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6.002x CIRCUITS AND ELECTRONICS
Damped Second-Order
Systems
Review
C A B
5V
+!!
5V
CGS large loop
2K 50
2K S
LvA 5
0
vB
0
vC
0
50
t
t
t
5
5
3
In the last lecture, we started by analyzing the simpler LC circuit to build intuition
4
And for initial conditions
We solved
For input
In the last lecture LC circuit
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In the last lecture LC circuit
Total solution
+!! C L +!
!)(tv
)(tvI
)(ti
6 See A&L Section 13.6
Today, we will close the loop on our observations in the demo by analyzing the RLC circuit
+!! C
L
+!
!
i(t)
v(t) vI (t) LC
I v
v(t) 2VI
t
VI
0
7
Recall element rules
Lets analyze the RLC network
L:
C:
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Set up the differential equation vA
+!! C L +!
!
R i(t)
v(t) vI(t) Need to get rid of vA
v
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Set up the differential equation differently
v +!!
C
L +!
!
R i
vI
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Solve IvLCvLCdtdv
LR
dtvd 112
2
=++
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Lets solve IvLC
vLCdt
dvLR
dtvd 112
2
=++
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1 Particular solution IvLC
vLCdt
dvLR
dtvd 112
2
=++
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1 Particular solution IvLC
vLCdt
dvLR
dtvd 112
2
=++
IPPP V
LCv
LCdtdv
LR
dtvd 112
2
=++
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Homogeneous solution 2 IvLC
vLCdt
dvLR
dtvd 112
2
=++
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Homogeneous solution 2 0vLC1
dtdv
LR
dtvd
HHH =++2
2
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0vLC1
dtdv
LR
dtvd
HHH =++2
2
Solution to Homogeneous solution 2
Assume solution of the form 2A
vH = AeST , A, S = ?
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Homogeneous solution 2 0vLC1
dtdv
LR
dtvd
HHH =++2
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Solution to
02 =++LC1s
LRs
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Homogeneous solution 2 0vLC1
dtdv
LR
dtvd
HHH =++2
2
Solution to
02 =++LC1s
LRs
Roots o1s22 +=
os 222 =
characteristic equation
LC1
o =
LR2
=
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Total solution 3
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v(t) =VI + A1ete22o( )t + A2ete
22o( )t
Lets stare at this a while longer 3 v(t) =VI + A1e+ 22o( )t + A2e
22o( )t
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v(t) =VI + A1ete22o( )t + A2ete
22o( )t3
Overdamped o
>
Underdamped o <
Critically damped o
=
Overdamped o >
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Underdamped o < 3 v(t) =VI + A1ete22o( )t + A2ete
22o( )t
Overdamped o
>
Underdamped o <
Critically damped o
=
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Underdamped contd o < 3 teKteKVtv d
td
tI
sincos)( 21 ++=
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Underdamped contd o <
Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7
v(t) =VI VIet cosdt VId
et sindt3
+=+
1
2122
2121 tancossincos A
AAAAA
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Critically damped o =
Section 13.2.3
3
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Remember this? We have now closed the loop
vA 5
0
vB
0
vC
0
50
t
t
t
5
5
27 See Sec. 12.7 of A&L textbook
ringing v(t)
VI 0
t
Easy way: Characteristic equation tells the whole story!
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Intuitive Analysis +!! C
L +!
!)(tvIv
)(ti
R
)0(i)0(v
given -ve +ve
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Next, introduce R: RLC Circuits
More in the next sequence! If you are impatient, see A&L Section 13.2
+!! C
L
+!!vI (t)
i(t)
v(t)
v(t)
t
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What about other variables?
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Parallel RLC Characteristic equation says it all