+∑Axn-1]+∑Bx2ln-1 *Ayi[n]+ By2[n]. Since p*0 hence the system is not linear Since there is output before the input hence the system is non-causal. Here, a bounded input produces bounded output hence the system is BIBO stable

11 (b) yn x n [] [] = 4 . For an input x1[n] the output is yi[n] = xi 4[n], i = 1, 2. Thus, for an input x3[n] = Ax1[n] + Bx2[n], the output y3[n] is given by y n A x n Bx n A x n A x n 3 12 4 4 1 4 4 2 4 [ ] ( [ ] [ ]) [ ] [ ] =+ ≠ + Hence the system is not linear. Since there is no output before the input hence the system is causal. Here, a bounded input produces bounded output hence the system is BIBO stable too. Let y[n] be the output for an input x[n], and let y1[n] be the output for an input x1[n]. If x n xn n 1 0 [] [ ] = − then y n x n x n n y n n 1 1 4 4 0 0 [ ] [ ] [ ] [ ]. = = −=− Hence, the system is time￾invariant. (c) yn xn [] [ ] =+ − = β ∑ l l 0 3 . For an input xi[n] the output is yn xn i i [] [ ] =+ − = β ∑ l l 0 3 , i = 1, 2. Thus, for an input x3[n] = Ax1[n] + Bx2[n], the output y3[n] is given by y n Ax n Bx n Ax n Bx n [] [ ] [ ] [ ] [ ] =+ − + − ( ) =+ − + − = == β β ∑ ∑∑ 1 2 0 3 1 0 3 2 0 3 ll l l l ll ≠ + Ay n By n 1 2 [ ] [ ]. Since β ≠ 0 hence the system is not linear. Since there is no output before the input hence the system is causal. Here, a bounded input produces bounded output hence the system is BIBO stable too. Also following an analysis similar to that in part (a) it is easy to show that the system is time￾invariant. (d) yn xn [] [ ] – =+ − = β ∑ l l 3 3 For an input xi[n] the output is yn xn i i [] [ ] – =+ − = β ∑ l l 3 3 , i = 1, 2. Thus, for an input x3[n] = Ax1[n] + Bx2[n], the output y3[n] is given by y n Ax n Bx n Ax n Bx n [] [ ] [ ] [ ] [ ] =+ − + − ( ) =+ − + − =− =− =− β β ∑ ∑∑ 1 2 3 3 1 3 3 2 3 3 ll l l l ll ≠ + Ay n By n 1 2 [ ] [ ]. Since β ≠ 0 hence the system is not linear. Since there is output before the input hence the system is non-causal. Here, a bounded input produces bounded output hence the system is BIBO stable