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LECTURE 8
TIME RESPONSES
M2Electric Circuits
1
Agenda2
� 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
�� � � ����� �� � �����
N. Mohan, Power Electronics: Converters, Applications and Design, 3 rd edition, Wiley
�� � � 1������
��� � � 1 � ����
�
First-Order Circuits4
• 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
• 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
• 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
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
5
6
Source-Free RC Circuit
� The capacitor is initially charged
(energy is already stored v(0) � V0)
� Solving the above equation
6
0 =+dt
dvC
R
v0 =+ CR iiBy KCL
1dvdt
v RC= −∫ ∫
0 0( )t / RC t /v t V e V e− − τ= =
Ohms lawCapacitor
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free RC Circuit
7
CR=τTime constant
10 0 00 368
t /V e V e . V− τ −= =
� t = �
t v(t)/V0
τ 0.36788
2τ 0.13534
3τ 0.04979
4τ 0.01832
5τ 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 Inductor9
• 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
• 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
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)
0 =+ RL vvBy KVL
0 =+ iRdt
diL
Inductor relationship
Ohms law
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)( )
(0) 0
i t t
i
di Rdt
i L= −∫ ∫
ln ( ) ln (0) 0Rt
i t iL
− = − +
τ−= /0)(teIti
R
L=τ
A general representation
where
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.
� Thy key: �� and τ
τ/0)(
teIti −=R
L=τwhere
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Summary: Natural Responses
13
A RL source-free circuit
where
A RC source-free circuit
where
� � � ���� �⁄� � �
� � � ���� �⁄� � � ��
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Initial Conditions
14
� 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
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
1
2
15
Step Response: RC Circuit
� 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:
• Applying KCL,
• Rearrange and integrate both sides
� 0� =� 0� � ��
���� �� � ��� � 0
0
( )
0
v t t
Vs
dv dt
v V RC= −
−∫ ∫C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step Response: RC Circuit
16
• Integration gives
• Combine and take the exponential
• Complete response
/
0
, t ts RC
s
v Ve e RC
V Vτ τ
− −− = = =−
0
/0
, 0
( ) , 0ts s
V tv
V V V e tτ−
( ) ( )0ln lns st
v V V VRC
− − − = −
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Step Response: RC Circuit17
Initially uncharged capacitor Initially charged capacitor
0
/0
, 0
( ) , 0ts s
V tv
V V V e tτ−
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Complete Response: RC
18
>−+
<=
− 0)(
0)(
/0
0
teVVV
tVtv
tss
τ
Final value
at t -> ∞
Initial value
at t = 0
Source-free
Response
Complete Response = Natural response + Forced Response
(stored energy) (independent source)
= V0e–t/τ + Vs(1–e
–t/τ)
Complete Response = transient response + steady-state response
(temporary) (permanent)
= ( V0 - Vs)e–t/τ + Vs
Summary: RC Step Response
19
� Typical form
� 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.
τ/ )]( )0( [ )( )( tevvvtv −∞−++∞=
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
20
• Initial current
• Final inductor current
• Time constant τ = L/R
� 0� =� 0� � ��
� ∞ � �� ��
� � �!"#�$%# � ��%"&'��"%C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
21
/L Rτ =
0sVI A
R= +
/0
ts sV Vi I eR R
τ− = − +
Step Response: RL Circuit
Will try this by inspection
� � �!"#�$%# � ��%"&'��"%� � (�� �⁄ � �� ��
The initial condition gives
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
22
Step Response: RL Circuit
� 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
τ/ )]( )0( [ )( )( teiiiti −∞−++∞=
/0
ts sV Vi I eR R
τ− = − +
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Second-Order Circuits23
• 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
0
0 0
1(0) , (0)
−∞= = =∫v idt V i IC
10
tdiRi L idt
dt C −∞+ + =∫2
20
d i R di i
dt L dt LC+ + =
• KVL
• 2 initial conditions needed
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
�� 0 � � ����)*� � �� � 0
Source-Free Series RLC Circuits
26
• Our experience with the first-order circuit suggests
sti Ae=
2 0st st stAR A
As e se eL LC
+ + = 2 1 0stR
Ae s sL LC
+ + =
(A & s are to be determined)
2
20
d i R di i
dt L dt LC+ + =
• Plugging in the differential equation
Characteristic equation
Source-Free Series RLC Circuits
27
� Roots of the characteristic equation
� 2 possible solutions
� Linear combination
21
2 2
R Rs
L L LC = − ± −
2 21 0
2 22 0
s
s
α α ω
α α ω
= − + −
= − − −
02 202
2
=++ idt
di
dt
idωα LC
andL
R 1
2 0== ωα
General 2nd order Form
where
(damping factor)
(undamped natural frequency)
1 21 1 2 2 &
s t s ti A e i A e= =1 2
1 2( )s t s ti t A e A e= +
The types of solutions for i(t) depend on the relative values of α and ω.
