Download Natural and Step Responses of RLC Circuits

Natural and Step
Responses of RLC Circuits
EE3301
Kamran Kiasaleh
Learning Objectives
1. Be able to determine the natural responses
of parallel and series RLC circuits
2. Be able to determine the step responses of
parallel and series RLC circuits
3. Be able to determine the responses (both
natural and transient) of second order
circuits with op amps
Parallel RLC circuit
The step response of a parallel RLC circuit.
A series RLC circuit.
The step response of a series RLC circuit.
The Natural Response of a Parallel RLC
1. Using KCL,
This will disappear
dv (t )
1 t
v (t ) /R + ∫ 0 v (s)ds + C
+ I0 = 0
L
dt
2
d v (t ) 1 dv (t ) v (t )
+
+
=0
2
dt
RC dt
LC
The Natural Response of a Series RLC
1. Using KVL,
This will disappear
di ( t )
1 t
Ri ( t ) + ∫ i ( s ) ds + L
+ V0 = 0
C 0
dt
2
d i ( t ) R di ( t ) i ( t )
+
+
=
0
2
dt
L dt
LC
The Natural Response of a
Parallel/Series RLC
v ( t ) = Ae st ⇐ parallel
i ( t ) = Ae st ⇐ series
1
1
s +
s+
=0⇒
RC
LC
R
1
=0⇒
s + s+
L
LC
2
2
1
1
 1 
s=−
± 
 −
2 RC
 2 RC  LC
2
2
1
R
 R 
± 
s=−
 −
2L
 2 L  LC
The Natural Response of a
Parallel/Series RLC
There are 3 distinct cases.
1
R
α=
orα =
Let
2 RC
2L
Then,
Neper
frequency
ω0 =
1
LC
s = −α ± α 2 − ω0 2
1)α = ω 0
Critically-damped
2)α < ω 0
Under-damped
3)α > ω 0
Over-damped
Resonant
radiant
frequency
The Natural Response of a Parallel RLC
There are 3 distinct cases.
1)α = ω 0 ⇒ s1 = s2 = −α
 s = −α + j ω 2 − α 2
1
0
2)α < ω 0 ⇒ 
 s1 = −α − j ω 0 2 − α 2
 s = −α + α 2 − ω 2
0
3)α > ω 0 ⇒  1
 s1 = −α − α 2 − ω 0 2
Critically-damped
Under-damped
Over-damped
The Natural Response of a
Parallel/Series RLC
How does the response look like?
v (t ) = A1e + A2e
s1 t
i(t ) = A1e + A2e
s1 t
s2 t
s2 t
⇐ parallel
⇐ series
Initial conditions must be used
to evaluate
How do we solve for the unknowns for
over-damped (parallel)?
v (0
)= A + A
Two equations with two
unknowns
dv (0 )
= As +A s ⇒
+
1
2
+
1 1
dt
2 2
+


dv (0 ) ic (0 ) 1 v (0 )
=
= −
− I0 
C 
dt
C
R

+
+
Initial voltage across the
capacitor
Initial current through the
inductor
How do we solve for the unknowns for
the over-damped case(series)?
i(0 + ) = A1 + A2
di(0
+
)= A s + A s
1 1
2 2
Two equations with two
unknowns
⇒
dt
+
+
di(0 ) v L (0 ) 1
=
= −Ri(0 + )− V0
dt
L
L
[
Initial current through the
inductor
]
Initial voltage across the
capacitor
Example 8.9
The circuit for Example 8.2.
I0 = 30mA;
v (0
+
) = 12
Figure 8.7
Example 8.2.
s1 = −5000;rad /sec
s2 = −20,000;rad /sec
The Natural Response of an underdamped Parallel/Series RLC
How does the response look like?
v (t ) = B1e −αt cos(ω d t )+ B2e −αt sin(ω d t ) ⇐ parallel
i(t ) = B1e
−αt
cos(ω d t )+ B2e
−αt
sin(ω d t ) ⇐ series
ωd = ω − α
2
0
Initial conditions must be used
to evaluate
2
How do we solve for the unknowns for
under-damped (parallel)?
v (0
)= B
dv (0 )
=ω
+
Two equations with two
unknowns
1
+
dt
d
B2 − αB1 ⇒
+


dv (0 ) ic (0 ) 1 v (0 )
=
= −
− I0 
C 
dt
C
R

+
+
Initial voltage across the
capacitor
Initial current through the
inductor
How do we solve for the unknowns for
under-damped (series)?
i(0 + ) = B1
di(0
+
dt
di(0
dt
+
)=ω
Two equations with two
unknowns
d
B2 − αB1 ⇒
) = v (0 ) = 1
+
L
L
Initial current through the
inductor
−Ri(0 )− V ]
[
L
+
0
Initial voltage across the
capacitor
The circuit for Example 8.4
I0 = −12.25mA;
v (0
+
)= 0
The voltage response for Example 8.4
s1 = −200 + j979.80
s2 = −200 − j979.80
The Natural Response of a Critically
Damped Parallel/Series RLC
How does the response look like?
v (t ) = D1te + D2e
st
st
Initial conditions must be used
to evaluate
s1 = s2 = s
How do we solve for the unknowns for
critically-damped (parallel)?
v (0
)= D
dv (0 )
= D − αD
+
2
+
1
dt
2
Two equations with two
unknowns
⇒
+


