r(t) u(t) y(t) g(s) r(t) e(t) u(t) y(t) - ut arlington – · pdf filefollows the...
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1
Open loop control vs. closed loop (feedback) control The next two figures present the structures of open loop and feedback control systems. Figure 1 shows an open loop control system whose function is to cause the output ( )y t to follow the reference ( )r t .
Figure 1. Open loop control system
Figure 2 shows a standard feedback control system with the same purpose that the output ( )y t follows the reference ( )r t .
Figure 2. Closed loop (feedback) control
The performances of a control system are prescribed in terms of
- the stability of the system - (range of) steady-state error lim ( ) lim ( ( ) ( ))ss t t
e e t r t y t
= =
- desired maximum allowed percent overshoot POV POV< - desired maximum allowed settling time s st t<
- desired maximum rise time r rt t< In this lecture we are only concerned with the stability and the steady state error performances. Design of controllers which satisfy the rest of the performances will be discussed in future lectures. Open loop control reduces to finding the desired control input trajectory which will produce a desired output of the controlled system. In effect it requires accurate knowledge on the model of the system (at least of the static characteristic of the system). Using as input of the model the desired output of the system, the model will output a signal which, when applied at the systems input, will produce that specified desired output. Take for example the steady state error for an open loop control system when the reference is a unit
step 1( )R ss
= then
0
00
lim ( ) lim ( ( ) ( )) lim ( ( ) ( ))
lim [1 ( ) ( )] ( ) lim[1 ( ) ( )] 1 (0) (0)
ss t t s
ss
e e t r t y t s R s Y s
s K s G s R s K s G s K G
= = = =
= = =
G(s)K(s)u(t) y(t) r(t)
G(s)K(s)u(t) y(t)r(t) e(t)
2
Thus the steady state error in this case is zero only if (0) 1/ (0)K G= (i.e. exact knowledge on the DC gain of the system is available). Lets for a moment consider that we know exactly the value of (0)G (even if this is not generally true). What if an external disturbance signal affects (is added to) the control input (like shown in Figure 3)?
Figure 3. Open loop control system with disturbance
Calculate the value of the steady state error if the disturbance is a unit step. See that in this case the steady state error will never be zero. Feedback control algorithms (or feedback controllers) calculate the control input based on the error differences between the desired system output (i.e. the reference) and the measured present system output in the sense of cancelling the error difference. Can such controllers cancel the steady state error in the presence of uncertainties on the system dynamics and/or in the presence of exogenous disturbances? If so, how can we design such controllers? Steady-state error The steady state error is defined as the value of the error signal ( ) ( ) ( )e t r t y t= as t . The steady state error is shown in the figures for a unit step and a unit ramp reference.
e s s
r(t)
y(t)
0 t
Figure 4 Nonzero steady-state error to a step reference
G(s)K(s)u(t) y(t) r(t)
d(t)
3
r(t)
y(t)
0 t
ess
Figure 5. Nonzero steady-state error to a ramp reference
Feedback can improve the tracking capabilities of a plant by making the steady-state error
smaller, preferably zero.
The closed-loop transfer function is given by (remember Masons formula) ( ) ( ) ( )( )( ) 1 ( ) ( )
Y s K s G sT sR s K s G s
= =+
.
However, to find the steady-state error lim ( )ss t
e e t
= , one determines
1( ) ( ) ( ) (1 ( )) ( ) ( )1 ( ) ( )
E s R s Y s T s R s R sK s G s
= = =+
.
To find the steady-state error in response to a unit step reference, select s
sR 1)( = and use the
Final Value Theorem to obtain
0 0
1 1lim ( ) lim ( ) lim1 ( ) ( )ss t s s
e e t sE s sK s G s s
= = =+
.
Example 1 Given the system in the next figure
G(s)K(s)
d(t)
y(t)r(t) e(t)
Figure 6
where the plant is 1
1)(+
=s
sG and the controller is the integral compensator sksK I=)( with kI the
integral gain, calculate the steady state error in response to a unit step reference and a unit step disturbance.
