The Laplace Transform - University of Toledo

Report
F (s) 


e
0
 st
f ( t ) dt  L  f ( t ) 
Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Laplace Transforms
Outline of Today’s Lecture
 Review
 Phase
 Phase Computations
 Full Bode Plot
 System Identification
 Using Bode Plots for System Identification
 Laplace Transform
 Inverse Laplace Transform
 Properties of the Laplace Transform
 Final Value Theorem
Phase
s  i  u ( t )  e
i t
 cos  i   i sin  i 
y ( t )  G ( i ) e
i t
 Me
 i    t
w here M  G ( i ) is the m agnitude or gain of G  i  a nd
  tan
1
 Im  G  i   

 is the phase angle or argum ent of G  i 
 R e  G  i   


 For a sinusoidal input, phase represents the lag of the system or,
alternatively, the processing time of the system to produce an output
from the input
 Phase is measured as an angle




A cycle of the input is consider to take 2p radians or 360 degrees
Phase is the angular distance it takes for the output to represent the input
Thus it is normal that as the frequency increases that the phase also increase
In the case where the phase exceeds 180 degrees, the output appears to
“lead” the input. This is particularly evident in the range of 270 to 360
degrees.
Phase
 G ( s )    tan
1
 Im  G  i   


 R e  G  i   


 As with magnitude there are 4 factors to consider which can be
added together for the total phase angle.
 We will consider, in turn,
K
s
n
(s  a)
 ( s  2  n s   n )
2
2
 The sign will be positive if the factor is in the numerator and
negative if the factor is in the denominator
Phase Computations
 K is the angle of a real num ber; therefor e it is alw ays 0 (or 180 if the num ber w ere negative)
 s   ( i n )
n
n
n
If n is an even num ber, then i w ill either be + 1 or -1, both real num ber s
for
n
an odd num ber, i w ill be -1:  s  180
n
n
 the angle of com plete turns (n )
4
2
for
n
an even num ber, i w ill be + 1:  s  0  the angle of com plete turn s (n )
n
n
4
2
If n is an o dd num ber, then i w ill either be  i or - i , b oth pure im aginary num bers
n
for
n 1
an odd num ber, i w ill be  i :  s  90
n
n
 the angle of com plete turns (n )
4
2
for
n 1
an ev en num ber, i w ill be - i:  s  270  the angle of com plete turns (n )
n
n
2
E xam ples:  s  90
s
1

1
  90
s
2
 -180
s
3
 s = -270
s
5
  90  360   450
4
Phase Computations
 
 ( s  a )   ( i  a )  arctan  
 a 
T his now involves the variable,  , w hich is plotted on the sem ilog sca le
W here 
 
a , arctan    0
 a 
W here 
 
0
a , arctan    90
 a 
W here
 
   10 a , arctan    a line from 0 to 90
10
 a 
a
 ( s  2  n s   n )   (   n  
2
2
2
2
 2  n  
2
2 


 n

  i 2   )  arctan  
n
T his now also involves the va riable, 
 
0
a
 
W here 
 n , arctan 
W here 
 n , arctan 
W here
 
0
  180
 a 
 
   10 a , arctan    a line from 0 to 180 w ith a correction for 
10
 a 
a
Matlab Command bode(sys)
System Identification
 It is not unusual for a field engineer to be shown a piece of
equipment and then asked if he can put a control system on it or
replace the control system for which there are no parts.
 The task of determining how an unknown structure responds is
called “System Identification”.
 To identify a system, there are many tools are your disposal
 First and foremost, what should the system structure look like?
 Motors are often first order transfer functions (G ( s )  K ) which you then
sa
attempt to identify the constants
 Perform step tests and see what the response looks like
 Perform tests with sinusoidal outputs and use the Bode plot to
identify the system
 Apply statistical/time series methods such as ARMAX and RELS
Using Bode Plots for
System Identification
 The overall order of the system will be the high frequency phase divided by 90 degrees
 The exponent of the “s” term will be the slope on the magnitude plot at the lowest frequency
divided by 20
 Alternatively, the exponent of “s” is the lowest frequency phase divided by 90 degrees.
 The system gain constant (Kt) in dB will be the height value at the extension of the “s” term
line on the magnitude plot to where it crosses1 rps
 Starting from the left (the lowest frequency) on the magnitude plot, determine the structural
components using the change in slopes in increments of 20 degrees either up or down
 Then by using the intersection of the lines at those places match to the test curve, determine
the break frequencies
 Write the transfer function in the form
  s2
  s2

 s
 s
  s
2 z1s
2 2 s

1

1
...

