13動態(tài)電路的復頻域分析

1、第十三章網(wǎng)絡的復頻域分析法拉氏變換運算法 網(wǎng)絡函數(shù)相量分析法：頻域(相量)時域時域復頻域分析法：復頻域時域時域拉氏變換拉氏逆變換解高階電路變換過程中已經(jīng)以某種形式計入原微分方程的初始條件。優(yōu)點：13.1變換變換的定義13.1.1在數(shù)學中，F(xiàn) (s) 正變換定義為：f (t)estdt0式中，s=+j為復數(shù)變量，通常稱為復頻率。f(t)：原函數(shù)；F(S)：f(t)在s域中的象函數(shù)。逆變換定義為： 1 2j jf (t) stF (s)e dt j典型函數(shù)的拉氏變換例1：求F (s) L沖擊函數(shù)(t)的象函數(shù)。 (t)est (t)f (t)estdtdt000 (t)dt例2：求 10階躍函數(shù)的

2、象函數(shù)。st1sF (s) L1(t)f (t)estdt1(t)edt001s est dtest 00例3：求指數(shù)函數(shù)e-at的象函數(shù)。 1F (s) Leat f (t)estdtes a0013.1.2 拉氏變換的基本性質(zhì)1 線性性質(zhì)：K1 f 1 ( t )+ K2 f 2 ( t )= K1F1 (s)+ K2F2 (s)例1：求cost的象函數(shù)。e jte jte jte jtsin t 由公式得：cost 2 j21212112Lcost jtjtjtjt LeeLeLe2 s 11 11 2s22 s j2 s j Lsin t 2s2LL d2微分性質(zhì)：f (t) sF (

3、s) f (0)dt d dtd f (t)dtLf (t)ststeedf (t)dt00f (t) s estf (t)dest f (0) sF (s)00微分定理說明：時域中的求導運算，對應復頻域中乘以s的運算，并以f(0-)計入原始條件。d 2ddd Lf (t) sLf (t) f (0)Lf (t)dtdtdtdt 2 s2F (s) sf (0) f (0) ssF (s) f (0) f (0) d n f n 1n 2(n 2)(n1)n S F (S ) Sf (0 ) Sf (0 ) Sf(0 ) f(0 )ndt3積分性質(zhì)：f 1(0)1tf (t)dt F (s)

4、sLs0(0) 1ff (t)dt d 是函數(shù)f(t)的積分式在t=0-時的值。tf (t)dt f (t)dt d dttf (t)dt F (s)Lttf (t)dt F (s)sLf (t)dt00f (t)dtf 1(0)11tf (t)dt F (s) sF (s) sLssf (t)dt 1 F (s)stL0例：L1(t) 1s1(t)dt t1(t)t0Lt1(t) L1(s) 1s2s4時域延遲定理 stL f (t t0 )1(t t0 ) e0 F (s)求f (t) e(t3)1(t 3)的象函數(shù)1f (t) e(t3)1(t 3) e3ss 113.1.3拉氏反變換：

5、sm1b bsmbF1 (s) m1F (s) m0sm1a asnF (s)a2m1n0F1 (s) K F10 (s)F (s) (n m)F2 (s)F2 (s)1單根的拉氏逆變換F1 (s) F1 (s)F (s) (s s1)(s s2 )(s sn )F2 (s)KnK1K2(s s1 )(s s2 )(s sn )KnK2(s s )F (s) K (s s )111(s s )(s s)2nK1 (s s )F (s)Ki (s si )F (s)(i 1,n)1s sssi12例1：求F (s) 的原函數(shù)。2 2s 5s22F (s) 2s 5s2(s 1)2 22f (t)

6、 L1F (s) et sin 2t(t 0)K12K2F (s) 2s 5s (1 j2)s (1 j2)s22 j 112K s (1 j2)K2 j1 2s 5s22s( 1 j 2)j 1j 1f (t) L1F (s) L12 2s (1 j2)s (1 j2)e j 2t e j 2t j 1 e( 1 j 2)t j 1 e( 1 j 2)t te2sin 2t2(t 0)2 j ets 3例2：求F (s) 的原函數(shù)(s 1)(s2 2s 5)F (s) s 3 as b 0.5(s 1)(s2 2s 5) 2s 5s 1s2 (a 0.5)s2 (a b 1)s b 2.5(