4
Note that if the feedback loop has a gain different that 1 then the error signal will not show up in the diagram. Thus even if in this case the error can be calculated using Masons formula in general it is better to use the definition on page 2. a. Find the Steady-State Error in Response to a Unit Step Disturbance d(t) To find the requested steady-state error, set R(s)=0 and D(s)= 1/s. The transfer function from d(t) to e(t) is given by
( )( ) ( )1 ( ) ( )
G sE s D sK s G s
=+
)()(
111
11
)( 2 sDkssssD
ssk
ssEII ++
=
++
+
= .
Using now the Final Value Theorem yields
20 0
1lim ( ) lim 0ss s s I
se sE s sss s k
= = = + +
.
Note that this means the output y(t) goes to zero, since the reference is r(t)=0. b. Find the Steady-State Error in Response to a Unit Step Reference r(t) To find the requested steady-state error, set R(s)=1/s and D(s)= 0. The transfer function from r(t) to e(t) is given by
1( ) ( )1 ( ) ( )
E s R sK s G s
=+
)()1()(
111
1)( 2 sRksssssR
ssk
sEII ++
+=
++
=
Using now the Final Value Theorem yields
20 0
( 1) 1lim ( ) lim 0ss s s I
s se sE s sss s k
+= = = + +
.
Note that this means the output y(t) goes to one, since the reference is r(t)=u-1(t). c. Find the POV For a Unit Step Reference. The closed-loop transfer function from the command r(t) to the output y(t) is given by
( ) ( )( ) ( )1 ( ) ( )
G s K sY s R sG s K s
=+
5
)()(
111
11
)( 2 sRkssksR
ssk
ssk
sYI
I
I
I
++=
++
+
= .
The closed-loop characteristic polynomial is
Iksss ++=2)( .
Comparing this to the standard form 22 2)( nn sss ++= one sees that
In
In
k
k
21
21
==
=
.
One can therefore select the integral gain to obtain any desired value of damping ratio, and hence of POV which is given by
%10021/ = ePOV .
d. Find the Output y(t) if )()(),()( 1
21 tuetdtuetr
tt
== . The transfer relation between the two inputs and the output is given by
( ) ( ) ( )( ) ( ) ( )1 ( ) ( ) 1 ( ) ( )
G s K s G sY s R s D sG s K s G s K s
= ++ +
or
)()(
)(
111
11
)(
111
11
)(
22 sDkssssR
kssk
sD
sskssR
ssk
ssk
sY
II
I
II
I
+++
++=
++
++
++
+
=.
Setting now 2
1)(,1
1)(+
=+
=s
sDs
sR one obtains
)2(1
)1(1)( 22 +
++
++
++
=skss
sskss
ksYII
I .
For the given value of kI one may now use the inverse Laplace transform to determine the output y(t).
6
Feedback control system design for zero steady state error
In general, to follow a reference R(s) with zero steady-state error, the path between the reference and the system output should contain a term like R(s). For instance, to follow a ramp (velocity) reference 2/1)( ssR = one requires at least two integrators in the path from the reference to the output. That is, for zero steady-state error in response to a given reference, the control system should contain a model of the desired reference trajectory.
Also, to reject the steady state component of a disturbance of nonzero mean the path between the reference and the disturbance input should contain a term like D(s).
These two rules express the internal model principle. Problem 1. Consider the system in the figure
G(s)K(s)
d(t)
y(t)r(t) e(t)
Figure 7.
Let the plant be given in the form
0 1
1
... ( )( )( )1 ...
mm
nn
b b s b s n sG sd sa s a s
+ + += =
+ + +
The compensator is given as ( ) NNkK ss
= , ( )( ) ( )( )
NN
kn sG s K sd s s
= , and assume that the plant does not have
any poles in zero.
A. Let ( ) 0D s = and 1( ) rR s s= , 0r .
What is the minimum value of N such that the steady state error is zero?
B. Let ( ) 0R s = and 1( ) dD s s= , 0d .
What is the minimum value of N such that the steady state error is zero?
C. Repeat points A and B considering that ( ) 1( )( ) g
n sG sd s s
= , 0g .
The advantages of feedback control
- a well designed controller can cancel the steady state error such that the output of the system will follow, after the transient response is finished, the desired specified output i.e. the reference, even in the presence of uncertainties relative to the model of the system or the appearance of exogenous disturbances
Some costs of feedback control
- increased complexity of the control system: sensors are required and sensors have their own dynamics; moreover one has to deal with measurement noise
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- in some cases feedback can make the system unstable for specific controller structure and certain values of the controller parameters