An introduction to impluse response.pptx

2-2 Transfer function and impulse-response
function
Definition: The transfer function of a linear
time-invariant system is defined as the ratio of
the Laplace transform of the output variable
to the Laplace transform of the input
variable when all initial conditions are zero.
LTI system
input
x(t)
output
y(t)
( )
( )
( )
Y s
G s
X s

2. Transfer function

1
0 1 1
1
n n
n n
n n
d d dy
a y a y a a y
dt dt dt



   

1
0 1 1
1
0 1 1
( )
: ( )
( )
m m
m m
n n
n n
bs bs b s b
Y s
G s
X s a s as a s a




   
 
   


By definition, the transfer function is
1
0 1 1
1
,
m m
m m
m m
d d dr
b x b x b b x n m
dt dt dt



     

Consider the linear time-invariant system
described by the following differential equation:

Example. Given a system described by the
following differential equation
The advantage of transfer function: It
represents system dynamics by algebraic
equations and clearly shows the input-output
relationship:
Y(s)=G(s)X(s)
G(s)
input
x(t)
output
y(t)
2
2
2 3 ( )
d d
y y y u t
dt dt
  
Find its transfer function.

Example. Spring-mass-damper
system:
Wall
friction
, b
r(t) Force
y
k
Let the input be the force
r(t) and the output be the
displacement y(t) of the
mass. Find its transfer
function.
Solution: The system
differential equation is
2
( ) ( )
( ) ( )
d y t dy t
M b ky t r t
dt dt
  
from which we obtain its transfer function
2
( ) 1
( )
Y s
R s Ms bs k

 

Comments on transfer function:
• is limited to LTI systems.
• is an operator to relate the output variable to
the input variable of a differential equation.
• is a property of a system itself, independent of
the
magnitude and nature of the input or driving
function.
• does not provide any information concerning
the
physical structure of the system. That is, the
transfer functions of many physically different
systems can be identical.

Wall
friction
, b
r(t) Force
y
k
L
R
C
u y

3. Convolution integral
( ) ( ) ( )
Y s G s X s

From
and by using the convolution theorem, we have
0
0
( ) ( ) ( )
( ) ( )
t
t
y t g x t d
g t x d
t t t
t t t
 
 


where both g(t) and x(t) are 0 for t<0.

Example. Given
1
( ) ( ) ( ) ( )
1
Y s G s R s R s
T s
 

with T >0. If R(s)=1/s, find y(t).
Solution: By the definition of convolution
integral, 1
1
1/ 1
( ) [ ] , 0
1/
1
( ) [ ] 1( )
t
T
T
g t e t
s T T
r t t
s



  

 
L
L
Hence,
1
( )
0 0
0
1
( ) ( ) ( ) 1
1 , 0
t t
t
T
t
t t
T T T
y t g t r d e d
T
e e d e t
T
t
t
t t t t
t
 
 
   
   
 

1
t

4. Impulse response function
Consider the output (response) of a system to a
unit-impulse input when the initial conditions
are zero:
G(s)
input
(t)
output
y(t)
Hence,
( ) ( ) ( ) ( ) ( ( ))
( ) 1 ( )
Y s G s R s G s t
G s G s
 
  
d
L
and 1
( ) [ ( )]: ( )
y t G s g t

 
L
where g(t) is called impulse response function.

An impulse response function g(t) is the
inverse Laplase transform of the system’s
transfer function!
Example. Given
( ) ( ) ( )
Y s G s R s

t
y
t
(t)
If
( ) , 0
t
y t e t

 
determine the system’s transfer
function.
R(s) Y(s)

It is hence possible to obtain complete
information about the dynamic characteristics of
the system by exciting it with an impulse input
and measuring the response (In practice, a pulse
input with a very short duration can be
considered an impulse).
Example. Let
Assume that the system is LTI. Determine its
transfer function.
t
(t)
R(s) Y(s)
t
y
1
1
e-t

2

Example. Let
3 2
1
( ) ( ) ( ) ( )
3 3 1
Y s G s R s R s
s s s
 
  
Determine its impulse response function.

2-3 Automatic control systems
1. Block Diagrams
 Block diagram of a system is a pictorial
representation of the functions performed by
each component and of the flow of signals.
G(s)
R C
( )
( )
( )
C s
G s
R s
 or ( ) ( ) ( )
C s G s R s


2. Block Diagram of a closed-loop system
A real physical system includes more than one
components. The following is a typical
feedback system represented by block
diagram:
G(s)
H(s)

R(s) C(s)
E
B(s)
where the summing point E(s)=R(s)B(s) and
the branch point are shown below.

