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 <1. From Table 3. 1, DTFT(x[n])= X(eo) a)x)=∑"un+lmn=∑a"em=a-+∑d"e -1n Jo (b)x2[n]=no u[n]. Note that x2[n]=nx[n]. Hence, using the differentiation-in-frequency property in Table 3. 2, we get X2(eJo)=j dX(e) ae Jo do -ce-1o)2 (c)x3叫= (M+1) +oMejoM1-a-Me-joMS-M (d)X4(eJo)=>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