1、第14章線性動(dòng)態(tài)電路的復(fù)頻域分析14.1拉普拉斯變換的定義14.2拉普拉斯變換的基本性質(zhì)本章重點(diǎn)14.3拉普拉斯反變換的部分分式展開14.4運(yùn)算電路14.5用拉普拉斯變換法分析線性電路首 頁 重點(diǎn)(1) 拉普拉斯變換的基本原理和性質(zhì)(2) 掌握用拉普拉斯變換分析線性電路的方法和步驟(3) 網(wǎng)絡(luò)函數(shù)的概念(4) 網(wǎng)絡(luò)函數(shù)的極點(diǎn)和零點(diǎn)返 回拉普拉斯變換的定義14.11. 拉氏變換法拉氏變換法是一種數(shù)學(xué)積分變換，其核心是把時(shí)間函數(shù)f(t)與復(fù)變函數(shù)F(s)聯(lián)系起來，把時(shí)域問題通過數(shù)學(xué)變換為復(fù)頻域問題，把時(shí)域的高階微分方程變換為頻域的代數(shù)方程以便求解。應(yīng)用拉氏變換進(jìn)行電路分析稱為電路的復(fù)頻域分析法，又

2、稱運(yùn)算法。返 回上 頁下 頁相量法i1 + i2 = i時(shí)域的正弦運(yùn)算變換為復(fù)數(shù)運(yùn)算正弦量= I&I&+ I&相量12運(yùn)算法拉氏變換對應(yīng)f(t)(時(shí)域原函數(shù))F(s)(頻域象函數(shù))返 回上 頁下頁2.拉氏變換的定義定義 0 , )區(qū)間函數(shù) f(t)的拉普拉斯變換式： 正變換 F(s) = L f (t) ，f (t) = L-1F(s)簡寫ss = s + jw復(fù)頻率返 回上頁下 頁反變換F (s) = +f (t)e-stdt0 -1c+ j f (t) =F (s)estds2jc- j3.典型函數(shù)的拉氏變換(1)單位階躍函數(shù)的象函數(shù)f (t) = e (t)=-stF (s) = Le

3、(t) =e (t)edtedt-st-00-0-= - 1 e-st= 1ss返 回上 頁下 頁+-stF (s) = 0f (t)edt-(2)單位沖激函數(shù)的象函數(shù)f (t) = d (t)0d (t) =d (t) ed-st+-stF (s) = L=dt(t)edt0 -0 -= e-s0= 1(3)指數(shù)函數(shù)的象函數(shù)f (t) = eate=0- 1s - aat-stF (s) = L= -at-( s-a )teeedt0 -1=s - a返 回上 頁下 頁14.2 拉普拉斯變換的基本性質(zhì)1.線性性質(zhì)L f1 (t) = F1(s),L f2 (t) = F2 (s)若LA1 f

4、1(t) + A2 f2 (t)= A1L f1(t)+ A2Lf2 (t)= A1F1 (s) + A2 F2 (s)則返 回上 頁下頁求:f (t) = K (1- e-at )的象函數(shù)例1F (s) = L K - LKe-at = K -KKa=s + as(s + a)s求:f (t) = sin(w t)的象函數(shù)例2F(s) = Lsin(t)= L 1 (e jw tw t- e- j) 解 2 j= 1 1- 1= w2 j s - jws + jw s+ w22返 回上 頁下頁解2. 微分性質(zhì)若： L f (t)= F(s)則：Ldf (t) = sF (s) - f (0)

5、-dtdf (t)e-stdt =Ldf (t) =-stedf (t)0-證dtdt0-0- = e-st f (t)f (t)(-se-st )dt0 -0-= - f (0- ) + sF (s)若s足夠大返回上 頁下 頁利用udv = uv - vdu例 利用導(dǎo)數(shù)性質(zhì)求下列函數(shù)的象函數(shù)(1)f (t) = cos(w t)的象函數(shù)dsin(wt) = wcos(wt) 解dtcos(wt) = 1 d(sinwt)wdtLcoswt = L 1d (sin(wt)w dtsw=1 s- 0=w + w2+ w 2s2s2返 回上 頁下 頁f (t) = ( t)的象函數(shù)(2) 解Le

6、(t) = 1d (t) = de (t)sdtLd (t)= L de (t)= s 1 - 0 = 1dts返 回上 頁下 頁3.積分性質(zhì)若：L f (t) = F(s)f (x )dx = 1 F (s)t則：Ls應(yīng)用微分性質(zhì)0-tf (t)dt = f(s)L證令-0L f (t) = Lf (t)dtdtf (t)dt-dt-00t =0 -tf(s)F (s) = s0-f(s) = F (s)s返 回上 頁下頁求:f (t) = te ( t)和f (t) = t2e (t)的象函數(shù)例 解Lte (t) = LLt 2e (t)=11s 1s20-e (t)dt=s=t2L2td

