1 module 14 - circuits with resistors, inductors, and capacitors (rlc)
TRANSCRIPT
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Module 14 - Circuits with Resistors, Inductors, and Capacitors (RLC)
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Capacitor and Resistor Exponential Forms
t
vC
t
vC
Inductor and Resistor Exponential Forms
vC = Vo et /RC
vC = Vo (1 et /RC)
t
iL
t
iL
iL = Io et R/L
iL = Io (1 et /RC)
One Reactive Element…
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• Behavior Very Different
Capacitor, Inductor, and Resistor
• Typical Response
decaying exponentialsinusoid
0 10 20 30 40 50 60 70 80 90 100-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
t
vC or iL
Let’s look at a circuit example…
RLC NETWORK:
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Series RLC Circuit
L
CR +
vC
–+vL
–
+ vR –
Vo
t = 0
i(t)
Find i(t)Objective:
Initial Conditions
vC = 0
iL = 0
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Governing Equation:
L
CR +
vC
–+vL
–
+ vR –
Vo
t = 0
i(t)
Find i(t)Objective:
Vo = vR + vC + vLKVL:
Ohm’s Law:Inductor Eqn:
L id
dtR i + vC + = Vo
Q: How do we take care of vC term?
A: Take of the entire equation!ddt
L id2
dt2+ + =R id
dt vCddt
dVo
dt
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L id2
dt2+ + =R id
dt iC
dVo
dt
Capacitor equation
Rearrange + Divide by L
L id2
dt2+ + =R id
dt vCddt
dVo
dt
Transform some more….
+ + =dVo
dti
LCid
dtLR id2
dt2 L1 zero
Note condition at t =
( and = 0 )ddt
d2
dt2
i = 0Makes sense:Capacitor is a dc open
Vo
L
C
R
+vC
–+vL
–
+ vR –
i(t)
= 0
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+ + =iLC
iddtL
R id2
dt2 L1
= 0
Let’s solve it!
Here’s our differential equation
est
(Same approach as before:)
Try solution of the form *
* (Suggested by mathematicians who came before us…)
i = A est
ddt
est
= s estd2
dt2
est
= s2 est
Note:
s2 Aest Aest
LR s
LC1
Aest+ + = 0
s2LC1
LR s+ + = 0
Will be determined by initial conditions• A can be anything…
• s must satisfy:
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x = –b – b2 – 4ac
2a x = –b + b2 – 4ac
2a
s2LC1
LR s+ + = 0
Characteristic Equation (of the circuit)
Vo
L
C
R
+vC
–+vL
–
+ vR –
i(t)
Quadratic Equation:
ax2 + bx + c = 0Root #1 Root #2
a 1 bLR
c LC1
Root #12LR
– +2LR
LC1
–2
Root #22LR
– –2LR
LC1
–2
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Root #1
2L 2LR
LC1R
– + –2
s1 =
Root #2
2LR
– –2LR
LC1
–2
s2 =
i = A est
L
CR +
vC
–+vL
–
+ vR –
Vo
t = 0
i(t)
Two solutions: Which one is correct?
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Root #1
2L 2LR
LC1R
– + –2
s1 =
Root #2
2LR
– –2LR
LC1
–2
s2 =
i(t) = A est
Two cases to consider:
2LR
LC1
> = REALPositive Number2LR
LC1
< = IMAGINARYNegative NumberLet’s look at these cases one at a time…
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Root #1
2L 2LR
LC1R
– + –2
s1 =
Root #2
2LR
– –2LR
LC1
–2
s2 =
2LR
LC1
> = REALPositive Number
L
CR
1 F
12 H
10 k
Vo
5 Vi(t)
2LR
= 417
LC1
= 83,300
s1 = –116 s2 = – 717
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L
CR
1 F
12 H
10 k
Vo
5 Vi(t)
s1 = –116 s2 = – 717
i(t) = A1es1t + A2e
s2t
Most General Solution: Linear Superposition
Must satisfy the Initial Conditions:
i(0) = 0
+vC
–
vC(0) = 0
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VoL
CR
i(t)
+vC
–
iL(0) = 0
vC(0) = 0
vL(0) = Vo
+vL
–
By KVL: (The voltage that’s left over after vC = 0 and vR = 0)
Initial Conditions:
i(0) = 0 vC(0) = 0 (The initial condition on vC translates into an IC for di/dt )
This is good!
2nd order differential equation Need I.C. for i and to solve. di dt
i(0) = 0 vR(0) = 0
Ohm’s Law:
t=0
d idt
= Vo L
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s1 = –116 sec–1 s2 = – 717 sec–1
i(t) = A1es1t + A2e
s2t
i(0) = 0 A1 + A2 = 0
t=0
d idt
= Vo L s1A1 + s2A2 = Vo
L
Solve for A1
s2A1 + s2A2 = 0( )
(s1–s2)A1 = Vo L
A1 = Vo
(s1–s2) L
s1 = –116 sec–1
s2 = – 717 sec–1
Vo = 5 V
L = 12 H
Find A1
= 0.7 mA
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s1 = –116 sec–1 s2 = – 717 sec–1
A1 + A2 = 0
Find A2
A1 = 0.7 mA A2 = –0.7 mA
i(t) = A1es1t + A2e
s2t
This term decays much more quickly
i(t) = 0.7 (e–166t– e –717t ) mA
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0.00
0.10
0.20
0.30
0.40
0.500.
