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 invariant
12 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 timeinvariance 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
229yn]=yn-1]+ Now for an input x[n]=a u[n], the ouput yIn] converges to some constant K as n- The above difference quatio as n=>o reduces to k=ik K+- which is equivalent to K K=a or in other words K It is easy to show that the system is non-linear. Now assume yi[n] be the output for an input xi[n]. Then y[n]=5yn-11+-,In y1n-1 xIn If x, [n]=x[n-nol. Then, y,[n]=ly[n-1+ y1[n-1 Thus y,n]=yIn-nol. Hence the above system is time invariant. 2.30 For an input xi [n i=l, 2, the input-output relation is given by yi[n]=Xi[n]-yiIn-1]+ yi[n-l]. Thus, for an input Axl[n]+ Bx2In], if the output is AyI[n]+ By2ln, then the input-output relation is given by A[n]+By2ln] Ax1m]+Bx1m]-(Ay1n-1+By2n-1)2+Ayn-1]+By2{n-1=Axm+Bxm A2yiIn-1]-B2[n-1]+2ABy In-lly2[n-1]+Ay[n-1]+By2[n-1] *AXI[n]-A yiIn-1]+AyIIn-1]+Bx2[n]-B-y2[n-1]+ By2ln-I]. Hence, the system is nonlinear Let yIn] be the output for an input x[n] which is related to x[n] through [n]=x[n]-y-In-1]+ yIn-1]. Then the input-output realtion for an input x[n-no] is given by yIn -no=xIn-no-yIn-no-1]+ yIn-no -l], or in other words, the system is time Now for an input x[n]=∝μn], the ouput yIn] converges to some constant K as n→∞ The above difference equation as n→∞ reduces to K=α-K2+K,orK2=a,ie. 231AsS[n]=n-μn-1,T6n]}=Tu[n]-T{u[n-1}→hn]=sn]-sn For a discrete Lti system y=∑地m-k=∑-k-1m+k=∑k-k-∑-1km-k k=-∞ 2.32 y[n]=2ms-g h(m]x[n-m]. Hence, y[n+kN]=2ms- h(mjx[n+kN-ml h[m]x[n-m]=yIn ]. Thus, yIn] is also a periodic sequence with a period N
13 2.29 yn yn x n y n [] [ ] [ ] [ ] = −+ − 1 2 1 1 Now for an input x[n] = α µ[n], the ouput y[n] converges to some constant K as n → ∞. The above difference equation as n → ∞ reduces to K K K = + 1 2 α which is equivalent to K2 = α or in other words, K = α . It is easy to show that the system is non-linear. Now assume y1[n] be the output for an input x1[n]. Then yn yn x n y n 1 1 1 1 1 2 1 1 [] [ ] [ ] [ ] = −+ − If x n x n n 1 0 [] [ ] = − . Then, yn yn xn n y n 1 1 0 1 1 2 1 1 [] [ ] [ ] [ ] = −+ . − − Thus y n y n n 1 0 [] [ ] = − . Hence the above system is time invariant. 