Laplace Transform

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Transcript Laplace Transform

LAPLACE TRANSFORM
AS AN USEFUL TOOL
IN TRANSIENT STATE ANALYSIS
Oana Mihaela Drosu
Dr. Eng. , Lecturer
POLITEHNICA University of Bucharest
Department of Electrical Engineering
LPP Erasmus+
PART I
INTRODUCTION TO
LAPLACE TRANSFORM
THEORY
Definition
The Laplace transform is a linear operator
that switched a function f(t) to F(s), where
s = s+wj.
(Go from time argument with real input to a
complex angular frequency input).
Note that the real part s of the complex
variable s must be large enough for the
integral to converge.
Restrictions
There are two governing factors that determine
whether Laplace transforms can be used:
•f(t) must be at least piecewise continuous
for t ≥ 0
•|f(t)| ≤ Meγt where M and γ are constants
Conditions
•to be limited and integrable on any interval (t1,
t2), where 0< t1< t2 ;
•to be absolute integrable on the interval [0, t0 ],
where t0 >0;
•at least one value s=s0, to exist for the integral
to have sense;
• if it is absolute convergent for s=s0, then it will
be generally absolute convergent :
es  es0 
•In these conditions we can find a minimum
value of Re{s}, denoted , for the Laplace
transform of f(t) to exist (this is simple
convergence abscissa);
•The definition domain for F(s) is the complex
semiplane at the right of :
es  
BIBLIOGRAPHY:
•Norman Balabanian: Electric Circuits; McGrawHill,Inc., USA, 1994
•K.E. Holbert: Laplace transform solutions of ODE’s,
2006
•Walter Green: The Laplace transform presentation;
University of Tennessee, Electrical and Computer
Engineering Dep. Knoxville, Tennessee
•E.Cazacu, O.Drosu, G.Epureanu, Theory and
applications of electric circuits: vol.1 Transient state
analysis Matrix Rom, Bucharest, 2005.
PART II
SOLVING TRANSIENT STATE
CIRCUITS
USING LAPACE TRANSFORM
METHOD
INVERSE LAPLACE TRANSFORM
Mellin-Fourier (or Bromwich-Wagner) formula is a general
case of Fourier integral transformation. It establishes that, for
each Laplace transform F(s) There is a coresppondance
original (time variable) function f(t) given by:
s + i
1
f (t )  L1F ( s ) 
F ( s) exp( st ) d s, t  0, i   1

2j s i
Most of the time in the common applications, the function F(s)
is expressed as the ratio of two polynomial functions, the
denominator having the higher degree:
M ( s)
F ( s) 
N ( s)
Performing the inverse transform is straightforward
when using partial fractions expansion .
mk
Ckl
F ( s)  
l
k l 1 ( s  sk )
Where sk are the multiple solutions of order mk for the
polynomial function N(s), and the coefficients Ckl are
given by:
mk l
1
d
Ckl  lim
s sk (m  l )! d s mk l
k
 (s  sk ) mk M ( s) 


N ( s)


Applying inverse Laplace transform we obtain the general
Heaviside formula for the original function:
 M ( s) 
Ckl k 1
f (t )  L 
 
t exp( sk t )

 N (s)  k l 1 (l  1)!
1
mk
If the denominator has only simple roots, then the
general Heaviside formula can be simplified for the
following cases:
1) If the denominator does not have also the solution
s=0, then the Laplace image F(s) can be written as :
M ( s)
F ( s) 
N ( s)
And the corresponding original function will be given by
Heaviside I formula:
M ( sk )
f (t )   '
exp( sk t )
k N ( sk )
N ( sk )  0 sk  poles (roots) of F ( s)
2) If the denominator has also the solution s=0, then the
Laplace image F(s can be written as :
M ( s)
F ( s) 
sP ( s)
In this case : N(s)=sP(s),
Then, the original function f(t) is given by Heaviside II
formula:
M ( sk )
M (0)
f (t ) 
+
exp( sk t )
'
P(0)
k sk P ( sk )
P(sk )  0
sk  poles (roots) of the function F (s)
Kirchhoff Equations in time-domain:
i
k( j )
k
0
j  1,2,, n  1


d ik
d ih 1
 Rk ik + Lk
   ek (t )
+
L
+
i
d
t


kh
k



d t h( k )
d t Ck
k( p ) 
 k( p )
p  1,2,, b
Kirchhoff Equations after Laplace transform is applied:
I
k( j )
k
 0, j  1,2,, n  1,


