(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