2.30 For an input x n i[ ], i = 1, 2, the input-output relation is given by yn xn y n yn iii i [ ] [ ] [ ] [ ]. = − −+ − 2 1 1 Thus, for an input Ax1[n] + Bx2[n], if the output is Ay1[n] + By2[n], then the input-output relation is given by A y n By n 1 2 [] [] + = Ax n Bx n Ay n By n Ay n By n 12 1 2 2 11 11 1 2 [] [] [ ] [ ] [ ] [ ] + − −+ − ( ) + −+ − = A x n Bx n 1 2 [] [] + − −− −+ − −+ −+ − A y n B y n ABy n y n A y n By n 2 1 2 2 2 2 12 1 2 [ ] [ ] [ ] [ ] [ ] [ ] 1 12 1 1 1 1 ≠ − −+ −+ − −+ − Ax n A y n Ay n Bx n B y n By n 1 2 1 2 1 2 2 2 2 11 11 2 [ ] [ ] [ ] [ ] [ ] [ ]. Hence, the system is nonlinear. Let y[n] be the output for an input x[n] which is related to x[n] through yn xn y n yn [] [] [ ] [ ] = − −+ − 2 1 1 . Then the input-output realtion for an input x n no [ ] − is given by y n n x n n y n n y n n oo o o [ ][ ] [ ][ ] − = − − − −+ − − 2 1 1 , or in other words, the system is timeinvariant. Now for an input x[n] = α µ[n], the ouput y[n] converges to some constant K as n → ∞. The above difference equation as n → ∞ reduces to K K K =− + α 2 , or K2 = α, i.e. K = α . 2.31 As δ µµ [] [] [ ] n nn = −−1 , ΤΤΤ { [ ]} { [ ]} { [ ]} δµµ nnn = −−1 ⇒ = −− hn sn sn [] [] [ ]1 For a discrete LTI system yn hkxn k k [ ] [ ][ ] = − =−∞ ∞ ∑ = −− ( ) − =−∞ ∞ ∑ sk sk xn k k [] [ ] [ ] 1 = −− − − =−∞ ∞ =−∞ ∞ ∑ ∑ skxn k sk xn k k k [ ][ ] [ ][ ] 1 2.32 yn hmxn m m [] [ ] = − ˜[ ] =−∞ ∞ ∑ . Hence, y n kN h m x n kN m m [ ] [] + = +− ˜[ ] =−∞ ∞ ∑ = −= =−∞ ∞ ∑ hmxn m yn m [ ]˜[ ] [ ]. Thus, y[n] is also a periodic sequence with a period N
2.33Nown-n-=∑m=列m-n-s-m]=n-r a)yn]=x1[nl⑧h1[n=(286n-1-0.58n-3(28m+8n-11-38n-3] 48n-11③8n-8n-3]④8n]+28n-1④8n-11-0.58n-3]③8n-1 -68n-1③8n-3]+1.58n-3]④8n-3]=48n-1-8n-3]+28n-1 0.5Sn-4]-6[n-4]+1.5δ[n-6 4δ[n-1+2δ[n-1-δn-3]-6.5δ-4]+1.5δ[n-6] (b)y2[m=xm③h2lnl=(-36n-18n+2)(-8n-2]-0.58n-11+38n-3 0.5δ[n+1]-δ[n]+3δn-1+1.5δ[n-2]+38n-3]-9δn-4] (c) y3[n]=x,[nOh[n] (28n-1]-0.58n-3(-8n-2]-0.586n-11+38n-3 δn-2]-2δ[n-3]-625δ[n-4]+0.5n-5]-1.58[n-6] (d)y4n]=x2{nl②h1n=(-386n-11+8n+2)(28n+8n-11-38n-3]= 2δ[n+2]+δn+1]-8n-1]-3n-2]-98n-4] 2.34 y[n]=2m2 N, glm]h(n-m]. Now, h(n-m] is defined for Misn-msM2.Thus,for m=NI, h[n-m] is defined for MiSn-NISM2, or equivalently, for M1+N1≤n≤M2+N1. Likewise,form=N2,h[n-m] is defined for M1≤n-N2≤M2,or equivalently,forM1+N2≤n≤M2+N2 (a) The length of yIn] is M2+n2-M1-NI+l (b) The range of n for yIn]≠0 Is mini(M1+N1,M1+N2)≤n≤max(M2+N1,M2+N2) M1+N1≤n≤M2+N2 235y=x1mn⊙x2=∑x+x2k Nwv=x1m-N1②x2n-N2]=∑以=x-N1-k1x2k-N2lLet k-n2=m. Then v[n]= 2ms- XIn-N1-N2-m]x2[m]=yIn-NI-N21 236gn=x1n]③x2{nx3n=ym③ X3In] where yIn=x1[m③x2lmn. Define v XI[n-N1lOx2In-N2]. Then, h(n]=v[n]Ox3In-N3]. From the results of Problem 2.32, v[n]= y[,-N2]. Hence, h(n]= yIn-N-N21Ox3In-N31.Therefore, using the results of Problem 2.32 again we get h[n]=gIn-N1-N2-N3I