1


1

 1  ...




 
2
2
 nz 1
 nz 2
 b1
  b2
  bm
   nz 1
   nz 2

G (s)  K t
2
2
 s
 s

2 p1s
2 p 2 s
 s
  s
p  s
s 
 1 
 1  ... 
 1 

 1 

 1  ...
2
2

 
 

 np 1
 np 2
 a1
  a2
  aq
  np 1
  np 2

Laplace Transform
 Traditionally, Feedback Control Theory was initiated by using
the Laplace Transform of the differential equations to develop
the Transfer Function
 The was one caveat: the initial conditions were assumed to be
zero.
 For most systems a simple coordinate change could effect this
 If not, then a more complicated form using the derivative
property of Laplace transforms had to be used which could lead
to intractable forms
 While we derived the transfer function, G(s), using the
convolution equation and the state space relationships, the
transfer function so derived is a Laplace Transform under
zero initial conditions
Laplace Transform
 CAUTION: Some Mathematics is necessary!
 The Laplace transform is defined as
For an analytic function f ( t )
(i.e., integrable everyw here and e veryw here less than e
F (s) 


e
 st
0
s0 t
for finite s 0 )
f ( t ) dt  L  f ( t ) 
F  s  is the Laplace transform of f ( t )
s is a com plex num ber
T he Inverse L aplace transform is defined as
f (t ) 
1
2p i 
a  i
a  i
e F ( t ) ds  L
st
1
 F (t ) 
Fortunately, we rarely have to use these integrals as there are other methods
Laplace
Transforms
Tables are available for
determining the Laplace
transform of most common
functions
This table which continues
on the next slide
is from Modern Control
Engineering by K. Ogata
4th ed., 2002
Laplace
Transforms
Laplace Transform

F (s) 

e
 st
0
f ( t ) dt  L  f ( t ) 
 Note that the index on the integral is 0:
 it is assumed that no dynamics are considered prior to t=0
f (t )  0
t0
 The Laplace is a linear transform:
L ( af ( t )) 
L  af  t   bg  t   





e
 st
0
af ( t ) dt  a  e
 st
0

e
 st
0

e
0

 st
f ( t ) dt  aL  f  t  
 af  t   bg  t   dt
af  t  dt 


e
 st
0
 aL  f  t    bL  g  t  
bg  t  dt
Some Common Laplace Transforms
 The Laplace Transform of the
Impulse Function
t0
 0

1
 (t )  
0  t< 


 0 t  0
L  (t )   1
 The Laplace Transform of the
Step Function
 0 t< 0
1( t )  
1 t  0
L 1  t   
1
Unit Ramp:
 0 t< 0
f (t )  
t t  0
1
s
2nd power of t:
 0

f (t )   t 2

2
t< 0
t0
1
L ( f ( t )) 
s
3
 The Place Transform of the
nth power of t:
 0

f (t )   t n

 n!
s
 The Laplace Transform of a
L ( f ( t )) 
 The Laplace Transform of the
2
t< 0
t0
1
L ( f ( t )) 
s
n 1
Some Common Laplace Transforms
 Laplace Trans Form of the
exponentials:
 0
f ( t )    at
e
L ( f ( t )) 
trigonometric functions:
t< 0
t0
1
sa
1
s  a
2
 0 t< 0
f ( t )   n  at
t0
t e
L ( f ( t )) 
n 1

s 
2
2
0 t< 0

f (t )  
 cos  t t  0
L ( f ( t )) 
s
s 
2
2
0 t< 0

f ( t )    at
 e sin  t t  0
L ( f ( t )) 
n!
s  a
 0 t< 0
f (t )  
 sin  t t  0
L ( f ( t )) 
 0 t< 0
f ( t )    at
t0
 te
L ( f ( t )) 
 Laplace Transforms of

s  a
2

2
Examples
 Find the Laplace Transform of
f ( t )  t  4 t  17
2
L  f  t    L  t  4 t  17   L  t
2
L t
2
  L  4 t   L 17  
2
s
3