7、t =s30返 回上 頁下 頁4.延遲性質(zhì)若： L f (t) = F(s)-st則：L f (t - t0 )e (t - t0 ) = e0 F(s)返 回上 頁下 頁5.拉普拉斯的卷積定理若： L f1(t) = F1(s)L f2 (t) = F2 (s)tf1 (t - x ) f2 (x )dx則： L f (t) * f(t) =L120= F1 (s)F2 (s)返 回上 頁下 頁常用時(shí)間函數(shù)及其象函數(shù)：Ad (t )AAe (t )A / sAe-atA s + at 1/ s 2sin(wt ) ws 2+ w 2cos(wt )ss 2+ w 214.3 拉普拉斯反變換的

8、部分分式展開由象函數(shù)求原函數(shù)的方法： 12jc+ j(1)利用公式f (t) =F (s)edsstc- j(2)對簡單形式的F(s)可以查拉氏變換表得原函數(shù)(3)把F(s)分解為簡單項(xiàng)的組合F (s) = F1 (s) + F2 (s) + Fn (s)f (t) =f1 (t) + f2 (t) + + fn (t)部分分式展開法返 回上 頁下頁（真分式）討論(1) 若D(s) = 0有n個(gè)單根分別為p1 pn利用部分分式可將F(s)分解為：待定常數(shù)K1K2KnF (s) =+ +s - p1s - p2s - pnf (t) =+Kep tp tp tK eK e12n12n返 回上 頁

9、下 頁N (s)a sm + a sm-1 + aF (s) = 01m (n m) D(s)b sn + b sn-1 + b01n待定常數(shù)的確定：K2Kn(s -p )F (s) =+ (s -+ +s - pKp )111s -pn 令s = p12返 回上 頁下頁K=N ( pi )iD ( p )i方法2Ki= F (s)(s - pi ) s= pi = 1、2、3L、ni方法1= lim N (s)(s - pi )求極限的方法KiD(s)spiN (s)(s - p ) + N (s)=lim iD (s)spi返 回上 頁下 頁K=N ( pi )iD ( p )i4s +

10、5求 F (s) =例的原函數(shù)+ 5s + 6s2 解法1- 3K1K274s + 5=+=+F (s) =s + 2s + 3s + 2s + 3+ 5s + 6s2= 4s + 5= 4s + 5= 7K= -3Ks=-3s + 22S =-2s + 31返 回上 頁下 頁 解法2N ( p1 )= 4s + 5K= -3s=-212s + 5D ( p )1= 4s + 5N ( p2 )K= 7s=-322s + 5D ( p)2- 37F (s) =+s + 2s + 3f (t) = -3e-2te (t) + 7e-3te (t)返回上 頁下 頁 p1 = a +jw(2) 若D

11、(s) = 0具有共軛復(fù)根 p= a -jw2F (s) = N (s) =N (s)(s - a - jw)(s - a + jw)D(s)K1s - a - jwK2s - a + jw=+K1、K2也是一對共軛復(fù)數(shù)注意返回上 頁下 頁K= F (s)(s -a m jw)=N (s)1，2s=a jwD(s)s=a jwsF (s) =例的原函數(shù)f (t)求+ 2s + 5s2= -1 j2 解p+ 2s + 5 = 0 的根s21,2s= 0.5 - j0.5 = 0.52 - 45oKs=-1+ j21s - (-1- 2 j)s= 0.5245oKs=-1- j22s - (-1+

12、 2 j)N (s)=s或：K= 0.52 - 45os=-1+ j212s + 2D (s)2e-tf (t) =cos(2t - 45o )返 回上 頁下 頁(3) 若D(s) = 0具有重根a sm + a sm-1 + aF (s) = 01m (s - p )n1K11K12K1n-1K1nF (s) =+ +(s - p )n-1s - p(s - p )2(s - p )n1111= (s - p )n F (s)Ks= p11n1d=(s - p )F (s)nK1n-1s= p11dsMdn-1(s - p1 )F (s)1=nK11s= pdsn-1(n -1)!1返回上

13、頁下頁s + 4求：F (s) =例f (t)的原函數(shù)s(s +1)2s + 4= K1K21K22 解F (s) =+s(s +1)2(s +1)(s +1)2sK= SF (S ) = s + 4= s + 4= 4 K= -3s=01(s +1)2s=-122s=d (s +1)2 F (s)s + 4dK=ds= -4 s=-1s=-121dssf (t) = 4 - 4e-t- 3te-t返 回上 頁下頁 由F(s)求f(t) 的步驟：小結(jié)jn =m 時(shí)將F(s)化成真分式和多項(xiàng)式之和F (s) = A + N0 (s)D(s) 求真分式分母的根，將真分式展開成部分分式K1K2KnF

