Lecture 01 --Introduction
1.1 Brief History
1.2 Steps to study a control system
1.3 System classification
1.4 System response
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Brief history of automatic control (I)
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1868
1877
1892
1895
1932
1945
1947
1948
1949
1955
1956
First article of control ‘on governor’s’ –by Maxwell
Routh stability criterion
Liapunov stability condition
Hurwitz stability condition
Nyquist
Bode
Nichols
Root locus
Wiener optimal control research
Kalman filter and controlbility observability analysis
Artificial Intelligence
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Brief history of automatic control (II)
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1957 Bellman optimal and adaptive control
1962 Pontryagin optimal control
1965 Fuzzy set
1972 Vidyasagar multi-variable optimal control and
Robust control
• 1981 Doyle Robust control theory
• 1990 Neuro-Fuzzy
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Three eras of control
• Classical control : 1950 before
– Transfer function based methods
• Time-domain design & analysis
• Frequency-domain design & analysis
• Modern control : 1950~1960
– State-space-based methods
• Optimal control
• Adaptive control
• Post modern control : 1980 after
– H∞ control
– Robust control (uncertain system)
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Control system analysis and design
• Step1: Modeling
– By physical laws
– By identification methods
• Step2: Analysis
– Stability, controllability and observability
• Step3: Control law design
– Classical, modern and post-modern control
• Step4: Analysis
• Step5: Simulation
– Matlab, Fortran, simulink etc….
• Step6: Implement
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Signals & systems
Output signals
Input signals
Time system
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Signal Classification
• Continuous signal
• Discrete signal
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System classification
• Finite-dimensional system (lumped-parameters system
described by differential equations)
– Linear systems and nonlinear systems
– Continuous time and discrete time systems
– Time-invariant and time varying systems
• Infinite-dimensional system (distributed parameters
system described by partial differential equations)
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Power transmission line
Antennas
Heat conduction
Optical fiber etc….
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Some examples of linear system
• Electrical circuits with constant values of circuit
passive elements
• Linear OPA circuits
• Mechanical system with constant values of k,m,b
etc
• Heartbeat dynamic
• Eye movement
• Commercial aircraft
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Linear system
A system is said to be linear in terms of the system input
x(t) and the system output y(t) if it satisfies the following
two properties of superposition and homogeneity.
Superposition：
y1 (t )
x1 (t )
x1 (t )  x2 (t )
y2 (t )
x2 (t )
y1 (t )  y2 (t )
Homogeneity：
x1 (t )
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y1 (t )
ax1 (t )
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ay1 (t )
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Example
x(t )
let
y (t )  x(t ) x(t  1)
y (t )
x(t )  x1 (t )
y1 (t )  x1 (t ) x1 (t  1)
let
x(t )  ax1 (t )
y (t )  ax1 (t )ax1 (t  1)  a 2 x1 (t ) x1 (t  1)  a 2 y1 (t )
y(t )  ay1 (t )
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Non linear system
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example
The system is governed by a linear ordinary differential equation (ODE)
y(t )  2 y(t )  y (t )  x(t )  3x(t )
x(t )
Linear time
invariant system
y (t )
y1(t )  2 y1 (t )  y1 (t )  x1 (t )  3x1 (t )
y2 (t )  2 y2 (t )  y2 (t )  x2 (t )  3x2 (t )
[ax1 (t )  bx2 (t )]  3[ax1 (t )  bx2 (t )]  ax1 (t )  bx2 (t )  a3 x1 (t )  b3x2 (t )
 a[ x1 (t )  3 x1 (t )]  b[ x2 (t )  3 x2 (t )]
 a[ y1(t )  2 y1 (t )  y1 (t )]  b[ y2 (t )  2 y2 (t )  y2 (t )]
 [ay1 (t )  by2 (t )]  2[ay1 (t )  by2 (t )]  [ay1 (t )  by2 (t )]
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linearity
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Examples : Circuit system
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Examples Discrete system
Time delay
u[k ]
u[ k  1]
A
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
y[k ]

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Properties of linear system :
(1)
(2)
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Time invariance
A system is said to be time invariant if a time delay or
time advance of the input signal leads to an identical time
shift in the output signal.
x(t )
y (t )
Time invariant
system
x(t  t0 )
y (t  t0 )
t0
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t0
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Example 1.18
x(t )
x(t )
y (t ) 
R(t )
y (t )
x1 (t )
y1 (t ) 
R(t )
x2 (t )  x1 (t  t0 )
x2 (t ) x1 (t  t0 )
 y2 (t ) 