4
s
2
2
  L  4 t   L 17 

17
s
 Find the Laplace Transform of f ( t )  4 sin(3t  2 )
N otice that this one has a phase (2) and there are no transform s in the tables for this
f ( t )  4 sin(3 t  2 )  4  sin  3 t  cos( 2 )  sin( 2 ) cos  3 t  


L  f ( t )   L 4  s in  3 t  cos( 2 )  sin( 2 ) cos  3 t    4 cos  2  L  sin  3 t    4 sin  2  L  cos  3 t  
L  f ( t )   4 cos  2 
3
s 9
2
 4 sin( 2 )
and then evaluate the constants
L  f (t )  
3.6372 s  4.9938
s 9
2
s
s 9
2
Lumped Parameter Model of an
Armature Controlled DC Motor
Find the transfer function for
this system with voltage as the
input and angular position as the
output using Laplace Transforms
R a ia  V b  V a
 V oltage L oop:

d
 B ack V oltage:
Vb  K b

dt

M otor T orque: T  K i a

2

d 
d
b
T
 R otations N S L : J
dt
dt

K d

i   b

 a
R a dt

 2
K ia
b
d 


 dt
J
J
Va
Ra
d
dt
 b
KK b  d
K Va
 


dt
JR a  dt
JR a
J
d 
2
 2  b
K Va  s 
KK b  
s  
 s  (s) 
JRa  
JR a
J

K
T . F .( s ) 
O utput ( s )
Input ( s )
 (s)
Va  s 


 (s)
Va  s 

K
JR a s   bR a  K K b  s
2
JR a
 b
KK b 
2
s  
s
J
JR
a 

Important Inverse Transforms
1 
 at
1 
L 
e
sa
 1

1
 at
 1  e 
L 

 s s  a  a


1


1
1
 bt
 at
e

e

L 


  s  a   s  b   b  a


1
2


n

L  2
2 
 s  2  n   n 
1
n
1
e
2
 i n t

sin  n t 1  
2

0<  < 1
Examples
 Find the Inverse Laplace Transform of
L
1
F (s) 
10
s s
2


10 
1
1 
1
t
F
(
s
)

L

10
L



  10  1  e 
 2

s s
 s  s  1 
 Find the Inverse Laplace Transform of
F (s) 
300
s  36 s
4


300
1


1

L  F (s)  L  4
 300 L  2 2
2 
 s  s  36  
 s  36 s 


1
1
3

 300
6

L  F (s) 
L  2 2
 6 t  sin  6 t  
2


216
216
s s 6 
 

1
L
1
300
1
 F ( s )   8.3333 t  1.3889 sin  6 t 
2
Properties of the
Laplace Transform
 Laplace Transforms have
several very import
properties which are useful
in Controls
 d

L
f ( t )   sF ( s )  f (0 )
 dt

 d2

d
2
L
f ( t )   s F ( s )  sf (0 ) 
f (0 )
dt
dt


L

t
0

f ( t ) dt 
F (s)
s
Now, you should see the advantage
of having zero initial conditions
Final Value Theorem
 If f(t) and its derivative satisfy the conditions for Laplace
Transforms, then
lim f ( t )  lim sF ( s )
t 
s 0
 This theorem is very useful in determining the steady state gain
of a stable system transfer function
 Do not apply this to an unstable system as the wrong
conclusions will be reached!
Example
 What is the steady state gain (DC Gain) of the system
G (s) 
3s  2
s  s  2 s  10 
2
lim g ( t )  lim sG ( s )  lim
t
s 0
s 0
3s  s  2 

s s  2 s  10
2

 0.6
Summary
 Laplace Transform
 Inverse Laplace Transform
 Properties of the Laplace Transform
 Final Value Theorem
Next Class: Using the Laplace Transform

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