Chapter 3(2e) 3.1 X(eo)=2xnJe-jon where x[n] is a real sequence. Therefore X(e)=Rl∑xnlo/。 ∑xR(-mu)=∑ x[n]cos(on),and xmm)=m∑刈nm∑刈mc-m)=-2 xn] sin(oon) Since cos(on)and sin(on)are, respectively, even and odd functions of o, Xre(eJo) is an even function of o and Xim( Jo)is an odd function of o lx()=vxr(ejo )+x m(ejo ) Now. X (ejo) is the square of an even function and Xim(eJo) is the square of an odd function, they are both even functions of o.Hence, X(eJo) is an even function of o ( X(eo)=tan-1(Xim(ejoy The argument is the quotient of an odd function and an even Xre(e) function, and is therefore an odd function. Hence, arg x(eJo ) is an odd function of a 1-ce-0 3.2X( 1-a cosO-Jasino 1-ae Jo 1-ae-Jo 1-ae- 1-2acoso +a1-2acos0+a Therefore, x(ejo)=-1-ocoso and Xim(e 1-20cos (+a- 1-2 a cos@+. X(e)=X(e0)x*(e) Therefore, X(e) Xim(e) a sIno tan e(o) Therefore, 0( O= tan Xre(eJo) 1-acos@ 1-ocoso 33(a)yn]=叫n=yem+ydm, where y=(n+y-n=叫m]+叫n)=+n, andyml=a]-y-n)=a--m)=山]-3-1an Now,Y(e)=1|2x>8(o+2rk)+=π>δ(o+2rk) Since yd[n]=u[n]-+=on], y[n]=un-1]-3+=Sn-1]. As a result
42 Chapter 3 (2e) 3.1 X( ejω ) = x n e j n n [ ] − =−∞ ∞ ∑ ω where x[n] is a real sequence. Therefore, X e xne xn e xn n re j jn n j n n n ( ) Re [ ] [ ]Re [ ]cos( ), ωω ω = ω = ( ) = − =−∞ ∞ − =−∞ ∞ =−∞ ∞ ∑∑ ∑ and X e xne xn e xn n im j jn n j n n n ( ) Im [ ] [ ]Im [ ]sin( ). ωω ω = ω = ( ) = − − =−∞ ∞ − =−∞ ∞ =−∞ ∞ ∑∑ ∑ Since cos( ) ωn and sin( ) ωn are, respectively, even and odd functions of ω , X e re j ( ) ω is an even function of ω , and X e im j ( ) ω is an odd function of ω . Xe X e X e j re j im j () () () ω ωω = + 2 2 . Now, X e re 2 j ( ) ω is the square of an even function and X e im 2 j ( ) ω is the square of an odd function, they are both even functions of ω . Hence, X ej ( ) ω is an even function of ω . arg ( ) tan ( ) ( ) X e X e X e j im j re j ω ω { } ω = −1 . The argument is the quotient of an odd function and an even function, and is therefore an odd function. Hence, arg ( ) X ejω { } is an odd function of ω . 