Control system topics, electronics and communication engineering

Control system topics

1. 5.4 Disturbance rejection
The input to the plant we manipulated is m(t).
Plant also receives disturbance input that we
do not control.
The plant then can be modeled as follow
C(s)= G
p
(s)M(s)+G
d
(s)D(s)
The control system should minimize G
d
(s)D(s).
D
HGG
G
R
HGG
GG
C
pc
d
pc
pc




11
C=T(s)R(s)+T
d
(s)R(s)
We want T
d
(s) as small as possible.
Let us use frequency approach.
T
d
(j) can be made small for specific
frequency.
Recall that G
c
(j) G
p
(j) H(j) must be
large to reduce the sensitivity to plant
variation.
)()()(1
)(
)(



jHjGjG
jG
jT
pc
d
d
)()()(
)(



jHjGjG
jG
pc
d
Methods to reduce T
d
(j)
1.make G
d(s) small
2.increase loop gain by increasing G
c
3.reduced D(s)
4.use feed forward compensation
D(s)
M(s) +
C(s)
G
p
(s)
G
d(s)
+
plant
G
c
+
–
H
G
d
(s)D(s)
R(s)

2. 5.4 Disturbance rejection
Feedforward compensation
Feedforward compensation can be applied if
the disturbance can be measured.
The addition of compensator G
cd
(s) does not
effect T(s) the TF from R(s) to C(s) but does for
T
d(s)
)()()(1
)()()()(
)(
sHsGsG
sGsGsGsG
sT
pc
pccdd
d



C(s)
D(s)
M(s) +
G
p
(s)
G
d(s)
+
plant
G
c
+
–
H
G
d
(s)D(s)
G
cd
(s)
–
R(s)
We choose G
cd
(s)G
c
(s) such that T
d
(s) will
as small as possible.

3. 5.5 Steady State Accuracy
We will examine the steady state error e
ss for
different system types. Assume H=1. hence
)(
)()(1
)()(
)( sR
sGsG
sGsG
sC
pc
pc


Step input, R(s) = 1/s hence
C(s)M(s)
G
p
(s)G
c
+
–
R(s)
E(s)
The steady state error e
ss
is defined as
pc
sst
ss
GG
sR
ssEtee


 1
lim)(lim)(lim
00
ppc
s
pc
s
s
ss
KGGGG
s
e







 1
1
lim1
1
1
lim
0
1
0
Ramp input, R(s) = 1/s
2
hence
vpc
s
pc
s
ss
KGsGGsGs
e
1
lim
11
lim
0
0





We can express
)(
)(
)()(
1sQs
sF
sGsG
Npc 
N is integer and the system is called system
type N.
N is also the number of integrator in G
c
G
p
we will demonstrate its importance
For N = 0 then K
p
= K
p
and e
ss
=1/(1+ K
p
)
For N > 0 then K
p
=  and e
ss
= 0.
For N = 0 then K
v = 0 and e
ss = 
For N = 1 then K
v = K
v  and e
ss = 1/ K
v
For N > 1 then K
v =  and e
ss = 0

4. 5.5 Steady State Accuracy
Parabolic input, R(s) = 1/s
3
hence
apc
s
pc
s
ss
KGGsGGss
e
1
lim
11
lim
2
0
22
0





For N < 2 then K
v
= 0 and e
ss
= 
For N = 2 then K
a
= K
a
 and e
ss
= 1/ K
a

For N > 2 then K
v
=  and e
ss
= 0
The steady state error e
ss
system
type
input
N 1/s 1/s
2
1/s
3
0 1/(1+K
P)  
1 0 1/K
v
 e
ss
2 0 0 1/K
a
K
p is called position error constant
K
v
is called velocity error constant
K
v is called acceleration error constant
Plot of ramp responses of different type system
type 0 system
type 1 system
type 2 system
e
ss
e
ss

