Transcript CHAPTER 2

Introduction to Control Systems
● Historical perspective
● Introduction to Feedback Control Systems
● Closed loop system examples
1
Historical Perspective
●
●
●
●
●
●
●
●
13.7B BC
13.4B
5B
3.8B
700M
200M
65M
600K
Big Bang
Stars and galaxies form
Birth of our sun
Early life begins
First animals
Mammals evolve
Dinosaurs extinct
First Trace of humans
2
Feedback Control Systems emerge rather recently
●
●
●
1600
1781
1765
Drebbel Temperature regulator
Pressure regulator for steam boilers
Polzunov water level float regulator
3
Closed loop example: Polzunov’s Water level
float regulator
4
Feedback Control Systems emerge rather recently
●
●
●
●
1600 Drebbel Temperature regulator
1681 Pressure regulator for steam boilers
1765 Polzunov water level float regulator
1769 James Watt’s Steam Engine and Governor
5
Closed loop example: James Watt’s flyball
governor
6
Open loop and closed loop control systems
Open Loop System
7
Open loop and closed loop control system
models
Open Loop System
Closed Loop System
8
Example: Feedback in everyday life
9
Multivariable Control System Model
10
Multivariable Control System
11
Robotics
A robot is a programmable computer integrated with a machine
12
Example: Disk Drive
13
Example: Automatic Parking Control
14
Feedback Control: Benefits and cost
Benefits:
Cost:
15
Feedback Control: Benefits and cost
Benefits:
•
•
•
•
Reduction of sensitivity to process parameters
Disturbance rejection
More precise control of process at lower cost
Performance and robustness not otherwise achievable
Cost:
16
Feedback Control: Benefits and cost
Benefits:
•
•
•
•
Reduction of sensitivity to process parameters
Disturbance rejection
More precise control of process at lower cost
Performance and robustness not otherwise achievable
Cost:
•
•
•
•
More mathematical sophistication
Large loop gain to provide substantial closed loop gain
Stabilizing closed loop system
Achieving proper transient and steady-state response
17
Ideal Control System Elements
Element Type
Physical Element
Describing Equation
Energy (E) or power (Þ)
Inductive storage
Electrical Inductance
v = L di/dt
E = (1/2) L i 2
Translational spring
dx/dt = (1/k) dF/dt
E = (1/2) F 2 / k
Rotational spring
ω = (1/k) dT/dt
E = (1/2) T 2 / k
Fluid inertia
P = I dQ/dt
E = (1/2) I Q 2
Electrical capacitance
i = C dv/dt
E = (1/2) C v 2
Translational mass
F = M d 2x/dt 2
E = (1/2) M ( dx/dt ) 2
Rotational mass
T = J dω /dt
E = (1/2) J ω 2
Fluid capacitance
Q = Cf d P/dt
E = (1/2) Cf P 2
Thermal capacitance
q = Ct dŦ/dt
E = Ct Ŧ 2
Electrical resistance
v = iR
Þ = i 2R = v 2 / R
Translational damper
F = b dx/dt
Þ = b ( dx/dt ) 2
Rotational damper
T = bω
Þ = bω 2
Fluid resistance
Q = ( 1/ Rf ) P
Þ = ( 1/ Rf ) P 2
Thermal resistance
q = ( 1/ Rt ) Ŧ
Þ = ( 1/ Rf ) Ŧ
Capacitive storage
Energy dissipators
18
Example: Mechanical system
Determine y(t)
19
Example: Mechanical system
Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0
r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2
r(t) = M d 2y(t)/dt 2 + b dy(t)/dt + k y(t)
L2CCDE
20
Example: Mechanical system
Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0
r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2
r(t) = M d 2y(t)/dt 2 + b dy(t)/dt + k y(t)
Homogenous solution: yH (t) r(t) = 0
Particular solution: yp(t) = f [r(t)]
Total solution: y(t) = yH (t) + yp(t)
21
Example: Mechanical system
Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0
r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2
r(t) = M d 2y(t)/dt 2 + b dy(t)/dt + k y(t)
22
Example: Mechanical system
Assume the system is initially at rest: y(0-) = v(0-) = 0
r(t) – k ∫0-tv(τ)d τ – b v(t) = M dv(t) /dt
r(t) = M dv(t) / dt + b v(t) + k ∫0-tv(τ)d τ
23
Example: Electrical system
Determine v(t)
Assume circuit initially at rest. What does this mean?
24
Example: Electrical system
Write node equation
r(t) = (1/L)
t
0-
v(τ)d τ + C dv(t)/dt + G v(t)
Homogenous solution vH (t) : r(t) = 0
Particular solution: vp(t) = f [r(t)]
Total solution: v(t) = vH (t) + vP (t)
25
How to describe a plant/system
mathematically?
if a dynamic system
Differential
Equation
L
V(t)
i
KVL
R
di
L
 Ri  V (t )
dt
26
Consider a plant that can be described by
an n-order differential equation
n
n 1
d x n1 d x
dx
an n  a
   ai  ao x  q(t )
n 1
dt
dt
dt
a1 , a2 ,, an
are constant coefficients, and
Input  q (t )
independent variable
Output  x (t )
dependent variable
an  0
There are 2 cases to be considered
(1) If q(t)=0  Homogeneous Case
(2) If q(t) ≠0  In-Homogeneous Case
27
Homogeneous Case
Let a 2nd order differential equation be:



