1、7.5 the zero-state response if all of the initial conditions have zero value, then the circuit is said to be in the zero-state, and the solution to nonzero inputs for the circuit is known as the zero-state response. (零狀態響應零狀態響應)zero-state response of an rc circuitk+ucacicabt=0t=0：uc(0)=0+ucuscicabr0

2、ccsdurcuutdtthe particular solution：it is usually called the forced response or the steady-state response.uc ()=us ，uc p = uc ()=us rusc+ucithe homogeneous solution：0ccdurcudtthe characteristic equation： rcs+1=0the characteristic value：1src trcstchuaeaethe homogeneous solution is usually called the

3、natural or transient response.uc (0+)=uc (0 )=0=us +a a= us0,)(tevvtvtsscteyyty)()()(0tccpchsuuuuaetthe general form of the step response of rc circuits：us-us usriuc0tuc=us(1- e )-rct（t 0）i=-rctusre（t 0）1 zero-state response of dc input:uc= us us e-rct（t 0）特解特解齊次解齊次解穩態分量穩態分量暫態分量暫態分量+-uc+-crius（2）dis

4、cussion7.6 step response circuitnwhen the dc source is suddenly applied, the signal of the source can be modeled as a step function, the response is known as a step response.the step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current s

5、ource.7.6.1 step response of rc circuitsk+ucacicabt=0+ucuscicabrus1(t)t=0：uc(0)=u00ccsdurcuutdt its solution includes two parts : the particular solution and the homogeneous solution.complete response: external source and initial condition of the storage element the particular solution：it is usually

6、 called the forced response or the steady-state response.uc ()=us ，uc p = uc ()=us rusc+ucithe homogeneous solution：0ccdurcudtthe characteristic equation： rcs+1=0the characteristic value：1src trcstchuaeaethe homogeneous solution is usually called the natural or transient response.uc (0+)=uc (0 )=u0=

7、us +a a= u0 us0,)()(0teuuututsscteyyyty)()0()()(0tccpchsuuuuaetthe general form of the step response of rc circuits：complete response例例ucusc12r+-u0us u0uc(0+)=u0rc=usducdt+（t 0）ucusuc=(u0 us)e-rct+there are two ways to excite the circuit:by initial conditions of the storage elements in the circuits.

8、by independent sources.1、two ways to determine complete response:（1）to linear circuit:teyyyty)() 0()()(teyyty)()()(complete responsezero_state responsezero_input responseteyty)0()(complete response = zero-input response + zero-state response.zero_input response is a special complete response when y(

9、 )=0;zero_state response is a special complete response when y( 0 )=0;y(t) = yh(t) + yp(t)自由自由分量分量強制強制分量分量 ty(t)= ke+ yp(t)暫態暫態響應響應穩態穩態響應響應(2) to linear circuit, complete response=natural response+forced responseteyyyty)()0()()(teyyyty)()0()()(thus, to find the step response requires three things:th

10、e initial value y(0+);the final value y( );1. the time constant ;（2） 與輸入無關，歸結為求由電容元件或電感元件觀與輸入無關，歸結為求由電容元件或電感元件觀 察的入端電阻察的入端電阻rlr=rc=關于關于uc(0-)，il(0-) 和和 uc(0+)，il(0+) （3） y(0+) uc(0-)，il(0-) uc(0+)，il(0+)確定其它確定其它變量的初始值變量的初始值（1） y( ) 歸結為求解電阻網絡（電容元件相當于開路，歸結為求解電阻網絡（電容元件相當于開路， 電感元件相當于短路）電感元件相當于短路）three f

11、actors:switch s is closed at t=0, uc(0-)=2v, calculate the response uc.+u-icccutucu ddciuiu 1212164dd4 ccutu1044 ppv5 . 14/6 cutcaeu tcau e5 . 1 0)( v e5 . 05 . 1 tutci12i1+2v+1 1 1 0.8fuc suc (v)t1.5oexample:method 1:reduce the circuit at first.s 18 . 0)25. 01( rc v 5 . 1）（cu0)( v e5 . 05 . 1 tutci

12、12i1+2v+1 1 1 0.8fuc s+1.5v+0.25 1 0.8fuc smethod 2: ty(t)= y(0+) - y( )e+ y( )+ul2usirl2l1k+ ul1 例：l1=1mh，l2=2mh，r=3k，us=3v，t=0 時 k 閉合，求零狀態響應 ul1與 ul2。l=l1+l2=3mh61 10lsr 61010tiemat 6610101122,20ttlldidiulevulevtdtdtmail1)(drill 1. calculate uc.3a100 50 0.3f+-ucuc(0)=150vuc(0+)=150v=0.3 5000/150=1

