N<n≤N 312(a)y1[n]= Then Y,(e.)-∑m=c0NcmN)=如mo小+ (b) yiN] <n<N Assume to be odd. Now y2[n]=yo[nOWhere [n] 2’ Thus Y2(e)=Y(eo) otherwise 2(o/2) Note: The above result also holds for n even cos(mn/2N,-N≤n≤N, )y3{n]= Then Y3(e10) i(πn/2 ∑ ej(In/2Ne-joor +1∑ca+p in(o-2N) n|(+)(N+ sin(@)/2) sin(+)/2 3.13 Denote xr (n+m-1)! a"[n],a<l. We shall prove by induction that DTFTXm[n])=Xm From Table 3. 1. it follows that it holds for m =1 (1-ae Jo) (n+1) Let m =2. Then xiN=n("un=(n+)XIIn=-nxIIn + xIIn) Theretore x-saoylbLe jo )2 1-ae (l-ae-jo,2 using the differentiation-in-frequency property of Table 3.2. Now asuume, the it holds for m. Consider next xm+lIn=(n+m) d"uln n!(m)! n+m)(n+m-1)! a2n],= In]=-.nxm[n]+xmIn]. Hence, m+1(e1) 1-ae-joom(-ae-jo ) m ae
47 3.12 (a) yn NnN otherwise 1 1 0 [ ] ,– , , . = ≤ ≤ Then Ye e e e e N j jn n N N j N j N 1 j 2 1 1 1 2 1 2 ( ) ( ) ( ) sin( ) sin( / ) ( ) ω ωω ω ω ω ω = = − − = + [ ] − =− − + ∑ − (b) y n n N NnN otherwise 2 1 0 [ ] , , , . = − −<< Assume N to be odd. Now y n N 2 0 y n 1 [] [] = * where y n N n N otherwise 0 1 1 2 1 2 0 [ ] , , , . = − − ≤ ≤ − Thus Y e N Y e N j j N 2 0 2 2 2 1 12 2 () () sin / sin ( / ) . ω ω ω ω =⋅ =⋅ ( ) Note: The above result also holds for N even. (c) y n n N NnN otherwise 3 2 0 [ ] cos( / ), , , . = −≤≤ π Then, Ye e e e e j j n N jn n N N j n N jn n N N 3 1 2 2 2 1 2 ( ) ω πω πω (/ ) (/ ) = + − − =− − =− ∑ ∑ = + − − =− + =− π π ∑ ∑ 1 2 1 2 e e 2 2 j n n N N j n n N N N N ω ω – = ( − + ) ( − ) + ( + + ) ( + ) π π π π 1 2 2 1 2 2 2 1 2 2 2 1 2 2 sin ( )( ) sin ( ) / sin ( )( ) sin ( ) / ω ω ω ω N N N N N N . 3.13 Denote x n n m n m m n n [ ] ( )! !( )! = [ ], . + − − < 1 1 αµ α 1 We shall prove by induction that DTFT x n { } m[ ] = X e e m j j m ( ) ( ) . – ω ω α = − 1 1 From Table 3.1, it follows that it holds for m = 1. Let m = 2. Then x n n n n n x n nx n x n n 2 1 11 1 [ ] 1 ( )! !( = [ ] ( ) [ ] [ ] [ ]. + α µ =+ = + Therefore, X e e e ee j j 2 j jj 2 2 1 1 1 1 1 ( ) () () – – –– ω ω ω ωω α α αα = − + − = − using the differentiation-in-frequency property of Table 3.2. Now asuume, the it holds for m. Consider next x n n m n m m n n + = + 1[ ] ( )! !( )! α µ[ ] = + + − − = + = ⋅⋅ + n m m n m n m n n m m x n m nx n x n n m mm ( )! !( )! [ ], [ ] [ ] [ ] 1 1 1 α µ . Hence, X e m j d d e e e e e m j jm jm j + jm jm + = − + − = − + − 1 1 1 1 1 1 1 1 1 1 ( ) ( ) ( )( ) ( ) – – – – – ω ω ω ω ω ω ω α α α α α = − + 1 1 1 ( ) . – α ω e j m
