Transcript Unit 70520 Review - University of Southern Queensland

3. Systems and Transfer function
Discrete-time system revision
• Discrete-time system
• A/D and D/A converters
• Sampling frequency and sampling theorem
• Nyquist frequency
• Aliasings
• Z-transform & inverse Z-transform
• The output of a D/A converter
3.1 Zero-order-hold (ZOH)
A Zero-order hold in a system
x(t)
x(kT)
Sampler
Zero-order h(t)
Hold
3.1 Zero-order-hold (ZOH)
How does a signal change its form in a discrete-time
system?
The input signal x(t) is sampled at discrete instants
and the sampled signal is passed through the zeroorder-hold (ZOH). The ZOH circuit smoothes the
sampled signal to produce the signal h(t), which is
a constant from the last sampled value until the
next sample is available. That is
h(kT  t )  x(kT ), for 0  t  T
3.1 Zero-order-hold (ZOH)
Transfer function of Zero-order-hold
The figure below shows a combination of a
sampler and a zero-order hold.
x(t)
x(kT)
Sampler
Zero-order h(t)
Hold
3.1 Zero-order-hold (ZOH)
Assume that the signal x(t) is zero for t<0,
then the output h(t)is related to x(t) as
follows:
h(t)
t
h(t )  x(0)[1(t )  1(t  T )]  x(T )[1(t  T )  1(t  2T )]  

  x(kT )[1(t- kT )- 1(t - (k  1)T )]
k 0
3.1 Zero-order-hold (ZOH)
As
 kTs
e
L[ f (t  kT )]  F ( s)e kTs , L[1(t  kT )] 
s
L[ (t )]  1, L (t  kT )  e kTs
The Laplace transform of the above equation
becomes

L[h(t )]  L[ x(kT )[1(t- kT )- 1(t - (k  1)T )]]
k 0
e  kTs e ( k 1)Ts
  x(kT)L[1(t - kT )- 1(t - (k  1)T )]  x(kT)[

]
s
s
k 0
k 0



e  kTs  e ( k 1)Ts
1  e Ts  kTs 1  e Ts
  x(kT)
  x(kT)
e

s
s
s
k 0
k 0


 kTs
x(kT)e

k 0
3.1 Zero-order-hold (ZOH)

 
X (s)  L x(kT ) (t  kT )   x(kT )ekTs
 k 0
 k 0
As
Therefore
1  e Ts
H (s)  L[h(t )] 
s

 x(kT)e
k 0
 kTs
1  e Ts

X ( s)
s
Finally, we obtain the transfer function of a ZOH as
H (s) 1  e Ts
Gh (s)  G0 (s) 

X ( s)
s
3.1 Zero-order-hold (ZOH)
There are also first-order-hold and high-order-hold
although they are not used in control system.
3.1 Zero-order-hold (ZOH)
A zero-order-hold creates one sampling interval
delay in input signal.
3.1 Zero-order-hold (ZOH)
First-order-hold
1 e
Gh1 ( s)  
 s
Ts
2
 Ts  1

 T
3.1 Zero-order-hold (ZOH)
First-order-hold and high-order-hold does not bring
us much advantages except in some special cases.
Therefore, in a control system, usually a ZOH is
employed. The device to implement a ZOH is a
D/A converter.
If not told, always suppose there is a ZOH in a
digital control system.
3.2 Plants with ZOH
Given a discrete-time system, the transfer function of
a combination of a ZOH and the plant can be
written as GHP(z) in Z-domain. HP, here, means
the ZOH and the Plant.
GHP(z)
ZOH
GP(s)
3.2 Plants with ZOH
The continuous time transfer function
GHP(s)=G0(s)GP(s)
The discrete time transfer function

 G P ( s) 
  sT G P ( s ) 
 sT G P ( s ) 
G HP ( z )  Z G HP ( s )  Z (1  e )
  Z
  Z e

s 
s 

 s 

 G P ( s )  1  G P ( s ) 
 G P (s) 
1
 Z
  z Z
  (1  z ) Z 

 s 
 s 
 s 
 G P (s) 
sT
G HP ( z )  (1  z ) Z 
,
z

e

s


1
3.2 Plants with ZOH
1
G P (s) 
s 1
Example 1: Given a ZOH and a plant
Determine their Z-domain transfer function.
 1 
 G (s) 
G HP ( z )  (1  z 1 ) Z  P   (1  z 1 ) Z 

 s 
 s ( s  1) 
1  1 
  1 
1   1 
 (1  z ) Z  
  (1  z ) Z    Z 
 
 s s  1
 s  1 
 s
1
1
 1

 (1  z )

