Second-Order Transient Response Analysis

Lecture 14
Second Order Transient Response

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Damped Harmonic Motion

Mechanical oscillator
2
d x dx
m 2 + b + kx = f m (t )
dt dt
Electrical
El t i l oscillator
ill t
d 2q
dq 1
L 2 +R + q = v s (t )
dt dt C


The Series RLC Circuit and its
Mechanical Analog

• Drive voltage vS(t) Applied force f(t)
• Charge q Displacement x
• Current i Velocity v
• Inductance L Mass m
• Inductive Energy U=(1/2)Li2 Inertial Energy (1/2)mv2
• Capacitance C Spring constant 1/k
• Capac. Energy U=(1/2)(1/C)q2 Spring energy U=(1/2)kx2
• Resistance R Dampening constant b

Second –Order Circuits

di (t )
t
+ Ri (t ) + ∫ i(t )dt + v (0) = v (t )
1
L C s
dt C 0

Differentiating with respect to time:
2
d i (t ) di (t ) 1 dv s (t )
L +R + i (t ) =
dt 2
dt C dt

Second –Order Circuits
d 2 i (t ) R di (t ) 1 1 dv s (t )
+ + i (t ) =
dt 2
L dt LC L dt
R Dampening
Define: α=
2L coefficient
1
ω0 = Undamped resonant
LC frequency

1 dv s (t )
f (t ) = Forcing function
L dt
2
d i (t ) di (t )
+ 2α + ω 0 i (t ) = f (t )
2
dt 2
dt

Solution of the Second-Order Equation

Particular solution
d i (t )
2
di (t )
+ 2α + ω 0 i (t ) = f (t )
2
dt 2
dt
Complementary solution
d i (t )
2
di (t )
+ 2α + ω 0 i (t ) = 0
2
dt 2
dt


Solution of the Complementary
Equation
d i (t )
2
di (t )
+ 2α + ω 2 i (t ) = 0 0
dt 2
dt

Try iC (t ) = Ke st :
s 2 Ke st + 2αsKe st + ω0 Ke st = 0
2

Factoring :
( s 2 + 2αs + ω 0 ) Ke stt = 0
2

Characteristic equation :
s 2 + 2αs + ω 0 = 0
2


Equation

Qaudratic Equation : ax 2 + bx + c = 0
− b ± b − 4ac 2
x=
2a
Characteristic Equation : s 2 + 2αs + ω0
2

− 2α ± (2α ) 2
− 4ω0
2
s= = −α ± α 2 − ω0
2
2


Equation
Roots of the characteristic equation:

s1 = −α + α − ω 2 2
0

s2 = −α − α − ω 2 2
0

α
ζ = Dampening ratio
ω0

1. Overdamped case (ζ > 1). If ζ > 1 (or
equivalently, if α > ω0), the roots of the
characteristic equation are real and distinct.
Then the complementary solution is:

ic (t ) = K1e s1t
+ K 2e s2t

In this case, we say that the circuit is
overdamped.


2. Critically damped case (ζ = 1). If ζ = 1 (or
equivalently, if α = ω0 ), the roots are real
and equal. Then the complementary solution
is

ic (t ) = K1e s1t
+ K 2te s1t

critically damped.
iti ll d d


3. Underdamped case (ζ < 1). Finally, if ζ < 1
(or equivalently, if α < ω0) the roots are
equivalently ),
complex. (By the term complex, we mean that
the roots involve the square root of –1.) In other
1)
words, the roots are of the form:

s1 = −α + jωn a n d s2 = −α − jωn

in which j is the square root of -1 and the
natural frequency is given by:
f q y g y

ωn = ω − α 2
0
2


I electrical engineering, we use j rather than i
In l t i l i i th th
to stand for square root of -1,because we use i
for
f current. F complex roots, the
t For l t th
complementary solution is of the form:

ic (t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin (ω n t )

underdamped.


Analysis of a Second-Order Circuit
with a DC Source

Apply KVL around the loop:

di (t )
L + Ri (t ) + vC (t ) = VS
dt

with a DC Source
di (t )
L + Ri (t ) + vC (t ) = VS
dt
dvC (t )
Substitute i (t ) = C
dt
d 2 vC (t ) dvC (t )
LC + RC + vC (t ) = Vs
dt 2
dt
Divide by
Di id b LC :
d 2 vC (t ) R dv(t ) C 1 Vs
+ + vC (t ) =
d2
dt L dt LC LC


with a DC Source
To find the particular solution, we consider the steady
solution
state. For DC steady state, we can replace the inductors
by a short circuit and the capacitor by an open circuit:

vCp (t ) = Vs = 10V

with a DC Source
To fi d the homogeneous solution, we consider three cases;
find h h l i id h
R=300Ω, 200Ω and 100Ω. These values will correspond to
over-damped, critically damped and under-damped cases.
p , y p p