x(t )  3 x(t )  2 x(t )  0, x(0)  0, x(0),  1
Define a differential operator D
d
D
dt
D
2

 3D  2 x(t )  0
2
d
D2  2
dt
D  2D 1x(t )  0
z (t )
28
Homogeneous Case (cont.)
D  2D 1x(t )  0
z (t )
D  1x(t )  z(t )
D  2z(t )  0
D

d
dt

x(t )  x(t )  z (t )
x(t )   x(t )  z (t )


z (t )  2 z (t )  0
z (t )  2 z (t )
29
Homogeneous Case (cont.)

x(t )   x(t )  z (t )
Re-write

z (t )  2 z (t )
    1
 x (t )   
 z (t )   0


1   x (t ) 
 2  z (t ) 
In general

x  Ax(t )
If one can solve a first order D.E, then it can be used to solve nth order D.E
30
Homogeneous Case (cont.)
Solve a 1st order differential equation

x(t )  ax(t )  p(t )
General eqn.
Notice that:
 
d at  at
xe  x e  x(ae at )
dt
Multiply both sides by eat

 at
at
x
(
t
)

ax
(
t
)
e

p
(
t
)
e





 ( x  ax)e at

d
at
at
x(t )e  p (t )e
dt
31
Homogeneous Case (cont)
Integrate both sides of equation
t

0

  p( )e
d
x( )e a 
d
t
a
d
Zero Input
Response
Zero State
Response
0
t
x (t )e at  x (0)e a ( 0 ) 

p ( )e a d
0
t
x (t )e at  x0 

p ( )e a d
0
x(t )  e
 at
t
xo   p( )e
 a ( t  )
d
0
due to initial condition
due to input
32
Homogeneous Case (cont.)

Consider


x(t )  3 x(t )  2 x(t )  0, x(0)  0, x(0),  1

x(t )  x(t )  z (t )
Initial
condition

z (t )  2 z (t )  0
a  2, p  0

x(t )  ax(t )  p(t )
z (t )  z (0)e

2t
 c1e
x(t )  x(t )  c1e
2t
 2t
p (t )
33
Homogeneous Case (cont.)

a 1
x(t )  x(t )  c1e
 2t
p (t )
t


x(t )  x(0)e t  e t  c1e  2 e d
0
e
t
 e t
 c e 
 c e  c   c e

1
t
0
t
1
1
1
t
 c1e  2t
t

x(t )  c1e  c2e
2t
34
Homogeneous Case (cont.)
What are C1 & C2???
Based on the initial condition:
x (t )  c1e  t  c2 e 2 t
x (0)  c1  c2  0
c1  c2