14、 (s) = A + +s - p1s - p2s - pn 求各部分分式的系數(shù) 對每個(gè)部分分式和多項(xiàng)式逐項(xiàng)求拉氏反變換返 回上 頁下頁+ 9s +11s2求：F (s) =例的原函數(shù)+ 5s + 6s24s + 5+ 9s +11s2= 1+s27F (s) =+ 5s + 6+ 5s + 6s2- 3= 1+s + 2s + 3f (t) = d (t) + (7e-3t- 3e-2t )返 回上 頁下 頁解14.4運(yùn)算電路1.基爾霍夫定律的運(yùn)算形式基爾霍夫定律的時(shí)域表示：i(t) = 0u(t) = 0根據(jù)拉氏變換的線性性質(zhì)得KCL、KVL的運(yùn)算形式 I (s) = 0U (s) = 0

15、對任一結(jié)點(diǎn)對任一回路1.基爾霍夫定律的運(yùn)算形式返 回上 頁下頁2.電路元件的運(yùn)算形式 電阻R的運(yùn)算形式i(t)I (s)電阻的運(yùn)算電路+RRuR(t)U (s)-時(shí)域形式： u=RiU (s) = RI (s)I (s) = GU (s)取拉氏變換返 回上 頁下 頁Z (s) = RY (s) = Gu = L di 電感L的運(yùn)算形式時(shí)域形式：dtLi(t)取拉氏變換,由微分性質(zhì)得+u(t)-U (s) = L(sI (s) - i(0)-= sLI (s) - Li(0 -)I (s) = U (s) + i(0 -)sLssLL的運(yùn)算電路i(0)s-I(s )U(s)+-返 回上頁下 頁Z

16、 (s) = sLY (s) = 1 sLI(s)sLLi(0 - )-+U(s)- 電容C的運(yùn)算形式時(shí)域形式：u = u(0) + 1Cti(t)i(x ) dx-C0-+u(t)-取拉氏變換,由積分性質(zhì)得u(0 - )1U (s) =I (s) +sCsI (s) = sCU (s) - Cu(0)-1/sCC的運(yùn)算電路Cu(0-)I(s )U(s)+-返 回上 頁下頁Z (s) = 1 sCY (s) = sCI(s)1/sCu(0 - )s+-+U(s)- 當(dāng)RLC元件初始儲能為零時(shí)：（初始值為0）iL(t)LIL(s)sL+uL(t)-+sLIL(s)-i (t)CI (s)1CCs

17、C+uC(t)-+UC(s)-iR(t)RIR(s)R+uR(t)-+UR(s)- 耦合電感的運(yùn)算形式時(shí)域形式：i2i1Mudi1 + M di2= L+*L11dtdtuLu2_112_di2 + M di1u= L22dtdt取拉氏變換,由微分性質(zhì)得U1 (s) = sL1I1 (s) - L1i1 (0 - ) + sMI2 (s) - Mi2 (0 -)U2 (s) = sL2 I2 (s) - L2i2 (0 - ) + sMI1 (s) - Mi1 (0 - ) 互感運(yùn)算阻抗返 回上 頁下頁ZM (s) = sM YM (s) = 1 sMU1 (s) = sL1I1 (s) -

18、L1i1 (0 - ) + sMI2 (s) - Mi2 (0 - )U2 (s) = sL2 I2 (s) - L2i2 (0 - ) + sMI1 (s) - Mi1 (0 -)I2 (s)sMI1 (s)+sL1-sL2-L2i2 (0 -+U1 (s)U2 (s)-L1i1 (0 - )+-+Mi1 (0- )Mi2 (0- )返 回上 頁下頁 受控源的運(yùn)算形式時(shí)域形式：i2= bi1取拉氏變換I2 (s) = bI1 (s)i1+u1_i2+u2_Rb i1I1 (s)+I2 (s)+RU1 (s)_U2 (s)_bI (s)1返 回上 頁下 頁3.運(yùn)算電路的畫法u 在換路前的電路中

19、計(jì)算及；L短路C開路開關(guān)未動(dòng)作u 將元件換為復(fù)頻域模型，繪運(yùn)算電路;Li(0 - )sLI(s)-+U(s)+-u(0 - )s1/sCI(s)+-+U(s)返回上 頁下 頁iL (0- )uC (0- )給出圖示電路的運(yùn)算電路模型。例50V -20W+5Wt0 運(yùn)算電路返 回上 頁下 頁+25/s-及；14.5應(yīng)用拉普拉斯變換法分析線性電路1. 運(yùn)算法的計(jì)算步驟u 在換路前的電路中計(jì)算u 將元件換為復(fù)頻域模型，繪運(yùn)算電路;u 據(jù)一般的電路分析方法對運(yùn)算電路進(jìn)行分析，計(jì)算響應(yīng)的象函數(shù)F(s)；u 按部分分式法展開F(S)；u 進(jìn)行反變換，得出時(shí)域響應(yīng)f(t)。返 回上 頁下頁iL (0- )uC (0- ) = 0 ，求uC(t)、iC(t)。is= d (t),圖示電路uc (0-例1+I(s) = 1sucUc(s)- 解畫運(yùn)算電路R1U(s) =I(s)CsR +1/ sCRsCu=1 e-t / RC (t 0)=cC返 回RC(s +1/ RC)上 頁下 頁IC (s)R1/sCisRC+I(s) = 1sUc(s)RsC-I(s