3.2 X e e e e e e j j j j ( ) cos cos sin cos – – – – – ω ω ω ω ω ω α α α α α α ωα α ωαω α ωα = − = − ⋅ − − = − − + = − − − + 1 1 1 1 1 1 1 1 2 1 1 2 2 2 Therefore, X e re j ( ) cos cos ω α ω α ωα = − − + 1 1 2 2 and X e im j ( )– sin cos ω α ω α ωα = 1 2 − + 2 . Xe Xe X e e e j jj j j ( ) ( ) *( ) cos . – ω ωω ω ω α α α ωα 2 2 1 1 1 1 1 1 2 =⋅ = − ⋅ − = − + Therefore, X ej ( ) cos . ω α ωα = − + 1 1 2 2 tan ( ) ( ) ( ) – sin cos θ ω . α ω α ω ω ω = = − X e X e im j re j 1 Therefore, θ ω α ω α ω ( ) tan – sin cos == . − −1 1 3.3 (a) yn n y n y n ev od [ ] [ ] [ ] [ ], == + µ where y n yn y n n n ev[] [] [ ] [] [ ] = +− ( ) = +− ( ) 1 2 1 2 µ µ = + 1 2 1 2 δ[ ] n , and y n yn y n n n n n od[] [] [ ] [] [ ] [] [] = −− ( ) = −− ( ) = − – . 1 2 1 2 1 2 1 2 µµ µ δ Now, Ye k k ev j k k () ( ) ( ). – – ω = π +π δω δω + =π + π + = ∞ ∞ = ∞ ∞ ∑ ∑ 1 2 1 2 1 2 22 2 Since yn n n od[ ] [ ] [ ], = −+ µ δ 1 2 1 2 yn n n od[ ] [ ] [ ]. = −−+ − µ δ 1 1 1 2 1 2 As a result
yod[n]-yodln-1 both sides we then get Yod(ejo)=jo Yod (ejo)=3(+e-jo )or 1+ (e)=Y(e)+Y(e1o) 兀>8+2rk) (b)Let x[n] be the sequence with the DTFt X(eJo)=>2Ib(o-O. 2ck). Its inverse DTFT is then given by x[n]=2nb(o-OoJeJondo 3.4 Let X(eJo)=k_ 2n 8(0+2nk). Its inverse DTFT is then given by xIn 2(o)e 3.5(a)Let yIn]=gIn -nol, Then Y(ejo)=>yinJe-jon= >.le-jon 0∑gnle-jomn=e-jon oG(e°). n=- (b) Let hn]=cogn, then H(eJ)=∑hl-10=∑ eJoongIn]e-m glnJe"J(@-@ n= G(ej(-oo) (c)G(eJo )=>gnJe-Joon.Hence d do2-jng[n]ejon d(G(ejo )) d(G(ejo ) Therefore,Jdon=-∞ Eng(nge -jon. Thus the DTFT of ngIn is J do (d)ynl=gh=∑skn-kl. Hence Y(eJ)=∑∑skhn-ke-mn n=-k=-0 ∑kH(co)elo=H(c)∑kle-o= H(eJo)G(e Jo (e)yIn]=gInjh(n]. Hence Y(ejo )=>ginjh[nge-jon
43 ynyn n n n n n n od od [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]. − −= − −+ −− = + − 1 11 1 1 2 1 2 1 2 1 2 µµ δ δ δ δ Taking the DTFT of both sides we then get Ye e Ye e od j j od j j () () . ωω ω ω – − =+ ( ) − 1 2 1 or Y e e e e od j j j j () . ω ω ω ω = + − = − − − − − 1 2 1 2 1 1 1 1 Hence, Ye Y e Y e e k j ev j od j j k ( ) ( ) ( ) ( ). ωωω ω =+= δ ω − +π + π − =−∞ ∞ ∑ 1 1 2 (b) Let x[n] be the sequence with the DTFT X e k j o k () ( ) ω = −+ πδ ω ω π =−∞ ∞ ∑2 2 . Its inverse DTFT is then given by xn e d e o jn j n [] ( ) . = −= o − ∫ 1 2 2 π πδ ω ω ω ω π π ω 3.4 Let X e k j k ( ) ( ). ω = π +π δ ω =−∞ ∞ ∑ 2 2 Its inverse DTFT is then given by xn e d j n [] ( ) . – = π π = π π = π π ∫ 1 2 2 2 2 δω ω 1 ω 3.5 (a) Let y[n] = g[n – no], Then Y e y n e j jn n ( ) [] ω ω = − =−∞ ∞ ∑ = − − =−∞ ∞ ∑gn n e o j n n [ ] ω = − − =−∞ ∞ e gn e ∑ jn jn n o ω ω [ ] = e G e jn j o −ω ω ( ). (b) Let