inc 112 basic circuit analysis
DESCRIPTION
INC 112 Basic Circuit Analysis. Week 10 RLC Circuits. RLC Circuits. Similar to RL and RC circuits, RLC circuits has two parts of responses, namely: natural response and forced response. Force Response: Similar to RL and RC, a step input causes a step output. Natural Response: - PowerPoint PPT PresentationTRANSCRIPT
INC 112 Basic Circuit Analysis
Week 10
RLC Circuits
RLC Circuits
Similar to RL and RC circuits, RLC circuits has two partsof responses, namely: natural response and forced response.
Force Response:Similar to RL and RC, a step input causes a step output.
Natural Response:Different and more difficult than RL, RC.
Source-free RLC Circuits
We study the natural response by studying source-free RLC circuits.
LR+
v(t)-
iR(t)C
iC(t)iL(t)
Parallel RLC Circuit
0)(1)(1)(
0)(
)(1)(
0
2
2
tvLdt
tdv
Rdt
tvdC
dt
tdvCdttv
LR
tv
iii CLR
Second-orderDifferential equation
This second-order differential equation can be solved by assuming solutions
The solution should be in form of
0)(1)(1)(
2
2
tvLdt
tdv
Rdt
tvdC
stAetv )(
If the solution is good, then substitute it into the equation will be true.
011
0)11
(
011
2
2
2
Ls
RCs
Ls
RCsAe
AeL
AseR
eCAs
st
ststst
which meanss=??
0112 L
sR
Cs
LCRCRCs
LCRCRCs
1
2
1
2
1
1
2
1
2
1
2
2
2
1
Use quadratic formula, we got
tseAtv 11)( Both and tseAtv 2
2)( are solution to the equation
Therefore, the complete solution iststs eAeAtv 21
21)(
LCRCRCs
1
2
1
2
12
2,1
From
Define resonant frequencyLC
10
Damping factorRC2
1
Therefore,20
22,1 s
in which we divide into 3 cases according to the term inside the bracket
Solution to Second-order Differential Equations
1. α > ω0 (inside square root is a positive value) Overdamped case
2. α = ω0 (inside square root is zero) Critical damped case
3. α < ω0 (inside square root is a negative value) Underdamped case
1. Overdamped case , α > ω0
7H6Ω+
v(t)-
iR(t)1/42f
iC(t)iL(t)
Initial conditionvc(0) = 0, iL(0) = -10A
Find v(t)
5.32
1
RC 6
10
LC
α > ω0 ,therefore, this is an overdamped case
20
22,1 s s1 = -1, s2 = -6
Therefore, the solution is in form oftt eAeAtv 6
21)(
Then, we will use initial conditions to find A1, A2
From vc(0) = 0 we substitute t=0
210
20
10)0( AAeAeAv
From KCLAt t=0
4206
0642
1)10(
0
0)(
)10()0(
0
21
0
621
0
AA
eAeAR
dt
tdvC
R
v
iii
t
tt
t
CLR
Solve the equation and we got A1 = 84 and A2 = -84
and the solution is
)(848484)( 66 tttt eeeetv
t
v(t)
)(84)( 6tt eetv
2. Critical damped case , α = ω0
7H8.573Ω+
v(t)-
iR(t)
1/42f
iC(t)iL(t)
Initial conditionvc(0) = 0, iL(0) = -10A
Find v(t)
45.21
2
10
LCRC
α = ω0 , this is an critical damped case s1 = s2 = -2.45
The complete solution of this case is in form of
stst eAteAtv 21)(
Then, we will use initial conditions to find A1, A2
From vc(0) = 0 we substitute t=0
20
20
1 )0(0)0( AeAeAv
Find A1from KCLat t=0
