M2
Electric Circuits
1
LECTURE 8
TIME RESPONSES
Agenda
2
Time response
First-order circuits
Natural response
Forced responses
Second-order circuits
Natural response
Forced responses
Responses in Time Domain
3
A first-order circuit is characterized by a first-order differential
equation
1
1
N. Mohan, Power Electronics: Converters, Applications and Design, 3 rd edition, Wiley
First-Order Circuits
4
• Source-free RC circuits
• Source-free RL circuits
• Step-response RC circuits
• Step-response RL circuits
Homogenous differential
equations (i.e. natural response)
Nonhomogenous differential
equations (i.e. forced response)
Pre-Charging the Capacitor
5
• The switch has been in
the “closed” position
(i.e. the switch is on)
for a long while
• The capacitor acts as an
open circuit path
• For this particular case, the capacitor voltage is equal to
the voltage across the resistor R2
• At t = 0 (the beginning of our interest), the switch is
opened and the capacitor is discharging
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RC Circuit
6
The capacitor is initially charged
(energy is already stored v(0) V0)
By KCL
v
dv
+C
=0
R
dt
iR + iC = 0
Ohms law
Capacitor
Solving the above equation
dv
1
∫ v = − RC ∫ dt
v(t ) = V0 e
− t / RC
= V0 e
−t / τ
6
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RC Circuit
7
t=
Time constant
τ=RC
V0 e − t / τ = V0 e −1 = 0.368V0
t
τ
2τ
3τ
4τ
5τ
v(t)/V0
0.36788
0.13534
0.04979
0.01832
0.00674
Decays more slowly
Decays faster
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RC Circuit
8
The natural response of a circuit = the behavior (in
terms of voltages and currents) of the circuit itself, with no
external sources of excitation
The time constant τ of a circuit is the time required for the
response to decay by a factor of 1/e or 36.8% of its initial
value.
decays faster for small and slower for large
The key to working with a source-free RC circuit is finding:
The initial voltage 0
across the capacitor.
The time constant
Energizing the Inductor
9
• The switch has been in
the “closed” position
(i.e. the switch is on)
for a long while
• The inductor acts as a
short-circuited path
• For this particular case, the inductor current is equal to the
current through the resistor
• At
0(the beginning of our interest), the switch is
opened and the inductor is de-energized
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RL Circuit
10
A first-order RL circuit consists of an inductor L (or
its equivalent) and a resistor R (or its equivalent)
By KVL
vL + v R = 0
di
L
+ iR = 0
dt
Ohms law
Inductor relationship
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RL Circuit
11
Solving for i(t)
i (t )
t
di
R
∫i (0) i = − ∫0 L dt
Rt
ln i (t ) − ln i (0) = − + 0
L
A general representation
i(t ) = I 0 e
where
−t / τ
L
τ =
R
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RL Circuit
12
The time constant of a circuit is the time required
for the response to decay by a factor of 1/e or 36.8% of
its initial value.
i(t) decays faster for small τ and slower for large The general form is very similar to a RC source-free
circuit.
i (t ) = I 0 e
Thy key:
− t /τ
L
where τ =
R
and τ
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Summary: Natural Responses
13
A RC source-free circuit
A RL source-free circuit
where
where
⁄
⁄
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Initial Conditions
14
1
The initial capacitor voltage
could be either zero or nonzero
When the switch is left at either
position for a long time, the
capacitor voltage builds up to a
specific value
2
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step Response: RC Circuit
15
The step response of a circuit is its behavior
when the excitation is the step function, which
may be a voltage or a current source.
• Initial condition:
0 = 0
• Applying KCL,
0
• Rearrange and integrate both sides
∫
v (t )
V0
t dt
dv
= −∫
0 RC
v − Vs
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step Response: RC Circuit
16
• Integration gives
t
ln ( v − Vs ) − ln (V0 − Vs ) = −
RC
• Combine and take the exponential
v − Vs
−t
= e RC = e−t /τ , τ = RC
V0 − Vs
• Complete response
,t <0
V0
v=
−t /τ
V
+
(
V
−
V
)
e
, t >0
 s
0
s
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step Response: RC Circuit
17
,t <0
V0
v=
−t /τ
Vs + (V0 − Vs )e , t > 0
Initially uncharged capacitor
Initially charged capacitor
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Complete Response: RC
18
t<0
V0
v(t ) = 
−t / τ
V
+
(
V
−
V
)
e
0
s
 s
Final value
at t -> ∞
Complete Response =
Complete Response
Initial value
at t = 0
t >0
Source-free
Response
Natural response
(stored energy)
+
=
V0e–t/τ
+
=
transient response
(temporary)
+
=
( V0 - Vs)e–t/τ
+
Forced Response
(independent source)
Vs(1–e–t/τ)
steady-state response
(permanent)
Vs
Summary: RC Step Response
19
Typical form
v (t) = v (∞) + [v (0+) − v (∞)]e−t /τ
To determine the step response of an RC circuit:
The initial capacitor voltage v(0)
The final capacitor voltage v(∞) — DC voltage across C
The time constant τ.
Step Response: RL Circuit
20
The step response of a circuit is its behavior when the excitation
is the step function, which may be a voltage or a current source
• Initial current
0 = 0
• Final inductor current
∞
• Time constant τ = L/R
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C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
20
Step Response: RL Circuit
21
Will try this by inspection
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The initial condition gives
⁄
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τ = L/ R
Vs
I0 = A +
R
Vs  −t /τ Vs

