Damped Second-Order
Systems
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Damped Second-Order Systems
5V
5V
2K
50
2K
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 the course
notes for next class]
6.002
Fall 03
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Damped Second-Order Systems
5V
5V
50
2K
2K
S
C
A
B
+
–
large
loop
Relevant circuit:
2K
CGS
L
B
5V
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+
–
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CGS
3
Observed Output
2k
5
vA
0
vB
2k
0
vC
0
Now, let’s try to speed up our inverter by
closing the switch S to lower the effective
resistance
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Observed Output
~50
5
vA
0
vB
50
0
vC
0
Huh!
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In the last lecture, we started by
analyzing the simpler LC circuit to
build intuition
L
+
–
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+
C
–
6
In the last lecture…
We solved
For input
And for initial conditions
v(0) = 0
i(0) = 0
[ZSR]
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In the last lecture…
Total solution
where
LC
vI
+
–
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L
+
C
–
8
Today, we will close the loop on our
observations in the demo by
analyzing the RLC circuit
R
L
+
+
–
C
–
LC
vI
add R
Damped sinusoids with R – remember demo!
See A&L Section 13.6
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Let’s analyze the RLC network
L
+
–
+
R
C
–
Node method:
Recall
element rules
L:
C:
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v, i state variables
10
Let’s analyze the RLC network
L
+
–
+
R
C
–
Node method:
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Solving
Recall, the method of homogeneous and
particular solutions:
1
Find the particular solution.
2
Find the homogeneous solution.
4 steps
3
The total solution is the sum of the
particular and homogeneous.
Use initial conditions to solve for the
remaining constants.
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Let’s solve
For input
And for initial conditions
v(0) = 0
i(0) = 0
[ZSR]
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1
Particular solution
is a solution.
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2
Homogeneous solution
Solution to
Recall, vH :
solution to homogeneous
equation (drive set to zero)
Four-step method:
A Assume solution of the form
B
Form the characteristic equation
f(s)
C Find the roots of the characteristic equation
D General solution
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2
Homogeneous solution
Solution to
A Assume solution of the form
so,
characteristic
equation
B
C Roots
D
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General solution
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3
Total solution
Find unknowns from initial conditions.
so,
Mathematically: solve for unknowns,
done.
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Let’s stare at this a while longer…
3 cases:
> o
Overdamped
< o
Underdamped
= o
Critically damped
Later…
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Let’s stare at underdamped a while longer…
< o
Underdamped contd…
Note: For
Same as LC as expected
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Let’s stare at underdamped a while longer…
< o
Underdamped contd…
Remember, scaled sum of sines (of the same
frequency) are also sines! -- Appendix B.7
LC
vI
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add R
20
Underdamped contd…
< o
Remember, scaled sum of sines (of the same
frequency) are also sines! -- Appendix B.7
LC
vI
add R
v
= o
Critically damped
underdamped
criticallydamped
overdamped
Section 13.2.3
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Remember this? Closed the
loop…
5
vA
0
vB
50
0
vC
0
See example 170 on page 898 for inverter
pair analysis
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Intuitive Analysis
Konquer it like Kolodziejski…
Sec. 13.8
Underdamped
“ringing”
Characteristic
equation
Oscillation frequency
Governs rate of decay
Final value
Initial value
Quality factor (approximately
the number of cycles of ringing)
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Intuitive Analysis
Sec. 13.8
Konquer it like Kolodziejski…
Ringing stops
after Q cycles
?
is –ve
so v(t)
must
drop
period
Characteristic
equation
L
+
–
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2
d
+
R
C
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–
given
-ve
+ve
24