7、s2 2s 5)(s 1) 0.5s 0.5 0.5s 0.5 0.50.5 2s 5s 1(s 1)2 22s 1s2 0.5(s 1)2 10.5(s 1)2 222(s 1)2 22 s 1 0.5et0.5et sin 2t 0.5et(t 0)cos 2t 0.5et 0.5et cos(2t )42重根的拉氏逆變換F1 (s)F (s) (s s )3 (s s )(s s)12nn K13K12K11Ki(s s )(s s )2(s s )3(s s )i2111in i2K13K12Ki(s s )3 F (s) K (s s )3(1)1111(s s )(s s )2(s

8、 s )11iK (s s )3 F (s)111ss1對(1)式兩邊求導得:n d ds d ds1Ki1(s K(s s )33s )F (s)2(ss )K111312(s s )i2idd 2(s 3Ks )F (s)(s s1) F (s)3K13121ds22 dss s1s s12s 3例3：求F (s) 的原函數(shù)。(s 2)3(s 1)2s 3K13K12K11K2F (s) (s 2)3 (s 1)s 2(s 2)2(s 2)3s 1K (s 2)3 F (s) 111s 2d 2s 31d (s 2)3 F (s)s2s 2 1K12s 1(s 1)2dsdss2d 21d

9、s2(s 2) F (s)3K13(s 1)3 1s2s2K (s 1)3 F (s) 12s11111f (t) L1s 2(s 2)2(s 2)3s 1 e2t te2t 1 t 2e2t et(t 0)213.2動態(tài)電路的復頻域分析法13.2.1KCL與KVL的復頻域模型 i 0 u 0L I (s) 0U (s) 0KCLKVLRiu (0 ) 0i (0 ) 0+uCLCu iR L di 1tidt-dtC013.2.21、電路元件的復頻域模型電阻元件I(S)i(t)RR-+U(S)+u(t)2、電感元件Li(0-)-i(t)LSLI(S)+-+u(t)-+U(S)u(t)=L d

10、i(t)微分性質(zhì)U(S)=SLI(S) Li(0-) 1 dtI(S)SLI(S)= 1U(S)+i(0-) i(0-) SSLS-+U(S)13.2.2電路元件的復頻域模型3、 電容元件U(S)= 1 I(S)+u(0-)I(S)=SCU(S) Cu(0-)SC 1 SCSSCI(S)u(0-)/SI(S)+cu(0-)+U(S)+U(S)4、線性時不變耦合電感元件didiU1(S)=SL1I1(S) + SMI2(S) L 1i1(0 )1+2u =LM11 dt Mi (0 )dt+2didi2U2(S)=SL2I2(S) + SMI1(S) L 2i2(0 )1+u =M+Ldt2 d

11、t2 Mi (0 )+113.2.2電路元件的復頻域模型 1 SCLi(0-)u(0-)/S-SLI(S)I(S)+-+U(S)+U(S):1）初具電源(附加電源)由uC(0-)、iL(0-)提供，參考方向，UL(S), UC(S)等的計算2）考慮零狀態(tài)情況運算阻抗與運算導納U(S)= 1 I(S)U(S)=RI(S)I(S)=GU(S)U(S)=SLI(S)SCI(S)=SCU(S)I(S)= 1 U(S)SL 1 U=jLIU= jC II=jCUU=RII=GU 1 I= jLU13.2.3動態(tài)電路的復頻域分析法1 運算電路模型RiI (S)1i21I2(S)RLRRL(t)C1/SCE

12、/SLSLuc (0 ) 0iL (0 ) 0運算電路時域電路電壓、電流用象函數(shù)形式元件用運算阻抗或運算導納電容電壓和電感電流初始值用附加電源表示例2050V+-IL(s)201/2siL50.5s-0.5H+ u-c102.5+25/s-2FUC(s)+105t=0時打開開關(guān)t 0復頻域模型時域電路uC(0-)=25ViL(0-)=5A總結(jié)：變換法分析電路步驟：由.換路前電路計算uc(0-) , iL(0-) 。畫運算電路模型應用電路分析方法求象函數(shù)。反變換求原函數(shù)。0.1例1：30已知：uc (0 ) 100Vt = 0時閉合s,求iL，uL.。-iLuc100010+200例1：30解：

13、(1)i (0 ) 5 A0.1HLuc (0 ) 100V(2)畫運算電路SL 0.1S-i Luc101000F+200V0.5V11S 1000 106300.1sSCI (S)I (S)L21000/S100010S200/S V100/S V0.5 V300.1sIL(S)I2(S)1000/SI1 (s)10100/S V200/SVI2 (s)I ( S )(40 0.1S ) 10I ( S ) 200 0.512S100(3)回路法1000 - 10I1( S ) (10 )I2( S ) SS5( S 2 700S 40000)I1 ( S ) S( S 200)25( S