h n e g n j n [] [] = o ω , then H e h n e e g n e j jn n j n jn n ( ) [] [] o ω ω ωω = = − =−∞ ∞ − =−∞ ∞ ∑ ∑ = − − =−∞ ∞ ∑gn e j n n [ ] o ( ) ω ω = G ej ( ) o ( ) ω ω− . (c) Ge gn e j jn n ( ) [] ω ω = − =−∞ ∞ ∑ . Hence dGe d jng n e j j n n ( ) [ ] ω ω ω ( ) = − − =−∞ ∞ ∑ . Therefore, j dGe d ng n e j j n n ( ) [ ] ω ω ω ( ) = − =−∞ ∞ ∑ . Thus the DTFT of ng[n] is j dGe d j ( ) ω ω ( ) . (d) y[n] = g[n] * h[n] = g k h n k k [][ ] − =−∞ ∞ ∑ . Hence Y e g k h n k e j jn n k ( ) [][ ] ω ω = − − =−∞ ∞ =−∞ ∞ ∑ ∑ = = − =−∞ ∞ − =−∞ ∞ ∑ ∑ gk He e He gk e j jk k j jk k [] ( ) ( ) [] ωω ω ω = H e G e j j ( )( ) ω ω . (e) y[n] = g[n]h[n]. Hence Y e g n h n e j jn n ( ) [][] ω ω = − =−∞ ∞ ∑
Since gIn 2r JG(e)eJende we can rewrite the above dtft as (y=∑h=∑ JH*(ejo jon do H*(e io[ 2einJe-jion do= 2]H*(ejo)G(ejo)do 3.6 DTFT(x[nJ)=X(ejo )=2x[nge=jon (a) DTFTIX-m]}=∑x-nemo=∑ x mEt (b)DTET(x"(-n )=2x"Inle= jom [2 xl-nleion using the result of Part( a) Therefore DTFT(x*[])= X*(eJo) (c)dTFT(Re(x[nD)= DTFT ∫x叫+x*[n I x(eio)+x"(e io o] using the result of Part (d)DTFT(j Im(x[n])= DTFrj ={xeo)-x*(e-10 ( DTFT xcs[n])=DTFT2x0)+x时x小=x (f DTFTxca[n=DTFT xIn]-x*H -n 2 x(eio -x-"(e ion)=jx imf(eln) 3.7 X(e Jo)=>x[n]e Jon where x[n] eal For a real X(ejo )=X*(e-jo ) and IDFT X*(
44 Since gn Ge e d j jn [] ( ) = − ∫ 1 2π θ θ θ π π we can rewrite the above DTFT as Ye hne Ge e d j j n j jn n ( ) [] ( ) ω ω θθ π π π = θ − =−∞ − ∞ ∑ ∫ 1 2 = − − =−∞ ∞ − ∫ ∑ 1 2π θ θ ωθ π π Ge hne d j jn n ( ) [] ( ) = − − ∫ 1 2π θ θ ωθ π π Ge He d j j ( )( ) ( ) . (f) yn gn h n gn H e e d n n j jn [ ] [ ] *[ ] [ ] *( ) = = =−∞ ∞ =−∞ ∞ − − ∑ ∑ ∫ 1 2π ω ω ω π π = 1 2π ω ω ω π π H e gn e d j n j n *( ) [ ] =−∞ ∞ − − ∫ ∑ = 1 2π ω ω ω π π H e Ge d j j *( ) ( ) − ∫ . 3.6 DTFT{x[n]} = X e x n e j jn n ( ) [] ω ω = − =−∞ ∞ ∑ . (a) DTFT{x[–n]} =− = = − =−∞ ∞ =−∞ ∞ − ∑ ∑ x ne xme Xe j n n j m m j [ ] [ ] ( ). ω ωω (b) DTFT{x*[-n]} = −= − − =−∞ ∞ =−∞ ∞ ∑ ∑ x ne x ne j n n j n n *[ ] [ ] ω ω using the result of Part (a). Therefore DTFT{x*[-n]} = X ej * ( ). ω (c) DTFT{Re(x[n])} = DTFT xn x n Xe X e j j [ ] *[ ] ( ) *( ) + = + { } − 2 1 2 ω ω using the result of Part (b). (d) DTFT{j Im(x[n])} = DTFT j xn x n j Xe X e j j [ ] *[ ] ( ) *( ) − = − { } − 2 1 2 ω ω . (e) DTFT x n DTFT xn x n Xe X e Xe X e cs jj j re j [ ] [ ] *[ ] { } = ( ) * ( ) Re ( ) ( ). + − = + { } = { } = 2 1 2 ωω ω ω (f) DTFT x n DTFT xn x n X e X e jX e ca j j im j [ ] [ ] *[ ] { } = ( ) * ( ) ( ). − − = − { } = 2 1 2 ωω ω 3.7 Xe xne j jn n ( ) [] ω ω = − =−∞ ∞ ∑ where x[n] is a real sequence. For a real sequence, Xe X e j j ( ) * ( ), ω ω – = and IDFT X e x n j * ( ), *[ ]. – ω { } = −
