dah presentation

1. Automatic control
Introduction to automatic control
Dr.-Ing.
Dr Ing Mohamed Saber Sokar

2. ▐ Objectives
 Review on elementary function
y
 Complex numbers
 Laplace transform
 Examples
 properties
 Inverse transform
 Partial fraction expansion
e pansion
 Partial fraction
 Common test input signals
 U d
Understand the Concept of Transfer Functions.
t d th C t fT f F ti
٢ Dr.-Ing. Mohamed Saber Sokar

3. ▐ Elementary functions
Exponential Function e x
e xe y  e x y , e x / e y  e x y
( e x ) y  e xy , e 0  1
Logrithmic F
L i h i Function ln( x ) i l (
ln (xy)  ln x  ln y
( y)
x
ln  ln x  ln y
y
ln (x a )  a ln x ,
( ln(1)  0
( )
1
 ln x
ln 1
e ln x
 x, e
, e x

x
٣ Dr.-Ing. Mohamed Saber Sokar

4. ▐ Elementary functions
Even Function: f(  x)  f(x), x
Odd Function: f(  x)   f(x), x
sin(  x)   sin x, cos(  x)  cos x
sin x  cos x  1
2 2
sin (x  y)  sin x cos y  cos x sin y
sin (x  y)  sin x cos y  cos x sin y
sin 2 x  2 sin x cos x
cos (x  y)  cos x cos y  sin x sin y
cos (x  y)  cos x cos y  sin x sin y
٤ Dr.-Ing. Mohamed Saber Sokar

5. ▐ Elementary functions
cos 2 x  cos 2 x  sin 2 x  1  2 sin 2 x  2 cos 2 x  1
1
cos x  ( 1  cos 2 x)
2
2
1
sin 2 x  ( 1  cos 2 x)
2
π π
sin x  cos ( x- )  cos ( -x)
2 2
π π
cos x  sin ( x  )  sin ( -x)
2 2
sin (π  x)  sin x
cos (π  x)   cos x
sin x cos x 1 1
tan x  , cot x  , sec x  , csc x 
cos x sin x
i cos x sin x
i
٥ Dr.-Ing. Mohamed Saber Sokar

6. ▐ Elementary functions
Simplifying
D  A cos   B sin 
B
then C  A  B , tan  
2 2
A
A cos x  B sin x  A2  B 2 cos (  )
C : magnitude,  : phase
٦ Dr.-Ing. Mohamed Saber Sokar

7. ▐ Complex Numbers
If z1  x1  i. y1 , and z 2  x2  i. y2
 adding z  z1  z 2  ( x1  iy1 )  ( x2  iy2 )  ( x1  x2 )  i ( y1  y2 )
 Multiplying z  z1 . z 2  ( x1  iy1 ) . ( x2  iy2 )
z  z1  z 2  x1 x2  y1 y2  i ( x2 y1  x1 y2 )
 Multiplying by conjugate of the denominator
z1 x1  iy1 ( x1  iy1 )( x2  iy2 )
z  
z 2 x2  iy2 ( x2  iy2 )( x2  iy2 )
x1 x2  y1 y2 x2 y1  x1 y2
z i
x2  y 2
2 2
x2  y 2
2 2
٧ Dr.-Ing. Mohamed Saber Sokar

8. ▐ Polar form of Complex Numbers
 z=x+i.y, if x=r . cos, y= r . sin
Then z = r(cos +i . sin = r 
 Th (  i 
 Absolute value (modulus) r2=x2+y2
 Magnitude
g r  x2  y2
 Argument = tan-1(y/x)
 Example z=2+i.4
z  20
4
  tan 1
2
٨ Dr.-Ing. Mohamed Saber Sokar

9. ▐ Laplace Transform
 D fi iti
Definition and examples
d l

F ( s )  L( f )   e  st f (t )dt
0
Inverse Lapalce f (t )  L1 ( F )
Example
Unit Step Function u(t)
f (t )  u (t )  1
 
1  st 1  s 1 0
F ( s )  L (1)   e  st
1 dt   e  e  e
0
s 0 s s
1
 F (s) 
s
٩ Dr.-Ing. Mohamed Saber Sokar

10. ▐ Laplace Transform (Example)
 Fi d Laplace transform for the f ll i
Find L l t f f th following f (t)  e
3t
function
 
