信號(hào)與系統(tǒng)習(xí)題3-1

1、第三章第三章拉普拉斯變換拉普拉斯變換一、公式及要點(diǎn)1.基本概念雙邊拉普拉斯變換（31）()()stFsftedt??????單邊拉普拉斯變換（32）0()()stFsftedt????拉普拉斯反變換（33）1()()2jstjftFsedsj?????????式中s復(fù)變量，亦稱“復(fù)頻率”，；的像函數(shù)。sj????()Fs()ft在S平面上，使積分（31）使絕對(duì)收斂，即（34）()stftedt???????的值的范圍陳為收斂域。?2.常

2、見函數(shù)的拉普拉斯變換公式序號(hào)原函數(shù)，()ft0t?像函數(shù)1()[()]Fsft???1()t?12()ut1s3t21s4ate1sa?5sint?22s???6cost?22ss??3.拉普拉斯反變換⑴部分分式展開法首先將展成部分分式，然后將各部分分式逐項(xiàng)進(jìn)行反拉斯變換，最后疊加起來即為()Fs原函數(shù)。()ft111012()()()()mmmmnnbsbsbsbFsaspspsp???????????1212()()()nnkkks

3、pspsp????????()()|iiispkspFs???(12)in??若的極點(diǎn)為階極點(diǎn)，則()Fs1pK1()()()()()()KNsNsFsDsspBs???111121111()()()()()kKKkkkAsspspspBs??????????⑴⑵1()ln()sFss??221()(1)(2)sseeFsss???????解：(1)S域微分性質(zhì)()dFsds??()tft???因?yàn)閷?duì)方程兩邊同時(shí)求導(dǎo)()ln(1)lnF

4、sss???()111dFsdsss???1()[](1)()tdFseutds?????所以1()(1)()tfteutt???⑵因?yàn)?111()(1)(2)12Fsssss???????所以21()[]()ttfteeut????根據(jù)時(shí)間平移特性=2()ft111()(1)(2)ftftft????(1)2(1)(2)2(2)2[](1)[](2)[]()tttttteeuteeuteeut??????????????????33用

5、部分分式展開法，求下列像函數(shù)的原函數(shù)⑴142()(7)(4)(2)Fssss????解：⑴142()(7)(4)(2)Fssss????147215(7)45(2)sss??????故111()[()]ftFs???7421421(7)()55ttteeeut??????34已知像函數(shù)，試求其原函數(shù)31()1sseFse?????()ft解：像函數(shù)中表示原函數(shù)中的時(shí)間延遲，而其分母中則表示時(shí)間函數(shù)()Fsse?31se??以3為周期的重

6、復(fù)函數(shù)1原函數(shù)?()t?像函數(shù)原函數(shù)1se???()(1)tt????所以31()1sseFse??????()()(1)(3)(4)(6)(7)fttttttt???????????????????或直接將展開()Fs31()1sseFse?????369(1)(1)sssseeee???????????34671ssssseeeee?????????????所以()()(1)(3)(4)(6)(7)fttttttt?????????