R(s)
E
B(s) Y(s)
Y(s)

The property of summing point:

 

1
R
2
R
3
R
C 1
R
2
R
3
R
C


1
R
2
R
3
R
C
1
E

Example. A network system is shown below,
where uc is the output and ur is the input. Draw
its block diagram.
i
1
i
2
i C
1
R
2
R
r
u c
u
1 1 ( )
r c
R i u u
= -
1 1
2 1
R
du di
i C CR
dt dt
= =
2 1 2
( )
c
u R i i
= +
Solution:
Sept 1. Write the input and output relationship of
each device:
ur
uc

1 1( ) ( ( ) ( ))
r c
R I s U s U s
= -
2 1 1
( ) ( )
I s R CsI s
=
2 1 2
( ) ( ( ) ( ))
c
U s R I s I s
= +
Sept 2. Taking Laplace transform of both sides of
the above equations yields:
Sept 3. Rearrange each above equation so that
its left-hand side is the output variable and the
right-hand side is the transfer function
multiplied by input signals:
1
1( ) [ (
1
) ( )] (1)
r c
I s U s U s
R
= -
2 1
1
( ) ( ) (2)
R Cs
I s I s
=
2 1 2
( ) [ ( ) ( )] (3)
c
U s I s I s
R
= +

i
1
i
2
i C
1
R
2
R
r
u c
u
1
R
1
)
(s
Ur
)
(s
Uc
)
(
1 s
I
Cs
R1
)
(
2 s
I
2
R
)
(s
Uc
Step 4. Based on (1)-(3), draw the block diagram:
1
2 1
1
1
2
1
2
( ) [ ( ) ( )] (1)
( ) ( ) (2)
( ) [ ( ) ( )] (3)
1
r c
c
I s U s U s
I s I s
U s I s
Cs
R I
R
R
s
ì
ï
ï = -
ï
ï
ï
ï
ï =
í
ï
ï
ï = +
ï
ï
ï
ï
î

Example. A system is described by the
following equations:
1
x r c
 
2 1 1 1
x x K x
t
 

3 2 2
x K x

4 3 5 5
x x x K c
  
5 3 4
x K x


4 5
K x T c c
 

Draw its block diagram,
where , Ki and T are
positive constants, the
input and output signals
are r and c, respectively,
and x1x5 are
intermediate variables. .
r c

Example. A system is described by the
following equations:
1 1
x r c n
  
2 1 1
x K x

3 2 5
x x x
 
4 3
T x x


5 4 2 2
x x K n
 
0 5
K x c c
 
 
Draw its block diagram,
where Ki and T are positive
constants, the input and
output signals are r and c,
respectively, n1, n2 are
disturbances, and x1x5 are
intermediate variables.
r c
n1 n2

3. Open-loop transfer function and feedforward
transfer function
Two important concepts:
Open-loop transfer function: The ratio of the
feedback signal B(s) to the actuating error signal
E(s) is called the open-loop transfer function. That
is,
G(s)
H(s)

R(s) C(s)
E
B(s)
( )
( ) ( )
( )
B s
G s H s
E s


Feedforward transfer function: The ratio of the
output C(s) to the actuating error signal E(s) is
called the feedforward transfer function, so that
( )
( )
( )
C s
G s
E s

G(s)
H(s)

R(s) C(s)
E
B(s)

4. Closed-loop transfer function
( ) ( )
( ) 1 ( ) ( )
C s G s
R s G s H s


G(s)
H(s)

R(s) C(s)
E
B(s)
( ) ( ) ( ) (1)
C s G s E s

( ) ( ) ( ) ( ) ( ) ( ) (2)
E s R s B s R s H s C s
   
Substituting (2) into (2) yields
( ) ( ) ( ) ( ) ( ) ( ) (3)
C s G s R s G s H s C s
 
from which we obtain the closed-loop transfer
function as

( )
R s
( )
m
m m
C
s J s f

( )
C s
b
K s
( )
B s
Example. A block diagram of a system is
shown below. Determine its closed-loop
transfer function C(s)/R(s).

a) Cascaded system
G1(s)G2(s)
R(s) C(s)
G1(s) G2(s)
R(s) C(s)
U
2 2 1
( ) ( ) ( ) ( ) ( ) ( )
C s G s U s G s G s R s
 