3.14(a)x(ejo)=28(@+2nk). Hence, x[n = &(o)jondo=l N+1) ∑ 0≤n≤N H )X(10)=1+2∑co(1)=2+∑ Hence xIn 0<n≤N, (d)xa(ejo)= u_io 2, la]<1. Now we can rewrite Xa (ejoy)as d Now x[n]=aun]. Hence, from Table 3.2, x,[n]=-jno un] 315(a)H1(e + Hence, the inverse of H1(eJo) is a length-5 sequence given by h1n=[1.51111.5 e20+e-20 )=3+ eJo/2 2 272ej20 +3ejo +4+4e" j0 +3e-j200 +2e-j3o). Hence, the inverse of M(eoy is a length-6 sequence given by h2[n]=[1 1.5 2 2 1.5 1].-2sns3 +eJo (c)H3(eJ0)=j3+ 6e7ei3o+2ej20+2ejo +0-2e Jou-2e-j20-e-j3o ) Hence, the inverse of H3(e u)is a stt-7sequence given by h3[n]=[0.5 1 10-1 -1 -05].-35ns3 e +e-j2o ejo 2 (d)H4(eJ0)=j4+ e jo/2 2 3+ Hence, the inverse of H4(eJo)is a length-6 sequence given by h4[n]=0.75 -0.25 1.5 -1.5 025-0.75.-3sns2 316(a)H2(e10)=1+2coso+(1+cos2o)=e10+ 5
48 3.14 (a) Xe k a j k () ( ) ω = + δω π =−∞ ∞ ∑ 2 . Hence, x[n] = 1 2 ( ) e d j n π = −π π ∫ δω ω ω 1. (b) X e e e e b j j N j j n n N ( ) ( ) ω ω ω ω = − − = + − − = ∑ 1 1 1 0 . Hence, x[n] = 1 0 0 , , , . ≤ ≤ n N otherwise (c) Xe e c j N j N N ( ) cos( ) ω ω =+ = + ω = − =− 12 2 ∑ ∑ 0 l l l l . Hence x[n] = 3 0 1 0 0 , , , , , . n n N otherwise = < ≤ (d) X e j e e d j j j ( ) ( ) , ω ω ω α α = α − − < − − 1 1 2 . Now we can rewrite X e d j ( ) ω as X e d d e d d X e d j j o j ( ) ( ) ( ) ω ω ω ω α ω = − − = ( ) 1 1 where X e e o j j ( ) ω ω α = − − 1 1 . Now x n n o n [] [] =α µ . Hence, from Table 3.2, x n jn n d n [] [] =− α µ . 3.15 (a) H e ee e e ee e e j jj j j jj j j 1 2 2 2 2 1 2 2 3 2 1 3 2 3 2 ( ) – – ω – – ωω ω ω ωω ω ω = + + + + =+ + + + . Hence, the inverse of H ej 1( ) ω is a length-5 sequence given by hn n 1[] . . , . = [ ] 15 1 1 1 15 2 2 −≤≤ (b) H e ee e e e e e j jj j j j j j 2 22 2 2 2 3 2 2 4 2 2 ( ) – – / –/ ω – / ωω ω ω ω ω ω = + + + + ⋅ + ⋅ = + ++ + + ( ) 1 2 2 3 44 3 2 2 23 ee ee e jj jj j ωω ω ω ω –– – . Hence, the inverse of H ej 2( ) ω is a length-6 sequence given by h n n 2[] . . , . = [ ] 1 15 2 2 15 1 2 3 −≤≤ (c) He j ee e e e e j j jj j j j j 3 22 2 2 3 4 2 2 2 2 ( ) – – / –/ ω ωω ω ω ω ω = + + + + ⋅ − = + + +− − − ( ) 1 2 2 2 02 2 32 23 eee eee jjj j jj ω ωω ω ω ω ––– . Hence, the inverse of H ej 3( ) ω is a length-7 sequence given by h n n 3[] . – – . , . = − [ ] 05 1 1 0 1 1 05 3 3 −≤≤ (d) He j ee e e e e j e j jj j j j j j 4 22 2 2 2 4 2 2 3 2 2 ( ) – – / –/ ω / ωω ω ω ω ω ω = + + + + ⋅ − ⋅ = − + −+ − 1 2 3 2 1 2 3 3 1 2 3 2 32 2 e ee e e j jj j j ω ωω ω ω – – . Hence, the inverse of H ej 4( ) ω is a length-6 sequence given by h n n 4[] . . . . . . , . =− − − [ ] 0 75 0 25 1 5 1 5 0 25 0 75 3 2 −≤≤ 3.16 (a) He e e e e j jj j j 2 2 2 1 2 3 2 1 2 3 4 5 2 3 4 ( ) cos ( cos ) . ω ωω ω ω – – =+ + + = + + + + ω ω
