CHAPTER 3.
DYNAMIC RESPONSE
3.1. Poles and Zeros
3.2. Inverse Laplace transform
3.3. Declare transfer function using simulation
software
3.4. Equivalent transfer function of the system
3.5. Modeling in state space
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CHAPTER 3.
DYNAMIC RESPONSE
3.6. Effect of poles and zeros
3.7. Transient qualities of system responses
3.8. Stability
3.9. Summary.
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OBJECTIVES
 Remember the definition and properties of
Laplace transform.
 Apply Laplace transform to derive the
transfer function.
 Find the equivalent transfer function using
block diagram method and signal-flow graph.
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OBJECTIVES
 Remember the mathematical description of a
system in the form of state space.
 Apply the basic methods to derive set of
state
variable
equations
from
different
differential equations and transfer function.
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OBJECTIVES
 Remember
the
concepts
of
percent
overshoot, setting time, rise time, …
 How to define whether a system is stable.
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3.1. POLES AND ZEROS
m 1
b0 s  b1s  ...  bm 1s  bm N ( s )
H (s)  n

n 1
s  a1s  ...  an 1s  an
D(s)
m
(
s

z
)

i
i 1
K
n
 i  ( s  pi )
m
 N: Numerator
 D: Denominator
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3.1. POLES AND ZEROS
 K: transfer function gain.
 z1, z2, … , zm: zeros of the system.
H (s) s  z  0
i
 p1, p2, … , pn: poles of the system
H (s) s  p  
i
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3.1. POLES AND ZEROS
Poles and zeros:
 The poles and zeros may be complex
quantities.
 Their locations are displayed in complex plane.
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3.2. INVERSE LAPLACE
TRANSFORM
Inverse Laplace transform:
m 1
b0 s  b1 s  ...  bm 1s  bm
H (s)  n
n 1
s  a1 s  ...  an 1s  an
m
( s  zi )

i 1
K
n
 i  ( s  pi )
Cn
C1
C2


 ... 
s  p1 s  p2
s  pn
m
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3.2. INVERSE LAPLACE
TRANSFORM
Inverse Laplace transform:
Determine the set of constant Ci:
Ci   s  pi  H ( s ) s  p
i
The time function:
n
h (t )   C i e
pi ( t )
i 1
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3.3. DECLARE TRANFER FUNCTION
USING SIMULATION SOFTWARE
Transfer function:
s2
Numerator polynomial
H (s)  2

s  2 s  10 Denominator polynomial
Using Matlab
The coefficients of the numerator polynomial are
displayed as row vector num
num  1 2
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3.3. DECLARE TRANFER FUNCTION
USING SIMULATION SOFTWARE
The coefficients of the denominator polynomial are
displayed as row vector den
den  1 2 10
Transfer function in Matlab is declared as:
H  tf  num, den 
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
3.4.1. Block diagram method
The block diagram:
 Represents the mathematical relationship
between the components.
 The interconnection of blocks include:
• Summing point.
• Pickoff point.
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
1. The block diagram algebra.
 Series system.
 Parallel system.
 Feedback system:
• Negative feedback.
• Positive feedback.
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
Example.
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
The interconnections of
diagram can be manipulated
without affecting the
mathematical relationships.
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
3.4.2. Block diagram using MATLAB.
 series
 parallel
 feedback
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
3.4.3. Mason’s rule and the signal flow graph
A signal-flow graph:
 Diagram consisting of nodes that are connected by
several directed branches.
 Graphical representation of a set of linear relations.
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
Definitions
Nodes: The input and output points or junctions
Branch: Line connecting two nodes.
Path: Branch or continuous sequence of branches that
can be traversed from one signal (node) to another
signal (node).
Loop (Self-loop): Closed path that originates and
terminates on the same node.
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
Mason’s signal-flow gain formula:
PD

