Elec 3202 Chap 7
Transcript of Elec 3202 Chap 7
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Chapter (7)Second Order Transient Circuits
•Here , we consider circuits with both capacitor , inductor , andresistors (RLC)•We expect the circuit to be descried by a second order differential equation.•Consider the parallel RLC circuit
iS (t)
iR (t) iL (t) iC (t)
L CR
+
-
VC (t)
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Take first derivative
( ) (t)idt
(t)dvCdττv
L1)(ti
R(t)v
sc
t
tc0L
c
0
=+++∴ ∫
dt(t)di
dt(t)vd
C(t)vL1
dt(t)dv
R1 s
2c
2
cc =++
Let’s assume that the voltage across the capacitor is VC(t)
(t)i(t)i(t)i(t)i CLRS ++=KCL :
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dt(t)di
C1(t)v
LC1
dt(t)dv
CR1
dt(t)vd s
cc
2c
2
=++
0dt
(t)diS =
0(t)vLC1
dt(t)dv
CR1
dt(t)vd
cc
2c
2
=++
If IS(t) = constant (DC)
Consider the series RLC :
VS(t)
+-
R L
Ci
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
KVL :
( ) (t)VdττiC1)(tV
dtdi(t)Li(t)R
0V(t)V(t)V(t)V
S
t
t0C
CLRS
0
=+++
=+++−
∫
dt(t)dV
(t)iC1
dti(t)dL
dtdi(t)R S
2
2
=++
0dt
(t)dVS =
Take first derivative :
If VS = constant
0(t)iCL
1dt
di(t)LR
dti(t)d
2
2
=++ Series RLC
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Hence, let’s assume that the differential equation we wish to solve is
A(t)xadt
dx(t)adt
(t)d212
X2
=++
It is known that the solution x (t) can be expressed as
(t)x(t)xx(t) CP +=
xP(t) : particular ( forced solution )xC(t) : complementary ( natural solution)
To find xP(t)
The only solution is the constant xP(t) = k0
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( ) ( )
2P
20
020102
2
aA(t)x
aAk
Akakdtdak
dtd
=⇒
=
=++
0(t)xadt
(t)dxadt
(t)dc2
c12
cX2
=++
To find xC(t) : ( natural solution )
For simplicity let’s define :
202
01
a
ξ2a
ω
ω
=
=
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Where :ξ : damping ratio.w0 : underdamped natural frequency
0(t)xdt
(t)dxξ2dt
(t)dc
20
c02
cX2
=++ ωω
Assume xc(t) = k e s t
[ ] [ ] [ ]
[ ] 0sξ2sek
0ekeskξ2esk
0ekekdtdξ2ek
dtd
200
2ts
ts20
ts0
ts2
ts20
ts0
ts2
2
=++
=++
=++
ωω
ωω
ωω
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k : non-zeroest : non-zero
0sξ2s 200
2 =++ ωω
1ξξs2
4ξ4ξ2s
200
20
20
20
−±−=
−±−=
ωω
ωωω
Characteristic polynomial
The roots :
1ξξs
1ξξs2
002
2001
−−−=
−+−=
ωω
ωω Complex frequencies[ rad / sec]
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ts2
ts1c
21 ekek(t)x +=⇒
ts2
ts10
CP
21 ekekkx(t)
(t)x(t)xx(t)
++=
+=Q
210 kkkx(0) ++=
k1 , k2 can be found from initial conditions
From x(0):
From dx(0) / dt:
2211
ts22
ts11
skskdt
dx(0)
eskeskdt
dx(t)21
+=
+=
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Note :There are 3 types of responses based on the values of (S)
1. Overdamped response ( ξ > 1)
S1 and S2 are real and distinct .
20
ts2
ts10
aAk
ekekkx(t) 21
=
++=
k1 and k2 can be found from initial conditions
2211
210
skskdt
dx(0)kkkx(0)
+=
++=
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2. Underdamped response ( ξ < 1 )
S1 & S2 are complex conjugate
2001,2
2001,2
2001,2
ξ-1ωjωξS
1ξ-1ωωξS
1ξωωξS
m
m
m
−=
−−=⇒
−−=⇒
−≡
≡2
0d
0
ξ1ωω
ωξσLet Damped radian frequency
[ rad / sec ]
d1,2 ωjσS m−=⇒
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( ) ( )
( )( ) ( )ΘsinjΘcose
ekekek
ekekk
ekekkx(t)
Θj
tωj2
tωj1
tσ0
tωjσ2
tωjσ10
ts2
ts10
dd
dd
21
±=
++=
++=
++=
±
−−
+−−−
Q
( ) ( )( ) ( ) ( )( )[ ]( ) ( ) ( )[ ]tωsinkjkjtωcos)k(kek
ωsinωcoskωsinωcoskekx(t)
d21d21tσ
0
dd2dd1tσ
0
−+++=
−+++=−
− tjttjt
( )212
211
kkjAkkALet−=
+=
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( ) ( )
20
dtσ
2dtσ
10
aAk
tωsinetωcoseAkx(t)
=
++= −− A
A1 & A2 can be found from initial conditions :
( ) ( )
( ) ( )
2d1
tσ-2dd
tσd2
tσ1
tσddd
tσ1
10
AωAσdt
dx(0)σeAtωsinωetωcosA
eσAetωcosωtωsineAdt
dx(0)Akx(0)
+−=
−+
−=
+=
−
−−
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+−=
+=
=
⇒
2d1
10
20
AωAσdt
dx(0)Akx(0)
aAk
021
2001,2
ωξSS1ξωωξSSince
−==⇒
−−= m
3. Critical damped response ( ξ = 1 )
*0(t)xαdt
(t)dxα2dt
(t)xd
ωξαLet
0(t)xωdt
(t)dxωξ2dt
(t)xd
c2
0c
2c
20
c20
c02
c2
=++⇒
≡
=++⇒
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Which means thattα
2tα
1c ekek(t)x −− +=
tα2
tα1c ekek(t)x −− += t
It is impossible for this solution to satisfy two initial conditions :
It can be shown that the following solution also satisfies the differential equation:
tα2
tα10
cP
etkekkx(t)
(t)x(t)xx(t)−− ++=
+=⇒
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
21
tα2
tα2
tα1
10
20
kαkdt
dx(0)
ekeαtkeαkdt
dx(t)kkx(0)
aAk
+−=
+−−=
+=
=
−−−
+−=
+=
=
⇒
21
10
20
kαkdt
dx(0)kkx(0)
aAk
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V (0)
+
-
RLC
Example :
A3(0)i,V12v(0)AssumeF1C
H0.5L
Ω31R
L ====
=
Find v (t) ?