5. 5.5 Steady State Accuracy
Non-unity-Gain feedback
The non-unity gain feedback above has
equivalent block diagram as follow provided
that H(s) is constant.
+
–
)(sG
p
H
R(s) )(sG
c
+
–
)()( sGsHG
pc
R
u(s) E(s) C(s)
Where R
u
(s) = R(s)/H.
Hence the position, velocity, and acceleration
error constant can be calculated in the same manner
HGGK
pc
s
p
0
lim


HGsGK
pc
s
v
0
lim


HGGsK
pc
s
a
2
0
lim



6. Disturbance Input Error
Let us consider steady state error due to
disturbance input.
Recall that the output of disturbed plant
C(s) =T(s)R(s)+T
d(s)C(s)=C
r(s)+C
d(s)
The system error is then
e(t)= r(t) – c(t) =[r(t) – c
r
(t)] – c
d
(t)
= e
r(t) + e
d(t)
where e
r
(t) is the error we’ve just considered
We see that e
d(t)= – c
d(t) and we want it to be
small. The steady state of e
d(t) is
where (assuming H = 1)
Let us model the disturbance input as
step function, D(s)=B/s. From (1) we have
)()(lim
0
sDssTe
d
s
dss

 (1)
To draw general conclusion we must consider
the type number of G
p
(s) G
c
(s) and G
d
(s).
pc
d
pc
d
d
GG
G
HGG
G
sT




11
)(
(2)
)()(1
)(
lim)(lim
00 sGsG
BsG
BssTe
pc
d
s
d
s
dss



(3)
Type number of
InputG
p(s)G
d(s)G
c(s)e
dss
Unit step0 0 0Finite
Unit step0 0 1 0
ramp 0 0 0 
ramp 0 0 2 0

7. TRANSIENT RESPONSE
The system TF can be written as
)(
)(
)(
)(
)(
1
i
n
i
n
psa
sP
sQ
sP
sT



The response for input R(s) is
)(
)(
)(
)()()(
1
sR
psa
sP
sRsTsC
i
n
i
n 


)(
1
1
1
1
1
1
sC
ps
k
ps
k
ps
k
r






 
Where C
r(s) is the part that
originate from the poles of R(s)
The inverse LT of this equation is
)()(
1
1
tcekektc
r
tp
n
tp
n
 
Here c
r(t) is the forced response
and the rest is the natural response
or the transient response
For stable system the natural
response will decay to zero. The
way it decays to zero is important.
The transient response c
tr(t) is
tp
n
tp
tr
n
ekektc  
1
1
)(
The factor
tp
i
eis called the modes
The nature or form of each term
is determined by the pole
location of p
i
.

8. TRANSIENT RESPONSE
The amplitude of each term is
j
ps
i
n
ji
i
n
j sR
psa
sP
k




 )(
)(
)(
1
Thus the amplitude of each term
determined by other pole location,
the numerator polynomial, and the
input function.
For higher order system it is
difficult to make general assertion,
however we can consider the nature
of each term of this response.
there is associated damping ratio ,
a natural frequency 
n
, and time
constant 
i =1/
n. If there is one
real dominant pole the system
responds essentially as 1
st
order
system. If there is dominant
complex poles then the system
respond essentially as 2
nd
order
system.
iipp
τ
11



For each real pole, a time constant
is associated with it with the value
2
ii ζ1ζ 
niip
For each set complex conjugate
poles of values

9. CLOSED LOOP FREQUENCY RESPONSE
The input output characteristics are determined by CL frequency response
T(0) is the system dc gain. This values determines the steady state error.
If we evaluate (1) for small  we see how the system will follow for slow varying
input. The bandwidth can be evaluated form (1). Large bandwidth indicate fast
system response.
The presence of any peaks, which denotes resonance, give indication of the
overshoot or decaying oscillation
)()()(1
)()(
)(



HGG
GG
T
pc
pc
(1).
|T()|


B