x(t )  c1e t  2c2 e  2t

x(0)  c1  2c2  1
 c1  2c2  1
c2  2c2  1
c2  2c2  1
c2  1
c1  1
35
Homogenous (cont.)
Thus
t
x(t )  e  e
2t
36
Homogeneous Case (cont.)
Easiest method:
Find the roots of the characteristic equations
an s  an 1s
n
n 1
   a1s  ao  0
For second order system:
as  bs  c  0
2
 b  b  4ac
s
 s1 , s2
2a
2
37
Homogeneous Case (cont.)
 b  b  4ac
s
 s1 , s2
2a
2
Therefore,
x(t )  c1e  c2e
s1t
So how to find these values?
s2t
Based on Initial Condition
38
Homogeneous Case (cont.)
Example:
Consider again
We’ll have


x(t )  3 x(t )  2 x(t )  0, x(0)  0, x(0),  1
s 2  3s  2  0
s  1,2
Thus,
So,

x(t )  c1e 1t  c2e 2t
By considering the initial conditions:
x(0)  c1e  c2e  0
0
We’ll get:
Therefore:
0
and

x(0)  c1e0  2c2e0  1
c1  1, c2  1
x(t )  e 1t  e 2t
39
4 cases to be considered
Case 1: Distinct real roots
Case 2: Equal roots & real
s1  s2
s1  s2  s
x(t )  c1e s1t  c2 e s 2t
x(t )  c1e st  c2test
GENERAL SOLUTION
Case 4: Complex conjugate roots
Case 3: Imaginary roots
s    j
s   j
x(t )  c1e
 jt
 c2 e
jt
x(t )  e t k1 sin t  k2 cos t 
x(t )  k1 sin t  k 2 cos t
40
In-Homogeneous Case
n 1
d nx
d
x
dx
n 1
an

a



a
 ao x  q(t )
i
n
n 1
dt
dt
dt
q(t )  0
The solution:
x(t )  xP (t )  xH (t )
Particular solution
Homogeneous solution
Determine the particular solution is
the most challenging!!!
Homogeneous solution can be found by setting q(t)=0,
and then find the solution as in homogeneous case
41
Finding the particular solution, xP(t)
Undetermined Coefficient Method
Assumption:
Various derivative of q(t) have a finite of functional form
Functional forms:
(1)exponential (eat)
(2)sinosoidal (sin ωt, cos ωt)
(3) polynomial (tm)
(4)combinational of these function and their products (teat, eatsinwt)
Then guess:


n
d q
xP (t )  Ao q(t )  A1 q(t )  A2 q (t )    An n
dt
42
In-Homogeneous Case (cont.)
Example:

Initial condition

x  3 x  2 x  5 sin t
x (0)  1

q(t )  5 sin t
(1) xH (t ) ?
x (0)  0
t
xH (t )  c1e  c2e
2t
(2) xP (t ) ?
(3) x(t ) ?
43
Finding Xp(t)
q(t )  5 sin( t )

q(t )  5 cos(t )

q (t )  5 sin( t )
x p (t )  A sin( t )  B cos(t )

q (t )  5 cos(t )



x p (t )  Ao q(t )  A1 q(t )  A2 q(t )  A3 q (t )
44
In-Homogenous case (cont.)
So, we’ll have
x(t )  xH (t )  xP (t )
x(t )  c1e t  c2e 2t  A sin t  B cos t
Now, we need to find c1,c2, A & B
First, how to find A & B? Let us consider xp(t) again and substitute it into the original
differential equation
x p (t )  A sin t  B cos t


x  3 x  2 x  5 sin t

x p (t )  A cos t  B sin t

x p (t )   A sin t  B cos t


x p  3 x p  2 x p  5 sin t
45


x p  3 x p  2 x p  5 sin t
 A sin t  3( A cos t  B sin t )  2( A sin t  B cos t )  5 sin t
Re-arrange:

:

A sin t

B cos t
3 x p (t )
:

3B sin t

3 A cos t
2 x p (t )
:
2 A sin t

2 B cos t
x p (t )

A  3B
A-3B = 5
B+3A = 0
So,
x p (t )  ½ sin t - 3 cos t
2
B  3A
A=½
B = -3/2
46
In-Homogenous case (cont.)
So,
x(t )  c1e t  c2e 2t  1 sin t  3 cos t
2
2
What about c1 & c2?
Again, you must find them by considering
the given INITIAL CONDITION
47
In Class exercise

In a group of two students, please solve the
following problem within 10 minutes. Please
submit them after right after 10 minutes ends.
Intellectual discussion is highly encouraged.