(a)xre(ejo)= X(ejo)+x*(e jo). Therefore, IDFT Xre(ejooy =DFX(e)+X*(1_1 ={xn]+x*[-n]}={xn]+x[-n]}=xen (b)jX )2/(eJo)-X*(e Jo) Therefore, IDFTjXim =2Dx)x*)=风x=一x=x 3.8 a) X(eJo)= ∑ x[ne Jon. Therefore, X (b) From Part (a), X(e)=X(e u Therefore, Xe(e)=Xr(e o) (e) From Part(a), X(eJ)=X*(e- Ju ). Therefore, Xm(eJ)=-Xim(e- Jo) X(e -arg X( Jo) X.(e 3.9 x/n = I X(eJo)eJondo. Hence, x*[n] on (a) Since x[n] is real and even, hence X(eJo)=X*(eJo).Thus x-n]=1「xe)e-0mdo, Therefore,xn)=2(xn +x-n)=2 JX(e io)cos(on do,. Now x[n] being even, X(e Jo)=X(e Jo). As a result, the term inside the above integral is even, and hence x[n]=-[X(ejo )cos( on )do (b) Since x[n is odd hence x[n]=-x[-n]. Thus x[n]=-(x[n]-x[-n)= 1/ X(eJ)sin(on )do. Again, since xn
45 (a) X e Xe X e re jj j ( ) ( ) *( ) . ωω ω – = + { } 1 2 Therefore, IDFT X e re j ( ) ω { } = + { } = +− { } = +− { } = 1 2 1 2 1 2 IDFT X e X e x n x n x n x n x n j j ev ( ) * ( ) [ ] *[ ] [ ] [ ] [ ]. ω ω – (b) jX e X e X e im jj j ( ) ( ) *( ) . ωω ω – = − { } 1 2 Therefore, IDFT jX e im j ( ) ω { } = − { } = −− { } = −− { } = 1 2 1 2 1 2 IDFT X e X e x n x n x n x n x n j j od ( ) * ( ) [ ] *[ ] [ ] [ ] [ ]. ω ω – 3.8 (a) Xe xne j jn n ( ) [] ω ω = − =−∞ ∞ ∑ . Therefore, X e x n e X e j jn n j * ( ) [ ] ( ), ω ωω – = = =−∞ ∞ ∑ and hence, X e xne Xe j jn n j * ( ) [ ] ( ). – – ω ωω = = =−∞ ∞ ∑ (b) From Part (a), X e X e j j ( ) * ( ). ω ω – = Therefore, X e X e re j re j ( ) ( ). ω ω – = (c) From Part (a), X e X e j j ( ) * ( ). ω ω – = Therefore, X e X e im j im j ( ) ( ). ω ω – = − (d) Xe X e X e j re j im j () () () ω ωω = + 2 2 = + Xe X e re j im 2 2 j () () – – ω ω = − X e j ( ) ω . (e) arg ( ) tan ( ) ( ) tan ( ) ( ) X e arg ( ) X e X e X e X e X e j im j re j im j re j ω j ω ω ω ω ω = =− =− − − − − 1 1 3.9 x[n] = 1 2π ω ω ω π π Xe e d j jn ( ) − ∫ . Hence, xn Xe e d j jn *[ ] *( ) – = − ∫ 1 2π ω ω ω π π . (a) Since x[n] is real and even, hence X e X e j j ( ) *( ) ω ω = . Thus, x[– n] = 1 2π ω ω ω