1
T 1 
1 e z 
1 z
T 1
1
T
1
1

e
z

1

z
(
1

e
)
z
 (1  z 1 )

1
T 1
(1  z )(1  e z )
1  e T z 1
1
3.2 Plants with ZOH
1
Example 2: Given a ZOH and a plant GP ( s) 
s ( s  1)
Determine their z-domain transfer function.
 1 
 GP ( s ) 
1
GHP ( z )  (1  z ) Z 
  (1  z ) Z  2

 s 
 s ( s  1) 
1
1 
1 1
1 
 1 
1   1 
 (1  z ) Z  2  
  (1  z ) Z  2   Z    Z 
 
s s  1
s
s
 s  1 
 s 
1
1


Tz
1
1
1

 (1  z )


1 2
1
T 1 
1 e z 
 (1  z ) 1  z
(T  1  e T ) z 1  (1  e T  Te T ) z  2

(1  z 1 )(1  e T z 1 )
3.2 Plants with ZOH
ab
Exercise 1: Given a ZOH and a plant GP (s) 
( s  a)(s  b)
Determine their z-domain transfer function.
Answer:
1  z 1  a  b
b
a

GHP ( z ) 



1
 aT 1
 aT 1 
a  b 1 z
1 e z
1 e z 
Assignment 1
You are required to implement a digital PID
controller which will enable a control object with
a transfer function of
Kn2
G( s) 
s 2  2 n s  n2
where K=0.2, n=10 rad/s, and =0.3.
to track a) a unit step signal, and b) a unit ramp
signal.
1) Simulate this control object and find the responses using
Matlab or other packages/computer languages.
Assignment 1
2) Choose a suitable sample period for a control loop for G(s)
and explain your choice.
3)* Derive the discrete-time system transfer function GHP(z)
from G(s).
4) Design a digital PID controller for the discrete-time
system, and optimize its parameters with respect to the
performance criterion below
using steepest descent
M
minimization process S   k .ek
k 0
5) Simulate the resulting closed-loop system and find the
responses. Swapping the input signals a) and b), discuss
the resulting responses.
3.3 Represent a system in difference
equation
1
G P (s) 
s 1
(1  e T ) z 1
GHP ( z ) 
1  e T z 1
For
we have
Let A=1-e-T and B=e-T, then the transfer function can
be rewritten as
(1  e T ) z 1
Az 1
Y ( z)
GHP ( z ) 


T 1
1
1 e z
1  Bz
X ( z)
Y ( z )(1  Bz 1 )  AX ( z )  Y ( z )  BY ( z ) z 1  AX ( z ) z 1
 y (k )  By(k  1)  Ax(k  1)
 y (k )  Ax(k  1)  By(k  1)
3.3 Represent a system in difference
equation
Simulate the above system
1) Parameters and input: A=1-e-T, B=e-T , x(k)=1
2) initial condition: x(k-1)=0, y(k)=y(k-1)=0, k=0
3) Simulation
While k<100 do
y(k)=Ax(k-1)+By(k-1); Calculate output
x(k-1)=x(k); y(k-1)=y(k); x(k)=1; k=k+1; Update data
print k, x(k), y(k);
Display step, input & output
End
3.3 Represent a system in difference
equation
Let T=1, we have A=0.6321 and B=0.3679
For a unit step input, the response is
y(k)=0.6321x(k-1)+0.3679y(k-1)
k= 0
1
2
3
4
5
x(k) 1
1
1
1
1
1
y(k) 0
0.6321
1
1
1
1
3.3 Represent a system in difference
equation
Step Response
1
0.9
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Time (sec.)
4
5
6
Assignment 1
1)* Simulate this control object and find the responses using
Matlab or other packages/computer languages.
Kn2
G( s)  2
s  2 n s  n2
Hints: Method 1
K n2
Y ( s)
G(s)  2

2
s  2 n s   n X ( s )
K n2
Y ( s)  G ( s) X ( s)  2
X ( s)
2
s  2 n s   n
2


K

1
1
n
y(t)  L [G ( s ) X ( s )]  L  2
X ( s)
2
 s  2 n s   n

Assignment 1
Hints: Method 2
K n2
Y (s)
Y ( z)
G (s)  2

,
G
(
z
)