Find the solution the given differential equation

x (t )  6 x(t )  8 x(t )  e 2t

provided that
x(0)  0, x(0)  1
48
Using Laplace Approach
x (0)  1
Consider the previous example again:


x  3 x  2 x  5 sin t

x (0)  0
Apply the Laplace transform to given diff. eqn
5
[ s X ( s )  sx (0)  x' (0)]  3[ sX ( s )  x(0)]  2 X ( s )  2
s 1
2
s3
5
X ( s)  2
 2
s  3s  2 s  1 s 2  3s  2
Simplify it:

X1(s)


X2(s)
49


x  3 x  2 x  5 sin t


Lap{ x  3 x  2 x}  lap{5 sin t}
5
s 1
5
5  ( s  3)( s 2  1)
2
( s  3s  2) X ( s )  2
 ( s  3) 
s 1
s2 1
5  ( s  3)( s 2  1)
As  B
K1
K2
X (s) 



( s 2  1)( s 2  3s  2)
s2 1
s 1 s  2
 (0)  3( sX ( s )  x (0))  2 X ( s ) 
s 2 X ( s )  sx (0)  x
5  ( s  3)( s 2  1)
K1 
( s 2  1)( s  2)
K2
5  ( s  3)( s 2  1)

( s 2  1)( s  1)

9
 4.5
2

10
 2
5
s  1
s  2

2



5  ( s  3)( s 2  1)   As  B s  1s  2   K1 s 2  1 s  2   K 2 s 2  1 s  1
s 3 coefficients on both sdes :1  A  K1  K 2  A  4.5  2  A  1.5
s  0  8  2 B  2 K1  K 2  2 B  9  2  B  0.5
 1.5s  0.5
4.5
2


2
s 1
s 1 s  2
x (t )  4.5e t  2e  2 t  1.5 cos(t )  0.5 sin( t )
X (s) 
50
Using Laplace Approach (cont.)
Partial fraction expansion:
A
B
s3
X 1 (s) 


s  1 s  2 ( s  1)( s  2)
and
C
D
Es  F
5
X 2 ( s) 

 2

s  1 s  2 s  1 ( s  1)( s  2)( s 2  1)
Determine the values for A,B,C,D,E & F
Then,
X ( s)  X 1 ( s)  X 2 ( s)
51
Using Laplace Approach (cont.)
Finally, x(t) can be found by applying the inverse Laplace transform of X(s)
1
x(t )  L
 X ( s )
52
Laplace Transforms


Def: F ( s)  L( f )   e  st f (t )dt for f (t ), t  0
0

Inverse:

Linearity:

f (t )  L1 ( F )
L{af (t )  bg (t )}  aL{ f (t )}  bL{g (t )}
Shifting Theorom:
L{e f (t )}  F ( s  a)
at
e at f (t )  L1{F ( s  a )}
53
The Laplace Transform
Laplace Transform of the unit step.

 1  st 
L[u(t )]   1e dt  e |
0
s
0
 st
1
L[u(t )] 
s
The Laplace Transform of a unit step is:
*notes
1
s
54
The Laplace Transform
The Laplace transform of a unit impulse:
Pictorially, the unit impulse appears as follows:
(t – t0)
f(t)
t0
0
Mathematically:
(t – t0) = 0 t
*note

t 0 
0
 (t  t )dt 1
0
 0
t0 
55
The Laplace Transform
The Laplace transform of a unit impulse:
An important property of the unit impulse is a sifting
or sampling property. The following is an important.
t2

t1
t1  t 0  t 2
 f (t 0 )
f (t ) (t  t 0 )dt  
t 0  t1 , t 0  t 2
0
56
The Laplace Transform
The Laplace transform of a unit impulse:
In particular, if we let f(t) = (t) and take the Laplace

