e)∫ t ldx(ejo do=2兀∑m对n2=378兀( Using Parseval's relation with differentiation-in 3.26(a)X(e10)=∑xm=-2+4-1+5-3-2+4+3=8 ∑xn-1=2+41+5+3-2 (c) (eo)do=2πx[0]=-4兀 do=2x∑m (e) do=2x∑xm12=12380c 3.27 Let G, (e Jo) denote the dTFt of g,[n] (b)g2[n]=g,[n]+g,[]. Hence, the DTFT of g,In] is given by (c) g3[n]=g [(n-3)]+g, [n-4]. Now, the DTFT of g,[n] is given by G(e Jo) Hence, the DTFT of g,[n] is given by G2( Jo)=e J>oG, (e Jo)+e J4oG, (e Jo) (d) gln]=g[n]+gl(n-7). Hence, the DTFT of gln] is given by G4(e0)=G1(e10)+e1G1(e1) 328Y(c10)=X1(e10),x,(e10)x3(eJo ∑-∑-m∑sm∑xm n=-oo (a) Therefore, setting 0=0 we get ∑15
52 (e) dX e d d n xn j n ( ) [] . ω ω ω 2 2 2 378 −π π =−∞ ∞ ∫ =π ⋅ = π ∑ (Using Parseval's relation with differentiation-infrequency property) 3.26 (a) Xe xn j n ( ) [] . 0 = =− + − + − − + + = 241532438 =−∞ ∞ ∑ (b) X e x n j n n ( ) [ ]( ) . π =−∞ ∞ = − = + ++ +− − + = ∑ 1 2 4 1 5 3 2 4 3 14 (c) Xe d x j ( ) [] . ω ω −π π ∫ = π =− π 20 4 (d) Xe d xn j n ( ) [] . ω ω 2 2 2 168 −π π =−∞ ∞ ∫ =π = π ∑ (Using Parseval's relation) (e) dX e d d n xn j n ( ) [] . ω ω ω 2 2 2 1238 −π π =−∞ ∞ ∫ =π ⋅ = π ∑ 3.27 Let G ej 1( ) ω denote the DTFT of g n 1 [ ]. (b) g n gn gn 2 11 [ ] [ ] [ ]. = +− 4 Hence, the DTFT of g n 2[ ] is given by G e Ge e Ge e Ge j jjj j j 2 1 4 1 4 1 ( ) ( ) ( ) ( ) ( ). 1 ω ω ωω ω ω = + =+ − − (c) gn g n gn 31 1 [ ] [ ( )] [ ]. = −− + − 3 4 Now, the DTFT of g n 1 [ ] − is given by G e j 1( ) − ω . Hence, the DTFT of g n 3[ ] is given by G e e G e e G e jj jjj 3 3 1 4 1 ( ) ( ) ( ). ω ω ω ωω = + − −− (d) g n gn g n 4 11 [ ] [ ] [ ( )]. = + −− 7 Hence, the DTFT of g n 4[ ] is given by G e Ge e Ge j jj j 4 1 7 1 ( ) ( ) ( ). ω ω ωω = + − − 3.28 Y e X e X e X e j jjj ( ) ( ) ( ) ( ), ω ωωω =⋅⋅ 123 i.e., yne x ne x ne x ne n j n n j n n j n n j n [] [] [] [] =−∞ ∞ − =−∞ ∞ − =−∞ ∞ − =−∞ ∞ − ∑∑∑∑ = ω ωωω 123 (a) Therefore, setting ω = 0 we get y n x n x n x n n nn n [] [] [] [] =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑∑∑∑ = 123 . (b) Setting ω=π we get ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] − =− − − =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑∑∑∑ 1 11 1 123 n n n n n n n n yn x n x n x n