Hence, the inverse of H,(e is a length-5 sequence given by h1n={07512510.75}-2≤n≤2 (b)H2(ejo)=1+3c0s0+-(1+cos 2o) cos( o/2)ejo 9 eox/de Jo +-e-j2o Hence, the inverse of H2(e) is a length-6 sequence given by h2[n]=0.5 1.25 2.75 2.75 1.25 0 5}-3≤n≤2 (c)H3(eJo)=j[3+4coso+(1+cos 2o)sin(o) Lejo +ej20 +ejo +0-le-jo_e- e330 inverse of H3(e )is a length-7 sequence given by h3[n]=0.25 1 1.75 0 -1.75 -1 -0.25)-3Sns3 (d)H4(ejo) =j4+2 cos 0+501+cos 2o) sin(@ /2)e-jo/2 22(4+ejo99f+103 length-6 sequence given by han N. jol Hence, the inverse of H4(e)is a 84488 3. 3.17 Y(ejo)=X(ejo)=X(ejo) 2). Now, X(ejo)=>-[nJe-jonHence Yc)=∑ylm=x(c1)=∑ x[n](e Jon)-=∑m×m/3-m Therefore, yin =xn1, n=0+3.6, K 318x(co)=∑xnls-m X(cm02)=∑x1), and X(eJ2)=∑xn(1e-o2).mhus n=- n=- ∑r =1xeo/2)+x(-cio/2=1 In]e-on I x[n]+x[n](-1)(o/2)n. Thus, yn =5(xIn+ x()m)=x0 . for n eve 3.19 From Table 3.3, we observe that even sequences have real-valued DTFTs and odd se have imaginary-valued DTFTs (a) Since Fn=In,, x, [n] is an even sequence with a real-valued DTFT (b)Since (n)=-n,x,In] is an odd sequence with an imaginary-valued DTFT
49 Hence, the inverse of H ej 1( ) ω is a length-5 sequence given by hn n 1[] . . . , . = [ ] 0 75 1 2 5 1 0 75 2 2 −≤≤ (b) He e j j 2 2 1 3 4 2 ( ) cos ( cos ) cos( / ) 12 2 ω ω/ =+ + + ω ωω = + + ++ + 1 2 5 4 11 4 11 4 5 4 1 2 32 2 ee e e e jj j j j ωωω ω ω – – . Hence, the inverse of H ej 2( ) ω is a length-6 sequence given by h n n 2[] . . . . . . , . = [ ] 0 5 1 25 2 75 2 75 1 25 0 5 3 2 −≤≤ (c) He j j 3( ) cos ( cos ) sin( ) 34 1 2 ω = + ++ [ ] ω ωω = + + +− − − 1 4 7 4 0 7 4 1 4 32 2 3 ee e e e e jj j j j j ωω ω ω ω ω –– – . Hence, the inverse of H ej 3( ) ω is a length-7 sequence given by h n n 3[] . . . . , . = − −− [ ] 0 25 1 1 75 0 1 75 1 0 25 3 3 −≤≤ (d) He j e j j 4 2 4 2 3 2 ( ) cos ( cos ) sin( / ) 12 2 ω ω – / =+ ++ ω ωω = + +− + − 1 2 3 4 1 4 9 2 9 2 1 4 3 4 2 23 ee e e e jj j j j ωω ω ω ω –– – . Hence, the inverse of H ej 4( ) ω is a length-6 sequence given by hn n 4 3 8 1 8 9 4 9 4 1 8 3 8 [] , . = −−− 2 3 −≤≤ 3.17 Ye Xe X e jj j ( ) ( ) ( ). ωωω = = ( ) 3 3 Now, X e x n e j jn n ( ) [] . ω ω = − =−∞ ∞ ∑ Hence, Ye yne X e xn e xm e j jn n j jn jm n m ( ) [ ] ( ) [ ]( ) [ / ] . ω ωω ω ω = = ( ) = = − =−∞ ∞ − − =−∞ ∞ =−∞ ∞ ∑ ∑ ∑ 3 3 3 Therefore, y n xn n otherwise [ ] [ ], , , , , . = = ±± { 036 0 K 3.18 Xe xne j jn n ( ) [] . ω ω = − =−∞ ∞ ∑ Xe xne j jn n ( ) [] , ω ω / ( /) 2 2 = − =−∞ ∞ ∑ and X e x n e j nj n n ( ) [ ]( ) . / ( /) −= − − =−∞ ∞ ∑ ω ω 2 2 1 Thus, Ye yne Xe X e xn xn e j n n j j nj n n ( ) [ ] ( ) ( ) [ ] [ ]( ) . ω ωωω ω / / ( /) = = +− { } = +− ( ) − =−∞ ∞ − =−∞ ∞ ∑ ∑ 1 2 1 2 1 22 2 Thus, yn xn xn x n for n even for n odd n [ ] [ ] [ ]( ) [ ], . = +− ( ) = { 1 2 1 0 . 3.19 From Table 3.3, we observe that even sequences have real-valued DTFTs and odd sequences have imaginary-valued DTFTs. (a) Since − = n n,, x n 1 [ ] is an even sequence with a real-valued DTFT. (b) Since ( ) , − =− n n 3 3 x n 2[ ] is an odd sequence with an imaginary-valued DTFT
