Chapter 9 Transient response
9.1 First-order transients
zero-input response, step response, pulse response, switched DC
transients
9.2 Switched AC transients
9.3 Second-order natural response
second-order circuit equations, overdamped response,
underdamped response, critically damped response
9.4 Second-order transients
initial conditions, switched DC transients
9-1
電路學講義第9章
9.1 First-order transients
Basics
1. Zero-input response
a1 y '+ ao y = 0,(or τ y '+ y = 0), i.c. y(to + ) = y(to − ) = Yo
s-domain solution:let y = Aest → characteristic equation a1s + ao = 0
s=−
ao
a
1
= − ,τ = 1 : time constant
a1
τ
ao
Ae = Yo → A = Yo e
sto
− sto
to
= Yo e τ
t-domain solution:zero-input response, y(t ) = Yo es (t −to ) = Yoe
+
o
y '(t ) = −
Yo
τ
: initial slope
9-2
−
−
t −to
τ
, t > to
t
= Yo e τ u(t − to )
電路學講義第9章
2. Step response: transient response with step function excitation
⎧0 t < 0
unit step function u(t ) ≡ ⎨
⎩1 t > 0
⎧ 0 t < to
Ku(t − to ) ≡ ⎨
⎩K t > t 0
a1 y' + ao y = Ku(t − to ), i.c. y(to+ ) = y(to− ) = 0
−
t
τ
step response:y(t ) = Ae + Yss
y(to+ ) = Ae
−
to
τ
+ Yss = 0 → A = −Yss e
y(t ) = Yss (1 − e
−
−
t −to
τ
to
τ
), t > to
t
τ
=Yss (1 − e )u(t − to )
rise time tr ≡ t90 − t10 ≈ 2.2τ
9-3
電路學講義第9章
3. Pulse response: transient response with pulse function excitation
pulse function : u (t ) − u (t − D)
a1 y' + ao y = u(t ) − u(t − D), y(0+ ) = y(0− ) = 0
−
t
pulse response:y(t ) = Ae τ + Yss , y(0) = A + Yss = 0 → A = −Yss
−
t
t
−
⎧
0<t ≤ D
⎪ Yss (1 − e τ ),
y(t ) = ⎨
D
t −D
−
−
⎪
τ
τ
,
t>D
⎩Yss (1 − e )t
y(t ) = Yss (1 − e ), t > 0
τ
−
D
y( D) = Yss (1 − e τ ), y(t ) = y( D)e
−
t −D
τ
,t > D
9-4
電路學講義第9章
4. Switched dc transients
given: DC source switched at t=to, steady-state response y(t)=Yss
find: transient response of first-order circuits
approach: for each time interval to write the corresponding
transient response based on the initial value, steady-state value,
and time constant
τ y '+ y = Yss , i.c. y (t o+ ) = Yo , t > t o
sol:y (t ) = Yss + Ae
+
o
y (t ) = Yss + Ae
−
to
τ
−
t
τ
to
= Yo → A = (Yo − Yss ) e τ (general expression)
⇒ solution : y (t ) = Yss + (Yo − Yss ) e
−
−
t − to
τ
, t > to
t
= Yss + (Yo − Yss ) e τ u (t − t o )
9-5
電路學講義第9章
Discussion
1. RC circuit (zero-input response)
continuous
vC (to+ )
V
iC (t ) = −
=− o
R
R
iC = CvC, , RiC + vC = 0 → RCvC, + vC = 0, (or τvC, + vC = 0), vC (to + ) = vC (to − ) = Vo
+
o
s-domain solution:let vC = Aest → characteristic equation RCs + 1 = 0,
s=−
1
1
= − , τ = RC : time constant
RC
τ
t-domain solution:vC (t ) = Vo e
−
t −to
τ
Vo −
, iC (t ) = − e
R
9-6
t −to
τ
, t > to
電路學講義第9章
(express using iC)
1
iC dt = 0
∫
C
vC (to − )
Vo
+
,
→ RCiC + iC = 0, iC (to ) = −
=−
R
R
t −t
Vo − τ o
t-domain solution:iC (t ) = − e
, t > to
R
D.E. is the same form as RCvC, + vC = 0, vC (to + ) = Vo
iC = CvC, , RiC + vC = 0 → RiC +
9-6
電路學講義第9章
2. RL circuit (zero-input response)