3.29(a) xn]=xev InJ+xodln]. Now, for a causal x[n], from the results of Problem 2.4(?),we x[n]=2x n]u[n]-x[OJS[n]=h(n]-x[O][n] x[n]=2x[[n]+x[OJo[n] Taking the dtFT of both sides of Eq(2)we get X(eJo)=H(eJo)-x[O] (3) where H(eJo )=DTFT2XevInlun-1xre(e lb)m(e//(o-0)de (4) Note: The DTFT of xev[n] is X.(eJo), and the dTFt of u[n] is m(eJo) Now, from Table 3.1, m(e)=I 1-e+∑8(o+2ml2-2co⑨ r∑0+2rk Substituting the above in Eq (4)we get e +兀 2 x)+x0x20 Substituting the above in Eq (3)we get X(eJo)=X.(eJo)+jX(eJo)=H(eJo)-x[O] x[0] 2π sInce X.(e Jo)de=x[O], as x[n] is real. Comparing the imaginary part of both sides of Eq(5)we therefore get Xim(ejo)=lxre(e lo)cof(@0\a0 (b) Taking the dTFf of both sides of Eq(2)we get where,G(ejo)=DTFT2xa[nJu(n]=J X (eje)m(ej(o-O)de (7) as jXi(eJu)is the DTFT of xod[n]. Substituting the expression for m( Jo) given above in k=-∞
53 3.29 (a) xn x n x n ev od [ ] [ ] [ ]. = + Now, for a causal x[n], from the results of Problem 2.4(??), we observe xn x n n x n hn x n ev [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ], = − =− 20 0 µδ δ (1) xn x n n x n od [ ] [ ] [ ] [ ] [ ]. = + 2 0 µ δ (2) Taking the DTFT of both sides of Eq. (2) we get Xe He x j j ( ) ( ) [ ], ω ω = − 0 (3) where H e DTFT x n n X e j ev re j ( ) [][] ( ) – ω θ = { } µ = π π π ∫ 2 1 m( ), ( ) e d j ω θ θ − (4) Note: The DTFT of xev[n] is X e re j ( ) ω , and the DTFT of µ[ ] n is m ( ) ejω . Now, from Table 3.1, m ( ) ejω = − +π + π = − +π + π − =−∞ ∞ =−∞ ∞ ∑ ∑ 1 1 2 1 22 2 2 e k j k j k k ω δ ω ω ( ) cot ( ). δ ω Substituting the above in Eq. (4) we get He X e j k d j re j k ( ) ( ) cot ( ) – ω θ θ = δθ θ π − +π + π π π =−∞ ∞ ∫ ∑ 1 1 22 2 2 = + π − π − π π π π ∫ ∫ Xe Xe d j Xe d re j re j re j ( ) ( ) ( )cot – – ωθ θ θ ω θ θ 1 22 2 . Substituting the above in Eq. (3) we get X e X e jX e H e x j re j im j j ( ) ( ) ( ) ( ) [] ω ω ωω = + =− 0 = + π − π − − π π π π ∫ ∫ Xe Xe d j Xe d x re j re j re j ( ) ( ) ( )cot [ ] – – ωθ θ θ ω θ θ 1 22 2 0 = − π − π π ∫ X e j Xe d re j re j ( ) ( )cot – ω θ ω θ θ 2 2 , (5) since 1 2 0 π = π π ∫ Xe d x re j ( ) [ ], – θ θ as x[n] is real. Comparing the imaginary part of both sides of Eq. (5) we therefore get X e Xe d im j re j ( ) – ( )cot . – ω θ ω θ = θ π − π π ∫ 1 2 2 (b) Taking the DTFT of both sides of Eq. (2) we get Xe Ge x j j ( ) ( ) [ ], ω ω = + 0 (6) where, G e DTFT x n n j X e j od im j ( ) [][] ( ) – ω θ = { } µ = π π π ∫ 2 m( ), ( ) e d j ω θ θ − (7) as j X e im j ( ) ω is the DTFT of xod[n]. Substituting the expression for m ( ) ejω given above in Eq. (7) we get G e j X e j k d j im j k ( ) ( ) cot ( ) – ω θ θ = δθ θ π − +π + π π π =−∞ ∞ ∫ ∑ 1 22 2 2
