(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 stable11 (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