14 2.33 Now δ[ ] n r − * δ[ ] n s − = δδ δ [ ][ ] [ ] mr nsm nrs m − −− = −− =−∞ ∞ ∑ (a) y n 1[ ] = x n 1[ ] * h n 1[ ] = (2 1 05 3 δ δ [ ] .[ ] n n −− − ) * (2 13 3 δδ δ [] [ ] [ ] nn n + −− − ) = 4 δ[ ] n − 1 * δ[ ] n – δ[ ] n − 3 * δ[ ] n + 2 δ[ ] n − 1 * δ[ ] n − 1 – 0.5 δ[ ] n − 3 * δ[ ] n − 1 – 6 δ[ ] n − 1 * δ[ ] n − 3 + 1.5 δ[ ] n − 3 * δ[ ] n − 3 = 4 δ[ ] n − 1 – δ[ ] n − 3 + 2 δ[ ] n − 1 – 0.5 δ[ ] n − 4 – 6 δ[ ] n − 4 + 1.5 δ[ ] n − 6 = 4 δ[ ] n − 1 + 2 δ[ ] n − 1 – δ[ ] n − 3 – 6.5 δ[ ] n − 4 + 1.5 δ[ ] n − 6 (b) y n 2[ ] = x n 2[ ] * h n 2[ ] = (− −+ + 31 2 δ δ [ ][ ] n n ) * (− − − −+ − δ δδ [ ] .[ ] [ ] n nn 2 05 1 3 3 ) = − +− + −+ − + −− − 05 1 3 1 15 2 3 3 9 4 . [ ] [] [ ] . [ ] [ ] [ ] δ δδ δ δ δ n nn n n n (c) y n 3[ ] = x n 1[ ] * h n 2[ ] = (2 1 05 3 δ δ [ ] .[ ] n n −− − ) * (− − − −+ − δ δδ [ ] .[ ] [ ] n nn 2 05 1 3 3 ) = − −− −− −+ − − δδ δ δ δ [ ] [ ] . [ ] . [ ]– . [ ] nn n n n 2 2 3 6 25 4 0 5 5 1 5 6 (d) y n 4[ ] = x n 2[ ] * h n 1[ ] = (− −+ + 31 2 δ δ [ ][ ] n n ) * (2 13 3 δδ δ [] [ ] [ ] nn n + −− − ) = 2 2 1 13 29 4 δδδ δ δ [ ] [ ] [ ] [ ]– [ ] nnn n n + + +− −− − − 2.34 yn gmhn m m N N [] [ ][ ] = − ∑ = 1 2 . Now, h[n – m] is defined for M n m M 1 2 ≤− ≤ . Thus, for m N= 1, h[n–m] is defined for M n N M 1 12 ≤− ≤ , or equivalently, for M N nM N 11 21 + ≤≤ + . Likewise, for m N= 2, h[n–m] is defined for M n N M 1 22 ≤− ≤ , or equivalently, for M N n M N 12 22 + ≤≤ + . (a) The length of y[n] is M N M N 2 2 11 +− + – . 1 (b) The range of n for y n[ ] ≠ 0 is min , max , (M NM N n M NM N 1 11 2 2 12 2 + + ) ≤≤ + + ( ), i.e., M N nM N 11 2 2 + ≤≤ + . 2.35 y[n] = x1[n] * x2[n] = x n k x k k 1 2 [ ] [] − =−∞ ∞ ∑ . Now, v[n] = x1[n – N1] * x2[n – N2] = x n N k x k N k 112 2 [ ][ ] −− − =−∞ ∞ ∑ . Let kN m − = 2 . Then v[n] = x n N N m x m y n N N m 1 12 2 12 [ ] [ ] [ ], −− − = −− =−∞ ∞ ∑ 2.36 g[n] = x1[n] * x2[n] * x3[n] = y[n] * x3[n] where y[n] = x1[n] * x2[n]. Define v[n] = x1[n – N1] * x2[n – N2]. Then, h[n] = v[n] * x3[n – N3] . From the results of Problem 2.32, v[n] = y n N N [ ] − −1 2 . Hence, h[n] = y n N N [ ] − −1 2 * x3[n – N3] . Therefore, using the results of Problem 2.32 again we get h[n] = g[n– N1– N2– N3]