c)Since sin(-ocn)=-sin(ocn)and oc-n)=-ocn,, xiN] is an even sequence with a real valued dtft ( d)Since xin] is an odd sequence it has an imaginary-valued DTFT (e) Since xsIn] is an odd sequence it has an imaginary-valued DTFT. 3.20 (a)Since Y,( Jo)is a real-valued function of o, its inverse is an even sequence (b) Since Y,(eJu) is an imaginary-valued function of ( its inverse is an odd sequence (e) Since Y3(eJ) is an imaginary-valued function of o, its inverse is an odd sequence 3. 21(a)HLLp(e Jo ) is a real-valued function of (. Hence, its inverse is an even sequence (b)HBLDIF(ejo)is a real-valued function of o. Hence, its inverse is an even sequence. 3.22 Let u[n]= x[I-n], and let X(eJo)and U(e Jo)denote the DTFTs of x[n] and u[n], respectively From the convolution property of the dtFt given in Table 3. 2, the DTFT of yIn]= x[n]Ou[n] is given by Y(eJo)= X(eJo)U(eJo). From Table 3.3, U(eJo)=X(e Jo). But from Table 3.4, (e-jo)=X*(ejo ) Hence, Y(ejo)=X(ejo )X*(ejo )=x(ejo) which is real-valued function of o 3.23 From the frequency-shifting property of the dtFt given in Table 3.2, the DTFT of x[n]ejn/ is given by X(ej(o+/3)). a sketch of this DTFT is shown below 兀2丌/3-丌/30/32/3兀 3.24 The DTFT of x[n]=-a"u[-n-l] is given by X(eJo)=>-ane-jon -ejon=-a-ejo ∑ For>1,X(e10)= /)1-e 1+a2-2ac From Parseval's relation, Jx(e)do∑kun
50 (c) Since sin( ) sin( ) − =− ω ω c c n n and ω ω c c () , − =− n n , x n 3[ ] is an even sequence with a realvalued DTFT. (d) Since x n 4[ ] is an odd sequence it has an imaginary-valued DTFT. (e) Since x n 5[ ] is an odd sequence it has an imaginary-valued DTFT. 3.20 (a) Since Y ej 1( ) ω is a real-valued function of ω , its inverse is an even sequence. (b) Since Y ej 2( ) ω is an imaginary-valued function of ω , its inverse is an odd sequence. (c) Since Y ej 3( ) ω is an imaginary-valued function of ω , its inverse is an odd sequence. 3.21 (a) He LLP j ( ) ω is a real-valued function of ω . Hence, its inverse is an even sequence. (b) He BLDIF j ( ) ω is a real-valued function of ω . Hence, its inverse is an even sequence. 3.22 Let u[n] = x[–n], and let X ej ( ) ω and U ej ( ) ω denote the DTFTs of x[n] and u[n], respectively. From the convolution property of the DTFT given in Table 3.2, the DTFT of y[n] = x[n] * u[n] is given by Y ej ( ) ω = X ej ( ) ω U ej ( ) ω . From Table 3.3, U e X e j j ( ) ( ). ω ω = − But from Table 3.4, Xe X e j j ( ) *( ). − = ω ω Hence, Y ej ( ) ω = X ej ( ) ω X ej *( ) ω = X ej ( ) ω 2 which is real-valued function of ω . 