Y s 
T s  

k
R s 
Where

k
k
D
D  1   Li   Li L j   Li L j Lm  ...
Pk: Gain of kth path.
i
i, j
i , j ,m
 D: Determinant of the graph.
 Dk: Cofactor of the path Pk.
 L: Self-loop
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3.4. EQUIVALENT TRANSFER
FUNCTION OF THE SYSTEM
Notes
 Two loops are said to be non-touching if they do not
have a common node.
 Two touching loops share one or more common
nodes.
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3.5. MODELING IN STATE SPACE
3.5.1.
Relationship
among
differential
equation, transfer function and state space
model
 Differential equation
 Transfer function
 State space model
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3.5. MODELING IN STATE SPACE
3.5.2. Derive the state space equation from the
differential equation
There is no derivative of input signal in right-hand side of
differential equations
d n y (t )
d n 1 y (t )
dy (t )
 a1
 ...  an 1
 an y (t )  b0 r (t )
n
n 1
dt
dt
dt
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3.5. MODELING IN STATE SPACE
x1 (t )  y (t )
Set
xi (t )  xi 1 (t ) (i  2, n )
Therefore, we have
 x1 (t )  x2 (t )
 x (t )  x (t )
3
 2

 x (t )  x (t )
n
 n 1
 xn (t )   an x1 (t )  an 1 x2 (t )  ...  a2 xn 1 (t )  a1 xn (t )  b0 r (t )
y (t )  x1 (t )
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3.5. MODELING IN STATE SPACE
 x (t )  Ax (t )  Br (t )

 y (t )  Cx (t )
In Matrix form
 x1 (t ) 
 0
 x (t ) 
 0
 2


x (t )     ; A   



x
(
t
)
 n 1 
 0
 x n (t ) 
  a n
1
0
0
1


0
0
 a n 1
 a n2
0 
0
0

0 

 

 ; B    ;

 

1 
0
b0 
 a1 

C  1 0  0 0
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3.5. MODELING IN STATE SPACE
3.5.2. Derive the state space equation from the
differential equation
There is derivative of input signal in right-hand side of
differential equations
d n y (t )
d n 1 y (t )
dy (t )
 a1
 ...  an 1
 a n y (t ) 
n
n 1
dt
dt
dt
d m r (t )
d m 1r (t )
dr (t )
b0
 b1
 ...  bm 1
 bm r (t )
m
m 1
dt
dt
dt
Condition: m = n - 1
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3.5. MODELING IN STATE SPACE
x1 (t )  y (t )
Set
xi (t )  xi 1 (t )   i 1r (t ) (i  2, n )
In Matrix form
 0
 0

A 

 0
  a n
1
0
0
1


0
0
 a n 1
 a n2
 x (t )  Ax (t )  Br (t )

 y (t )  Cx (t )
0 
 1 
 

0 

 2 

 ; B    ; C  1 0  0 0




1 

 n 1 
  n 
 a1 

 1  b0 ;  2  b1  a1  1 ;
 3  b2  a1  2  a 2  1 ; ;  n  bn 1  a1  n 1    a n 1  1
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3.5. MODELING IN STATE SPACE
3.5.3. Derive the state space equation from the
transfer function
Phase coordinated method
System transfer function
Y ( s ) b0 s m  b1 s m 1  ...  bm 1s  bm
T (s) 
 n
R(s)
s  a1s n 1  ...  an 1s  an
Conditions: m=n-1, a0 = 1
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3.5. MODELING IN STATE SPACE
Set the auxiliary variable Z(s) that satisfies
 Y ( s )   b0 s m  b1s m 1  ...  bm 1s  bm  Z ( s )


n
n 1
 R ( s )   s  a1s  ...  an 1s  an  Z ( s )
Inverse Laplace transform

d m z (t )
d m 1 z (t )
dz (t )
 b1
 ...  bm 1
 bm z (t )
 y(t)  b0
m
m 1
dt
dt
dt

n
n 1
d
z
(
t
)
d
z (t )
dz (t )
 r (t ) 

a

...