14、 2 700S 40000)I1 ( S ) S( S 200)25( S 2 700S 40000) lims 5i(0 ) lim SF ( S )s22 400S 200S5( S 2 700S 40000)i() lim SF ( S ) 5lims022 400S 200s0S(4)反變換求原函數(shù)F2 ( S ) 0有3個根S1 0，S2 S3 200I ( S ) K1 K21S 200K221( S 200)2SI ( S ) K1 K21S 200K221( S 200)2S2 5( S 700S 40000)K F ( S )S5S 01S 0S 2 400S 2002 F

15、( S )( S 200)2 1500K22S 200 d ( S 200)2 F ( S )0KS 20021ds5 0( S 200)1500I ( S ) 1( S 200)2Si1(t ) (5 1500te200t ) (t ) A0.5 V300.1sIL(S)I2(S)1000/SUL(S)10200/SV100/SVU L ( S ) ? I1 ( S ) SL求UL(S) 30000150S 200U ( S ) I ( S )SL 0.5 ( S 200)2L1ete200tVtuL (t ) is (t ), uc (0 ) 0例2：求沖激響應+ucI (s)issU (

16、S)Rc1/SCCRIs (s) 1R1RU (s) I(s)RC ( S 1 / RC )CsR 1 / SCSC RSC 1 1RSC1I ( S ) U( S )CCRSC 1RSC 1RSC 1SCu 1 et / RC (t 0)1et / RC (t 0)i (t ) ccCRC例3 求圖示電路的沖激響應+ 11F+S1S1(t)U(S)u111F時域分析的節(jié)點方程(2S+1)U(S) =SU(S)= S = 112S+124(S+1/2) t 211u(t)= U(S)=1(t) e1(t)24例4(見13-9) 0.1 +1000.4SIL(S)0.4H100100 iL+50

17、S 25S+uc100F50vk104/SuC(0-)=25viL(0-)=0.25AA* S+125+j96.850(S) =A+0.1 25I+IL(S)= SSS+125 j96.8L100+0.4s+104/SA= 0.25S+62.5 | 0.25S+62.5IL(S)=S2+250SS= 125+j96.8S+125+j96.8=0.204 52.2 = 0.25S+62.5(S+125)2(S) = 0.204 52.2 + 0.20452.2 0.4H100I100 iS+125 j96.8LS+125+j96.8L+uc100F50vkiL(t)=0.408e125t cos

18、(96.8t 52.2 )(t0)u (0-)=25vi (0-)=0.25ACL例5圖示電路在開關(guān)閉合前處于穩(wěn)態(tài)，t=0時將開關(guān)閉合，求開關(guān)閉合后uC(t)和iL(t)的變化規(guī)律。+ 40V-iL25H 40 50i (0-)=0.8 AL+uC-20300.01FuC(0-)= 0.820 =16 V+ 40V-iL25H 40 50i (0-)=0.8 A+LuC-20300.01FuC(0-)= 0.820 =16 VIL-+-2016 + 4025SS -UC-20S100S(S)= 16S2+80S +160 1 1U(0.01S+20 + 25S )UCCS(S+1)(S+4)2

19、0 + 40/S U C40 +20I (S)=0.16 + SL25S25S(S)= 20S2+124S +200I(S +5S +4S)UC32L25S(S+1)(S+4)= 16S2+80S +160(S)= 16S2+80S +160(S)= 20S2+124S +200IUS(S+1)(S+4)L25S(S+1)(S+4)CA2 S+1A3 S+4A1 SU =+C= 16S2+80S +160A =U (S)S=401CS=0(S+1)(S+4)S=0= 16S2+80S +160A =(S+1)U (S)= 322CS(S+4)S= 1S= 1= 16S2+80S +160= 8

20、A =(S+4)U (S)S= 43CS(S+1)S= 4t 0t 0uC(t)=40 32e t + 8e 4tiL(t)=2 1.28e t +0.08e 4t13.3網(wǎng)絡函數(shù)(networkfunctions)13.3.1 網(wǎng)絡函數(shù)的定義性時不變零狀態(tài)電路中，網(wǎng)絡函數(shù) H(S) 定義為響應 R(S)與輸入激勵 E(S)之比。即：S域中的網(wǎng)絡函數(shù)=零狀態(tài)響應象函數(shù)/激勵象函數(shù)R(S)=Lr(t)，E(S)=L e(t)，網(wǎng)絡函數(shù)為：R(S ) E(S )H (S )網(wǎng)絡函數(shù)有六種類型：I1(S)I2(S)+U2(S)U1(S)NZ0I2(S)+U2(S)I1(S)+U1(S)N0ZI1(S