Substituting the above in Eq (6)we get X(eJo)=X(e Jo)+jX(eJo)=G(e Jo)+x[O] =c)+xm9x+0 Xim(e j )cot de+x[0 (8) as Xim(eje)d0=0 since Xim(ejoy) is an odd function of 0. Comparing the real parts of 330s=∑w-m=∑2mk+1)N n=0 Ifk-l≠ rn then s 1-e2mn(k-1) In(k-1/N Ifk-l=rn then s 罗e-]2r Hence, W(k-D)n-N, for k-l=rN r an Integer, otherwise 331=∑对面n-mhen+kN=∑对面+kN- r]. Since h[n] is periodic in I with a period N, h(n+kN-r]= h(n-r]. Therefore yin+kN]=>[r]h[n-r]=yIn],hence yIn] is also periodic in n with a period N 332xn]={010-23} and hn]={2010-2 Now50=∑对0-n=0+面4+和1+对+刘4= Similarly yll=∑对-n=和+和面0+823+82=5 Continuing the process we can show that y[2]=4, y(3]=-9, and y[4]=6
54 = + π + π − π π π π ∫ ∫ jX e j Xed Xe d im j im j im j ( ) ( ) ( )cot – – ωθ θ θ ω θ θ 2 1 2 2 Substituting the above in Eq. (6) we get X e X e jX e G e x j re j im j j ( ) ( ) ( ) ( ) [] ω ω ωω = + =+ 0 = + π + π − + π π π π ∫ ∫ jX e j Xed Xe d x im j im j im j ( ) ( ) ( )cot [ ] – – ωθ θ θ ω θ θ 2 1 2 2 0 = + π − + π π ∫ jX e X e d x im j im j ( ) ( )cot [ ] – ω θ ω θ θ 1 2 2 0 (8) as 1 2 0 π = π π ∫ Xed im j ( ) – θ θ since X e im j ( ) ω is an odd function of ω. Comparing the real parts of both sides of Eq. (8) we finally arrive at Xe X e d x re j im j ( ) ( )cot [ ]. – ω θ ω θ = θ π − + π π ∫ 1 2 2 0 3.30 SW e N k ln n N j nk l N n N = = − − = − − = − ∑ ∑ ( ) ( )/ 0 1 2 0 1 π If k – l ≠ rN then S = 1 1 1 1 1 0 2 2 2 − − = − − = − − − e e e j nk l j nk l N j nk l N π π π ( ) ( )/ ( )/ . If k – l = rN then S = W e N N rnN n N j nr n N n N − = − − = − = − ∑∑∑ = == 0 1 2 0 1 0 1 1 π . Hence, W N for N k ln n N − − = − ∑ = ( ) , 0 1 k – l = rN, r an integer, 0, otherwise. 3.31 ˜[ ] ˜[ ]˜ yn xr hn r [ ] r N = − = − ∑ 0 1 . Then ˜[ ] ˜[ ]˜ y n kN x r h n kN r [ ] r N + = +− = − ∑ 0 1 . Since ˜ h n is periodic in n [ ] with a period N, ˜ h n kN r [ ] + − = ˜ hn r [ ] − . Therefore ˜[ ] ˜[ ]˜ y n kN x r h n r [ ] r N += − = − ∑ 0 1 = y n , hence ˜[ ] y n is also periodic in n with a period N. ˜[ ] 3.32 x n ˜[ ] = − { } 0 1 0 2 3 and ˜ h n[] . = − { } 2010 2 Now, ˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜ y xr h r x h x h x h x h x h[ ] r 0 0 00 14 23 32 41 0 4 = −= + + + + = ∑ = –4. Similarly ˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜ y xr h r x h x h x h x h[ ] r 1 1 01 10 23 32 0 3 = −= + + + = ∑ = 5. Continuing the process we can show that y2 4 ˜[] , = y3 9 ˜[] , = − and y4 6 ˜[] . =
