Download Laplace form of solution

CIRCUITS and
SYSTEMS – part II
Prof. dr hab. Stanisław Osowski
Electrical Engineering (B.Sc.)
Projekt współfinansowany przez Unię Europejską w ramach Europejskiego Funduszu Społecznego. Publikacja dystrybuowana jest bezpłatnie
Lecture 11
Transient states in electrical
circuits – Laplace transformation
approach
Laplace representation of basic
elements
3
Resistor
Inductor
Capacitor
Any real circuit element has its Laplace model valid in complex
frequency space (s-space).
Kirchhoff’s laws for transforms
Current law
n
I
k 1
k
(s)  0
k
(s)  0
Voltage law
n
U
k 1
These laws are formed identically as for real time currents and
voltages.
4
Transient in the circuit using Laplace
transforms
1) Determine the initial values iL(0-) and uC(0-)
2) Determine the steady state in circuit after commutation iLu(0+) and
uCu(0+)
3) Calculate the natural responses ucp and iLp of the circuit deprived
of external excitations (voltage source - short circuit, current source
– open circuit)
4) Final solution is the superposition of both states
iL (t )  iLu (t )  iLp (t )
uC (t )  uCu (t )  uCp (t )
This is so called method of superposition of states (necessary at
sinusoidal
excitations).
5
Calculation of natural response
• Eliminate the external sources the RLC circuit
• Determine the initial conditions for natural response
iLp (0  )  iL (0  )  iLu (0  )
uCp (0  )  uC (0  )  uCu (0  )
• Form the Laplace model of the RLC circuit deprived of external
sources
• Using Kirchhoff’s laws find the solution of this circuit in s-space
(operator form)
• Calculate the inverse Laplace transforms (original fuctions) of the
currents of inductors and voltages of capacitors.
6
Example
Determine the transient of inductor current after commutation.
Assume: R=2 , L=1H, C=1/4F,
e(t )  10 2 sin( 4t  45o )
Solution:
Initial conditions
  4, I L 
7
j 45o
10e
2,5

, iL (t )  2,5 sin( 4t )
4  j4
2
iL (0 )  0, uC (0 )  0
Steady state after commutation
j 45o
I Lu
10e
 j11, 31o

 2.77e
,
2  j 4  j1
U Cu   j1 I Lu  2,77e
 j101, 31o
iLu (t )  2,77 2 sin( 4t  11,31o ),
uCu (t )  2,77 2 sin( 4t  101,31o )
iLu (0  )  0,76, uCu (0  )  3,84
8
Natural response
Laplace model of the circuit for natural response
Initial conditions for natural response
iLp (0 )  iL (0 )  iLu (0 )  0,76, uCp (0 )  uC (0 )  uCu (0 )  3,84
Solution as Laplace transform
I Lp ( s) 
9
LiLp (0  ) 
uCp (0  )
s2
s
4
s

0,76s  3,84
s 2  2s  4
Final solution
Because of complex poles we apply the table of trasforms
1
0,76( s  1)  4,6 
3
3
I Lp ( s) 
2
2
s  1  3
 
Natural response in time form
iLp (t )  0,76e t cos( 3t )  2,67e t sin( 3t )
Total current of the inductor
iL (t )  iLu (t )  iLp (t ) 
2,77 2 sin( 4t  11,31o )  0,76e t cos( 3t )  2,67e t sin( 3t )
10
11
Transient state in RLC circuit at DC
excitation
Zero initial conditions
uC (0  )  0, iL (0  )  0
Laplace model of the circuit
Laplace form of solution
Current in Laplace form
E/s
E/L

sL  R  1 / sC s 2  R s  1
L
LC
Characteristic equation
I (s) 
s2 
R
1
s
0
L
LC
Poles
2
R
1
 R 
s1  
   
,
2L
 2 L  LC
2
12
R
1
 R 
s2  
   
2L
 2 L  LC
Three cases of general solution
• Overdamped (aperiodic) case:
L
R2
C
• Critically damped case
L
R2
C
• Oscillatory (periodic) case
L
R2
C
• Critical resistance
R2
13
L
C
Overdamped case
Both poles are real and single. The time form of current
i (t ) 
E
2
1
 R 
2L   
 2 L  LC
Damping coefficient
e
s1t
e
s2t


E
2
1
 R 
L   
 2 L  LC
R
2L
Voltages of capacitor and inductor
uC (t )  E 
E
2
s e
2
s1t
 s1e s2t

1
 R 
2   
 2 L  LC
di
E
s1t
s2t
u L (t )  L 
s
e

s
e
1
2
2
dt
1
 R 
2   
 2 L  LC

14

e

R
t
2L
  R 2 1 
sh   
t
  2 L  LC 


Graphical form of solution
Examplary transients in RLC circuit for R = 2,3, C = 1F i L = 1H
at E = 1V.
15
Transients of capacitor voltage and
current in RC and RLC circuits
16
Critically damped case
Double pole
R
s1  s2    
2L
Laplace form of current solution
I (s) 
17
E/L
R 

s 

2L 

2
18
Time form of solution
Current of inductor
R
E 2Lt
i(t )  te
L
Voltage of inductor
 t
di
R 
uL (t )  L  Ee 2 L 1  t 
dt
 2L 
R
Voltage of capacitor
uC (t )  E  Ri L (t )  uL (t )  E  Ee

R
t
2L
R 

t
1 
 2L 
Comparison of uC(t) at overdamped
and critically damped cases
19
Oscillatory case
Both poles are complex.
Laplace form of solution
1
R2
 2
E/L
LC
4L
I ( s) 

2
R
1
R   1
R2
s2  s 

 2
 
L
LC  s 
2 L   LC 4 L

Self-oscillation frequency

20
1
R2
 2
LC 4 L




2

E/L
1
R2
 2
LC 4 L
21
Time solution
• Current of inductor
R
E  2Lt
i (t ) 
e
sin( t )
L
• Voltage of inductor
R
 t
di
E
u L (t )  L  
e 2 L sin( t   )
dt
 LC
  arctg

R / 2L
• Voltage of capacitor
R

E  2Lt 
L
uC (t )  E  uL (t )  Ri (t )  E 
e
sin( t   )
 R sin( t ) 
L
C


Graphical form of solution
22
Transient uC(t) at different resistances
in oscillatory case
23