L[ (t )]    (t )e dt  e
 st
0 s
1
0
57
The Laplace Transform
An important point to remember:
f (t )  F ( s)
The above is a statement that f(t) and F(s) are
transform pairs. What this means is that for
each f(t) there is a unique F(s) and for each F(s)
there is a unique f(t). If we can remember the
Pair relationships between approximately 10 of the
Laplace transform pairs we can go a long way.
58
The Laplace Transform
Building transform pairs:

e
L[e u(t )]   e e dt   e
at
0
at  st
L(e
( s  a ) t
dt
0
 st
e
1

L[e u( t )] 
|0 
( s  a)
sa
 at
A transform
pair
 at
e u( t )

1
sa
59
The Laplace Transform
Building transform pairs:

L[tu(t )]   te dt
 st
0



 udv  uv |   vdu
0
0
u=t
dv = e-stdt
0
tu(t )

1
2
s
A transform
pair
60
The Laplace Transform
Building transform pairs:

(e jwt  e  jwt )  st
L[cos( wt )]  
e dt
2
0
1 1
1 
 


2  s  jw s  jw 
s
 2
s  w2
cos( wt )u(t )

s
s2  w2
A transform
pair
61
The Laplace Transform
Time Shift

L[ f (t  a )u(t  a )]   f (t  a )e  st
a
Let x  t  a , then dx  dt and t  x  a
As t  a , x  0 and as t  , x  . So,


0

f ( x )e  s ( x  a ) dx  e as  f ( x )e  sx dx
0
L[ f (t  a)u(t  a)]  e
 as
F ( s)
62
The Laplace Transform
Frequency Shift
L[e
 at

f (t )]   [e
 at
 st
f (t )]e dt
0


 f ( t )e
( s  a ) t
dt  F ( s  a )
0
L[e
 at
f (t )]  F ( s  a)
63
The Laplace Transform
Example: Using Frequency Shift
Find the L[e-atcos(wt)]
In this case, f(t) = cos(wt) so,
s
F ( s)  2
s  w2
(s  a)
and F ( s  a ) 
(s  a)2  w 2
L[e
 at
( s  a)
cos( wt )] 
( s  a)2  (w)2
64
The Laplace Transform
Time Integration:
The property is:

 t
  st
L   f (t )dt      f ( x )dx e dt
0
 0 0

Integrate by parts :
t
Let u   f ( x )dx , du  f (t )dt
0
and
 st
dv  e dt ,
1  st
v  e
s
65
The Laplace Transform
Time Integration:
Making these substitutions and carrying out
The integration shows that

 1
L  f (t )dt    f (t )e  st dt
0
 s0
1
 F ( s)
s
66
The Laplace Transform
Time Differentiation:
If the L[f(t)] = F(s), we want to show:
df (t )
L[
]  sF ( s )  f (0)
dt
Integrate by parts:
u  e , du   se dt and
 st
*note
 st
df ( t )
dv 
dt  df ( t ), so v  f ( t )
dt
67
The Laplace Transform
Time Differentiation:
Making the previous substitutions gives,
 df 
L    f ( t )e
 dt 

|  f (t ) se dt
 st 
 st
0
0

 0  f (0)  s  f (t )e  st dt
0
So we have shown:
 df (t ) 
L

sF
(
s
)

f
(
0
)
 dt 
68
The Laplace Transform
Time Differentiation:
We can extend the previous to show;
 df (t ) 2  2
L
 s F ( s )  sf (0)  f ' (0)
2 
 dt 
 df (t ) 3 
3
2
L

s
F
(
s
)

s
f (0)  sf ' (0)  f ' ' (0)
3 
 dt 
general case
 df (t ) n 
n
n 1
n2
L

s
F
(
s
)

s
f
(
0
)