1
1
I k + uCk (0)   Ek
Rk I k + Lk [sI k  ik (0)] +  Lkh [sI h  ih (0)] +

sC k
s
k( p ) 
h(k )
 k( p )
The ideal resistor
R iR (t )  u R (t )  Z R I R ( s)  U R ( s),
where Z R  R
The inductor
d iL (t )
u L (t )  L
 U L ( s)  Z L ( s) I L ( s)  LiL (0)
dt
where Z L ( s)  sL
The ideal capacitor
1
1
uC (t )   iC (t ) d t  U C ( s )  Z C ( s ) I C ( s ) + uC (0)
C
s
1
where Z C ( s ) 
sC
The coupled coils
Ideal current and voltage sources
The algorithm for solving transient circuits
using Laplace transform
•We determine the initial conditions of the circuit (current
through the inductors, voltages on capacitors) before
switching action.
•We draw the equivalent operational circuit, containing the
Laplace transform of given sources, the sources
corresponding to initial conditions, operational impedances
corresponding to the R, L,C elements.
•We calculate the images of the given time variable
functions (usually voltage of current sources) using direct
transformation formula or transform pairs tables. The
expressions will depend on the complex variable s.
•We apply the operational expresions of Kirchhoff
Equations (or different methods ussually applied
to circuits: loop method, node potentials methods,
etc). These equations will be solved with respect to
Laplace images of the unknown functions.
•After finding the Laplace transforms of the
unknown quantities, the original functions (time
dependent) are determined using inversion
methods (Mellin-Fourier, Heaviside formulas or
tables with transform pairs);
Example 1 Let’s consider the circuit given in Fig. 1-a, with
the following values of the circuit elements : R1 = 6 , R2 = 3 ,
inductance L = 0.8 mH, and constant value of the input voltage
source E = 36V. At moment t =0 the switch K closes. Using
Laplace transform method determine the variations of the
currents through the three elements of circuit and variation of the
inductor voltage.
Fig. 1-a
Fig. 1-b
The initial condition of the circuit is determined by the current
through the inductor and it can be easily calculated from the circuit
presented in the Fig. 1-b:
i L(0–) = 0
After switching (closing K), the equivalent operational circuit is
given in the Fig. 1-c
Fig. 1-c Equivalent circuit after switching
Taking into consideration the equivalent circuit, we can
determine the total equivalent operational impedance and, then,
using Kirchhoff II Theorem, we can calculate the current through
the resistor R1.
R2  s L
Z (s)  R1 +
R2 + s L
E  R2 + s L 
E s
I1 s  

R2  s L
s R1  R2 + s L R1 + R2 
R1 +
R2 + s L
Using current divider formula, we can determine the current
through resistor R2, respectively through the inductor L:
R2
E R2 + s L
R2
E  R2
I L s   I1 s  



R2 + s L s R1  R2 + s L R1 + R2  R2 + s L s R1  R2 + s L R1 + R2 
E R2 + s L  s L
sL
EL
I 2 s   I1 s  


R2 + s L s R1  R2 + s L R1 + R2  R2 + s L R1  R2 + s L R1 + R2 
In order to determine the voltage on the inductor, we apply Ohm’s
law on the respective branch:
U L s   I L s   s L 
E R2  s L
E R2 L

s R1  R2 + s L R1 + R2  R1  R2 + s L R1 + R2 
The original (time dependent functions) are calculated using
Heaviside II formula for currents i1 and iL, respectively Heaviside I
formula for current i2 and voltage uL:

 R1R2 
R1 R2 
E  R2 + L 

  R1 R2 t

t



L
R
+
R
E  R2
R
E
1
2 
i1 t  
+ 
 e LR1 + R2    1  2 e LR1 + R2  
R1R2
R1  R2
R1  R1


 L R1 + R2 


L R1 + R2 
EL
i2 t  
e
L R1 + R2 
R1 R2

t
LR1 + R2 
E

e
R1 + R2 
R1 R2

t
LR1 + R2 
E R2
E R2
i L t  
+
e
 R1R2
R1R2
 LR1 + R2 
2 R1 + R2 
E R2 L
u L t  
e
LR; 1; ;+ R2 
R1 R2

t
LR1 + R2 
R1 R2

t
LR1 + R2 
R1 R2



t
E
LR1 + R2  
 1  e

R1 



E R2

e
R1 + R2 
R1 R2

t
LR1 + R2 
Replacing also the numerical values, the final solution is:

i1 t   6 1  0,5  e

i L t   6 1  e
2500 t
2500 t


i2 t   4  e 2500 t
u L t   12  e
2500 t
The representations of these functions with respect to the time
, is presented in the Fig. 1-d, respectively Fig. 1-e
Fig. 1-d The variation of the circuit currents
LR1 + R2 
t is the time-constant of the circuit: t 
 0.0004 s
R1R2
Fig. 1-e The variation of of the inductor voltage
Example2 Let’s consider the circuit given in Fig. 2-a, with the
following values of the circuit elements: R = 5 , capacitance
C = 1 mF and constant value of the input voltage source E = 20V.
At t =0 the switch K moves from position a to position b. Using
Laplace transform method determine the variations of the currents
through the capacitor, respectively the voltage drop on this element.
Fig. 2-a
Fig. 2-b The circuit before changing the position of switch K.
The initial condition for the given circuit is determined by the
voltage at the terminal of the capacitor, as represented in the Fig. 2b:
uC 0    E  20 V
After switching from a to b, the equivalent operational circuit
is presented in Fig. 2-c:
Fig 2-c The circuit after switching from a to b
If we apply Kirchhoff II Theorem on the loop that contains the
capacitor, we obtain:
E
1 

  I c s  R +

s
sC 



 E s s C
I c s  

and
 EC
E

1 
CR s + 1
s CR + 1

R s +

RC 

U c 0  
1
E
E
U c s  
 I c s  +

+
1  s
sC
s

sCR s +

RC 

Taking into consideration the operational expressions of the two
functions with respect to s, we apply Heaviside I to determine the
time variable current, respectively Heaviside II to determine the time
variable voltage on the capacitor:
.
Hence we obtain the time-domain expressions for the current,
respectively voltage on the capacitor:
t
 EC  R C
ic t  
e
R


t 
t
 E


R
uC (t )  E  
+
e RC   E e RC
 1 
 RC 1


RC




RC
RC




Replacing the numerical values, we obtain:
ic t   4  e
200t
A
uC (t )  20 e -200 t V
.
Time variations of these two functions are presented in Fig. 2-d,
respectively Fig. 2-e:
Fig. 2-d Variation with respect to time of current through capacitor
t is the time constant of the circuit:
t  RC  0.005 s
Fig. 2-e Variation with respect to time of voltage on capacitor
Example 3 Let’s consider the circuit given in Fig. 3-a, with the
following values of the circuit elements: R1 = R2 = 1 k, .
inductance L1 = 1 H, capacitanceC2 = 100 F and constant value of
the input voltage source E = 1kV. At moment t =0 the switch K
closes. Using Laplace transform method determine the variations of
the currents through the inductor and of the voltage on the capacitor.
The initial condition for the given circuit is determined by the
voltage on the terminals of the capacitor, as represented in the Fig.
2-b:
iL(0–) = 0;
uC(0–) = E =1000 V
After switching (closing K), the circuit is made by two distinctive
circuits: the first one includes the voltage source, the resistor R1 and
inductor L2, the second includes the resistor R2 and capacitor C2.
Fig. 3-c presents the circuit after switching and the equivalent
operational circuits:
Fig. 3-c
Using Kirchhoff II on the respective loops, we can easily calculate
the Laplace images of the current through the inductor and the
voltage on the capacitor.
E

+ Li L (0  )
E + Li L (0  )
1000 
s
I L ( s) 


  A + B  0
A  1
R1 + s
s( s + R1 )
s( s + 1000 )   

1000 A  1000 B  1


s( A + B) + 1000 A
A
B
I L ( s)  +



s s + 1000
s( s + 1000 )
From the Laplace expression of the current (that was decomposed
using partial fractions) we deduce the original function (timevariable current):
1
1
I L ( s)  
s s + 1000
 i L (t )  1  exp(1000t ) A
In order to determine the voltage on the capacitor we have to
calculate firstly the current through the loop IC(s) and then, taking
into consideration Kirchhoff II applied on the loop, the voltage can
be written as: UC(s) = R2IC(s).
u C (0  )
u C (0  )C 2
1
s
I C (s) 


1
1 + sR2 C 2
s + 10
R2 +
sC 2
U C (s)  R2 I C (s) 
1000
s + 10
 u C (t )  1000 exp( 10t ) V
i L (t )  1  exp(1000 t ) A

u C (t )  1000 exp(10t ) V
The time variations of the current and voltage are represented in the
figures Fig. 3-d, respectively Fig. 3-e.
L1
t1 
 0.001 s
R1
Fig. 3-d
t 2  R2 C2  0.1 s
Fig. 3-e
BIBLIOGRAPHY:
•Norman Balabanian: Electric Circuits; McGrawHill,Inc., USA, 1994
•K.E. Holbert: Laplace transform solutions of ODE’s,
2006
•Walter Green: The Laplace transform presentation;
University of Tennessee, Electrical and Computer
Engineering Dep. Knoxville, Tennessee
•E.Cazacu, O.Drosu, G.Epureanu, Theory and
applications of electric circuits: vol.1 Transient state
analysis Matrix Rom, Bucharest, 2005.