i =  I0 −  e +
R
R

C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step Response: RL Circuit
22
Vs  −t /τ Vs

i =  I0 −  e +
R
R

i (t ) = i (∞) + [i (0+ ) − i (∞)] e − t /τ
Three steps in finding out the step
response of an RL circuit:
The initial inductor current i(0) at t = 0+
The final inductor current i(∞)
The time constant τ
Note: This technique applies only for the step responses
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Second-Order Circuits
23
• Source-free series RLC circuits
• Source-free parallel RLC circuits
• Step-response series RLC circuits
• Step-response series RLC circuits
What is a 2nd order circuit?
24
• A second-order circuit is characterized by a second-order
differential equation. It consists of resistors and the
equivalence of two energy storage elements.
Series RLC circuit
Parallel RLC circuit
RL T-configuration
RC Pi-configuration
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free Series RLC Circuits
25
• KVL
di 1 t
Ri + L + ∫ idt = 0
dt C −∞
d 2i R di
i
+
+
=0
2
dt
L dt LC
• 2 initial conditions needed
1 0
v(0) = ∫ idt = V0 , i (0) = I 0
C −∞
0
)
*
0
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free Series RLC Circuits
26
• Our experience with the first-order circuit suggests
i = Ae st
(A & s are to be determined)
• Plugging in the differential equation
d 2i R di
i
+
+
=0
2
dt
L dt LC
AR st
A st
As e +
se +
e =0
L
LC
2 st
1 
 2 R
Ae  s + s +
=0
L
LC 

st
Characteristic equation
Source-Free Series RLC Circuits
27
Roots of the characteristic equation
2
R
1
 R 
s=−
± 
 −
2L
 2 L  LC
 s = −α + α 2 − ω 2
0
1

 s2 = −α − α 2 − ω02
(undamped natural frequency)
General 2nd order Form
d 2i
di
2
+
2
α
+
ω
0i =0
2
dt
dt
where
R
α=
2L
and ω0 =
(damping factor)
2 possible solutions i1 = A1e s t & i2 = A2 e s t
Linear combination i(t ) = A1e s t + A2 e s t
1
2
1
2
The types of solutions for i(t) depend on the
relative values of α and ω.
1
LC
Source-Free Series RLC Circuits
28
• 3 possible solutions
1. If α > ωo, over-damped case
i (t ) = A1e s1t + A2 e s2t where s1, 2 = − α ± α 2 − ω0 2
2. If α = ωo, critically damped
case
i (t ) = ( A2 + A1t ) e − α t where s1 = s 2 = − α
3. If α < ωo, under-damped
i (case
t ) = e −αt ( B1 cos ω d t + B 2 sin ω d t )
where ω d = ω 02 − α 2
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Critically Damped Response
29
Critically damped case (repeated roots; α = ωo)
i (t ) = A1e −α t + A2 e −α t = A3e −α t
i (t ) = A1e s1t + A2 e s2t
Recall
Something wrong!! 2 ICs cannot satisfy a single constant
df
Go back to the general form
+α f = 0
dt
2
d  di

 di

d i
di
2
+
α
+
α
+
α
i
i
+ 2α + α i = 0
−α t



 =0
2
f
=
A
e
dt  dt
1
dt
dt

 dt

Thus,
di α t
d αt
e
+ e α i = ( e i ) = A1
dt
dt
di
+ α i = A1e −α t
dt
αt
Integrating both sides
eα t i = A1t + A2
i = ( A1t + A2 ) e −α t
Underdamped Response
30
Underdamped case (α < ωo)
Damping frequency
 s = −α + α 2 − ω 2 = −α + jω
0
d
1