1 1
ω0 = = = 10 4 rads / sec
LC (10mH )(1μF )


with a DC Source
Case I : R = 300Ω
R 300Ω α 1.5 x10 4
α= = = 1.5 x10 4 ζ = = = 1.5 → Overdamped
2 L 2(10mH ) ω0 10 4

s1 = −α + α 2 − ω0 = −1.5 x10 4 − (1.5 x10 4 ) 2 − (10 4 ) 2 = −2.618 x10 4
2

s 2 = −α − α 2 − ω0 = −0.382 x10 4
2

Adding the particular and complementary solutions :

vC (t ) = 10 + K1e s1t + K 2 e s2t


with a DC Source
To find K1 and K2 we use the initial conditions:
vc (t ) = 10 + K1e s1t + K 2 e s2t
vc (0) = 0 = 10 + K1e 0 + K 2 e 0 = 10 + K1 + K 2

i (0) = 0
dvc (t ) dvc (0)
i (t ) = C i (0) = 0 → =0
dt dt
dvc (t )
= s1 K1e s1t + s 2 K 2 e s2t
dt
dvc (0)
= 0 = s1 K1e 0 + s 2 K 2 e 0 = s1 K1 + s 2 K 2
dt

10 + K + K = 0
1
with a DC Source
2
s1 K1 + s 2 K 2 = 0

K1 + K 2 = −10
s1 K1 + s 2 K 2 = 0

− 10 1
0 s2 (−10)( s 2 ) − (0)(1) − 10 s 2 3.82 x10 4
K1 = = = = = 1.71
1 1 (1)( s 2 ) − ( s1 )(1) s 2 − s1 − 0.3820 x10 4 − (−2.618 x10 4 )
s1 s2
1 − 10
s1 0 (1)(0) − ( s1 )(−10) 10 s1 − 26.18 x10 4
K2 = = = = = −11.7
1 1 (1)( s 2 ) − ( s1 )(1) s 2 − s1 − 0.3820 x10 − (−2.618 x10 )
4 4

s1 s2

vc (t ) = 10 + 1.708e s1t − 11.708e s2t

with a DC Source
Case II : R = 200Ω
R 200Ω α 10 4
α= = = 10 4 ζ = = 4 = 1 → Critically damped
2 L 2(10mH ) ω0 10

s1 = s 2 = −α + α 2 − ω 0 = −α = −10 4
2


vC (t ) = 10 + K1e s1t + K 2 te s1t


with a DC Source

vc (t ) = 10 + K1e s1t + K 2 te s1t
vc (0) = 0 = 10 + K1e 0 + K 2 (0)e 0 = 10 + K1

i (0) = 0
dvc (t ) dvc (0)
i (t ) = C i (0) = 0 → =0
dt dt
dvc (t )
= s1 K1e s1t + s1 K 2 te s1t + K 2 e s1t
dt
dvc (0)
= 0 = s1 K1e 0 + s1 K 2 (0)e 0 + K 2 e 0 = s1 K1 + K 2
dt

with a DC Source
10 + K1 = 0 → K1 = −10
s1 K1 + K 2 = 0 → K 2 = − s1 K1 = 10s1 = 10(−10 4 ) = −10 5

vc (t ) = 10 − 10e s1t − 10 5 te s2t


with a DC Source
Case III : R = 100Ω
R 100Ω α 0.5 x10 4
α= = = 0.5 x10 4 ζ = = = 0.5 → Underdamped
2 L 2(10mH ) ω0 10 4

ω n = ω0 − α 2 = (10 4 ) 2 − (0.5 x10 4 ) 2 = 8660
2


vC (t ) = 10 + K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
dvC (t )
= −αK1e −αt cos(ω n t ) − K1e −αt ω n sin(ω n t ) − αK 2 e −αt sin(ω n t ) + K 2 e −αt ω n cos(ω n t )
dt


with a DC Source
v c ( 0) = 0
dvc (0)
=0
dt

10 + K1 = 0 → K1 = −10
α 5000
− α K1 + ω n K 2 = 0 → K 2 = K1 = (−10) = −5.774
ωn 8660

vc (t ) = 10 − 10e −αt cos(ω n t ) − 5.774e −αt sin(ω n t )

Overdamped
O d d


Critically d
C iti ll damped
d


Underdamped
U d d d


Normalized Step Response of a Second
Order System

u (t ) = 0 t < 0
u (t ) = 1 t > 0


Order System
Example: Closing a switch at t=0 to apply a DC voltage A
t 0

v(t ) = Au (t )


Order System

What value of ζ
should be used to get
to the steady state
position quickly
without overshooting?