continuous
iL(t)
Io
vL(t)
vL (to+ ) = −iL (to+ ) R = − I o R
-RIo
vL = LiL, , RiL + vL = 0 → LiL, + RiL = 0, (or τiL, + iL = 0), iL (to + ) = iL (to − ) = I o
s-domain solution:let iL = Ae st → characteristic equation Ls + R = 0,
s=−
R
1
L
= − , τ = : time constant
L
τ
R
t-domain solution:iL (t ) = I o e
−
t − to
τ
, vL (t ) = − RI o e
9-7
−
t − to
τ
, t > to
電路學講義第9章
3. Ex.9.1 find i (t) , v (t) for t > 0
vL = LiL, ,(40 + 10)iL + vL = 0 → LiL, + (40 + 10)iL = 0
L 60 ×10−3
25
= 1.2 ×10−3 , iL ( 0 ) =
= 2.5
τ= =
R
50
10
−
t
iL (t ) = 2.5e τ , v(t ) = −40 × iL (t ) = −100e
−
t
τ
4. RC circuit (step response)
voc = Vssu(t − to ), vC (to+ ) = vC (to− ) = 0
t > to , ReqiC + vC = Vss → ReqCvC, + vC = Vss ,τ = ReqC
vC = Vss + Ae
−
t
τ
to
→ A = −Vss e τ → vC = Vss (1 − e
−
t −to
τ
), t > to
RL circuit (step response): dual circuit
vC → iL , C → L, Req → Geq ,τ = ReqC →
−
L ,
→
iL + iL = I ss → iL = I ss (1 − e
Req
t −to
τ
L
Req
), t > to
9-8
電路學講義第9章
5. Ex. 9.2 vc(0-)=0, C=50uF, find step response i (t) , v (t) for t > 0
t > 0 → Req = 3 // 6 = 2 k , voc =
v (t ) = Vss (1 − e
i (t ) =
−
t − to
τ
6
× 12 = 8 = Vss , τ = RC = 0.1
3+ 6
−
t
) = Vss (1 − e τ )u (t − t o ) = 8(1 − e −10 t )u (t )
8 − v (t )
= 4 e −10 t u (t ) mA
2000
9-9
電路學講義第9章
6. Ex. 9.3 iL(0-)=0, v s (t) = 5[u (t ) - u(t - 30ms)] to find the relay last
time with iL reaching 150mA till falling to 40mA
Req = 10 + 15 = 25,τ =
L 400m
5
=
= 16ms, I ss =
= 200mA
Req
25
25
0 < t < 30ms : iL (t ) = 200(1 − e
−
t
16
)mA → iL (t1 ) = 200(1 − e
−
t1
16
) = 150 → t1 = 22.2ms
t = 30ms : iL (30) = 169mA
t > 30ms : iL (t2 ) = 169e
−
t2 −30
16
= 40 → t2 = 53.1ms, t2 − t1 = 30.9ms
9-10
電路學講義第9章
7. Ex.9.4 Sequential switched transients, to find v(t), i(t)
Step 1: for the circuit t = 0
find y (0), vC (0), iL (0)
vC (0 − ) = 0, i (0 − ) = 0
Step 2: for the circuit t > 0 find
y (0 + ) = Yo from the continuity
of capacitor voltage and
inductor current
Vo = vC (0+ ) = vC (0− ) = 0
C is SC
−16 − v(0+ )
= −2mA
I o = i (0 ) =
8
+
Step 3: find the steady
state value of Yss
Vss = (−16)
I ss =
24
= −12 C is OC
8 + 24
−16
= −0.5mA
8 + 24
9-11
電路學講義第9章
Step 4: find the time constant τ
Step 5: find the response for
each time interval
(1) 0 < t ≤ 1
τ = (8k // 24k )100µ = 0.6 s
vC (t ) = Vss + (Vo − Vss )e
= −12 + 12e
−
−
t
τ
t
0.6
vC (1) = −9.73
i (t ) = I ss + ( I o − I ss )e
= −0.5 − 1.5e
−
t
0.6
−
t
τ
mA
iC (1) = −0.783mA
i=
9-12
− 16
= − 2 mA
8
電路學講義第9章
(2) t > 1
Rth = (12k // 8k ) // 24k = 4.8k // 24k = 4k
60 + 16
× 12 = 14.4
12 + 8
τ = 4k × 100µ = 0.4
Vth = 60 −
Vo = vC (1+ ) = −9.73
14.4 − vC (1+ )
Io =
= 5.03mA
4.8
24
Vss = 14.4 ×
= 12
4.8 + 24
14.4
I ss =
= 0.5mA
4.8 + 24
vC (t ) = Vss + (Vo − Vss )e
= 12 − 21.73e
−
= 0.5 + 4.53e
t − to
τ
t −1
0.4
i (t ) = I ss + ( I o − I ss )e
−
−
t −1
0.4
mA
−
t − to
τ
i=
9-13
14 . 4 − ( − 9 . 73 )
= 5 . 03 mA
4 .8
電路學講義第9章
9.2 Switched AC transients
Basics