3.23 From the frequency-shifting property of the DTFT given in Table 3.2, the DTFT of xne j n [ ] − π / 3 is given by X ej ( ) ( /) ω+π 3 . A sketch of this DTFT is shown below. X(e j(ω+π/3)) – π π –2 π /3 2 – π / 3 0 π / 3 π / 3 ω 1 3.24 The DTFT of x n n n [ ] – [– – ] = αµ 1 is given by Xe e e e e j n n jn n n jn j j n n ( ) – ω ω ωω – –– ω α αα α = − =− =− = ∞ − = ∞ = ∞ ∑∑ ∑ 1 1 1 0 . For α >1, Xe e e e j j j j ( ) ( /) . ω ω – ω ω α α α = − − = − − 1 1 1 1 1 X ej ( ) cos . ω α αω 2 2 1 1 2 = + − From Parseval's relation, 1 2 2 2 π = −π π =−∞ ∞ ∫ Xe d xn ∑ j n ( ) []. ω ω
Hence, O=-2. Therefoe, x[n]=-(2)u[-n-1 5+4 Now, 4 x(ejo f don0e-don=∑k2y ∑(4)=x∑ 4 n=0 Hence, a=1. 5 and therefore, x[n]=-(1.5) ul-n-1]. Now 3.25-3cos0 K)do2kn)dor∑a=x15y= n=-c0 ∑(台 (c)Using the differentiation-in-frequency property of the dtFt, the inverse dTFt of xsiigotl-ae jo (1-ae o)2 Is[n]=-noul-n-I].Hence, the inverse DTFT of (1-ae-jis-(n+1)a2-n-1 Hence, a=2 and yIn]=-(n+1)2 u[-n-1] Now °do=xNn2=4x∑a+2-2 9/16 325(a)X(e0) xn]=3+1-2-3+4+1-1=3 n=-∞ (b)X(e)=∑xnle=-3-1-2+3+4-1+1=1 (c)|x(e1°)do=2兀x0=-4兀 k(ejo y do =2T >[n]=82T.(Using Parseval,'s relation
51 (a) X ej ( ) cos . ω ω 2 1 5 4 = + Hence, α=−2. Therefoe, x n n n [ ] ( ) [ ]. =−− − − 2 1 µ Now, 42 4 2 0 2 2 Xe d Xe d xn j j n ( ) ( ) [] ω ω ω ω π −π π =−∞ ∞ ∫ ∫ = =π ∑ =π − =−∞ − 4 2 ∑ 2 1 ( )n n = π = π = π = ∞ = ∞ 4 ∑ ∑ 1 4 1 4 4 3 1 0 n n n n . (b) X ej ( ) . cos . ω ω 2 1 3 25 3 = − Hence, α = 1 5. and therefore, x n n n [ ] ( . ) [ ]. =− − − 15 1 µ Now, Xe d Xe d xn j j n ( ) ( ) [] ω ω ω ω 2 0 2 1 2 2 π −π π =−∞ ∞ ∫ ∫ = =π ∑ = π =−∞ − ∑(.) 1 5 2 1 n n = π = ∞ ∑ 4 9 1 n n = π = π ⋅ = π = ∞ ∑4 9 4 9 4 9 9 5 4 5 0 n n . (c) Using the differentiation-in-frequency property of the DTFT, the inverse DTFT of Xe j d d e e e j j j j ( ) ( ) ω ω ω ω ω α α α = − = − − − − 1 1 1 2 is x n n n n [ ] [ ]. =− − − α µ 1 Hence, the inverse DTFT of 1 1 2 ( ) − − α ω e j is − + −− ( ) [ ]. n n n 1 1 α µ Y ej ( ) ( cos ) . ω ω 2 2 1 5 4 = − Hence, α = 2 and y n n n n [ ] ( ) [ ]. =− + − − 12 1 µ Now, 42 4 2 0 2 2 Xe d Xe d xn j j n ( ) ( ) [] ω ω ω ω π −π π =−∞ ∞ ∫ ∫ = =π =π + ⋅ ∑ =−∞ − 4 12 ∑ 2 2 1 ( ) n n n = π ⋅ =π = π = ∞ ∑ 1 4 9 4 9 16 4 0 2 n n n / / .. 3.25 (a) Xe xn j n ( ) [] . 0 = = +− −+ +−= 312 3 411 3 =−∞ ∞ ∑ (b) Xe xne j n j n ( ) [] . π =−∞ ∞ π = =− − − + + − + = ∑ 312 3 4111 (c) Xe d x j ( ) [] . ω ω −π π ∫ = π =− π 20 4 (d) Xe d xn j n ( ) [] . ω ω 2 2 2 82 −π π =−∞ ∞ ∫ =π = π ∑ (Using Parseval's relation)