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1、第三章作業(yè)題答案作業(yè):%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3.2設(shè)X(ej<0)是序列的離散時(shí)間傅里葉變換,利用離散時(shí)間傅里葉變換的定義及性質(zhì),求下列各序列的離散時(shí)間傅里葉變換。(4)g(n)=x(2ti)解:利用DFT的定義進(jìn)行求解。G(嚴(yán))=乞8(小嚴(yán)(這是一種錯(cuò)誤的解法,正確的如下所示。)=£x(2n)e~j^+8?(0加=一8=x(嚴(yán))G(R")=》g(n)e~J(0,tm=2n=為x(2n)e-j(,At=為”=Y>加=一8(注意,此處n為奇數(shù)的項(xiàng)為零。)二血)+(j)“班訃s?=-<*>/=£丄[兀⑺)+嚴(yán)兀
2、何]嚴(yán)咗二丄x(嚴(yán)勺+丄x(小必詞)=-X(e^2)+-X(-^2)%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3.3試求以下各序列的離散時(shí)間傅里葉變換。8]“兀5(〃)=》(丁)力(斤一3加)〃戶(hù)04解:利用DTFT的定義和性質(zhì)進(jìn)行求解。X(嚴(yán))二工兀5(〃0泅4-oooo1=工工G)和-3加0曲?:=-
3、1,2,0,-3,2,1;料=0丄2,3,4,5咖%其它它的離散時(shí)間傅里葉變換為X(R")。不具體計(jì)算X(£^),試直接確定下列表達(dá)式的值。(3)解:不計(jì)算X(R0),解法如下:X(n)=-LrX(ejM}ejMtdco令n=0,貝!J:x(0)=_L^X(eJa})dco=-l因此,?:xW?)da)=-2兀%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3.11證明:(1)若序列兀⑺)是實(shí)偶函數(shù),則其離散時(shí)間傅里葉變換X(ejM)是血的實(shí)偶函數(shù)。(2)若序列兀⑺)是實(shí)奇函數(shù),則其離散時(shí)間傅里葉變換X(ei(0)是純虛數(shù),且是。的
4、奇函數(shù)。解:此題求解需要利用DTFT的性質(zhì)DTFT[x(-n)]和IDTFT^(嚴(yán))]首先,(1)當(dāng)班〃)為實(shí)偶序列時(shí):x(-n)=x(n)根據(jù)DTFT的性質(zhì),可知:DTFT[x(-n)=X(e-j(0)因此:X(e-jM)=X(ej(,))因此,X(R”)為血的偶函數(shù)。此外,DTFT性質(zhì),IDTFT^X^(嚴(yán))]=x(-n)=x(n)=IDTFT^X(嚴(yán))]因此,X(ej(0)為實(shí)函數(shù)。綜上,X(R”)為Q的實(shí)偶函數(shù)。(2)利用同樣的性質(zhì)可以證明若序列兀⑺)是實(shí)奇函數(shù),則其離散時(shí)間傅里葉變換X(RQ)是純鹿數(shù),且是e的奇函數(shù)。%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
5、%%%%%%%%%%%%%%%%%%%%%%%3.16若序列x(h)是因果序列,己知其離散時(shí)間傅里葉變換X(£溝)的實(shí)部Xr(丘溝)為Xr("")=1+COS69求序列x(n)及其離散時(shí)間傅里葉變換X(eja))o解:此處的條件為:班〃)是因果序列。因此此題的求解必然使用因果序列的對(duì)稱(chēng)性。注意:此處并沒(méi)有提及班冊(cè)為實(shí)序列,因此,此題需加如條件兀5)為實(shí)序列。XR^)=DTFT[xe^=1+COS(69)=1+[嚴(yán)+嚴(yán)]/2注意,在常見(jiàn)序列DTFT中,A(e>)=DTFT[^(n)]=lo根據(jù)位移特性,DTFT[x(n-n())]=X。因此,=5(比)+[5(比一1)+5(比+1)]/2因此可得
6、:0n<01n=0x(n)=<訃)n=0=<1n=12xjn)n>00elsex(嚴(yán))=£兀(町廠劭=1+0%H=_8%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3.17若序列兀⑺)是實(shí)因果序列,x(0)=1,已知其離散時(shí)間傅里葉變換X(嚴(yán))的虛部X/(R")=-sinp求序列%(?)及其離散時(shí)間傅里葉變換x(R")。解:X,(ejM)=—sine二一丄「嚴(yán)一嚴(yán)八)2八1FT[_x(f(町]=兀(嚴(yán))=-十加-嚴(yán)]=£x°(町嚴(yán)乙打=Y>0,Z2<01,n=0n=1else2xo(n),n>0[0,-re^z'x+re'x(n
7、)=<£(〃),n=0=<,X(£溝)二£兀(町廠加=l+£%斤二YO%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3.21試計(jì)算下列各序列的z變換和相應(yīng)的收斂域,并畫(huà)出各自相應(yīng)的零極點(diǎn)分布圖。(5)x5(n)=Arncos(?〃+(p)u(n),0