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 linear9 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