28
• 3 possible solutions1. If α > ωo, over-damped case
tsts eAeAti 21 21)( +=2
02
2,1 ωαα −±−=swhere
2. If α = ωo, critically damped case tetAAti α−+= )()( 12 1 2s s α= = −where
3. If α < ωo, under-damped case )sincos()( 21 tBtBeti dd
t ωωα += −
where 220 αωω −=d
Source-Free Series RLC Circuits
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Critically Damped Response
29
1 21 2( )
s t s ti t A e A e= + 1 2 3( )t t ti t A e A e A eα α α− − −= + =
22
22 0
d i dii
dt dtα α+ + = 0
d di dii i
dt dt dtα α α + + + =
0df
fdt
α+ =
1tf A e α−=
1tdi i A e
dtαα −+ = ( ) 1t t tdi de e i e i Adt dt
α α αα+ = =
( )1 2 ti A t A e α−= +1 2te i A t Aα = +
� Critically damped case (repeated roots; α = ωo)
� Recall
� Something wrong!! 2 ICs cannot satisfy a single constant
� Go back to the general form
� Thus,
� Integrating both sides
Underdamped Response
30
� Underdamped case (α < ωo)
� Euler’s identities
2 21 0
2 22 0
d
d
s j
s j
α α ω α ω
α α ω α ω
= − + − = − +
= − − − = − −2 20dω ω α= −
Damping frequency
( )( ) ( )1 2 1 2( ) d d d dj t j t j t j tti t A e A e e A e A eα ω α ω ω ωα− − − + −−= + = +
[ ]( )
1 2
1 2 1 2
( ) (cos sin ) (cos sin )
cos ( )sin
td d d d
td d
i t e A t j t A t j t
e A A t j A A t
α
α
ω ω ω ω
ω ω
−
−
= + + −
= + + −
[ ]1 2( ) cos sint d di t e B t B tα ω ω−= +
Source-Free Parallel RLC Circuits
31
� The damping effect is due to the presence of ‘R’
� 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.
LCand
L
R 1
2 0== ωα
Source-Free Parallel RLC Circuits
32
0
0
1(0) ( )i I v t dt
L −∞= = ∫Let
v(0) = V0Apply KCL to the top node:
∫∞−
=++t
dt
dvCvdt
LR
v0
1
011
2
2
=++ vLCdt
dv
RCdt
vdThe 2nd order expression
Characteristic equation
2 2 20
1 12s s s s
RC LCα ω+ + = + +
0
1 1
2and
RC LCα ω= =
C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edition, McGraw Hill
Source-Free Parallel RLC Circuits
33
The solutions of the characteristic equation are
2 21,2 0= − ± −s α α ω
1. If a > wo, overdamped casetsts eAeAtv 21 )( 21 +=
20
22,1 ωαα −±−=swhere
2. If a = wo, critically damped case
tetAAtv α−+= )( )( 12 α−= 2,1swhere
3. If a < wo, underdamped case
)sincos()( 21 tBtBetv ddt ωωα += − where 220 αωω −=d
1,2 ds jα ω= − ±
Step-Response: Series RLC Circuits
34
� The step response is obtained by the sudden application of a dc source.
The 2nd order of expression LC
v
LC
v
dt
dv
L
R
dt
vd s=++2
2
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
� 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
� The steady-state response is the final value of � �� ��� � � �+∞,
Step-Response: Series RLC Circuits
35
(over-damped)
(critically damped)
(under-damped)
� � � �! � � ��� �
�! � � (���- � (.��/�! � � +(��(.�,��0�! � � ��0 (�1234&� � (.3�54&�
Step-Response: Series RLC Circuits
36
� The complete response becomes:
(over-damped)
(critically damped)
(under-damped)
� The values of A1 and A2 are obtained from the initial conditions:
� � � �� � (���- � (.��/� � � �� � +(��(.�,��0� � � �� � ��0 (�1234&� � (.3�54&�
� 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 LC
I
LC
i
dt
di
RCdt
id s=++1
2
2
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
(over-damped)
(critical damped)
(under-damped)
• The steady-state response is the final value of i(t) o iss(t) = i(∞) = Is
� � � �! � � ��� �
�! � � (���- � (.��/�! � � +(��(.�,��0�! � � ��0 (�1234&� � (.3�54&�
Step-Response: Parallel RLC Circuits
39
� The complete response:
tstss eAeAIti 21 21)( ++= (over-damped)
ts etAAIti
α−++= )()( 21 (critically damped)
)sincos()( 21 tAtAeIti ddt
s ωωα ++= − (under-damped)
� The values of A1 and A2 are obtained from the initial conditions:
� 0 , ����)*