5. Obtaining cascaded, parallel, and feedback
transfer functions

a
u
( )
m
m m
C
s J s f

m

1
a a
L s R

b
K s
b
E
Example. A block diagram of DC motor is
shown below. Determine its open-loop transfer
function and feedforward transfer function.

b) Parallel Blocks
G1(s)
G2(s)

R(s) C(s)
1 2
( ) ( )
G s G s

R(s)
C(s)
1 2 1 2
( ) ( ) ( ) ( ) ( ) ( ( ) ( )) ( )
C s G s R s G s R s G s G s R s
   

c) Feedback loop
( )
1 ( ) ( )
G s
G s H s

R(s) C(s)
G
H

R(s) C(s)
E
( )
( ) ( )
1 ( ) ( )
G s
C s R s
G s H s



Diagram simplification: Moving a summing point
ahead of a block:
G s
( )
1
X s
( )
3
X s
( )
2
X s
( )
1
G s
( )
+
G s
( )
1
X s
( )
3
X s
( )
2
X s
( )
+

Diagram simplification : Moving a summing
point behind a block:
1
X s
( ) 3
X s
( )
2
X s
( )
G s
( )
G s
( )
-
G s
( )
1
X s
( )
3
X s
( )
2
X s
( )
-

Example. Simplify the following block diagram.
Then obtain the closed-loop transfer function
Eo(s)/Ei(s).
i
E
2 ( )
G s
o
E
1( )
G s
ab
U
1
I ab
U
3 ( )
G s
2
I
4 ( )
G s
If there exists a branch point between two
summing points, do not move summing point.

Diagram simplification: Moving a branch
point ahead of a block
1
X s
( )
3
X s
( )
2
X s
( )
G s
( )
G s
( )
G s
( )
1
X s
( )
3
X s
( )
2
X s
( )

Diagram simplification: Moving a branch
point behind of a block
G s
( )
1
X s
( )
3
X s
( )
2
X s
( )
1
G s
( )
G s
( )
1
X s
( )
3
X s
( )
2
X s
( )

Example. Simplify the following block diagram.
Then obtain the closed-loop transfer function
Eo(s)/Ei(s).
i
E
2 ( )
G s
o
E
1( )
G s
ab
U
1
I ab
U
3 ( )
G s
2
I
4 ( )
G s

-
R s
( ) C s
( )
1
G s
( ) 2
G s
( )
1
H s
( )
-
-
Diagram simplification: Examples
Example. The block diagram of a given system is
shown below. Obtain C(s)/R(s).
Note that only the movement between two
summing points (two branch points) is valid.

-
R s
( ) C s
( )
2
G s
( )
2
H s
( )
3
G s
( )
1
H s
( )
4
G s
( )
1
G s
( )
- -
+
shown below. Obtain C(s)/R(s).
If there exists a branch point (summing point)
between two summing points (two branch
points), do not move them.

shown below. Obtain the transfer function that
relates the output C(s) in function of the input
R(s), i.e., C(s)/R(s).
G1
G2
G3
H1
R(s)
C(s)
Note that only the movement between two
summing points (two branch points) is valid.

G2
H1
G1
G3
R(s) C(s)
G2
H1
G1
G3
G1
R(s)
C(s)

G2
H1
G1
G3
An alternative way to simplify the diagram is
G2
H1
G1
G3
1/G1

1
G 2
G 3
G 4
G
1
H
2
H
3
H
R C
Example. The block diagram of a given
system is shown below. Obtain C(s)/R(s).

1
G 2
G 3
G 4
G
1
H
2
H
3
H
R C
1
G 2
G 3
G 4
G
1
H
4
2
G
H
3
H
R C

1
G 2
G 3
G 4
G
1
H
4
2
G
H
3
H
R C
1
G 2
G
4
2
G
H
3
H
R C
1
4
3
4
3
1 H
G
G
G
G


1
G 2
G
4
2
G
H
3
H
R C
1
4
3
4
3
1 H
G
G
G
G

1
G
3
H
R C
2
3
2
1
4
3
4
3
2
1 H
G
G
H
G
G
G
G
G



1
G
3
H
R C
2
3
2
1
4
3
4
3
2
1 H
G
G
H
G
G
G
G
G


R C
3
4
3
2
1
2
3
2
1
4
3
4
3
2
1
1 H
G
G
G
G
H
G
G
H
G
G
G
G
G
G




Example. Block reduction: Redrawing block
diagram. Consider the following diagram. With
redrawing the diagram, the simplification can
be proceeded.
G1
G4
H3
G2 G3
H1

G1
G4
H3
G2 G3
H1
H1
H3
G1
G4
G2 G3
H3
H1

1 ( )
G s
3 ( )
G s
2 ( )
G s
( )
R s ( )
C s
+
Example. The block diagram of a system is shown
below. Obtain C(s)/R(s).