0
5.4
10.8
16.2
21.6
27.0
32.4
37.8
43.2
48.6
54.0
59.4
Time (ms)
Cur
rent
(m
A)
0 10 20 30 40 50 60 t (ms)
i(t) = 0.7 (e–166t
– e–717t
) mA
Overdamped Response
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-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
0.0
3.6
7.210
.814
.418
.021
.625
.228
.832
.436
.039
.643
.246
.850
.454
.057
.60 10 20 30 40 50 60
0
0.20
0.40
0.60
0.80
– 0.60
– 0.40
–0.20
0
– 0.8
i(t) = 0.7 (e–166 t
– e–717 t
) mA
i1(t) + i2(t)
i1(t) = 0.7 e–166t
i2(t) = – 0.7 e–717 t
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Root #1
2L 2LR
LC1R
– + –2
s1 =
Root #2
2LR
– –2LR
LC1
–2
s2 =
2LR
LC1
> = REALPositive Number
L
CR
1 F
12 H
10 k
Vo
5 Vi(t)
2LR
= 417
2LR
LC1
< = IMAGINARYNegative Number1 k
41.7
LC
1
= 83,300(same)
s1 = – 41.7 + –81,600 s2 = – 41.7 – –81,600
Huh?
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–81,600
Huh?
–81,600
How to deal with this term…
–81,600
–1 ij (Mathematicians) (Engineers)
s1 = – + j o
s2 = – – j o
= 42 units: sec–1
–1 –81,600
s1 = – 42 + j 81,600
s2 = – 42 – j –81,600
(41.7)
o = 81,600 = 286
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i(t) = A1es1t + A2e
s2t
e s1t = e(– jo)t+
Huh?(again)
= e–t jote
xe = 1 + x x2
2!x3
3!x4
4!x5
5!x6
6!+ + + + + + …
xe = 1 + x x2
2!x3
3!x4
4!x5
5!x6
6!+ + + + + + …
j 2 = –1 j 4 = +1 j 6 = –1j x
e = 1 – xx2
2!x3
3!x4
4!x5
5!x6
6!+ – + … + j + – …–
s1 = – + j o s2 = – – j o
n=0 n=1 n=2
n! = n(n–1)(n–2)(n–3)..(1)
j 2 j 4 j 6
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j xe = 1 – xx2
2!x3
3!x4
4!x5
5!x6
6!+ – + … + j + – …–
cos x = 1 – x2
2!x4
4!x6
6!+ – + …
x x3
3!x5
5!+ – …–sin x =
Basic definitions ofsine and cosine
j xe = cos x + j sin x “Euler’s Equation”
j ote = cos ot + j sin ot Back to the circuit…
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s1 = – + j o s2 = – – j o
= 42 o = 16.9
A1es1t = A1e
(– jo)t+ cos ot + j sin ote
–t= A1
–
Soln #1: cos te–t2 (A1+ A2) e–t cos ot= K1
sin ote–t= K2
sin ote–t+ K2 cos ote–t K1i(t) =General Solution:
Do some linear algebra…
Vo
L
C
R
+vC
–+vL
–
+ vR –
i(t)
Soln #2: sin te–t2 j (A1– A2)
A2es2t = A2e
(– jo)t cos ot – j sin ote–t
= A2
–A2e
s2t = A1e(– jo)t cos ot – j sin ote
–t= A2+--
A1es1t = A1e
(– jo)t+ cos ot+ j sin ote
–t= A1
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sin ote–t+ K2 cos ote–t K1i(t) =
General Solution:
Initial Condition #1: i(0) = 0 K1 = 0
Equal 0 at t = 0Equals 1 at t = 0
Initial Condition #2:
t=0
d idt
= Vo L
Equals 0 at t = 0
cos oto e–t
+ sin ot– e–t
K2
Equals 1 at t = 0Equals 1 at t = 0
K2o = Vo L
K2 = Vo o L
Find derivative of i(t)
Equals 1 at t = 0 Equals 1 at t = 0
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-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
0.0 10 20 30 40 50 60 70 80 90 100 t (ms)
1.5
1.0
0.5
–0.5
–1.0
0
K2e–t envelope
Period 2o
sin ote–ti(t) = Vo o L
Underdamped Response
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Root #1
2L 2LR
LC1R
– + –2
s1 =
Root #2
2LR
– –2LR
LC1
–2
s2 =
R 1
2L LC= 0Two Double Roots
2LR –s1 = s2 =
Critically Damped Response
(Boundary between Overdamped and Underdamped Responses)
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End of This Module
Homework