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0CL
1sCR
1s
0(t)vLC1
dt(t)dv
CR1
dt(t)vd
2
cc
2c
2
=
++
=++
Characteristic polynomial is :
11.0622
3ω23ξ
3ωξ22ω2ω
0CL
1SCR
1S
0
0
02
0
2
>===
==⇒=
=++
Overdamped
From before we know
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A363
112(0)i12v(0)Since
(1)kk12v(0)ekekv(t)
ekekv(t)
2s,1s
R
21
t2-2
t-1
tS2
tS1
21
21
==⇒=
+==+=
+=
−=−=
LL
39dt
dv(0)
0dt
dv(0)139
0dt
dv(0)C336
0(0)i(0)i(0)i CLR
−=
=+
=++
=++
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
t2t2
1
21
21
t22
t1
e27e15-v(t)
27k15-k
(2)39k2k
39k2kdt
dv(0)
ek2ekdt
dv(t)
−−
−−
+=
==
=+
−=−−=
−−=
KK
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24 V
+-
RL
Ct = 0
F4.0CH0.1LΩ802R
µ===
Find vc (t) , t > 0 ?
Example :
If vc(0) = 0iL(0) = 0
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
KVL :
( )dττiC1(0)Vi(t)R
dtdi(t)L24
V(t)V(t)V24t
tC
CRL
0
∫+++=
++=
0(t)i10*25dt
di(t)2800dt
i(t)d
0(t)iCL
1dt
di(t)LR
dti(t)d
(t)iC1
dtdi(t)R
dti(t)dL0
62
2
2
2
2
2
=++
=++
++=
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Char. Poly. is
128.0100002800
ω228008002ωξ2
rad/sec0005ω10*52ω
010*52s8002s
00
062
0
62
<===⇒=
=⇒=
=++
ξ
Underdamped
4800j1400ωjσ
ξ-1wjwξs
d
20021,
m
m
m
−=−=
−=
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( ) ( )0k
tωsineAtωcoseAki(t)
0
dtσ
2dtσ
10
=++= −−
( ) ( )t4800sinet4800coseAi(t) t14002
t14001
−− += A
A1 & A2 can be found from initial conditions
420R(0)i(0)V42(0)V(0)V(0)V
L
CRL
=++=++
i(0) = A1 = 0
Since homogenous
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t)sin(4800e0.05i(t)
0.05AA4800Aω402
4800ω
AωAσ402dt
di(0)
42dt
di(0)L(0)v
t1400-2
22d
d
2d1
L
=
===⇒
=
+−==
==
dtdi(t)0.1i(t)28042(t)v
(t)Vi(t)Rdt
di(t)L24
V(t)V(t)V24
c
C
CRL
−−=
++=
++=From KVL
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( )[ ]( )
( ) ( )
+−
−=
−
−
−
t1400
t1400
t1400c
e1400-0.05*t4800sin4800t4800cose0.05
0.1
t4800sine0.0528042(t)v
( )( ) ( )
( ) ( )t4800cose42t4800sine742(t)vt4800sine7t4800cose42
t4800sine1442(t)v
t1400t1400c
t1400t1400
t1400c
−−
−−
−
−−=
+−
−=
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+
-
R
L Ci L(0) vc(0)
Example :
F91C
H1LΩ6R
=
==
Find vc (t)?
vc(0) = 1 V , iL(0) = 0
0(t)i9dt
di(t)6dt
i(t)d
0(t)iCL
1dt
di(t)LR
dti(t)d
2
2
2
2
=++
=++
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α−==⇒−==⇒=+
=++
21
212
2
ss3ss03)(s
09s6s
0ki(0)etkeki(t)
1
t32
t31
==+= −−
1dt
di(0)
010dt
di(0)
01R(0)idt
di(0)L
0(0)V(0)V(0)V CRL
−=
=++
=++
=++From KVL :
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t32
21
eti(t)
1k
1kkdt
di(0)
−−=
−=
−=+−= α
dtdi(t)L(t)iR(t)V
0(t)V(t)V(t)V
C
CRL
−−=
=++From KVL :
Now, we need to find Vc(t):
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[ ] ( ) ( )[ ]
t3t3c
t3t3t3
t3t3t3c
eet3(t)veet3et6
1e3et-et-6(t)v
−−
−−−
−−−
+=
+−=
−+−−−=