Let yIn] be the output for an input x[n], and let 1 [n] be the output for an input xi[n]. If x1n]=xn- nol then yiIn]=β+∑xn-1=β+∑x1n-no-1=yn-nol. Hence the 1=-3 system is time-invariant (e)yIn]=ax[-n] The system is linear, stable, non causal. Let yIn] be the output for an input x[n] and y,n be the output for an input x,[n]. Then yIn]=ax[-n] and yin]=ax1[-n]. Let x, In]=xn-nol, then yiIn]=ax1l-nj=axl-n-nol, whereas yIn -nol=ax[no-n] Hence the system is time-varying (fy[m]=x[n-5] The given system is linear, causal, stable and time-invariant 2.27yln]=x[n+11-2x[n]+x[n-1] Let y,ln] be the output for an input x,ln] and y2Ln] be the output for an input x,In]. Then for an input x3[n]=ax, [n]+Bx, In] the output y3[] is given by y3]=x3n+1-2x3n+x3{n- =0x1n+1-20xn]+0x1n-1+Bx2[n+1-2阝x2{n+阝x2{n-1 ay1[n]+βy2In Hence the system is linear. If x,[n]=xn-noI then y,n]=yIn-noJ. Hence the system is time-inavariant. Also the systems impulse response is given by n hIn]={1,n=1,-1, 0. elsewhere Since h[n]#0 n<o the system is non-causal 2.28 Median filtering is a nonlinear operation. Consider the following sequences as the input to a median filter: X,[n]=3,4,5]and x2In]=2, -2,-2). The corresponding outputs of the median filter are y,[n]=4 and y2ln]=-2 Now consider another input sequence xiN xInJ+ x2n]. Then the corresponding output is y3 In]=3, On the other hand, y,In]+y2[n]=2. Hence median filtering is not a linear operation. To test the time- invariance property, let x[n] and xiIn] be the two inputs to the system with correponding outputs yIn] and yi[n]. If x, [n]=xIn-noI ther 1m]= median(x1{n-k,……x1{n,……X1{n+k] median xn-k-nol……xn-nol……xn+k-nol=yn-nol Hence the system is time invariant12 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 yn n 1 1 3 3 1 0 3 3 0 [ ] [ ] [ ] [ ]. – – =+ − =+ − − = − = = β β ∑ ∑ l l l l Hence the system is time-invariant. (e) yn x n [] [ ] = − α The system is linear, stable, non causal. Let y[n] be the output for an input x[n] and y n 1 [ ] be the output for an input x n 1 [ ]. Then y n x n [] [ ] = − α and y n x n 1 1 [] [ ] = − α . Let x n x n n 1 0 [] [ ] = − , then y n x n x n n 11 0 [] [ ] [ ] = − = −− α α , whereas y n n x n n [ ][ ] −= − 0 0 α . Hence the system is time-varying. (f) y[n] = x[n – 5] The given system is linear, causal, stable and time-invariant. 2.27 y[n] = x[n + 1] – 2x[n] + x[n – 1]. Let y n 1 [ ] be the output for an input x n 1 [ ] and y n 2[ ] be the output for an input x n 2[ ]. Then for an input x n x n x n 3 12 [] [] [] = + α β the output y n 3[ ] is given by yn xn xn xn 33 33 [] [ ] [] [ ] = +− + − 12 1 = +− + −+ +− + − α αα β ββ xn xn xn x n x n x n 1 11 2 22 [ ] [] [ ] [ ] [] [ ] 12 1 12 1 = + α β yn y n 1 2 [ ] [ ]. Hence the system is linear. If x n x n n 1 0 [] [ ] = − then y n y n n 1 0 [] [ ] = − . Hence the system is time-inavariant. Also the system's impulse response is given by h n n [ ] , , , , = − =     2 1 0 0 n = 1,-1, elsewhere. Since h[n] ≠ ∀ 0 n < 0 the system is non-causal. 2.28 Median filtering is a nonlinear operation. Consider the following sequences as the input to a median filter: x n 1 [ ] {, , } = 345 and x n 2[ ] {, , } = −− 222 . The corresponding outputs of the median filter are y n n 1 [] [] = =− 4 2 and y2 . Now consider another input sequence x3[n] = x1[n] + x2[n]. Then the corresponding output is y n 3[ ] = 3, On the other hand, yn yn 1 2 [] [] + = 2. Hence median filtering is not a linear operation. To test the time￾invariance property, let x[n] and x1[n] be the two inputs to the system with correponding outputs y[n] and y1[n]. If x n x n n 1 0 [] [ ] = − then y n median x n k x n x n k 1 1 11 [ ] { [ ],......., [ ],....... [ ]} =− + = −− − +− = − median x n k n x n n x n k n y n n { [ ],......., [ ],....... [ ]} [ ] 0 0 00 . Hence the system is time invariant