F ( s )  L ( e 3 t )   e  st e 3 t dt   e  ( s  3 ) t dt
0 0

1 ( s 3)t 1
 e 
s3 0 s3
1
 F (s) 
s3
1
L (e ) 
at
sa
١٠ Dr.-Ing. Mohamed Saber Sokar

11. ▐ Laplace Transform

L ( e sin  t )   e  st e at sin  tdt
at
0
 
1 1
 e  ( s  a ) t sin  t   ( ) e  ( s  a ) t  cos  tdt
sa 0 0
sa



sa 
0
e  ( s  a ) t cos  tdt
 
  1
cos  t  ( ) e  ( s  a ) t   2 sin  tdt
( s  a )t
 e 
(s  a)2 0
sa 0
sa

 2
e sin  tdt
( s  a )t
 
(s  a) 2
(s  a) 2
0

2 
2 
(1  ) e  st sin  tdt 
(s  a) 0
(s  a)2



0
e  st sin  tdt 
(s  a)2   2

sa

0
e  st cos  tdt 
(s  a)2   2
١١ Dr.-Ing. Mohamed Saber Sokar

12. ▐ Laplace transform table
١٢ Dr.-Ing. Mohamed Saber Sokar

13. ▐ Laplace transform theorem-1
١٣ Dr.-Ing. Mohamed Saber Sokar

14. ▐ Laplace transform theorem-2
١٤ Dr.-Ing. Mohamed Saber Sokar

15. ▐ Laplace Transform
 Find Laplace transform
y
y”+9y=0, y(0)=0, y’(0)=2
y y( ) y( )
1- L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2
2- L(y)=Y(s)
Solution
Sol tion
(s2+9)Y(s)-2=0
+9)Y(s) 2 0
Y(s)=2/ (s2+9) = (2/3)[3/(s2+9)]
y(t)=(2/3) sin 3t
١٥ Dr.-Ing. Mohamed Saber Sokar

16. ▐ Laplace Transform
Solve: y”+2y’+5y=0 y(0)=2 y’(0)= 4
y”+2y’+5y=0, y(0)=2, y’(0)=-4
 L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2s+4
(y ) ( ) y( ) y ( ) ( )
 L(y’)=sY(s)-y(0)=sY(s)-2
 L(y)=Y(s)
 (s2+2s+5)Y(s)=2s
 Y(s)=2s/
(s2+2s+5)=2(s+1)/[(s+1)2+22]-
2/[(s+1)2+22]
 y(t)= e-t(2cos 2t –sin 2t)
١٦ Dr.-Ing. Mohamed Saber Sokar

17. ▐ Inverse Laplace Transform
1 1 1 1 1 1 3 1
L ( 2 )  L ( 2 2 )  L ( 2 2 )  sin 3t i
s 9 s 3 3 s 3 3
1 s3 1 s2 1
L ( )L (  )
( s  2)  3
2 2
( s  2)  3 ( s  2)  3
2 2 2 2
 2t 1  2t  2t 1
 e cos 3t  e sin 3t  e (cos 3t  sin 3t )
3 3
1 3s  2 1 3s 2
L ( 2 2 )  L ( 2 2  2 2 )  3 cos t  2 sin t
s 1 s 1 s 1
Find
1 3s  2
L ( 2 )?
s  2s  5
١٧ Dr.-Ing. Mohamed Saber Sokar

18. ▐ Partial Fraction
s 1
Y (s)  3 2 , What is y(t)?
s  s  6s
s 1 s 1 A B C
Y (s)  3 2    
s  s  6 s s ( s  2)( s  3) s s  2 s  3
multiply by s, and put s  0
s 1 Bs Cs 1
 A   A
( s  2)( s  3) s 0 s  2 s 0 s  3 s 0 6
multiply by s - 2, and put s  2
s 1 A( s  2) C ( s  2) 3
 B   B
s ( s  3) s  2 s s 2 s  3 s 2 10
١٨ Dr.-Ing. Mohamed Saber Sokar

19. ▐ Partial Fraction
multiply by s  3, and put s  3
s 1 B ( s  3) A( s  3)
C 
s ( s  2) s  3 ( s  2) s  3 s s  3
2
C
15
s 1 11 3 1 2 1
Y (s)  3 2   
s  s  6s 6 s 10 s  2 15 s  3
1 3 2 t 2  3t
y (t )    e  e
6 10 15
١٩ Dr.-Ing. Mohamed Saber Sokar