333m={2-123-2} and hn]={12-30-3 Now,y[0]=>对h[0-r]=对0O]+刘4+对2[3]+3h2]+刘4]h[l=-8 Similarly=∑对-]=80面+刘面0+刘23+刘32=3 Continuing the process we can show that y[2]=-15, y3]=16, and y[4]=-8\ 3.34 Since Vr[n+rN]=vKInI, hence all the terms which are not in the range 0, I, N-I can be accumulated to y, Inl, where OsksN-1. Hence in this case the Fourier series representation involves only N complex exponential sequences. Let n=∑双2Nun k=0 ∑ Trn/N ∑ SXkJej2r(k-r)n/N=I >XIk ej2ri(k-r)n/N n=0 n=0k=0 k=0 Now from Eq. (3.28), the inner summation is equal to N if k r, otherwise it is equal to 0 Thus ∑e-PmN=xr n=0 XN=∑和k+1N小N=∑刘mNe-12=∑对-P小N=x 3.35(a)x,In]=cos 4/2fejrn/4+e-jxrn 14\. The period of i, [n] is N=8 X1k]= j2πn/8-j2πkn/8 n /.-j2Tkn/8 AABe-j2In(k-1)18+2e-j2rn(k+1)/8. Now, from Eq(3.28)we observe ∑ for= l, and >e-j2rn(k+/8=8 for k=7, ence otnerwise n=0 otnerwise 3 e-jrn/3 2tejrcn/4+e-jin/4. The period of 吨()6mp)mc购间gmr (6,8) and is24.X,k]= n=0
55 3.33 x n ˜[ ] =− − { } 2 1 2 3 2 and ˜ h n[] . = −− { } 12 30 3 Now, ˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜ y xr h r x h x h x h x h x h[ ] r 0 0 00 14 23 32 41 0 4 = −= + + + + = ∑ = –8. Similarly ˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜[ ] ˜[ ]˜ y xr h r x h x h x h x h[ ] r 1 1 01 10 23 32 0 3 = −= + + + = ∑ = 3. Continuing the process we can show that y 2 15 ˜[] , = − y 3 16 ˜[] , = and y4 8 ˜[] . = − \ 3.34 Since ψ ψ ˜ [ ] ˜ [ ] k k n rN n + = , hence all the terms which are not in the range 0,1,....N-1 can be accumulated to ψ˜ [ ], k n where 0 k N –1 ≤ ≤ . Hence in this case the Fourier series representation involves only N complex exponential sequences. Let ˜[ ] ˜ [ ] / x n N Xkej kn N k N = = − ∑ 1 2 0 1 π then ˜[ ] ˜ [ ] / ( )/ xne N Xke n N j rn N j k r n N k N n N = − − − = − = − ∑ ∑ = ∑ 0 1 2 2 0 1 0 1 π π 1 = 1 2 0 1 0 1 N Xk ej k rn N n N k N ˜ [ ] π − ( )/ = − = − ∑ ∑ . Now from Eq. (3.28), the inner summation is equal to N if k = r, otherwise it is equal to 0. Thus ˜[ ] / xne n N j rn N = − − ∑ 0 1 2π = ˜ X r[ ]. ˜ [ ] ˜[ ] ˜[ ] ˜[ ] ( )/ / / Xk N xne xne e xne n N j k Nn N n N j kn N j n n N j kn N += = = = − − + = − − − = − − l ∑∑ ∑ l l 0 1 2 0 1 2 2 0 1 π ππ π2 = ˜ X k[ ]. 3.35 (a) ˜ [ ] cos . / / x n n e e jn jn 1 4 4 4 1 2 = π = + { } π −π The period of x n ˜ [ ] 1 is N = 8. ˜ [ ] // // Xk e e e e j n n j kn j n n j kn 1 2 8 0 7 2 8 28 0 7 1 2 8 2 = + π = −π −π = − π ∑ ∑ = + −π − = −π + = ∑ ∑ 1 2 2 18 0 7 2 18 0 7 e e j nk n j nk n ( )/ ( )/ . Now, from Eq. (3.28) we observe e for k otherwise j nk n −π − = ∑ = = { 2 18 0 7 8 1 0 ( )/ , , , , and e for k otherwise j nk n −π + = ∑ = = { 2 18 0 7 8 7 0 ( )/ , , , . Hence, ˜ [ ] , ,, , . X k k 1 otherwise 4 17 0 = = { (b) ˜ [ ] sin cos . // // x n n n j ee ee jn jn jn jn 2 33 44 3 3 4 1 2 3 2 = π + π = − { } + + { } π −π π −π The period of sin π n 3 is 6 and the period of cos π n 4 is 8. Hence, the period of x n ˜ [ ] 2 is the gcm of (6,8) and is 24. ˜ [ ] // // X k j ee e e j n n j kn j n n j kn 2 8 24 0 23 2 24 8 24 0 23 1 2 24 2 = − π = −π −π = − π ∑ ∑