Circuits with Parallel L and C
t
V dq dv
∫
1
iR = iL = vdt + iL (0) ic = =C
R L dt dt
0

We can replace the circuit with it’s Norton equivalent
and then analyze the circuit by writing KCL at the top
y y g p
node:
t
dv(t ) 1
∫
1
C + v(t ) + v(t )dt + i L (0) = in (t )
dt R L
0

t
d (t ) 1
dv
∫
1
C + v(t ) + v(t )dt + i L (0) = in (t )
dt R L
0
differentiating :
d 2 v(t ) 1 dv(t ) 1 din (t )
C + + v(t ) =
dt R dt L dt
d 2 v(t ) 1 dv(t ) 1 1 din (t )
+ + v(t ) =
dt RC dt LC C dt


d 2 v(t ) 1 dv(t ) 1 1 din (t )
+ + v(t ) =
dt RC dt LC C dt
1
Dampening coefficient α =
2 RC
1
Undamped resonant frequency ω 0 =
LC
1 din (t )
Forcing function f (t ) =
C dt
d 2 v(t ) dv(t )
+ 2α + ω0 v(t ) = f (t )
2
dt dt


d 2 v(t ) dv(t )
+ 2α + ω0 v(t ) = f (t )
2
dt dt
This is the same equation as we found for the series LC
circuit with the following changes for α:

1
Parallel circuit α =
2 RC
R
Series circuit α =
2L


Example Parallel LC Circuit: RF-ID
Tag
L

C



Find v(t) for R=25Ω, 50Ω and 250Ω



Find v(t) for R=25Ω, 50Ω and 250Ω

1 1
ω0 = = = 1x10 5 rad / sec
LC (1x10 −3 H )(1x10 −7 F )
1
α=
2 RC


Vp(t) = 0

To find the particular solution vP(t) (steady state
response) replace C with an open circuit and L with a
short circuit
circuit.


Case I : R = 25Ω
1 1 5 −1 α 2 x10 5 s −1
α= = = 2 x10 s ζ = = = 2 → Overdamped
2 RC 2(25Ω)(1x10 −7 F ) ω0 10 5 rad / s

s1 = −α + α 2 − ω 0 = −2 x10 5 + (2 x10 5 ) 2 − (10 5 ) 2 = −0.268 x10 5
2

s 2 = −α − α 2 − ω 0 = −2 x10 5 − (2 x10 5 ) 2 − (10 5 ) 2 = −3.73 x10 5
2


vC (t ) = K1e s1t + K 2 e s2t
v P (t ) = 0
vT (t ) = K1e s1t + K 2 e s2t


Initial conditions :
v (0 −) = 0 → v (0 + ) = 0 Since the voltage across a capacitor
and the current through an inductor
i L (0 −) = 0 → i L (0 + ) = 0 cannot change instantaneously

KCL at t = 0 +
v (0 + ) dv(0+)
0.1 = + i L (0 + ) + C
R dt
dv(0+ ) dv(0+ ) 0.1 0.1
0.1 = C → = = = 10 6 v / s
dt dt C 0.1x10 −6

v(0+ ) = K1e 0 + K 2 e 0 → K1 + K 2 = 0
d (0 + )
dv
= K1s1e 0 + K 2 s 2 e 0 = K1s1 + K 2 s 2 = 10 6
dt


K1 + K2 = 0
s1 K1 + s 2 K 2 = 10 6

0 1
10 6 s2 (0)( s 2 ) − (10 6 )(1) − 10 6 − 10 6
K1 = = = = = 2.89
1 1 (1)( s 2 ) − ( s1 )(1) s 2 − s1 − 0.268 x10 − (−3.73 x10 )
5 5

s1 s2
K 2 = − K1 = −2.89

− 0.268 x105 t −3.73 x105 t − 0.268 x105 t −3.73 x105 t
v(t ) = 2.89e − 2.89e = 2.89(e −e )



Case II : R = 50Ω
1 1 5 −1 α 1x10 5 s −1
α= = = 1x10 s ζ = = = 1 → Critically damped
2 RC 2(50Ω)(1x10 − 7 F ) ω0 10 5 rad / s

s1 = −α = −1x10 5


vC (t ) = K1e s1t + K 2 te s1t
v P (t ) = 0
vT (t ) = K1e s1t + K 2 te s1t


v(t ) = K1e s1t + K 2te s1t
v(0) = K1e0 + K 2 (0)e0 = K1 = 0

v(t ) = K 2te s1t
dv(t )
= K 2e s1t + s1K 2te s1t
dt
dv(0 + )
= K 2e0 = 106 → K 2 = 106
dt
6 −105 t
v(t ) = 10 te

Case III : R = 250Ω
1 1 5 −1 α 0.2 x10 5 s −1
α= = = 0.2 x10 s ζ = = = 0.2 → Underdamped
2 RC 2(250Ω)(1x10 −7 F ) ω0 10 5 rad / s

ω n = ω 0 − α 2 = (1x10 5 ) 2 − (0.2 x10 2 ) 2 = 0.98 x10 5
2

s1 = −α + jω n = −0.2 x10 5 + j 0.98 x10 5
s 2 = −α − jω n = −0.2 x10 5 − j 0.98 x10 5


vC (t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
v P (t ) = 0
vT (t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
( i (



v(t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
dv(t )
= −αK1e −αt cos(ω n t ) − ω n K1e −αt sin(ω n t ) − αK 2 e −αt sin(ω n t ) + ω n K 2 e −αt cos(ω n t )
dt

v(0) = 0 = K1e −αt → K1 = 0
dv(0) 10 6
= 10 = −αK1 + ω n K 2 = ω n K 2 → K 2 =
6
= 10.2
dt 0.98 x10 5

− 0.2 x105 t
v(t ) = (10.2)e sin(0.98 x10 5 t )


Second-Order Transient Response Analysis

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