s
f ' (0)
n 
 dt 
 ...  f ( n 1) (0)
69
The Laplace Transform
Transform Pairs:
f(t)
F(s)
 (t )
1
1
u( t ) ____________________________________
s
1
 st
e
sa
1
t
s2
n!
n
t
s n 1
f (t )
F ( s)
70
The Laplace Transform
Transform Pairs:
f(t)
te
 at
n  at
t e
sin( wt )
cos( wt )
F(s)
1
s  a 2
n!
( s  a )n 1
w
s2  w2
s
s2  w2
71
The Laplace Transform
Transform Pairs:
f(t)
e at sin( wt )
e
 at
cos( wt )
sin( wt   )
cos( wt   )
F(s)
w
(s  a)2  w 2
sa
2
2
(s  a)  w
s sin   w cos 
s2  w2
s cos   w sin 
2
2
s w
Yes !
72
The Laplace Transform
Common Transform Properties:
f(t)
F(s)
f ( t  t 0 )u( t  t 0 ), t 0  0
f ( t )u( t  t 0 ), t  0
e  at f ( t )
d n f (t )
dt n
e
e
 to s
 to s
F ( s)
L[ f ( t  t 0 )
F (s  a)
s n F ( s )  s n  1 f ( 0)  s n  2 f ' ( 0)  . . .  s 0 f n  1 f ( 0)
dF ( s )
ds
tf ( t )

t
1
F ( s)
s
 f ( )d
0
73
The Laplace Transform
Using Matlab with Laplace transform:
Example
Use Matlab to find the transform of
te  4 t
The following is written in italic to indicate Matlab code
syms t,s
laplace(t*exp(-4*t),t,s)
ans =
1/(s+4)^2
74
The Laplace Transform
Using Matlab with Laplace transform:
Example
F ( s) 
Use Matlab to find the inverse transform of
s ( s  6)
( s  3)( s  6 s  18)
2
prob .12.19
syms s t
ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))
ans =
-exp(-3*t)+2*exp(-3*t)*cos(3*t)
75
The Laplace Transform
Theorem:
Initial Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
Has the Laplace transform F(s), and the lim sF ( s ) exists, then
s
lim sF ( s )  lim f ( t )  f (0)
s
t 0
Initial Value
Theorem
The utility of this theorem lies in not having to take the inverse of F(s)
in order to find out the initial condition in the time domain. This is
particularly useful in circuits and systems.
76
The Laplace Transform
Example: Initial Value Theorem:
Given;
F ( s) 
( s  2)
( s  1)2  5 2
Find f(0)


s 2  2s
f (0)  lim sF ( s )  lim s
 lim  2

2
2
s
s   ( s  1)  5
s
 s  2 s  1  25 
( s  2)
s2 s2  2 s s2
 lim
s
s
2
s  2 s s  ( 26 s )
2
2
2
1
77
The Laplace Transform
Theorem:
Final Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
has the Laplace transform F(s), and the lim sF ( s ) exists, then
s
lim sF ( s )  lim f ( t )  f ( )
s0
t 
Final Value
Theorem
Again, the utility of this theorem lies in not having to take the inverse
of F(s) in order to find out the final value of f(t) in the time domain.
This is particularly useful in circuits and systems.
78
The Laplace Transform
Example: Final Value Theorem:
Given:
F ( s) 
( s  2) 2  3 2
( s  2)
2
 32

note F 1 ( s )  te  2 t cos 3t
Find f ( ) .
f ( )  lim sF ( s )  lim s
s0
s0
( s  2) 2  3 2
( s  2)
2
3
2

0
79
List of Laplace Transforms
f(t)
L(f)
1
1
1/s
2
t
3
4
5
f(t)
L(f)
7
cos t
1/s2
8
sin t
t2
2!/s3
9
cosh at
tn
n!
10
sinh at
11
eat cos t
s
s2   2

s2   2
s
s2  a2
a
s2  a2
sa
( s  a) 2   2
12
eat sin t
(n=0, 1,…)
s n 1
ta
(a  1)
s a 1
1
sa
(a positive)
6
eat

( s  a) 2   2
80
Review: Partial Fractions
• Case I: unrepeated real factor
Case II: repeated real factor
s m  a m 1 s m 1   a0
( s  p1 ) k ( s  p2 )  ( s  pn )

A1k
An
A11
A12
A2






( s  p1 ) k ( s  p1 ) k 1
( s  p1 ) ( s  p2 )
s  pn
• Case III: unrepeated complex factor
• Case mIV: repeated
complex factor
m 1
s  a m 1 s   a0
( s 2  bs  c ) k ( s  p2 )  ( s  pn )