6m/24e-]2mkm/24+e-jm/24e-2n/24 /emk-324-Se-pmk+)m2413/3 e-pm(k-4)124+e-j2mk+4)124 =0 n=0 12,k=3, Hence X2[k]= 36, k=4, 20, 0. otherwise 3.36 Since pIn] is periodic with period N, then from Eq. (3.168a)Of Problem 3.34, P[n]=N2PIkJe-j2Tkn/N where using Eq (3.168b)we get P[k]=>PnJe-j2Tkn/N=I e-j2rckn/N 337Xk]=X(e 2πk/N X(ein /N)= 2 xInJe-j2rkn/N, -o ck< oo (a)k+N]=X(2(k+N)N)=X(e2/Ne21)=X(e2/N)=k] (b)刘2mNN20 ∑对1]-PnNp2nN k=0 ∑∑x]e2n(n-l)N.Let=n+rN. Then x[n]= [n+r·N] But -j2rckr=N. Hence, x[n] x[n +r N k=0 3.38(a)G[k]=>]e-j2rkn/N- KI ∑ 对n]]kmN.Now,还]= ∑ Therefore, G=∑∑xrn-2k X[] yIn]e-j2r(k-r)n/N n=0r=0 ∑X (b)n=∑kY ∑∑Ykpn k=0 k=0r=0
56 + + π = −π −π = − π ∑ ∑ 3 2 6 24 0 23 2 24 6 24 0 23 2 24 ee e e j n n j kn j n n // // j kn = − + + −π − = −π + = −π − = −π + = ∑∑ ∑∑ 1 2 3 2 2 3 24 0 23 2 3 24 0 23 2 4 24 0 23 2 4 24 0 23 j ee ee j nk n j nk n j nk n j nk n ( )/ ( )/ ( )/ ( )/ . Hence ˜ [ ] , , , , , ,, , . X k j k j k k otherwise 2 12 3 12 21 36 4 20 0 = − = = = 3.36 Since p n ˜[ ] is periodic with period N, then from Eq. (3.168a) 0f Problem 3.34, ˜[ ] ˜[ ] / p n N Pke j kn N k N = − = − ∑ 1 2 0 1 π where using Eq. (3.168b) we get ˜[ ] ˜[ ] / Pk pne j kn N n N = − = − ∑ 2 0 1 π = 1. Hence ˜[ ] / p n N e j kn N k N = − = − ∑ 1 2 0 1 π . 3.37 ˜ [] ( ) ( ) [] , – . / / / Xk Xe Xe xne k j k N j k N j kn N n = = = ∞< <∞ = π π −π =−∞ ∞ ∑ ω ω 2 2 2 (a) ˜ [ ]( )( )( ) ˜ [ ]. ( )/ / / Xk N Xe Xe e Xe Xk j k N N j kNj j kN += = = = π+ π π π l 2 22 2 l l (b) ˜[ ] ˜ [ ] / x n N Xkej kn N k N = = − ∑ 1 2 0 1 π = 1 2 2 0 1 N xe e j k N j kn N k N [ ] / / l l l − π =−∞ ∞ = − ∑ ∑ π = π − =−∞ ∞ = − ∑ ∑ 1 2 0 1 N x ej kn N k N [ ] ( )/ l l l . Let l= +⋅ n r N . Then x n ˜[] [ ] N xn r N e j kr k N r = +⋅ − π = − =−∞ ∞ ∑ ∑ 1 2 0 1 . But e N j kr k N − π = − ∑ = 2 0 1 . Hence, xn xn r N ˜[ ] [ ]. r = +⋅ =−∞ ∞ ∑ 3.38 (a) ˜ [ ] ˜[ ] ˜[ ]˜[ ] / / Gk gn e xn yn e j kn N n N j kn N n N = = − = − − = − ∑ ∑ 2 0 1 2 0 1 π π . Now, ˜[ ] ˜ [ ] / x n N Xrej rn N r N = = − ∑ 1 2 0 1 π Therefore, ˜ [ ] ˜ [ ]˜[ ] ( )/ G k N Xr yne j k rn N r N n N = − − = − = − ∑ ∑ 1 2 0 1 0 1 π = 1 2 0 1 0 1 N Xr yne j k rn N n N r N ˜ [ ] ˜[ ] − − ( )/ = − = − ∑ ∑ π = 1 0 1 N XrYk r r N ˜ [ ]˜ [ ] − = − ∑ . (b) ˜[ ] ˜ [ ] ˜ [ ] / h n N XkYkej kn N k N = = − ∑ 1 2 0 1 π = 1 0 1 2 0 1 N xrYk e r N j kn r N k N ˜[ ] ˜ [ ] ( )/ = − − = − ∑ ∑ π