R(t )
R(t )
x1 (t  t0 )
but y1 (t ) 
R(t  t0 )
y1 (t  t0 )  y2 (t ),
for t0  0
Time varying system
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di (t ) dL
vi (t )  L
i
 Ri (t )
dt
dt
di (t )
vi (t )  L0 (1  cos t )
 (L0 sin t )i (t )  Ri (t )
dt
di (t )
vi (t )  L0 (1  cos t )
 ( R  L0 sin t )i (t )
dt
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LTI System representations
Continuous-time LTI system
1. Order-N Ordinary Differential equation
2. Transfer function (Laplace transform)
3. State equation (Finite order-1 differential equations) )
Discrete-time LTI system
1. Ordinary Difference equation
2. Transfer function (Z transform)
3. State equation (Finite order-1 difference equations)
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Continuous-time LTI system
d 2 y(t )
dy (t )
LC
 RC
 y (t )  u (t )
2
dt
dt
Order-2 ordinary differential equation
constants
LCs 2Y ( s )  RCsY ( s )  Y ( s )  U ( s ) Linear system  initial rest
Y (s)
1

Transfer function
2
U ( s ) LCs  RCs  1
U (s )
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LCs 2  RCs  1
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Y (s )
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let
x1 (t )  y (t )
dy (t )
x2 (t ) 
dt
x1 (t )  x2 (t )
R
1
x2 (t )   x2 (t ) 
x1 (t )  u (t )
L
LC
 x1 (t )   0
 x (t )   1
 2   LC
u (t )
x (t )

1   x1 (t )  0
  u (t )

R 
 L   x2 (t ) 1
x(t )
A
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System response: Output signals due to inputs and ICs.
1. The point of view of Mathematic:
Homogenous solution y h (t ) ＋
Particular solution y p (t )
2. The point of view of Engineer:
Natural response y n (t )
＋
Forced response
y f (t )
3. The point of view of control engineer:
Zero-input response y zi (t ) ＋
Transient response
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Zero-state response y zs (t )
Steady state response
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Example: solve the following O.D.E
d 2 y (t )
dy (t )
 2t

4

3
y
(
t
)

e
, t  0,
2
dt
dt
y (0)  1,
dy (0)
1
dt
(1) Particular solution: [ y p (t )]  u (t )
d 2 y p (t )
dt 2
dy p (t )
4
dt
 3 y p (t )  e  2t
y p (t )  e2t
let
then
y ' p (t )  2e2t
yp (t )  4e2t
4e2t  4(2)e2t  3e2t  e2t    1
we have
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(2) Homogenous solution:
[ yh (t )]  0
yh(t )  4 yh (t )  3 yh (t )  0
t
yh (t )  Ae  Be
3t
y (t )  y p (t )  yh (t ) have to satisfy I.C.
y (0)  1
y (0)  1 ,
dy (0)
1
dt
yh (0)  y p (0) 1
dy (0)
 1
dt
yh (0)  yp (0)  1
5  t 1  3t
yh (t )  e  e
2
2
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(3) zero-input response: consider the original differential equation with no input.
y zi (t )  4 y zi (t )  3 y zi (t )  0,
t0
y zi (0)  1,
y zi (0)  1
y zi (t )  K1e t  K 2 e 3t , t  0
y zi (0)  K1  K 2
y zi (0)   K1  3K 2
K1  2
K 2  1
y zi (t )  2e  t  e 3t , t  0
zero-input response
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(4) zero-state response: consider the original differential equation but set all I.C.=0.
y zs (t )  4 y zs (t )  3 y zs (t )  e 2t ,
t0
y zi (0)  0 ,
y zi (0)  0
y zs (t )  C1e t  C 2 e 3t  e 2t
y zs (0)  C1  C 2  1  0
y zs (0)  C1  3C 2  2  0
1
2
1
C2 
2
C1 
1  t 1  3t
y zs (t )  e  e  e  2t
2
2
zero-state response
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(5) Laplace Method:
d 2 y (t )
dy (t )
 2t

4

3
y
(
t
)

e
, t  0,
2
dt
dt
y (0)  1,
dy (0)
1
dt
1
s Y ( s )  sy (0)  y (0)  4sY ( s )  4 y (0)  3Y ( s ) 
s2
2
1
1
5
s5

1
s

2
2
Y ( s)  2


 2
s  3 s  2 s 1
s  4s  3
 1  3t
5 t
 2t
y (t )   [Y ( s )] 
e e  e
2
2
1
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Complex response
Zero state response
y zs (t ) 
1  t 1  3t
e  e  e  2t
2
2
Forced response
(Particular solution)
 1  3t
5
e  e  2 t  e t
2
2
Zero input response
y zi (t )  2e  t  e 3t , t  0
Natural response
(Homogeneous solution)
y p (t )  e2t
Steady state response
y (t ) 
yh (t ) 
5  t 1  3t
e  e
2
2
Transient response
 1  3t
5 t
 2t
y (t ) 
e e  e
2
2
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