A1k s  B1k
An
A11s  B11
A12 s  B12
A2






( s 2  bs  c ) k ( s 2  bs  c ) k 1
( s 2  bs  c ) ( s  p2 )
s  pn
81
Transform of Derivatives

THEORM 1
 Laplace of f(t) exists
 f’(t) exists and piecewise continuous for t>=0
L( f ' )  sL( f )  f (0)

THEOREM 2
L( f ( n) )  s n L( f )  s n1 f (0)  s n2 f ' (0)    f ( n1) (0)
82
Differential Equations
y"ay'by  r (t )
y(0)  K 0
y' (0)  K1
1st step
[ s 2Y  sy (0)  y' (0)]  a[ sY  y(0)]  bY  R( s)
2nd step
1
Q( s)  2
s  as  b
Y ( s)  [( s  a ) y(0)  y' (0)]Q( s)  R( s)Q( s)
3rd
step
y(t )  L1 (Y )
83
Transform of Integrals
t
L( f ) F ( s)
L{ f ( )d } 

s
s
0
t

0
F ( s)
f ( )d  L {
}
s
1
84
Unit Step Function
0 if
u( t  a )  
1 if
u(t )
t a
t a
u( t  a )
85
t-Shifting
f (t  10)
f ( t ) u( t )
f (t  10)u(t  10)
86
Applications – t-shifting

THEOREM
L{ f (t  a)u(t  a)}  e as F ( s)
f (t  a)u(t  a)  L1{e as F ( s)}
e  as
L{u(t  a )} 
s
87
Dirac Delta (unit impulse) Function

Impulse of f(t), for
t 0  t  t1
t

“Generalized function”

 if t  a
 (t  a )  
0 otherwise
Laplace transform
t

and
  (t  a )  1
0
L{ (t  a)}  e as
88
Convolution

Definition
t
h(t )  ( f  g )( t )   f ( ) g (t   )d


0
Property
Application
( f * g)(t )  L1{F ( s)G ( s)}
1. Solve y"ay'by  r (t )
1
Q ( s)  2
s  as  b
Y ( s)  R( s)Q ( s)
y(0)  0
y' (0)  0
t
y(t )   q(t   )t ( )d
0
2. Calculate integral
89
Solution of Partial Fraction Expansion

The solution of each distinct (non-multiple)
root, real or complex uses a two step
process.


The first step in evaluating the constant is to
multiply both sides of the equation by the factor
in the denominator of the constant you wish to
find.
The second step is to replace s on both sides of
the equation by the root of the factor by which
you multiplied in step 1
90
K2
K3
8( s  3)( s  8) K1
X ( s) 



s( s  2)( s  4)
s s2 s4
8( s  3)( s  8)
K1 
( s  2)( s  4)
8( s  3)( s  8)
K2 
s( s  4 )
s0
s 2
8(0  3)(0  8)

 24
(0  2)(0  4)
8( 2  3)( 2  8)

 12
2 ( 2  4)
91
8( s  3)( s  8)
K3 
s( s  2 )
s 4
8( 4  3)( 4  8)

 4
4( 4  4 )
The partial fraction expansion is:
24
12
4
X ( s) 


s s2 s4
92

The inverse Laplace transform is found from
the functional table pairs to be:
x(t )  24  12e
2 t
 4e
4 t
93
Repeated Roots


Any unrepeated roots are found as before.
The constants of the repeated roots (s-a)m
are found by first breaking the quotient into a
partial fraction expansion with descending
powers from m to 0:
Bm
B2
B1


m
2
( s  a)
( s  a)
( s  a)
94

The constants are found using one of the
following:
m i
1
d
Bi 
(m  i )! ds m i
Bm 


P( s)

m
 Q( s ) /( s  a1 )  s  a1
P(a )
Q( s) / ( s  a )
m
sa
95
K1
K2
8( s  1)
Y ( s) 


2
s  2 ( s  2) 2
( s  2)
8( s  1)( s  2)
K2 
2
( s  2)
2
 8( s  1) s2  8
s 2
96
1 d
Bi 
(2  1)! ds