Z
[
G
(
s
)]

s  2 n s   n2 X ( s )
X ( z)

 b0  b1 z 1  b2 z  2
K n2
Y ( z)
 Z 2

2
1
2
X ( z)
s

2

s


1

a
z

a
z
n
n 
1
2

Y ( z )(1  a1 z 1  a2 z  2 )  X ( z )(b0  b1 z 1  b2 z  2 )
Y ( z )  X ( z )(b0  b1 z 1  b2 z  2 )  Y ( z )(a1 z 1  a2 z  2 )

y(kT)  L1[Y ( z )]  L1 X ( z )(b0  b1 z 1  b2 z  2 )  Y ( z )(a1 z 1  a2 z  2 )
k  0,1,2 

3.4 System stability
We can rewrite the difference equation as
y (k )  Ax(k  1)  By(k  1)  A( x(k  1) 
 A( x(k  1)  y (k  1)),   B A
B
y (k  1))
A
If A=1 and =0.9, for an impulse input we have
y(k )  x(k  1)  0.9 y(k  1))
k
0
1
2
3
4
...
x(k) 1
0
0
0
0
...
y(k) 0
1
0.9 0.81 0.729 …
It decreases exponentially, a stable system.
3.4 System stability
If K=1 and =1.2, we have
y(k )  x(k  1)  1.2 y(k  1))
k
0
1
2
3
4
...
x(k) 1
0
0
0
0
...
y(k) 0
1 1.2 1.44 1.728 2.074…
It increases exponentially, an unstable system.
3.4 System stability
If K=1 and = -0.8, we have
y(k )  x(k  1)  0.8 y(k  1))
k
0
1
2
3
4
...
x(k) 1
0
0
0
0
...
y(k) 0
1
-0.8 0.64 -0.512
…
It decays exponentially, and alternates in sign,
a gradual stable system.
3.4 System stability
It is clear that the value of  determines the system
stability. Why  is so important?
First, let A=1, we have
1
1
y(k )  x(k  1)  y(k  1)  Y ( z )  X ( z ) z  Y ( z ) z
Y ( z)
z 1
1
G( z ) 


1
X ( z ) 1  z
z 
From the transfer function, we can see that z= is a
pole of the system. The pole of the system will
determine the nature of the response.
3.4 System stability
For continuous system, we have stable,
critical stable and unstable areas in s
domain.
Critical stable area
Stable area
Unstable area
3.4 System stability
What is the stable area, critical stable area and
unstable area for a discrete system in Z
domain ?
Stable area:
unit circle
Critical stable: on the unit circle
Unstable area: outside of the unit circle
3.4 System stability
 GP ( s ) 
sT
GHP ( z)  (1  z )Z 
,
z

e

 s 
1
As
For the critical stable area in s domain s=j,
z  e sT  e j  cos  j sin 
 (cos ) 2  (sin  ) 2   1
As  is from 0 to , then the angle will be
greater than 2. That is the critical area
forms a unit circle in Z domain.
3.4 System stability
If we choose a point from the stable area at S
domain, eg s=- a + j, we have
z  e sT  e  a  j  e  a (cos  j sin  )
 e a (cos ) 2  (sin  ) 2   e a 
Let eg s=-  + j
z  e sT  e   j  e  (cos  j sin  )
 e  (cos ) 2  (sin  ) 2   e    0
The stable area in Z domain is within a unit circle
around the origin.
3.4 System stability
Exercise 2: Prove that the unstable area in Z
domain is the area outside the unit circle.
Hint: Follow the above procedures.
3.4 System stability
Z domain responses
1
0
3.5 Closed-loop transfer function
Computer controlled system
GHP(z)
R(z)
M(z)
E(z)
Gc(z)
Computer system
C(z)
ZOH
GP(s)
Plant
3.5 Closed-loop transfer function
Let’s find out the closed-loop transfer function
C ( z )  M ( z )G HP ( z )  E ( z )GC ( z )G HP ( z )
E ( z )  R( z )  C ( z )
C ( z )  ( R ( z )  C ( z ))GC ( z )G HP ( z )
C ( z )  C ( z )GC ( z )G HP ( z )  R ( z )GC ( z )G HP ( z )
GC ( z )G HP ( z )
C ( z)
T ( z) 