5. The Mason’ Formula (supplement material)
• Li=transfer function of the ith loop;
1
( ) 1
( )
N
k k
k
C s
P
R s 
 


• LiLj=product of transfer functions of two
non-touching loops;
• LiLjLk=product of transfer functions of three
non-touching loops;
• LiLjLkLl…..;
• = the characteristic polynomial of the
system
=1 Li+LiLj  LiLjLk+;

1
( ) 1
( )
1
N
k k
k
i i j i j k
C s
P
R s
L L L L L L

 

     

   
• N=total number of forward paths (from R(s)
to C(s) without visiting a point more than
once);
• Pk=transfer function of kth forward path;
• k= the cofactor of the kth forward path with
the loops touching the kth forward path
removed.

i
E
2 ( )
G s
o
E
1( )
G s
ab
U
1
I ab
U
3 ( )
G s
2
I
4 ( )
G s
Example. Determine the characteristic
polynomial of the following block diagram:

G 1
H 1
H 2
G 4
G 3
G 2
R C
shown below. Determine its characteristic poly-
nomial, forward paths and cofactors.

-
R s
( ) C s
( )
1
G s
( ) 2
G s
( )
4
H s
( )
3
G s
( )
3
H s
( )
2
H s
( )
1
H s
( )
4
G s
( )
-
-
+
-
Example. For the following block diagram, find
C(s)/R(s).
Solution: By using Mason’s formula,

1 1 2 3 4
=
P G G G G
1 2 3 3
2 1 2 3 2
3 1 2 3 4 1
4 3 4 4


 




 

L G G H
L G G G H
L G G G G H
L G G H
1
1 2 3 4
2 3 3 1 2 3 2 3 4 4 1 2 3 4 1
=
1




   

n
k k
k
P
C
R
G G G G
G G H G G G H G G H G G G G H
1 2 3 4
1 ( )
L L L L
     
1 1
 
The cofactor is:
Only one forward
path:
Four individual loops:
No nontouching
loops, therefore,
Consequently,

By principle of superposition, let,
respectively, R(s)=0 and D(s)=0 and calculate
the corres-ponding outputs CD(s) and CR(s).
Then,
C= CD(s) +CR(s)
G1(s) G2(s)
H(s)
R(s) C(s)
D
2 1 2
1 2 1 2
( ) ( ) ( )
( ) ( ) ( )
1 ( ) ( ) ( ) 1 ( ) ( ) ( )
G s G s G s
C s D s R s
G s G s H s G s G s H s
= +
+ +
6. Closed-loop systems subjected to a disturbance

Example. The block diagram of a given system
is shown below. Obtain C(s)/R(s) and C(s)/D(s).
H
_ ( )
c
G s
( )
D s
( )
f
G s
( )
C s
( )
R s
1( )
G s ( )
p
G s

Summary of Chapter 2
In this chapter, we mainly studied the
following issues:
1. Laplace transformation and the related
theorems.
2. Transfer function:
1
0 1 1
1
0 1 1
( )
( ) :
( )
m m
m m
n n
n n
bs bs b s b
C s
G s
R s a s as a s a






   
 
   

from which we know that
• C(s)=G(s)R(s);
• In time domain, the above equation
represents a convolution integral:
0
( ) ( ) ( )
t
c t g t r d
t t t
 

where g(t)=1
[G(s)].
• In particular, if r(t)=(t),
c(t)=1
[G(s)],
that is, the impulse response is the inverse
Laplase transform of its transfer function.

3. Block diagram:
Note that only one block is less meaning.
However, with a block diagram, the
interrelationship between components and the
signal flows can be revealed pictorially
compared with the mathematical expression of
a set of equations; for instance:
H
_ ( )
c
G s
( )
D s
( )
f
G s
( )
C s
( )
R s
1( )
G s ( )
p
G s

4. Block diagram simplification (reduction):
• feedback;
• cascade;
• parallel;
• moving between two summing (branch)
points;
• redrawing the block diagram for some special
cases.

By using diagram simplification techniques,
one can finally obtain, no matter how
complex a system may be,
Cr(s)=G(s)R(s);
Cd(s)=(s)D(s).
Based on which, system analysis can be
proceeded.

An introduction to impluse response.pptx

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An introduction to impluse response.pptx

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