dv (0 ) ic (0 ) 1 v (0 )
=
= −
− I0 
dt
C
C 
R

+
+
Initial voltage across the
capacitor
Initial current through the
inductor
How do we solve for the unknowns for
the critically-damped (series)?
i(0 + ) = D2
di(0
+
) = D − αD
1
2
Two equations with two
unknowns
⇒
dt
+
+
di(0 ) v L (0 ) 1
=
= −Ri(0 + )− V0
dt
L
L
[
Initial current through the
inductor
]
Initial voltage across the
capacitor
Example 8.5 (critically-damped)-R has been changed to
make this happen
ω 0 = 10
α = ω 0 ⇒ R = 4kΩ
6
s = −1000
v (t ) = 98000te
−1000t
A circuit used to describe the step response
of a parallel RLC circuit
The Step Response of a Parallel RLC
1. Using KCL,
vL ( t )
dvL ( t )
+ iL ( t ) + C
=I
R
dt
2
d iL ( t )
L diL ( t )
+ iL ( t ) + LC
=I
2
R dt
dt
2
d iL ( t ) 1 diL ( t ) 1
I
+
+
iL ( t ) =
2
dt
RC dt
LC
LC
The Step Response of a Parallel RLC
(direct method)
1. First find the natural response
2. Add to the natural response the final value
3. Use the initial conditions to solve for
coefficients
d iL ,n (t ) 1 diL ,n (t ) 1
+
+
iL ,n (t ) = 0
2
dt
RC dt
LC
iL (t ) = I f + iL,n (t )
2
The Step Response of a Parallel RLC
(direct method)
iL (0 ),
+
diL (0
+
) ⇐ known
dt
' s1 t
' s2 t
iL (t ) = I f + A1e + A2e
iL (t ) = I f + B e
' −αt
1
iL (t ) = I f + D te
'
1
cos(ω d t )+ B e
−αt
' −αt
2
' −αt
2
+De
sin(ω d t )
A circuit used to illustrate the step response
of a series RLC circuit.
The Step Response of a Series RLC
1. Using KVL,
dvc ( t )
d vc ( t )
RC
+ vc ( t ) + LC
=V
2
dt
dt
diL ( t )
1 t
'
'
L
+ RiL ( t ) + ∫ iL (t )dt = V
dt
C
2
d vc ( t ) R dvc ( t ) 1
V
+
+
vc ( t ) =
2
dt
L dt
LC
LC
2
The Step Response of a Series RLC
(direct method)
v c (0 ),
+
dv c (0
+
) ⇐ known
dt
' s1 t
' s2 t
v c (t ) = V f + A1e + A2e
v c (t ) = V f + B e
' −αt
1
v c (t ) = V f + D te
'
1
cos(ω d t )+ B e
−αt
' −αt
2
' −αt
2
+De
sin(ω d t )
The circuit for Example 8.12
v c (0 + ) = 0,
dv c (0 + )
=0
dt
v c (t ) = 48 + B1'e −αt cos(ω d t )+ B2'e −αt sin(ω d t )
α = −1400;ω d = 4800
Second order circuits with op amps
d 2v 0 (t )
1
1
=
v g (t )
2
dt
R1C1 R2C2
Second order circuits with op amps
•
•
•
•
This is a variation of the second order system
The output is the double integration of the input
Depending on the initial charges on the capacitors,
the response will vary
For a constant input, the output will increase
indefinitely
d 2v 0 (t )
1
1
=
v g (t )
2
dt
R1C1 R2C2
v g (t ) = V0
V0
v 0 (t ) =
t2
2R1C1R2C2
Second order circuits with op
amps-imperfect integrator
dv0 (t ) RbC2
v 01(t ) = −RbC2
−
v (t )
dt
τ2 0
dv01(t ) Ra C1
v g (t ) = −RaC1
−
v 01(t )
dt
τ1
τ1 = R1C1;τ2 = R2C2
v g (t )
d 2v 0 (t )  1 1  dv0 (t )  1 
+
+
+
v
t
=



 0( )
dt 2
τ
τ
dt
τ
τ
RaC1RbC2
 1 2
 1 2
−1
−1
s1 = ;s2 =
τ1
τ2
v g (t ) = V0U (t )
v 0 (t ) = V f + A1e s1 t + A2e s2 t ⇔ v 0 (0+ ) =
V0τ1τ 2
R1R2V0
=
Vf =
Ra C1RbC2 Ra Rb
dv0 (0+ )
dt
=0