 s2 = −α − α 2 − ω02 = −α − jωd
ωd = ω02 − α 2
(
i (t ) = A1e− (α − jωd )t + A2e − (α + jωd )t = e −α t A1e jωd t + A2e − jωd t
)
Euler’s identities
i (t ) = e −α t [ A1 (cos ωd t + j sin ωd t ) + A2 (cos ωd t − j sin ωd t ) ]
= e −α t ( A1 + A2 ) cos ωd t + j ( A1 − A2 ) sin ωd t 
i (t ) = e−α t [ B1 cos ωd t + B2 sin ωd t ]
Source-Free Parallel RLC Circuits
31
The damping effect is due to the presence of ‘R’
R
α=
2L
and
ω0 =
1
LC
If R = 0, the response is not only undamped but also
oscillatory
By adjusting R, the response can be varied
Oscillatory response is possible due to the storage elements.
It is difficult to tell the difference between overdamped and
critically damped responses.
Critically damped response approaches the final values most
rapidly.
Source-Free Parallel RLC Circuits
32
0
1
Let i (0) = I 0 = ∫ v(t ) dt
L −∞
v(0) = V0
Apply KCL to the top node:
t
v 1
dv
+ ∫ vdt + C = 0
R L −∞
dt
The 2nd order expression
d 2 v 1 dv 1
+
+
v=0
2
RC dt LC
dt
Characteristic equation
1
1
2
= s 2 + 2α s + ω02
s +
s+
RC
LC
1
α=
2 RC
and ω0 =
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
1
LC
Source-Free Parallel RLC Circuits
33
The solutions of the characteristic equation are
s1,2 = −α ± α 2 − ω02
1. If a > wo, overdamped case
v(t ) = A1 e s1t + A2 e s2t where
s1, 2 = − α ± α 2 − ω0
2
2. If a = wo, critically damped case
v ( t ) = ( A 2 + A1 t ) e − α t where
s1 , 2 = − α
3. If a < wo, underdamped case
v(t ) = e −αt ( B1 cos ω d t + B2 sin ωd t ) where ω =
d
s1,2 = − α ± jωd
ω 02 − α
2
Step-Response: Series RLC Circuits
34
The step response is
obtained by the sudden
application of a dc source.
The 2nd order
of expression
vs
v
d 2 v R dv
+
+
=
2
L dt LC LC
dt
The above equation has the same form as the equation
for source-free series RLC circuit.
• The same coefficients (important in determining
the frequency parameters).
• Different circuit variable in the equation.
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step-Response: Series RLC Circuits
35
The solution of the equation have 2 components:
Transient response ! + ,
Steady-state response + ,
!
The transient response vt is the same as that of the
source-free case
!
!
!
(
+(
-
0
(.
(. ,
0
/
( 1234&
(over-damped)
(critically damped)
(. 3 54&
(under-damped)
The steady-state response is the final value of
+∞,
Step-Response: Series RLC Circuits
36
The complete response becomes:
(
-
+(
0
(.
(. ,
/
(over-damped)
0
(critically damped)
( 1234&
(. 3 54&
(under-damped)
The values of A1 and A2 are obtained from the initial
conditions:
0 ,
)
*
Step-Response: Parallel RLC Circuits
37
• The step response is
obtained by the sudden
application of a dc source.
The 2nd order
of expression
d 2i 1 di
i
Is
+
+
=
2
dt
RC dt LC LC
This is the same form as the equation for source-free
parallel RLC circuit.
• The same coefficients (important in determining the
frequency parameters).
• Different circuit variable in the equation.
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step-Response: Parallel RLC Circuits
38
•
The solution of the equation have two components:
o Transient response: ! + ,
!
o Steady-state response: + ,
•
The transient response it is the same as the source-free case
!
!
!
•
(
-
+(
0
(.
(. ,
/
0
( 1234&
(over-damped)
(critical damped)
(. 3 54&
(under-damped)
The steady-state response is the final value of i(t)
o
iss(t) = i(∞) = Is
Step-Response: Parallel RLC Circuits
39
The complete response:
i (t ) = I s + A1e s1t + A2 e s2t
(over-damped)
i ( t ) = I s + ( A1 + A 2 t ) e − α t
(critically damped)
i (t ) = I s + e −αt ( A1 cos ω d t + A2 sin ω d t )
(under-damped)
The values of A1 and A2 are obtained from the initial
conditions:
0 ,
)
*