1. First-order circuit with a sinusoidal excitation
given: sinusoidal source switched at t=to, initial value y(to)=Yo
find: transient response of first-order circuit
approach: for each time interval, write the corresponding transient
response based on the initial value, time constant, and resulting
phasor value
−
t
τ
y (t ) = y F (t ) + Ae , t > to , y F (t ) = Ym cos( wt + φ), i.c. y (to+ ) = Yo
+
o
+
o
y (t ) = y F (t ) + Ae
−
to
τ
+
o
= Yo → A = [Yo − y F (t )]e
+
o
⇒ solution y (t ) = y F (t ) + [Yo − y F (t )]e
9-14
−
t − to
τ
to
τ
(general expression)
電路學講義第9章
Discussion
6 cos 15t , t < 0
1. Ex. 9.5 find i(t), v(t), vs (t ) = ⎧⎨
⎩12 cos 15t , t > 0
(1)t < 0
I=
6
= 0.2∠ − 30 o
26 + j15
i (t ) = 0.2 cos(15t − 30 o )
V = j15 I = 3∠ 60 o , v (t ) = 3 cos(15t + 60 o )
(2)t = 0
i (0 + ) = I o = i (0 − ) = 0.2 cos( −30 o ) = 0.173, v (0 + ) = Vo = vs (0 + ) − 26i (0) = 12 − 26 × 0.173 = 7.5
(3)t > 0
12
I=
= 0.4∠ − 30°, V = j15 I = 6∠ 60°
26 + j15
iF (t ) = 0.4 cos(15t − 30°), vF (t ) = 6 cos(15t + 60°), τ =
+
i (t ) = iF (i ) + [ I o − iF (0 )]e
+
−
t
τ
v (t ) = vF (i ) + [Vo − vF (0 )]e
−
L
1
=
s , iF (0 + ) = 0.346, vF (0 + ) = 3
R 26
= 0.4 cos(15t − 30°) − 0.173e −26 t
t
τ
= 6 cos(15t + 60°) + 4.5e −26 t
9-15
電路學講義第9章
9.3 Second-order natural response
Basics
1. A second-order circuit contains two independent energy-stored
elements.
2. vC(t) and iL(t): state variables
3. General second-order differential form
1R
1 1
1
y "+ 2αy '+ wo2 y = f (t ), α =
, or
, wo =
: resonant frequency
2L
2 RC
LC
series RLC circuit
parallel RLC circuit
KVL → L(CvC' )'+ RCvC' + vC = vs
vC" +
1
1
R '
vC +
vC =
vs
L
LC
LC
L '
i L + i L = is
R
1 '
1
1
iL" +
iL +
iL =
is
RC
LC
LC
KCL → C ( LiL' )'+
9-16
電路學講義第9章
4. Zero-input response
y"N + 2αy N' + wo2 y N = 0
let y N = Ae st → y N' = sAe st , y"N = s 2 Ae st ⇒ s 2 Ae st + 2αsAe st + wo2 Ae st = 0
characteristic equation:s 2 + 2αs + wo2 = 0 = ( s − p1 )( s − p2 )
eigen values (characteristic value):p1 = −α + α 2 − wo2 , p2 = −α − α 2 − wo2
solution:y N (t ) = A1e p1t + A2 e p2t
5. Case 1: overdamped response α 2 > wo2
p1 , p2 : real, negative
τ1 = −
1
1
, τ 2 = − : time constants
p1
p2
y N (t ) = A1e
−
t
τ1
+ A2 e
−
τ 1 > τ 2 , A1 < 0, A2 > 0
t
τ2
9-17
電路學講義第9章
6. Case 2: underdamped response α 2 < wo2
p1 , p2 = −α ± jwd : complex
wd ≡ wo2 − α 2
α : damping coefficient
wd : damped frequency
y N (t ) = A1e − p1t + A2 e − p2t = e −α t ( A1e jwd t + A2 e − jwd t ), A1 = A1 e j∠A1 , A2 = A1* = A1 e− j∠A1
⇒ y N (t ) = 2 A1 e −α t cos( wd t + ∠A1 )
ifα = 0, y N (t ) = 2 A1 cos( wo t + ∠A1 ) : oscillation
7. Case 3: critically damped response α 2 = wo2
p1 , p2 = −α
⇒ y N (t ) = A1e −αt + A2te −αt
9-18
電路學講義第9章
Discussion
1. Procedure to write the second-order D.E.
• Apply ohm’s law and KVL, KCL to express iC and vL in terms of
vC, iL and source
• Obtain two first-order D.E.s, then combine them to be a secondorder D.E.