π π Xe e d j jn ( ) − − ∫ , Therefore, xn xn x n Xe nd j [ ] [ ] [ ] ( )cos( ) . = +− ( ) = − ∫ 1 2 1 2π ω ω ω π π Now x[n] being even, X e X e j j () ( ) ω ω – = . As a result, the term inside the above integral is even, and hence xn Xe nd j [ ] ( )cos( ) = ∫ 1 0 π ω ω ω π (b) Since x[n] is odd hence x[n] = – x[– n]. Thus x[n] = 1 2 ( ) xn x n [] [ ] − − = j Xe nd j 2π ω ω ω π π ( )sin( ) − ∫ . Again, since x[n] = – x[– n]
X(ejo)=-X(e-jo ) The term inside the integral is even, hence x[n =3[X(ejo )sin( on )do 3.10 x[n]=a"cos(oo n+)u[n]=Aa"/eooejo +e-joo"ejo [n] Cejp(aejoo )un+re-o(ae-joo u[n].Therefore j 1-ae -JOe- jo 3.11 Let x[n]=a H[n], a ane" jon=>a"e" jon-1-ae Joo-a2e Jc (e)xs(e)=∑na"em=∑ Joon-20 ae Jo (x6(e1)=∑a"e0=∑
46 Xe Xe j j () ( ) ω ω – = − . The term inside the integral is even, hence x n j Xe nd j [ ] ( )sin( ) = ∫ π ω ω ω π 0 3.10 xn n n A eee e n n o n j nj j n j [ ] cos( ) [ ] [ ] = += + − − α ω φµ α µ ωφ ω φ 0 0 2 = A ee n A ee n j j n j j o o 2 2 φω φ ω (αµ αµ ) + ( ) − − [ ] [ ]. Therefore, X e A e e e A e e e j j j j j j j o o ( ) ω φ ω ω φ ω ω α α = − + − − − − − 2 1 1 2 1 1 . 3.11 Let x[n] = αµ α n [ ], . n <1 From Table 3.1, DTFT{x[n]} = X e e j j () . – ω ω α = − 1 1 (a) Xe n e e e e j n jn n n jn n j n jn n 1 1 1 0 () [ ]1 ω ω ωω ω = + = =+ αµ α α α − =−∞ ∞ − =− ∞ − − = ∞ ∑∑ ∑ = + − = − − − α α α α α ω ω ω ω 1 1 1 1 1 e e e e j j j – – j . (b) xn n n n 2[ ] [ ]. = αµ Note that x n n x n 2[ ] [ ]. = Hence, using the differentiation-in-frequency property in Table 3.2, we get Xe j dX e d e e j j j 2 j 2 1 ( ) ( ) ( ) ω ω ω ω ω α α = = − − − . (c) x n n M otherwise n 3 0 [ ] , , , . = ≤ α Then, Xe e e j n jn n M n jn n M 3 0 1 ( ) ωω ω = + α α − = − − =− − ∑ ∑ = 1 1 1 1 1 1 1 − − + − − +− + − − − − − α α α α α ω ω ω ω ω M jM j MjM M jM j e e e e e ( ) . (d) Xe e e e e j n n jn n n jn j j 4 3 0 2 2 ( ) 1 ω ω ω ωω – – = = −− − α α αα = ∞ = ∞ − − ∑ ∑ = − −− − − 1 − − 1 1 2 2 α α α ω ω ω e e e j j j . (e) Xe n e n e e e j n n jn n n jn j j 5 2 0 22 1 ( ) 2 ω ω ω ωω – – = = −− α α αα =− ∞ = ∞ − − ∑ ∑ = − − − − − α − − α α α ω ω e ω ω e e e j j j j ( ) . 1 2 2 22 1 (f) Xe e e e e j n n jn m m jm m m j m 6 j 1 1 0 1 1 1 1 ( ) 1 ω ωω ω – ω αα α α = = = −= − − =−∞ − − = ∞ − = ∞ ∑∑ ∑ − = − e e j j ω ω α
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)