0)(42
110
0)45.2(42
1)10(
0
0)(
)10()0(
0
1
0
45.21
45.21
0
A
eAetAR
dt
tdvC
R
v
iii
t
tt
t
CLR
Therefore A2 =0 and the solution is reduced totteAtv 45.2
1)(
Solve the equation and we got A1 = 420 and the solution is
t
v(t)
ttetv 45.2420)(
ttetv 45.2420)(
3. underdamped case , α < ω0
from20
22,1 s
The term inside the bracket will be negative and s will be a complex number
define 220 d
Then djs 2,1
and
)()(
)(
21
)(2
)(1
tjtjt
tjtj
dd
dd
eAeAetv
eAeAtv
)()( 21tjtjt dd eAeAetv
Use Euler’s Identity sincos je j
)sincos()(
sin)(cos)(()(
)sincossincos()(
21
2121
2211
tBtBetv
tAAjtAAetv
tjAtAtjAtAetv
ddt
ddt
ddddt
)sincos()( 21 tBtBetv ddt
3. Underdamped case , α < ω0
7H10.5Ω+
v(t)-
iR(t)
1/42f
iC(t)iL(t)
Initial conditionvc(0) = 0, iL(0) = -10A
Find v(t)
22
1
RC 6
10
LC
α < ω0 ,therefore, this is an underdamped case
2220 d
and v(t) is in form )2sin2cos()( 212 tBtBetv t
Then, we will use initial conditions to find B1, B2
From vc(0) = 0 we substitute t=0
Find B2from KCLat t=0
0)2(42
110
02sin22cos242
1)10(
0
0)(
)10()0(
0
2
0
22
22
0
B
teBteBR
dt
tdvC
R
v
iii
t
tt
t
CLR
Therefore B1 =0 and the solution is reduced to
1210 )0sin0cos()0( BBBev
)2sin()( 22 tBetv t
we got and then the solution is
t
v(t)
29722102 B
)2sin297)( 2 tetv t
)2sin297)( 2 tetv t
Natural responseat different time
Mechanical systems are similar to electrical systems
A pendulum is an example of underdamped second-order systems in mechanic.
t
displacement(t)
Series RLC
0)(1)()(
0)(1)(
)(
0
2
2
tiCdt
tdiR
dt
tidL
dttiCdt
tdiLRti
vvv CLR
LR +
vL(t)-
+ vR(t) -
C
i(t)
-vC(t)
+
0)(1)()(
2
2
tiCdt
tdiR
dt
tidL
Series RLCParallel RLC
0)(1)(1)(
2
2
tvLdt
tdv
Rdt
tvdC
tsts eAeAtv 2121)(
LCRCRCs
1
2
1
2
12
2,1
20
22,1 s
RC2
1
LC
10
tsts eAeAti 2121)(
LCL
R
L
Rs
1
22
2
2,1
20
22,1 s
L
R
2
LC
10
Procedure for Solving RLC Circuits
1. Decide that it is a series or parallel RLC circuit. Find α and ω0. Then, decide which case it is (overdamped, critical damped, underdamped).
2. Assume the solution in form of (natural response+ forced response)
fstst VeAteA 21
ftsts VeAeA 21
21
fddt VtBtBe )sincos( 21
Overdamped
Critical damped
Underdamped
3. Find A, B, Vf using initial conditions and stable conditions
Example
S1 30
t = 0 sec
+
Vc
-
1/27f3H4A 5AiL(t)
Find vc(t)
52
L
R 31
0 LC
9,120
22,1 s
This is overdamped case, so the solution is in form
ftt VeAeA 9
21
ftt
C VeAeAtv 921)(
Consider the circuit we found that the initial conditions will bevC(0) = 150 V and iL(0) = 5 A and the condition at stable point will bevC(∞) = 150 V and iL(∞) = 9 A
fVAA 21150Using vC(0) = 150 V , we got
Using vC(∞) = 150 V , we got fV 00150
Therefore, Vf = 150, A1+A2 = 0
Consider the circuit, we find that iC(0) = 4A
21
02
01
921
9108
)9(27
14)0(
)9(27
1)()(
AA
eAeAi
eAeAdt
tdvCti
C
ttC
A1 = 13.5, A2 = -13.5
Therefore, 1505.135.13)( 9 ttC eetv