6.002
CIRCUITS AND
ELECTRONICS
Damped Second-Order
Systems
6.002
Fall 03
1
Damped Second-Order Systems
5V
5V
2KΩ
50Ω
2KΩ
S
C
A
B
+
–
large
loop
CGS
Remember this Demo
Our old friend, the inverter, driving another.
The parasitic inductance of the wire and
the gate-to-source capacitance of the
MOSFET are shown
[Review complex algebra appendix in Agarwal & Lang
for next class]
6.002
Fall 03
2
Damped Second-Order Systems
5V
5V
50Ω
2KΩ
2KΩ
S
C
A
B
+
–
large
loop
Relevant circuit:
5V +
–
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Fall 03
2KΩ
CGS
L
B
CGS
3
Observed Output
2kΩ
5
vA
0
t
vB
2kΩ
t
0
vC
0
t
Now, let’s try to speed up our inverter by
closing the switch S to lower the effective
resistance
6.002
Fall 03
4
Observed Output
~50Ω
5
vA
0
t
vB
50Ω
0
t
vC
0
t
Huh!
6.002
Fall 03
5
In the last lecture, we started by
analyzing the simpler LC circuit to
build intuition
i (t )
L
vI (t )
6.002
+
–
Fall 03
C
+
v(t )
–
6
In the last lecture…
We solved
d 2v 1
1
+
v
=
vI
2
dt
LC
LC
For input
VI
vI
0
t
And for initial conditions
v(0) = 0 i(0) = 0 [ZSR]
6.002
Fall 03
7
In the last lecture…
Total solution
v(t ) = VI − VI cosω t
o
where
1
LC
ωo =
v(t )
2VI
LC
VI
vI
0
t
i (t )
L
v I (t )
6.002
+
–
Fall 03
C
+
v (t )
–
8
Today, we will close the loop on our
observations in the demo by
analyzing the RLC circuit
R
L
vI (t ) +
–
i (t )
C
+
v(t )
–
v(t )
2VI
LC
VI
vI
0
add R
t
Damped sinusoids with R – remember demo!
See A&L Section 13.6
6.002
Fall 03
9
Let’s analyze the RLC network
vA
vI (t )
L
i (t )
+
v(t )
–
R
+
–
C
Node method:
vA :
1 t
vA − v
∫ (vI − v A ) dt =
L −∞
R
vA − v
dv
=C
R
dt
v:
Recall
element rules
L:
vL = L
t
di
dt
1
vL dt = i
∫
L −∞
d 2 v R dv 1
1
+
+
v=
vI
2
dt
L dt LC
LC
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Fall 03
C:
dvC
iC = C
dt
v, i state variables
10
Let’s analyze the RLC network
vA
vI (t ) +
–
L
i (t )
R
C
+
v(t )
–
Node method:
1 t
vA − v
∫ (v I − v A ) dt =
L −∞
R
vA :
vA − v
dv
=C
R
dt
v:
1
d 2v
( vI − v A ) = C 2
L
dt
1
d 2v
( vI − v A ) = 2
dt
LC
dv
v A = RC + v
dt
dv
1
d 2v
( vI − RC − v ) = 2
LC
dt
dt
d 2 v R dv 1
1
+
+
v
=
vI
2
dt
L dt LC
LC
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Fall 03
11
Solving
Recall, the method of homogeneous and
particular solutions:
1
Find the particular solution.
2
Find the homogeneous solution.
L
4 steps
3
The total solution is the sum of the
particular and homogeneous.
Use initial conditions to solve for the
remaining constants.
v = vP (t ) + vH (t )
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Fall 03
12
Let’s solve
d 2 v R dv 1
1
+
+
v
=
vI
2
dt
L dt LC
LC
For input
VI
vI
0
t
And for initial conditions
v(0) = 0 i(0) = 0 [ZSR]
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Fall 03
13
1
Particular solution
d 2 vP R dvP
1
1
+
+
vP =
VI
2
dt
L dt LC
LC
vP = VI
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Fall 03
is a solution.
14
2
Homogeneous solution
Solution to
1
d 2 vH R dvH
+
+
vH = 0
2
dt
LC
dt
L
Recall, vH : solution to homogeneous
equation (drive set to zero)
Four-step method:
A Assume solution of the form
vH = Ae st
, A, s = ?
B
Form the characteristic equation
f(s)
C Find the roots of the characteristic equation
s1 , s2
D General solution
vH = A1e s1t + A2 e s2t
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Fall 03
15
2
Homogeneous solution
1
d 2 vH R dvH
+
+
vH = 0
2
dt
LC
dt
L
Solution to
A Assume solution of the form
vH = Ae st
, A, s = ?
so,
As2est +
R
1
Asest +
Aest = 0
L
LC
characteristic
equation
R
1
s + s+
=0
L
LC
B
2
s + 2αs + ω
2
2
o
ωo =
=0
α=
C Roots
1
LC
R
2L
s1 = −α + α 2 − ω 2 o
s2 = −α − α 2 − ω 2 o
D
General solution
vH = A1e
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Fall 03
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t
⎝
⎠
+ A2 e
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t
⎝
⎠
16
3
Total solution
v(t ) = vP (t ) + vH (t )
v(t ) = VI + A1e
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t
⎝
⎠
+ A2 e
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t
⎝
⎠
Find unknowns from initial conditions.
v(0) = 0 : 0 = VI + A1 + A2
i (0) = 0 :
dv
i (t ) = C
dt
(
CA (− α −
)
)e
= CA1 − α + α 2 − ω 2 o e
2
so,
(
α −ω
2
2
o
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t
⎝
⎠
⎛⎜ −α − α 2 −ω 2 o
⎝
) (
+
⎞⎟ t
⎠
0 = A1 − α + α 2 − ω 2 o + A2 − α − α 2 − ω 2 o
)
Mathematically: solve for unknowns,
done.