2. Ex. 9.6 write its second-order D.E.
state variables: i1 , i2
⎧v1 = Ri + v2 = R(is − i1 ) + Rx (is − i1 − i2 )
KVL ⎨
⎩v2 = Rx ix = Rx (is − i1 − i2 )
⎧⎪v1 = L1i1'
⎧ L1i1' = ( R + Rx )is − ( R + Rx )i1 − Rx i2
→⎨
⎨
'
'
L
i
⎪⎩v2 = L2i2
2 2 = Rx (is − i1 − i2 )
⎩
R
R R + Rx ' RRx
R
)i2 +
i2 = x is' , i2" + 2α i2' + wo2i2 = x is'
⇒ i2" + ( x +
L2
L2
L1
L1 L2
L2
9-19
電路學講義第9章
3. Ex. 9.7 phase-shift oscillator (EX.6.10), write its second-order
D.E. of vout
state variables : vC , iL
Kv x − vC
v
⎧
⎪iC = i1 + iL = 1 + iL =
+ iL
⎨
R
R
⎪⎩
vL = v1 − v x = Kv x − vC − v x
1
⎧ '
⎪CvC = KiL − vC + iL
'
'
v x = RiL , iC = CvC , vL = LiL → ⎨
R
'
⎪⎩ LiL = ( K − 1) RiL − vC
L
⇒ LCiL" + ( RC − KRC + )iL' + 2iL = 0
R
v
v
R L
2
"
'
iL = x = out ⇒ vout
+ ( 2 + 1 − K )vout
+
vout = 0
R KR
L RC
LC
"
'
vout
+ 2αvout
+ wo2 vout = 0
9-20
電路學講義第9章
4. Ex. 9.8 find the natural response of a series RLC circuit with
L=0.1H, R=14Ω, C=1/400F
state variables : vC , iL
0
iC = iL
⎧
⎨
⎩v L = − v R − vC = − Ri L − vC
⎧ Cv C' = iL
1
iC = Cv , v L = Li → ⎨ '
⇒ Li L" = − Ri L' − vC' = − Ri L' − iL
C
⎩ Li L = − Ri L − vC
R
R
1
1
i L = 0, α =
⇒ iL" + iL' +
= 70, wo2 =
= 4000
L
LC
LC
2L
'
C
'
L
p1 , p 2 = −70 ± 70 2 − 4000 = −40,−100 : overdamped response
τ 1 , τ 2 = 25 ms ,10 ms
iL (t ) = A1e − 40 t + A2 e −100 t
9-21
電路學講義第9章
5. Ex. 9.9 phase-shift oscillator, find its natural response with
L=10mH, R=100Ω, C=2uF
"
'
+ 2αvout
+ wo2vout = 0
vout
2
R L
'
+
( 2 + 1 − K )vout
vout = 0
L RC
LC
R L
( 2 + 1 − K ) = 5000(2.5 − K )
α=
2L R C
2
(1) properly design amplification
= 108
wo2 =
factor K and L, C
LC
(2) results with different frequencies
⎧(1) K = 2.5 → α = 0
⎪
⎪vout (t ) = 2 A1 cos(10000t + ∠A1 ) : oscillation
⎪
2
2
⎪(2)K = 2 → α = 2500, wd = wo − α = 9680, p1 , p2 = −α ± jwd
gain selection⎨
− 2500t
=
v
(
t
)
2
A
e
cos(9680t + ∠A1 ) : decaying oscillation
1
out
⎪
⎪
2
2
⎪(3)K = 3 → α = −2500, wd = wo − α = 9680, p1 , p2 = α ± jwd
⎪v (t ) = 2 A e 2500t cos(9680t + ∠A ) : growing oscillation
1
1
⎩ out
"
+
vout
9-22
電路學講義第9章
9.4 Second-order transients
Basics
2
2
+
+
1. switched DC transient y "+ 2αy '+ wo y = wo Yss , t > 0, i.c. y(0 ), y '(0 )
solution y (t ) = Yss + yN (t )
2. Case 1: overdamped response
α 2 > wo2 → p1 , p2 = −α ± α 2 − wo2
+
⎧
y
(
0
) = Yss + A1 + A2
y (t ) = Yss + A1e p1t + A2 e p2t , t > 0, i.c.⎨
→ A1 , A2
+
⎩ y ' (0 ) = p1 A1 + p2 A2
3. Case 2: underdamped response
α 2 < wo2 → p1 , p2 = −α ± jwd , wd = wo2 − α 2
