6.002
CIRCUITS AND
ELECTRONICS
Second-Order Systems
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Second-Order Systems
5V
5V
Demo
2KΩ
50Ω
2KΩ
S
A
+
–
C
B
large
loop
CGS
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 for next class]
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Second-Order Systems
5V
5V
Demo
50Ω
2KΩ
2KΩ
S
C
A
+
–
Relevant circuit:
B
large
loop
2KΩ
CGS
L
5V +
–
B
CGS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Observed Output
2kΩ
5
vA
t
0
vB
2kΩ
t
0
vC
t
0
Now, let’s try to speed up our inverter by
closing the switch S to lower the effective
resistance
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Observed Output
~50Ω
5
vA
t
0
vB
0
50Ω
t
vC
t
0
Huh!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
First, let’s analyze the LC network
i (t )
L
vI (t )
+
–
C
+
v(t )
–
Node method:
dv
i (t ) = C
dt
Recall
di
vI − v = L
dt
t
dv
1
v
v
dt
C
(
−
)
=
∫ I
L −∞
dt
1
(v I − v )
L
1 t
(vI − v) dt = i
∫
L −∞
d 2v
=C 2
dt
d 2v
LC 2 + v = vI
dt
time2
v, i state variables
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
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 )
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 Fall 2000
Lecture
15
Let’s solve
d 2v
LC 2 + v = vI
dt
For input
V0
vI
t
0
And for initial conditions
v(0) = 0 i(0) = 0 [ZSR]
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
1
Particular solution
d 2 vP
LC 2 + vP = V0
dt
is a solution.
vP = V0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
2
Homogeneous solution
Solution to
d 2 vH
LC 2 + vH = 0
dt
Recall, vH :
solution to homogeneous
equation (drive set to zero)
Four-step method:
A Assume solution of the form*
vH = Ae st
, A, s = ?
so,
B
LCAs 2 e st + Ae st = 0
s2 = −
characteristic
equation
1
LC
1
s=±j
LC
C Roots
j = −1
1
ωo =
LC
s = ± jω o
General solution,
D
vH = A1e jωot + A2 e − jωot
Differential equations are commonly
solved by guessing solutions
*
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
3
Total solution
v(t ) = vP (t ) + vH (t )
v( t ) = V0 + A1e jωot + A2 e − jωot
Find unknowns from initial conditions.
v(0) = 0
0 = V0 + A1 + A2
i ( 0) = 0
dv
i (t ) = C
dt
i( t ) = CA1 jωo e jωot − CA2 jωo e − jωot
so,
0 = CA1 jωo − CA2 jωo
or,
A1 = A2
− V0 = 2 A
V0
A1 = −
2
so,
V0 jωot
v( t ) = V0 − (e + e − jωot )
2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
3
Total solution
Remember Euler relation
e jx = cos x + j sin x
(verify using Taylor’s
expansion)
e jx + e − jx
= cos x
2
so,
v( t ) = V0 − V0 cos ωot
where
i( t ) = CV0ωo sin ωot
1
ωo =
LC
The output looks sinusoidal
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
v(t )
Plotting the Total Solution
2V0
V0
0
π
π
3π
2
2
CV0ωo
0
2π
ωo t
i (t )
π
π
2
3π
2
2π
ωo t
− CV0ωo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Summary of Method
1
Write DE for circuit by applying
node method.
2
Find particular solution vP by guessing
and trial & error.
3
Find homogeneous solution vH
A Assume solution of the form Aest .
B Obtain characteristic equation.
C Solve characteristic equation
for roots si .
D Form vH by summing Ai esit
terms.
4
Total solution is vP + vH ,
solve for remaining constants using
initial conditions.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Example
What if we have:
L
iC
+
C vC
–
vC (0) = V
iC (0) = 0
We can obtain the answer directly from
the homogeneous solution (V0 = 0).
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Example
L
iC
+
C vC
–
vC (0) = V
iC (0) = 0
We can obtain the answer directly from
the homogeneous solution (V0 = 0).
vC ( t ) = A1e jωot + A2 e − jωot
vC (0) = V
V = A1 + A2
iC (0) = 0
0 = CA1 jωo − CA2 jωo
or
V
or A1 = A2 =
2
V jω o t
vC = (e + e − jωot )
2
vC = V cos ωot
iC = −CV ωo sin ωot
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Example
vC
V
2π
ωo t
CVωo iC
2π
ωo t
− CVωo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Energy
EC
C:
1
CV 2
2
1
2
CvC
2
2π
ωo t
EL
1
2
L : LiC
2
1
CV 2
2
2π
Notice
ωo t
1
1
1
2
2
CvC + LiC = CV 2
2
2
2
Total energy in the system is a constant,
but it sloshes back and forth between the
Capacitor and the inductor
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
RLC Circuits
R
L
vI (t ) +
–
i (t )
C
+
v(t )
–
v(t )
no R
add R
t
Damped sinusoids with R – remember demo!
See A&L Section 12.2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15