20. ▐ Partial fraction; repeated factor
y"3 y '2 y  4t  e 3t , y(0)  1, y' (0)  -1
y( ) , y ( )
4 1
s Y  s  1  3( sY  1)  2Y  2 
2
s s3
s  7 s  13 s  4 s  12 G ( s )
4 3 2
Y (s)  
s ( s  3)( s  3s  2)
2 2
H (s)
A2 A1 B C D
y (t )  2    
s s s  3 s  2 s 1
A 2 , B, C, D can be obtained as before, but A 1?
s 4  7 s 3  13 s 2  4 s  12
A 2  Q ( s ) s 0  2
( s  3)( s 2  3s  2) s 0
٢٠ Dr.-Ing. Mohamed Saber Sokar

21. ▐ Partial fraction; repeated factor
s 4  7 s 3  13s 2  4 s  12
Y (s)  2 
s ( s  3)( s  3s  2)
2
A2 A1 B C D
 2    
s s s  3 s  2 s 1
Multiply by s , and put s  0, A 2 can be obtained
py y 2
p ,
s  7 s  13s  4 s  12
4 3 2
Bs 2
Cs 2
Ds 2
 A2  A1s   
( s  3)( s  3s  2)
2
s  3 s  2 s 1
Differenti ate with s and put s  0
s,
A1  Q ' ( s ) s  0  3
٢١ Dr.-Ing. Mohamed Saber Sokar

22. ▐ Test signals used in control systems
٢٢ Dr.-Ing. Mohamed Saber Sokar

23. ▐ Test signals used in control systems
٢٣ Dr.-Ing. Mohamed Saber Sokar

24. ▐ Free & Forced Responses
 Forced Response [zero ICs & nonzero u(t)]
 The response of a system to nonzero input and zero initial conditions.
 Can be obtained by
 Assume zero ICs and use LT to solve for the forced
response {replace differentiation with (s) in the
Input/Output (I/O) ODE model}.
 Free Response [u(t) = 0 & nonzero ICs]
 The response of a system to zero input and nonzero initial conditions.
 Can be obtained by
Let u(t) = 0 and use Laplace transform (LT) considering
Initial condition (IC) values to solve for the free
response.
response
٢٤ Dr.-Ing. Mohamed Saber Sokar

25. ▐ Transient and Steady State Response
Ex: Find the total response of a stable first order system:
p y
y  5 y  10u

to a ramp input: u  t   5t with I.C.: y 0  2
10 5 1
Y s   2

- Total response s5 s 2
s  5 y0
Transfer Function U  s   
 
   Free Response
- Partial fraction eq. Forced Response
a1 a2 a3 2
Y s   2  
s s s5 s5
d 21 2
a1 
1
 2  1 ! ds 21
s Yforced ( s) s0  ds  s505    s 505 2
d


 2
   s 0    s 0
a2   s 2Yforced ( s ) 
  s0  10 a3   s  5 Yforced ( s ) 
  s5  2
 
 
y t   2 
  10t
   a 3 e 5 t
  2 e 5 t
 
Steady state response
d  Transient response Transient response
T i t

from Forced response  from Forced response free response

٢٥ Dr.-Ing. Mohamed Saber Sokar

26. ▐ Transient and Steady State Response
Ex: Find the response of a given second order system:
p g y
  4 y  3 y  6u
y 
to a sinusoidal input: u  t   5sin  3t  with I.C.: y  0   0, y  0   2

6 53 2s  4  2
Y s   2
- Total response s2  4s  3 s2  
 
 32    3
s 4 s 

Forced Response Free Response
- Partial fraction equation
 a1 a2 a3 a4   b1 b2 
Y s  s 3      
 s 1 s3j s 3j s 3 s  1

5 9 1 1
a1   a2  a 3  1  j a 4  1  j b1  1 b2  3
2 2 2 2
y  t   2 Re a3e 3 jt     a1  b1  e 3t   a2  b2  e  t 
 
       
Steady state response Transient response
 7 15  t 
 5 sin  3t  tan 1  2      e 3t  e 
 2 2 
٢٦ Dr.-Ing. Mohamed Saber Sokar

27. ▐ Transfer function
The transfer function H() of a system is the frequency
dependent ratio of the system output Y(s) to a system
input U(s)
U(s).
U(s) Y (s) Y(s)
H(s) =
U ( s)
(s
Vo ( )
H( )  Voltage gain 
Vi ( )
I o ( )
H( )  Current gain 
I i ( )
٢٧ Dr.-Ing. Mohamed Saber Sokar

28. ٢٨ Dr.-Ing. Mohamed Saber Sokar