85-4 The driven RL circuit r at t0 Ri+ d t In(v-Ri=t+k dt V-Ri R L i(0)=0→i(0)=0∴k Inv i V/R--forced-response R V/R V-Ri IIn(v-ri)-Inv=t R -complete-resnnn or e (t> R R ( L )u(t) (for all t) R -V/R natural- response
§5-4 The driven RL circuit (1 e )u(t) ( for all t) R V i t L R − = − complete response e R V t L R − − − − − (1 ) −V / R natural response e R V t L R − − − − − i t 0 V / R V / R− − forced − response at t 0 i(t) = 0 V Ri t k R L dt V Ri Ldi V dt di for t Ri L = − − = + − 0 + = ln( ) V R L i(0 ) = 0 → i(0) = 0 k = − ln − t L R e V V Ri V Ri V t R L − = − − [ln( − ) − ln ] = = − = (1− ) ( 0) − − e t R V e R V R V or i t L R t L R
The complete response-the natural response+forced response I=L+ i=Ae (Ri+L==0 R (t→> i=i +i=ae+ Ri(0)=0→ 0=4+ R R i=n(1-e)=r(1-eo (t=L/R R R V/R 0.632V/R t=T=L/R, →i(z)=0.632/R
The complete response=the natural response+forced response n f i = i + i (1 ) (1 e ) ( L/ R) R V e R V i t t L R = − = − = − − = ( + = 0) − dt di i Ae Ri L t L R n = (t → ) R V i f R V i i i Ae t L R = n + f = + − i t 0 V / R 0.632V / R i V R t L R ( ) 0.632 / / , → = = = R V A R V i(0) = 0 → 0 = A+ = −
Example: Find i(t) Solution 50uctNv 2 I=L+ Ae 3/1.5=2s R 5 H 75/15=50 ∴i=Ae2+50(t>0) 1.5 i(0+)=i(0)=50/2=25 (t) .25=A+50→A=-25 i=50-25e2=50-25e4(t>0) 75y i=25A(t<0 st or i=25+ 25(1-e u()a (for all f)
Example: Find i(t). 25 25(1 ) ( ) ( ) 0.5 or i e u t A for all t − t = + − Solution: n f i = i + i s R L i Ae eq t n = = = 3/1.5 = 2 − = 75 / 1.5 = 50 f i 50 ( 0) 2 = + − i Ae t t i = 25A (t 0) 50 25 50 25 ( 0) 2 0.5 = − = − − − i e e A t t t 25 50 25 (0 ) (0 ) 50 / 2 25 = + → = − = = = + − A A i i
V1 3H 50V 0 +-+ 40A 30A 6 10s 口I(L1)
Time 0s 2s 4s 6s 8s 10s I(L1) 20A 30A 40A 50A V1 I 2 6 3H 1 2 50V 0
i=50-25e057(t>0) ori=25e0+50(1-e05)(t>0) zero-input zero-state response response Zero input response--term due to the excitation of the network by a nonzero initial condition Zero state response--terms due to the excitation of the network by sources
25 50(1 ) ( 0) 50 25 ( 0) 0.5 0.5 0.5 = + − = − − − − or i e e t i e t t t t response zero − input response zero − state Zero input response--term due to the excitation of the network by a nonzero initial condition. Zero state response--terms due to the excitation of the network by sources
50u(tV 50V 50u(tV 2 6 i(0)=i(0-)=25A i(0)=i(00)=0Uan=751Ra0=1.52 i(t)=25e2A i(t)=50(1-e2)A i()=25e2+50(1-e2)4 The complete response=zero input response+zero state response
i t e A i i A t 2 ( ) 25 (0 ) (0 ) 25 − + − = = = i t e A i i V R t eq eq ( ) 50(1 ) (0 ) (0 ) 0 75 1.5 2 − + − = − = = = = i t e e A t t ( ) 25 50(1 ) 2 2 − − = + − The complete response=zero input response+zero state response
I. with all independent sources killed, simplify the circuit to determine Reg, Leg, and the time constant T=L/ 2. Viewing Leg as a short circuit, use dc-analysis methods to findi,(0"), the inductor current just prior to the discontinuity 3. Again viewing r as a short circuit, use dc analysis methods to find the forced response. This is the value approached by f(t as t)00; we represent it y∫(∞)
1. With all independent sources killed, simplify the circuit to determine Req , Leq , and the time constant τ = Leq / Req. Leq (0 ) − L i 2. Viewing as a short circuit, use dc-analysis methods to find , the inductor current just prior to the discontinuity. Leq → 3. Again viewing as a short circuit, use dcanalysis methods to find the forced response. This is the value approached by f(t) as t ; we represent it by f ( )
4. Write the total response as the sum of the natural and forced responses f(t)=Ae-"/T+f(oo) 5. Find f(o)by using the condition that i, (0t)=iL(0"). If desired, Leg may be replaced by a current source ii(o*) lan open circuit if i(ot)=0 for this calculation. with the exception of inductor currents(and capacitor voltages), other currents and voltages in the circuit may change abruptly. 6. Then, f(0)=A+f(oo), and f(t)=If(0)-f(oole+f(oo)
(0 ) ( ), ( ) [ (0 ) ( )] ( ) / = + = − + + + − f A f and f t f f e f t 6. Then, ( ) ( ). / = + − f t Ae f t 4. Write the total response as the sum of the natural and forced responses (0 ) + f (0 ) + L i (0 ) + L i (0 ) + L i (0 ) − L i Leq 5. Find by using the condition that = . If desired , may be replaced by a current source [an open circuit if = 0] for this calculation. With the exception of inductor currents (and capacitor voltages), other currents and voltages in the circuit may change abruptly
f(t)=f(0)-f(∞)le+f(∞) initial-value forced-value time-cons tant DP1: Find i and U L Solution 10 z=4/4=1s 6 i1(0+)=i1(0)=0 4H) i1(∞)=8/4=2A i1(t)=2-2e-4(t>0) U2()=L=8ev(t>0) dt oI 十 4i1(t)+U1(t)=8 i ∴U,(t)=8eV(t>0
initial−value forced − value time −constant ( ) [ (0 ) ( )] ( ) / = − + + − f t f f e f t DP 1: Find iL and . L ( ) 8 ( 0) 4 ( ) ( ) 8 = + = − t e V t i t t or t L L L ( ) 2 2 ( 0) ( ) 8 / 4 2 (0 ) (0 ) 0 4 / 4 1 : = − = = = = = = − + − i t e A t i A i i s Solution t L L L L ( ) = = 8 ( 0) − e V t dt di t L L t L
DP 2 (1):let:i.=2n()A,fnd:l,i1(1):z=0.0ls (2):lert:i=-2()4,fnd:l2,i1(o)=60/30+2=4A (3): let: i =-2A, find: i,, i i1(0)=i1(0)=0 100t 30 e 301+O.22z(1) 60 d t 60u(ty p2H (2-26 7o-10ot e
DP 2: 1 1 1 (3): : 2 , : , (2): : 2 ( ) , : , (1): : 2 ( ) , : , let i A find i i let i u t A find i i let i u t A find i i s L s L s L = − = − = i ( . e )u(t )A dt di ( t ) i . i (t ) e u(t )A i ( ) i ( ) i ( ) / A ( ) : . s t L t L L L L 100 1 1 100 2 2 67 30 0 2 60 4 4 0 0 0 60 30 2 4 1 0 01 − − + − = − + = = − = = = + = =