a
 a n z (t )
1
n 1
n
n

1

dt
dt
dt
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3.5. MODELING IN STATE SPACE
Set
x1 (t )  z (t )
x2 (t )  x1 (t )  z (t )
d n 1 z (t )
xn (t )  xn 1 (t ) 
dt n 1
 x (t )  Ax (t )  Br (t )
Set of state variable equations

 y (t )  Cx (t )
1
0
0 
 0
0 
 0
0 
0
1
0 


 
 ; B    ; C   bm bm 1
A
b1 b0 


 
0
0
0
1


0 
  an  an 1  an  2
 1 
 a1 
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3.5. MODELING IN STATE SPACE
3.5.4. Derive the characteristic equation from
the state space model
state space
State space model
state model
space model
 x (t )  Ax (t )  Br (t )

 y (t )  Cx (t )
Characteristic equation
det  sI  A   0
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3.6. EFFECT OF POLES AND
ZEROS
 Depending on the transfer function, we
analyze the response of the system.
 Response of the system:
• Impulse response (natural response).
• Step response.
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3.6. EFFECT OF POLES AND
ZEROS
 From the partial fraction expansion: Slide 6.
1
H (s) 
s 
 The impulse response is:
h(t )  e
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 t
1(t )
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3.6. EFFECT OF POLES AND
ZEROS
- When σ > 0, impulse response
is stable. If σ < 0, unstable.
- The time constant is the time
when the response is 1/e times
the initial value, measure the
time of decay.

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1

403036 - Chapter 3. Dynamic Response
3.6. EFFECT OF POLES AND
ZEROS
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3.6. EFFECT OF POLES AND
ZEROS
Example:
2s  1
H (s)  2
s  3s  2
Poles: s = -1; s = -2
 Poles farther to the left in the s – plane decay
faster than poles closer to the imaginary axis.
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3.6. EFFECT OF POLES AND
ZEROS
 The impulse response:
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3.6. EFFECT OF POLES AND
ZEROS
 Complex poles:
s    jd
 The second order transfer
function:
2
n
H (s)  2
s  2 n s   n2
 The denominator of H(s):
  n ; d  n 1   2
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3.6. EFFECT OF POLES AND
ZEROS
 Where:
 : damping ratio
n : undamped natural frequency
 The poles are located
at a radius  n and at an
angle   sin 1 
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3.6. EFFECT OF POLES AND
ZEROS
 When   1 , we have no damping,   0 , the
damped natural frequency d  n
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3.6. EFFECT OF POLES AND
ZEROS
Example:
2s  1
H (s)  2
s  2s  5
n  ? 5(rad/s)
  ? 0.447
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3.6. EFFECT OF POLES AND
ZEROS
 Effects of zeros and additional poles.
 Effect of zero.
s / 
H (s)  2
s  2 s  1
  0.5
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3.6. EFFECT OF POLES AND
ZEROS
 Effects of zeros and additional poles.
 Effect of zero.
s / 
H (s)  2
s  2 s  1
  0.707
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3.6. EFFECT OF POLES AND
ZEROS
 Effects of zeros and additional poles.
 Effect of zero to the pole locations
24  s /  z 
H (s) 
z ( s  4)( s  6)
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3.6. EFFECT OF POLES AND
ZEROS
 Effects of zeros and additional poles.
 Effect of complex zero to the pole locations
2
s






2
H (s) 
2

( s  1)  ( s  0.1)  1
 1
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3.6. EFFECT OF POLES AND
ZEROS
 Effects of zeros and additional poles.
 Effect of extra pole.
H ( s) 
1
( s / n  1)  ( s / n ) 2  2 ( s / n )  1
  0.5
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
 Performance specifications for a control system
design involve certain requirements associated
with the time response of the system.
• The rise time t r
• The settling time t s
• The overshoot M p
• The peak time t p
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
3.7.1. The rise time t r :
• The time it takes the system to reach the vicinity
of its new set point.
• The rise time from y = 0.1 to y = 0.9:
tr 
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1.8
n
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
3.7.2. The percent overshoot M p:
• Maximum amount the system overshoots its
final value divided by its final value (expressed
in percentage)
• The percent overshoot:

Mp e
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
1 2
 100%, 0    1
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
 The peak time t p :
• The time it takes the system to reach the
maximum overshoot point.
• The peak time:
tp 
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
n 1  
2
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
 Overshoot versus damping ratio for the 2nd
order system.
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
3.7.3. The setting time t s :
• The time it takes the system transient to decay
to a small value so that y(t) is almost in the
steady state.
• The settling time (1%):
ts 
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4.6
 n
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
Graph of regions in the s plane:
(a): rise time (b): overshoot (c): settling time
(d): composite of all three requirements
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
Example:
n  3(rad/s)
  0.6
  1.5(s)
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
3.7.4. Analyzing the system qualities using
simulation software
 impulse.
 step.
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
 Effects of zeros and additional poles.
 For the second order system with no finite
zeros, the transient response parameters are
approximated as follows:
tr 
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1.8
n
5%


0.7

4.6

M p  16%   0.5 t s 
 n

35%   0.3
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3.7. TRANSIENT QUALITIES OF
SYSTEM RESPONSES
 Effects of zeros and additional poles.
 A zero in the LHP will increase overshoot.
 A zero in the RHP will depress the overshoot.
 An additional pole in the LHP will increase the
rise time.
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3.8. STABILITY
 An LTI system is said to be stable if all the
roots of the transfer function denominator
polynomial have negative real parts (that is,
they are all in the left hand s – plane) and is
unstable otherwise.
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3.8. STABILITY
1. Bounded Input Bounded Output Stability.
 If every bounded input results in a bounded
output.
 If input u(t) is bounded, u  M   , the
output with impulse response is BIBO stable if

and only if
 h( ) d  

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3.8. STABILITY
Stability of LTI Systems:
Consider the LTI system whose transfer function
denominator polynomial leads to the
characteristic equation:
s  a1s
n
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n 1
 ...  an  0
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3.8. STABILITY
 The transfer function:
m 1
Y ( s ) b0 s  b1s  ...  bm
T (s) 
 n
n 1
R(s)
s  a1s  ...  an
m
(
s

z
)

i
i 1
K
n
 i 1 ( s  pi )
m
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3.8. STABILITY
 The solution to the differential equation:
n
y (t )   K i e
pi t
i 1
 The system is stable if and only if (necessary
and sufficient condition) every term goes to zero
as t  
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Re  pi   0
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3.8. STABILITY
 If all poles in the LHP, the system is stable.
 If any pole in the RHP, it is unstable.
 If the system has nonrepeated jω axis pole, it
is neutrally stable.
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3.8. STABILITY
2. Routh’s Stability Criterion.
Consider the characteristic equation of an nth
order system.
s  a1s
n
n 1
 ...  an  0
A necessary (but not sufficient) for stability is
that all the coefficients of the characteristic
polynomial be positive.
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3.8. STABILITY
2. Routh’s Stability Criterion.
 A system is stable if and only if all the
elements in the first column of the
Routh array are positive.
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3.8. STABILITY
 Routh array.
• Arrange the coefficients of the characteristic
polynomial in two rows, beginning with the first
and second coefficients and followed by the even
numbered and odd numbered coefficients.
• Compute the elements from the (n-2)th rows as:
cij  ci  2 , j 1   i  ci 1, j 1
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
ci  2 ,1 
 i 



c
i 1,1 

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3.8. STABILITY
Example 1.
- The polynomial:
A( s )  s 6  4 s 5  3s 4  2 s 3  s 2  4 s  4
- There are two poles in the RHP because there
are two sign changes.
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3.8. STABILITY
 Routh’s method is also useful in determining
the range of parameters for which a feedback
system remains stable.
 Example:
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3.8. STABILITY
 The transient response for the system:
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3.8. STABILITY
Example:
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3.8. STABILITY
 Region for stability and the transient response
for the system:
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3.8. STABILITY
3. Analyzing the stability using MATLAB.
 step
 roots
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3.9. SUMMARY
 The Laplace transform.
 The key property of Laplace transform.
 The final value theorem.
 Block diagrams.
 Modeling in state space.
 Effect of poles and zeros .
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3.9. SUMMARY
 Transient qualities of system responses:
rise time, settling time, overshoot, peak time.
 Stability conditions.
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ASSIGNMENT
 Review questions : from 3.1 to 3.12, page 179
 Problems: from 3.1 to 3.61, pages from 179 to
199
 Read [1]: 138-150, [4]: 76-81, [5]: 103-120
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