Lecture #2
Introduction to Systems
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system
A system is an entity that manipulates one or more
signals to accomplish a function, thereby yielding new
signals.
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Example of system
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System interconnection
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System properties
•
•
•
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Causality
Linearity
Time invariance
Invertibility
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Causality
A system is said to be causal if the present value of the
output signal depends only on the present or past values
of the input signal.
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Causal and noncausal system
Example: distinguish between causal and noncausal systems
in the following:
u (t )
1
t
2
(1) Case I y (t )  u (t )
y (t )
when t  1
but
y (t )  0
2
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t
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u (t )  0
Noncausal system
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(2) Case II y (t )  u (t   )
y (t )
Delay system
1  
(3) Case III
t
2 
causal system
y (t )  u (t )  u (t  2)
causal system
At present
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past
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(4) Case IV
y (t )  u (t )  u (t  2)
noncausal system
At present
(5) Case V
future
y(t )  u(t 2 ) if
u(t ) is unit
y (t )
when t  0
but
y (t )  0
t
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u (t )  0
noncausal system
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Linearity
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 )
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ax1 (t )
ay1 (t )
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Example 1.19
x[n ]
y[n]  nx[n]
y[n]
y[n]  nx[n]
let
x[n]  x1[n]  y1[n]  nx1[n]
let
x[n]  x2 [n]  y2 [n]  nx2 [n]
let
x[n]  ax1[n]  bx2 [n]
 y[n]  n{ax1[n]  bx2 [n]}
 anx1[n]  bnx2 [n]
 ay1[n]  by2 [n]
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Example 1.20
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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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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Invertibility
A system is said to be Invertible if the input of the system
can be recovered from the output.
x(t )
y (t )
x(t )
H
Hinv
y (t )  H {x(t )}
x(t )  H { y(t )}
inv
H { y(t )}  H {H {x(t )}}
inv
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inv
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Example 1.15
x(t )
y(t )  x(t  t0 )
y (t )
H  x(t  t0 )
H inv  x(t  t0 )
HH
inv
Example 1.16
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I
Inverse system
x(t )
y (t )  x (t )
2
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y (t )
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LINEAR TIME-INVARIANT (LTI) SYSTEMS:
A basic fact: If we know the response of an LTI
system to some inputs, we actually know the
response to many inputs
System identification
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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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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
4
dy p (t )
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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y p (t )  e2t
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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
dy (0)
 1
dt
y (0)  1 ,
dy (0)
1
dt
yh (0)  y p (0) 1
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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