y (t ) = Yss + 2 A1 e −αt cos( wd t + ∠A1 ), t > 0, i.c. → A1 ∠A1
4. Case 3: critically damped response
α 2 = wo2 → p1 = p2 = −α
y (t ) = Yss + A1e −αt
+
⎧
y
(
0
) = Yss + A1
+ A2te −αt , t > 0, i.c.⎨
→ A1 , A2
+
⎩ y ' (0 ) = −αA1 + A2
9-23
電路學講義第9章
Discussion
1. Procedure to write the initial conditions of the second-order circuit
• determine vC(0+), iL(0+) based on continuity property
• write iC(t), vL(t), y(t) in terms of state variables and input
• calculate v’C(0+)= iC(0+)/C, i’L(0+)= vL(0+)/L
• calculate y’(0+) from v’C(0+), i’L(0+)
2. Ex. 9.10 Find the initial values and slopes of vC, iL, vR, vL with a
switched source vs=V1, t<0 and vs=V2, t>0
(1)vC (0+ ) = vC (0− ) = V1 , iL (0+ ) = iL (0− ) = 0
(2)t > 0, iC (t ) = iL (t ), vL (t ) = V2 − RiL (t ) − vC (t )
vR (t ) = RiL (t )
(3)iC (0+ ) = 0, vL (0+ ) = V2 − RiL (0+ ) − vC (0+ ) = V2 − V1
iC (0+ )
vL (0+ ) V2 − V1
+
'
= 0, iL (0 ) =
=
vR (0 ) = RiL (0 ) = 0, v (0 ) =
C
L
L
R(V2 − V1 ) ' +
R(V2 − V1 )
(4)vL' (0+ ) = − RiL' (0+ ) − vC' (0+ ) = −
, vR (0 ) = RiL' (0+ ) =
L
L
+
+
'
C
+
9-24
電路學講義第9章
3. Ex. 9.11 Find the zero-input response for t>0, L=0.1H, R=5Ω,
C=1/640F
R '
1
iL +
iL = 0
L
LC
R
1
= 25, wo2 =
= 6400
α=
2L
LC
iL" +
⎧30, t < 0
vs (t ) = ⎨
⎩ 0, t > 0
p1 , p2 = −25 ± 252 − 6400
= −25 ± j 76 : underdamped response
iL (t ) = I ss + 2 A1 e−25t cos(76t + ∠A1 ), t > 0, I ss = 0
iL (0+ ) = 0, LiL' (0+ ) = V2 − V1 = −30 → A1 = 1.974∠90°
iL (t ) = 3.95e−25t cos(76t + ∠90°)
9-25
“ringing”
電路學講義第9章
4. Ex. 9.12 Find the step response for t>0, L=0.1H, C=1/640F
1
1
1
R '
R
vC +
vC =
vs , α =
= 5 R, wo2 =
= 6400
2L
L
LC
LC
LC
iC (0 + ) iL (0 + )
'
+
−
vC (0 ) = vC (0 ) = 0, vC (t ) =
=
= 0, Vss = 30
C
C
vC" +
(1) R = 34Ω → α = 170, p1 , p2 = −170 ± 170 2 − 6400 = −20,−320
overdamped response vC (t ) = Vss + A1e − 20 t + A2 e −320 t
⎧ vC (0 + ) = 0 = 30 + A1 + A2
→ A1 = −32, A2 = 2
⎨ '
v
(
t
)
=
0
=
−
20
A
−
320
A
1
2
⎩ C
⎧ 0, t < 0
vs (t ) = ⎨
⎩30, t > 0
(2) R = 5Ω → α = 25, p1 , p2 = −25 ± 25 2 − 6400 = −25 ± j 76
underdampe d response vC (t ) = Vss + 2 A1 e − 25t cos( 76t + ∠A1 )
⎧
vC (0 + ) = 0 = 30 + A1 + A2
→ A1 = 15.8∠161.8° = A2∗
⎨ '
⎩vC (t ) = 0 = ( −25 + j 76) A1 + ( −25 − j 76) A2
(3) R = 16Ω → α = 80 = wo , p1 , p2 = −80
critically damped response vC (t ) = Vss + A1e −80 t + A2te −80 t
⎧ vC (0 + ) = 0 = 30 + A1
→ A1 = −30, A2 = −2400
⎨ '
v
(
t
)
=
0
=
−
80
A
+
A
1
2
⎩ C
9-26
電路學講義第9章