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Second-Order Systems
6.002x CIRCUITS AND ELECTRONICS
[Review complex algebra Appendix C in textbook]
Second-Order Systems
+!–!5V CGS
2KΩ B
L
Relevant circuit: C A B
5V
+!–!
5V
CGS
large loop
2KΩ 50Ω
2KΩ S
L
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02
2
=+ HH v
dtvdLCSolution to
Homogeneous solution 2
Recall, vH : solution to homogeneous equation (drive set to zero)
Four-step method:
Assume solution of the form* 2A ?s,A,Aev st
H ==*Differential equations are commonly solved by guessing solutions
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Total solution 3 Find unknowns from initial conditions.
tjtjI
oo eAeAVtv ωω −++= 21)(v(t) = vP(t) = vH (t)
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Summary of Method Write DE for circuit by applying node method.
Find particular solution vP by guessing and trial & error.
Find homogeneous solution vH
Total solution is vP + vH , then solve for remaining constants using initial conditions.
Assume solution of the form Aest .
Obtain characteristic equation.
Solve characteristic equation for roots si .
Form vH by summing Ai esit terms.
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2 3
4
D
C
A
B
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We can obtain the answer directly from the homogeneous solution (VI = 0).
tj2
tj1C
oo eAeA)t(v ωω −+=
Example C L
+!
–!iC
vC
vC (0) = V
iC (0) = 0
vC (0) = V
iC (0) = 0