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Fall 03
17
Let’s stare at this a while longer…
⎛ α 2 −ω 2 o
−αt ⎜⎝
v(t ) = VI + A1e e
⎞⎟ t
⎠
⎛ − α 2 −ω 2 o
−αt ⎜⎝
+ A2 e e
⎞⎟ t
⎠
3 cases:
α > ωo
Overdamped
v(t ) = VI + A1e
α < ωo
−α1t
v(t ) = VI + A1e e
−αt
= VI + A1e e
α = ωo
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+ A2 e
−α 2 t
v
t
Underdamped
⎛ j ω 2 o −α 2 ⎞⎟ t
−αt ⎜⎝
⎠
= VI + K1e
VI vI
−αt
jω d t
⎛⎜ − j ω 2 −α 2 ⎞⎟ t
o
−αt ⎝
⎠
+ A2 e e
−αt − jωd t
+ A2e e
cosωd t + K 2e
−αt
sin ωd t
ωd = ω 2 o − α 2
e jωd t = cosωd t + j sin ωd t
Critically damped
Later…
Fall 03
18
Let’s stare at underdamped a while longer…
α < ωo
Underdamped contd…
v(t ) = VI + K1e−αt cosωd t + K 2e−αt sin ωd t
v(0) = 0 : K1 = −VI
dv
i (0) = 0 : i (t ) = C
dt
= −CK1αe−αt cosωd t − CK 2ωd e−αt sin ωd t
− CK1αe−αt sin ωd t + CK 2ωd e−αt cosωd t
0 = − K1α + K 2ωd
Vα
K2 = − 1
ωd
v(t ) = VI − VI e
−αt
α −αt
cosωd t − VI
e sin ωd t
ωd
Note: For R = 0
⇒α = 0
v(t ) = VI − VI cosωot
Same as LC as expected
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Fall 03
19
Let’s stare at underdamped a while longer…
α < ωo
Underdamped contd…
v(t ) = VI − VI e
−αt
α −αt
cosωd t − VI
e sin ωd t
ωd
Remember, scaled sum of sines (of the same
frequency) are also sines! -- Appendix B.7
ωo −αt ⎛
α
v(t ) = VI − VI
e cos⎜⎜ ωd t − tan −1
ωd
ωd
⎝
⎞
⎟⎟
⎠
v(t )
2VI
LC
VI
vI
0
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add R
t
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20
α < ωo
Underdamped contd…
v(t ) = VI − VI e
−αt
α −αt
cosωd t − VI
e sin ωd t
ωd
Remember, scaled sum of sines (of the same
frequency) are also sines! -- Appendix B.7
ωo −αt ⎛
−1 α
v(t ) = VI − VI
e cos⎜⎜ ωd t − tan
ωd
ωd
⎝
⎞
⎟⎟
⎠
v(t )
2VI
LC
VI
vI
0
add R
t
v
α = ωo
Critically damped
underdamped
criticallydamped
overdamped
t
Section 13.2.3
6.002
Fall 03
21
Remember this? Closed the
loop…
5
vA
0
t
vB
50Ω
0
t
vC
0
t
See example 12.9 on page 664 of the A&L textbook
for inverter-pair analysis
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Fall 03
22
Intuitive Analysis
See Sec. 12.7 of A&L textbook
ωo −αt ⎛
−1 α
⎜
v
(
t
)
V
V
e
cos
t
tan
ω
=
−
−
Underdamped
I
I
⎜ d
ωd
ωd
⎝
v(t )
e −αt
⎞
⎟⎟
⎠
“ringing”
VI
0
2π
t
ωd
Characteristic
equation
s2 +
R
1
s+
=0
L
LC
s 2 + 2αs + ω 2 o = 0
ωd : Oscillation frequency
α : Governs rate of decay
ωd = ω 2 o − α 2
VI : Final value
v(0) : Initial value
Q=
6.002
ωo
: Quality factor (approximately
2α
the number of cycles of ringing)
Fall 03
23
Intuitive Analysis
See Sec. 12.7
of A&L textbook
Ringing stops
after Q cycles V I
v(t )
VI
v(0)
i (0)
is –ve
so v(t)
must
drop
?
0
period
2π
t
ωd
Characteristic
equation
s2 +
R
1
s+
=0
L
LC
s 2 + 2αs + ω 2 o = 0
ωd = ω 2 o − α 2
i(t )
L
vI +–
6.002
Q=
R
C
Fall 03
+
v(t )
–
ωo
2α
given
i (0) -ve
v(0) +ve
24