13、0suc=100 + 50e0.1t (t 0)uc( )=3 5000/150=100v ty(t)= y(0+) - y( )e+ y( )10ma1k +-1k 2k 10v+-ulil1hdrill 2. calculate ul.il(0)=5mail(0+)=5mail( )=5+5=10maul(0+)=5vul( )=0 =103sil=10 5e1000t maul=5e1000t v5ma+-2k 2k 10v+-ul+-10v(t=0+) ty(t)= y(0+) - y( )e+ y( )+-15 10 30v+-ucil3 fik2mh5 drill 3. calcu

14、late uc, il, ik.iuc(0)=20vil(0)=1auc(0+)=20vil(0+)=1ai (0+)=20/15=4/3auc( )=1.2 15=18vi( )=30/25=1.2ail( )=0 c=3 106 150/25=18 106 s l=2 103/5=1/2500 suc=18+2e v106t18il=e2500t ai k=1.2+ e e2500t a215106t18 ty(t)= y(0+) - y( )e+ y( )例：us=6v，r1=r2=1，c =0.5f，原k1、k2打開，c 上無電荷；t=0 時k1閉合，求uc (t)，又當 t=2s 時

15、，k2又閉合，求 uc(t)。usk1r1c+ uc k2r2圖(a)usr1c+ uc r2圖(b)0t2, equivalent circuitsolution：uc( )=us=6v，uc (0+)=uc (0)=0，1=c(r1+r2)=0.52=1susk1r1c+ uc k2r2圖(a)usr1c+ uc r2圖(b)02),1 (6)()0()()(teeuuututtcccct=2s：uc (2)=66e2=5.19 vuc ( )=6v，uc (2+)=uc (2)=5.19 v，2=r1c=0.5susk1r1c+ uc k2r2圖(a)usr1c+ uc 圖(c)2,81

16、. 06)()2()()()2(2teeuuututtcccc+u-cn+us-+u*-ln+us-fig.afig.bn is a linear resistive circuit, in fig.a, c=1f, uc(0)=0, us=u0, u=(0.75-e-2t)v;in fig.b, l=1h, il(0 )=0, calculate u*.+-rc1c2+-u1(0) u2(0)u1u2+-i1usrsrl2l1i2special cases:已知：已知：vurrhlhls6,1,2,02. 0,01. 02121 并并定定性性畫畫出出波波形形。求求：)(),(21titi解：

17、此題為初始值躍變情況解：此題為初始值躍變情況aiai0)0(,3)0(21 )0()0(21 ii電流發生躍變電流發生躍變s(t =0)r1r2l1l2i1i2us)0()0()0()0(22112211 ilililil回路磁鏈守恒回路磁鏈守恒allililii103. 0301. 0)0()0()0()0(21221121 srrllaii01. 0,2126)()(212121 aetitit100212)()( 12i20ti13vettdidlutl100111)(02. 0 vettdidlutl1002222)(02. 0 slluuuirir 21221162)(02. 0)(

18、02. 0224100100100100 ttttetetee 電容電壓初值一定會發生躍變。電容電壓初值一定會發生躍變。解解 v1)0(1 euc0)0(2 cu合合k前前)0()0()0(21 cccuuu合合k后后 已知圖中已知圖中e=1v , r=1 , c1=0.25f , c2=0.5f 。求：求： uc1 、 uc2 、 ic1 和和 ic2 并畫出波形。并畫出波形。 )0()0()0()0(22112211 ccccucucucuc)0()()0()0(212211 cccuccucuc節點電荷守恒節點電荷守恒q(0+)= q(0-)erc1c2+-uc1+-uc2k(t=0)i

19、ic1ic2+v315 . 025. 0125. 0)0(211 ccecuc可解得可解得uc( ) = 1v 0321)131(1)(3434teetuttc = r (c1+c2) s43 0)0(1)0(021 ccuut)(92)(61)(98)()321()(41)(98)()321()(41dd34343434111tettetttetettucittttc )()321()()(341tettutc )()321()(342tetutc uc2ut01/31uc1-1/6 2/9ic1it 4/91/6ic2tucicdd222 )()321()(342tetutc )()321

20、()()(341tettutc )(92)(61341tetitc )(94)(6134tett )(98)()321(2134tett iic1ic2i 無沖激無沖激7.7 first_order op amp circuitsan op amp circuit containing a storage element will exhibit first_order behavior.examples: differentiators and integrators c+_uo_+ +_uirici- -u- -irturcud1io r+_uo_+ +_uiciri- -u- -icdtdvrcvioanalyze first_order op amp circuits nclassical way: write the circuit equation in the first_order differential form at first, get the solution.nuse the general form of the complete response of the first_order circuits. teyyyty)()0()()(drill