21、)I2(S)+U2(S)U1(S)NZ0I1(S )H (S ) =Y11(S)Driving-poadmittanceU1(S )策動點導納型H (S ) I2 (S ) =Y21(s)Transform admittanceU1 (S )轉(zhuǎn)移導納型H (S ) U2 (S ) =HU(S)Voltage gain轉(zhuǎn)移電壓比U1(S )I2(S)+U2(S)I1(S)+U1(S)N0ZH (S ) U1(S )=Z(S)Driving-poimpedance11I (S )1策動點阻抗型H (S ) U2 (S ) =Z (s)Transformimpedance21I (S )1轉(zhuǎn)移阻抗型

22、I 2 ( S )H ( S ) 轉(zhuǎn) 移 電 流 比 型=H ( S )C urrent gainII1 ( S ) 1 , R 1 , L 1 H , C 1F ，電路初始狀態(tài)為零。求網(wǎng)絡函數(shù)例： R12224H (S ) U1(S ) , H(S ) I1(S )R1R2I2(S)I1(S)12U (S )U (S )+U1(S)U(S)LC1解：(R1+SL)I1(S)SLI2(S)=U(S)1SLI (S ) (R SL (S ) 0)I122SCS 2S 2 2S 4I1(S ) 2U (S ) ,S 2S 2I2 (S ) 2SU (S ) 2S 21I (S )2(S ) I1(

23、S ) S 2S 4 H (S ) U1(S ) 2SSC, H12S 2S 2 2S 2 2S 2U (S )U (S )U (S )213.3.2網(wǎng)絡函數(shù)H(S)與沖擊響應h(t)之間的關(guān)系： H (S ) R(S ) , R(S ) H (S )E(S ) .E(S )當輸入 e(t)=(t)時H (S ) R(S ) h(t) h(t) .E(S)= 1 and (t)E(S )或h(t) L1H (S )網(wǎng)絡函數(shù)等于 對應沖激響應的拉氏變換。1例：R =8，R =4，L=2H， C F ，求沖激響應。1216LR1H (S ) h(t)+ (t)(t)RCh(t)21 R12SC11

24、(R SL)SL R22SC1SCR1R SL 2SC222(S 2)2 (22)2 h(t) L1H (S ) (2e2t2t) I (t)sin 2零點、極點與零極點圖：bm Sm bm1S m1 b1S b0P(S )H (S ) Q(S )Sn1 a S aSnaan 1n10m(S zi )H (S ) H (0) (S z1 )(S z2 )(S zm ) H (0) i1(S p1)(S p2 )(S pn )n(S p j )j1H (0) bm是實系數(shù)。an當S等于 z1, z2, ., zm時， 網(wǎng)絡函數(shù)等于零。 所以Zi稱為網(wǎng)絡函數(shù)的零點。當S等于 p1p, ., pn時

25、， 網(wǎng)絡函數(shù)趨于無窮。 所以pj稱為網(wǎng)絡函數(shù)的極點。上以s的實部為橫軸、虛部j 為縱軸，在其上標出極零點的位置。用符號 表示零點，用符號 表示極點。稱該圖為網(wǎng)絡函數(shù)的零、極點圖。S 2 (S 3)例： H (S ) (S 1)(S 2 j)(S 2 j)j1解：該網(wǎng)絡函數(shù)有三個零點，三個極點。Z1=Z2=0，Z3= 3，P1= 1，P2= 2+j，P3= 2j31113.3.3 網(wǎng)絡函數(shù)的極點與網(wǎng)絡的穩(wěn)定性的關(guān)系0H (S ) (S )2 20 j0p12h(t) e0t1h(t) etH (S ) S 若網(wǎng)絡函數(shù)的極點在s平面的左半開平面，則當t趨于無窮時，沖擊響應將趨于零，對應的網(wǎng)絡是漸近穩(wěn)定的。距離虛軸越遠，穩(wěn)定性越好。1h(t) etH (S ) S a0h(t) et sin t, p a jH (S ) 1200(S a)2 202 若網(wǎng)絡函數(shù)含有在s平面的右半開平面的極點，則當t趨于