R ( z ) 1  GC ( z )G HP ( z )
3.5 Closed-loop transfer function
C(z): output;
E(z): error
R(z): input;
M(z): controller output
GC(z): controller
GP(z)/G(z): plant transfer function
GHP(z): transfer function of plant + ZOH
T(z): closed-loop transfer function
GC(z)GHP(z): open-loop transfer function
1+ GC(z)GHP(z)=0: characteristic equation
3.6 System block diagram
C(z)
+
R(s)
G(s)
C ( z)
G( z )

R( z ) 1  G ( z ) H ( z )
H(s)
C(s)
+
R(s)
G(s)
H(s)
C(z)
C ( z)
G( z )

R( z ) 1  GH ( z )
3.6 System block diagram
The difference between G(z)H(z) and GH(z)
G(z)H(z)=Z[G(s)]Z[H(s)]
GH(z)=Z[G(s)H(s)]
Usually, G(z)H(z)  GH(z)
G(z)H(z) means they are connected through a
sampler. Whereas GH(z) they are connected
directly.
3.6 System block diagram
Example: Find the closed-loop transfer
function for the system below.
C(z)
+
R(s)
G1(s)
G2(s)
H(s)
Solution: The open-loop is G1(z)G2H(z).
The forward path is G1(z)G2(z).
3.6 System block diagram
C(z)
+
R(s)
G1(s)
G2(s)
H(s)
E ( z )  R( z )  E ( z )G1 ( z )G2 H ( z ); R( z )  E ( z )  E ( z )G1 ( z )G2 H ( z )
E ( z )  R( z ) /(1  G1 ( z )G2 H ( z ))
C ( z )  E ( z )G1 ( z )G2 ( z )  G1 ( z )G2 ( z ) R( z ) /(1  G1 ( z )G2 H ( z ))
C ( z ) R( z )G1 ( z )G2 ( z )
T ( z) 

R( z ) 1  G1 ( z )G2 H ( z )
3.6 System block diagram
*Exercise 3: Find the output for the closedloop system below.
C(s)
+
R(s)
G(s)
H(s)
C(z)
GR( z )
C ( z) 
1  GH ( z )
3.6 System block diagram
*Exercise 4: Find the output for the closedloop system below.
C(z)
+
R(s)
G1(s)
H(s)
RG1 ( z)G2 ( z)
C( z) 
1  G1G2 H ( z)
G2(s)
Reading
Study book
• Module 3: Systems and transfer functions
(Please try the problems on page 3.46-47)
Textbook
• Chapter 3 : Z-plane analysis of discretetime control system (pages 74-83 & 104114).
Tutorial
ab
Exercise 1: Given a ZOH and a plant GP (s) 
( s  a)(s  b)
Determine their z-domain transfer function.


ab
 G ( s) 
GHP ( z )  (1  z 1 ) Z  P   (1  z 1 ) Z 

s
s
(
s

a
)(
s

b
)




k 
k
k
 1 b (a  b) a (b  a) 
 (1  z 1 ) Z  1  2  3   (1  z 1 ) Z  


sa
sb 
s
 s s  a s  b
 1 
b
a
 1 
 1 
 (1  z 1 ) Z   
Z
Z

 
s
a

b
s

a
b

a
s

b




  
b
1
a
1
 1

 (1  z 1 )


1
 aT 1
bT 1 
1

z
a

b
1

e
z
b

a
1

e
z 

(1  z 1 )  a  b
b
a






a  b  1  z 1 1  e  aT z 1 1  e  aT z 1 
Tutorial
You are required to implement a digital PID
controller which will enable a control object with
a transfer function of
Kn2
G( s) 
s 2  2 n s  n2
where K=0.2, n=10 rad/s, and =0.3.
to track a) a unit step signal, and b) a unit ramp
signal.
1) Simulate this control object and find the responses using
Matlab or other packages/computer languages.
Tutorial
2) Choose a suitable sample period for a control loop for G(s)
and explain your choice.
3) Derive the discrete-time system transfer function GHP(z)
from G(s).
4) Design a digital PID controller for the discrete-time
system, and optimize its parameters with respect to the
performance criterion below
using steepest descent
M
minimization process S   k .ek
k 0
5) Simulate the resulting closed-loop system and find the
responses. Swapping the input signals a) and b), discuss
the resulting responses.
Tutorial
2) Choose a suitable sample period for a control loop
for G(s) and explain your choice.
a.
b.
c.
d.
e.
f.
Sampling theorem
Input signal
Bandwidth of a system
Bold plots
Applying sampling theorem
Sampling frequency