8( s  1)
8
 ( s  2) 2 /( s  2) 2 

 s  2
The partial fraction expansion yields:
8
8
Y ( s) 

2
s  2 ( s  2)
97
The inverse Laplace transform derived from the functional
table pairs yields:
y(t )  8e
2 t
 8te
2 t
98
A Second Method for Repeated Roots
K1
K2
8( s  1)
Y ( s) 


2
s  2 ( s  2) 2
( s  2)
8(s  1)  K1 (s  2)  K2
8s  8  K1s  2K1  K2
Equating like terms:
8  K1 and 8  2K1  K2
99
8  K1 and 8  2K1  K2
8  2  8  K2
8 16  8  K2
Thus
8
8
Y (s) 

s  2 s  22
y(t )  8e 2t  8te2t
100
Another Method for Repeated Roots
8( s  1)
K1
K2
Y ( s) 


2
2
( s  2)
s  2 s  2
As before, we can solve for K2 in the usual manner.
8( s  1)( s  2) 2
K2 
2
( s  2)
 8( s  1) s2  8
s 2
101
8( s  1)
8
2 K1
2
( s  2)
 ( s  2)
 ( s  2)
2
2
( s  2)
s2
s  2
2
d 8( s  1) d s  2 K1  8

ds
ds
8  K1
8( s  1)
8
8
Y ( s) 


2
2
( s  2)
s  2 s  2
y(t )  8e 2t  8te2t
102
Unrepeated Complex Roots


Unrepeated complex roots are solved similar
to the process for unrepeated real roots.
That is you multiply by one of the
denominator terms in the partial fraction and
solve for the appropriate constant.
Once you have found one of the constants,
the other constant is simply the complex
conjugate.
103
Complex Unrepeated Roots
5.2
K
K*
Z ( s)  2


s  2 s  5 ( s  1  j 2) ( s  1  j 2)
5.2( s  1  j 2)
K
( s  1  j 2)( s  1  j 2)
  j13
.
s 1 j 2
K  j13
.
*
104
-
5.2
 j1.3
j1.3
Z ( s)  2


s  2s  5 ( s  1  j 2) ( s  1  j 2)
e  j1.3
e j1.3
Z ( s) 

( s  1  j 2) (s  1  j 2)
105
-
5 .2
5 .2
5 .2
2


2
2
2
s  2 s  1  4 ( s  1)  2
2 ( s  1) 2  2 2
5 .2  t
e sin( 2t )
2
106
X ( s) 
k1 
k2 
s 1

( s  2)( s  3)( s 2  s  1)
s 1
( s  3)( s  s  1)
2
 1
s  2
s 1
( s  2)( s  s  1)
2
k1
k
As  B
 2  2
s  2 s  3 s  s 1
4
s  3
7
s  1  k1 ( s  3)( s 2  s  1)  k 2 ( s  2)( s 2  s  1)  ( As  B)( s  2)( s  3)
0  k1  k 2  A  A  (k1  k 2 )  3
7
s  0  1  3k1  2k 2  6 B  B  13
42
1
4 / 7 3 / 7 s  13 / 42
1
4/7
3 / 7 s  13 / 42





s2 s3
s  2 s  3 s 2  s  0.25  0.75
s2  s 1
1
4/7
3 / 7 s  13 / 42



s  2 s  3 ( s  0.5) 2  (0.75 0.5 ) 2
1
4 / 7 3 / 7( s  0.5)  (( (3 / 7)0.5  13 / 42 ) / 0.75 0.5 ) * 0.75 0.5


s2 s3
( s  0.5) 2  (0.75 0.5 ) 2
 e  2t  (4 / 7)e 3t  (3 / 7)e 0.5t cos 0.75 0.5 t  (( (3 / 7)0.5  13 / 42 ) / 0.75 0.5 ) * e 0.5t sin 0.75 0.5 t
As  B
( B  Ab / 2) b / 2t
 b / 2t
2 0.5
2 0. 5

Ae
cos((
c

(
b
/
2
)
)
t

e
sin((
c

(
b
/
2
)
) t
2
2 0.5
s  bs  c
(c  (b / 2) )
107