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 ) .
r(t)
u(t)
K(s)
y(t)
G(s)
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 ) .
r(t)
e(t)
u(t)
K(s)
G(s)
y(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 ess = lim e(t ) = lim (r (t ) - y (t ))
t ®¥
t ®¥
-
desired maximum allowed percent overshoot POV < POV
desired maximum allowed settling time ts < ts
-
desired maximum rise time tr < tr
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 system’s 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
1
step R( s ) = then
s
ess = lim e(t ) = lim (r (t ) - y (t )) = lim s ( R( s ) - Y ( s )) =
t ®¥
t ®¥
s ®0
= lim s[1 - K ( s )G ( s )]R( s ) = lim[1 - K ( s )G ( s )] = 1 - K (0)G (0)
s ®0
s ®0
1
Thus the steady state error in this case is zero only if K (0) = 1/ G (0) (i.e. exact knowledge on the DC
gain of the system is available).
Let’s for a moment consider that we know exactly the value of G (0) (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)?
d(t)
r(t)
K(s)
u(t)
y(t)
G(s)
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.
es s
y(t)
r(t)
0
t
Figure 4 Nonzero steady-state error to a step reference
2
ess
y(t)
r(t)
0
t
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 Mason’s formula)
Y ( s)
K ( s )G ( s )
T (s) =
=
.
R( s ) 1 + K ( s )G ( s )
However, to find the steady-state error ess = lim e(t ) , one determines
t ®¥
E ( s ) = R( s ) - Y ( s ) = (1 - T ( s )) R( s ) =
1
R( s) .
1 + K ( s )G ( s )
To find the steady-state error in response to a unit step reference, select R( s ) =
1
and use the
s
Final Value Theorem to obtain
ess = lim e(t ) = lim sE ( s ) = lim s
t ®¥
s ®0
s ®0
1
1
.
1 + K ( s )G ( s ) s
Example 1
Given the system in the next figure
d(t)
r(t)
e(t)
y(t)
K(s)
G(s)
Figure 6
1
k
and the controller is the integral compensator K ( s ) = I with kI the
s
s +1
integral gain, calculate the steady state error in response to a unit step reference and a unit step
disturbance.
where the plant is G ( s ) =
3
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 Mason’s 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
-G ( s )
E ( s) =
D( s)
1 + K ( s )G ( s )
æ 1 ö
-ç
÷
-s
s +1ø
è
E ( s) =
D( s) = 2
D( s) .
s + s + kI
æ kI 1 ö
ç1 + ×
÷
s s +1ø
è
Using now the Final Value Theorem yields
æ
ö1
-s
ess = lim sE ( s ) = lim s ç 2
÷ =0.
s ®0
s ®0 ç s + s + k ÷ s
è
I ø
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
E ( s) =
R( s)
1 + K ( s )G ( s )
1
s ( s + 1)
E ( s) =
R(s ) = 2
R(s)
s + s + kI
æ kI 1 ö
ç1 + ×
÷
s s +1ø
è
Using now the Final Value Theorem yields
æ s ( s + 1) ö 1
ess = lim sE ( s ) = lim s ç 2
÷ =0.
s ®0
s ®0 ç s + s + k ÷ s
è
I ø
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
G( s) K (s )
Y (s) =
R(s)
1 + G( s) K ( s)
4
æ kI 1 ö
ç ×
÷
k
s s +1ø
è
Y (s) =
R( s) = 2 I
R( s) .
s + s + kI
æ kI 1 ö
ç1 + ×
÷
s s +1ø
è
The closed-loop characteristic polynomial is
D( s) = s 2 + s + k I .
2
Comparing this to the standard form D( s ) = s 2 + 2zw n s + w n one sees that
w n = kI
1
1 .
=
2w n 2 k I
One can therefore select the integral gain to obtain any desired value of damping ratio, and hence of
POV which is given by
z =
POV = e -pz /
1-z 2
´ 100% .
d. Find the Output y(t) if r (t ) = e - t u -1 (t ), d (t ) = e -2t u -1 (t ) .
The transfer relation between the two inputs and the output is given by
G ( s) K (s )
G( s)
Y (s) =
R(s) +
D( s)
1 + G( s) K (s )
1 + G( s) K (s )
or
æ kI 1 ö
æ 1 ö
ç ×
÷
ç
÷
s s +1ø
s +1ø
è
è
Y (s) =
R( s) +
D( s )
æ kI 1 ö
æ kI 1 ö
ç1 + ×
÷
ç1 + ×
÷
.
s s +1ø
s s +1ø
è
è
k
s
= 2 I
R(s ) + 2
D( s )
s + s + kI
s + s + kI
1
1
Setting now R( s ) =
, D( s) =
one obtains
s +1
s+2
k
1
1
s
Y (s) = 2 I
×
+ 2
×
.
s + s + k I ( s + 1) s + s + k I ( s + 2)
For the given value of kI one may now use the inverse Laplace transform to determine the output y(t).
5
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 R( s ) = 1 / s 2 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
d(t)
r(t)
e(t)
y(t)
K(s)
G(s)
Figure 7.
Let the plant be given in the form
b + b s + ... + bm s m n( s )
G (s) = 0 1
=
d ( s)
1 + a1s + ... + an s n
The compensator is given as K ( s ) =
kN
s
N
, G (s) K ( s) =
n( s ) k N
, and assume that the plant does not have
d ( s) s N
any poles in zero.
A. Let D( s ) = 0 and R( s ) =
1
, r ³0.
sr
What is the minimum value of N such that the steady state error is zero?
1
B. Let R( s ) = 0 and D( s ) = d , d ³ 0 .
s
What is the minimum value of N such that the steady state error is zero?
n( s ) 1
C. Repeat points A and B considering that G ( s ) =
, g ³0.
d (s) s g
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
6
-
in some cases feedback can make the system unstable for specific controller structure and
certain values of the controller parameters
Stable feedback control
Bad design of feedback controllers can lead to an unstable closed loop system. Thus while choosing a
structure for the controller you must always check to see if the closed loop system will remain stable
(you can use Routh test on the characteristic equation or calculate the poles of the system and specify
the conditions such that all the poles are in the left half s-plane) and for which values of controller
parameters. The next example illustrates this idea.
Problem 2
Take the feedback control system in the figure
d(t)
r(t)
e(t)
y(t)
K(s)
G(s)
Figure 8
where D( s ) = 0 , K ( s ) = k a proportional controller, and the system has the transfer function
s -1
G (s) =
.
s+2
a. Calculate the poles and the zeros of the system.
b. Is the system minimum phase?
Y (s)
c. Calculate the transfer function of the closed loop system H ( s ) =
.
R(s )
d. For which values of k will the closed loop system remain stable? Note that here the controller is a
proportional controller.
e. Calculate the steady-state error in response to a unit step reference.
k
f. In order to obtain zero steady-state error we take the controller to be K ( s ) = I . This is an integral
s
controller. (According to the internal model principle we know that we need an integrator in the
controller. But will it be able to also stabilize the system?).
f.1. For which values of k I will the system be stable?
f.2. Calculate the steady state-error in response to a unit step reference.
k
g. Let the controller be K ( s ) = k p + I . This controller is a PI (proportional and integrator).
s
g.1. For which values of k I will the system be stable?
g.2. Calculate the steady state error in response to a unit step reference.
7