12第四节傅里叶变换的性质设本节中,各函数的傅氏变换均存在.基本性质线性性质设F1()=F[f1(t)],F2()=F[f2(t)],,为常数,则F[f1(t)+f2(t)]=F1()+F2()F1[F1()+F2()]=f1(t)+f2(t)3平移性质若F()=F[f(t)],t0,0为实常数,则)(e0iFtF[f(tt0)]F1[F(0)])(e0itft线性性质可直接由积分的线性性质推出.证明tttftde)(i0F[f(tt0)]4uuftuttude)()(i00)(ede)(e00iiiFuuftuttttftde)(i0F[f(tt0)]),0()i(1)(00为实数G例求G()的傅氏逆变换g(t).解由平移性质F1[F(0)])(e0itft5F1[F(0)]F1[F()]tie0g(t)=F1[G()]=F1])(i1[0t0ieF1]i1[i1)(,0,e0,0)(Ftttft时又P120例2.1.0,0,0,210,,e)()i(0ttttgt所以6伸缩性质(相似性质)若F[f(t)]=F(),a为非零实常数,则)(||1)]([aFaatfF证明tatftde)(iF[f(at)]令u=at,则有当a0时F)(1de)(1)]([iaFauufaatfua7F[f(t)]=iF[f(t)]当a0时F)(1de)(1)]([iaFauufaatfua).(||1)]([aFaatf综上F微分性质则若,0)(lim||tft则若,)1,,1,0(0)(lim)(||nktfkt8F[f(n)(t)]=(i)nF[f(t)]根据傅氏变换公式,并利用分部积分得证明得由0)(lim||tft0|)(|lim|e)(|lim||||tftfttitttftftde)()]([iFttftfttde)(ie)(ii9ttftfttde)(ie)(ii0|e)(|limi||tttfttftde)(ii=iF[f(t)]反复利用分部积分可证F[f(n)(t)]=(i)nF[f(t)]傅氏逆变换的微分性质F()=F[itf(t)]10F(n)()=(i)nF[tnf(t)]积分性质则若设,0)(lim,d)()(tgftgtti1)]([tgFF[f(t)].11)(]d)([dd)(tffttgt由于证明F[f(t)]=F[g(t)]=iF[g(t)]F[g(t)]=F[f(t)]i1F[f(t)]=iF[f(t)]i1]d)([tf即FF[f(t)]12例求解微积分方程)(d)()()(thttxctbxtxat其中t+为常数,为a,b,c为常数,h(t)为已知函数,且傅氏变换存在.解记F[x(t)]=X(),F[h(t)]=H(),对方程两边取傅氏变换,得)()(i)()(iHXcbXXa13)(i)()(cabHX解得求上式的傅氏逆变换,得de)(i)(π21)(itcabHtx这就是所求的微分方程的积分形式的解.14卷积与卷积定理卷积定义设函数f1(t),f2(t)在(,+)内绝对可积,则积分d)()(21tff称为f1(t)与f2(t)的卷积(也叫褶积),记为d)()()()(2121tfftftf卷积的性质15根据定义,可以验证卷积满足如下性质:交换律成立:f1(t)f2(t)=f2(t)f1(t)结合律成立:f1(t)[f2(t)f3(t)]=[f1(t)f2(t)]f3(t)分配律成立:f1(t)[f2(t)+f3(t)]=f1(t)f2(t)+f1(t)f3(t)16求f1(t)与f2(t)的卷积.例.0,e0,0)(,0,10,0)(21tttftttft若解由卷积的定义,有d)()()()(2121tfftftf若f1()f2(t)0,只要求下式成立:ttt00,00即17当t0时,f1()f2(t)=0,有f1(t)f2(t)=0当t0时,f1()f2(t)0,因此ttfftftf02121d)()()()(tttee1d10)(卷积定理若F[f1(t)]=F1(),F[f2(t)]=F2(),则有F[f1(t)f2(t)]=F1()F2()18或F1[F1()F2()]=f1(t)f2(t)证明tdtfftde])()([i21ttfftdde)(e)()(i2i1d]de)([e)()(i2i1ttfft=F1()F2()ttftftftftde)]()([)]()([i2121F19这个定理说明,两个函数卷积的傅氏变换等于这两个函数傅氏变换的乘积.同理可得即两个函数乘积的傅氏变换等于这两个函数傅氏变换的卷积除以2.例求下列函数的卷积(0,0)tttftttfπsin)(,πsin)(21)()(π21)]()([2121FFtftfF20解查附录1傅氏变换简表第3式,得到设F[f1(t)]=F1(),F[f2(t)]=F2().||,0||,1)(1F其他,0||,1)(]πsin[00FttF||,0||,1)(2F21由卷积定理有ttπsinF1[F1()F2()]=f1(t)f2(t)=.||,0,||,1)()(21FF(其中=min|,|)因此22