Steady-State Error
Steady-State Error for Closed-Loop Systems
Steady-State Error for Unity Feedback Systems
Steady-State Error for Disturbances
Steady-State Error for Nonunity Feedback Systems
Unit 6: Steady-State Error
Engineering 5821:
Control Systems I
Faculty of Engineering & Applied Science
Memorial University of Newfoundland
March 3, 2010
ENGI 5821
Unit 6: Steady-State Error
Steady-State Error
Steady-State Error for Closed-Loop Systems
Steady-State Error for Unity Feedback Systems
Steady-State Error for Disturbances
Steady-State Error for Nonunity Feedback Systems
1
Steady-State Error
1
Steady-State Error for Closed-Loop Systems
1
Steady-State Error for Unity Feedback Systems
Examples
System Type
Integral Control
1
Steady-State Error for Disturbances
1
Steady-State Error for Nonunity Feedback Systems
ENGI 5821
Unit 6: Steady-State Error
Steady-State Error
Control systems are judged according to their performance in three
areas:
Transient response
Stability
Steady-state error
Clearly we desire systems with zero steady-state error. We will
obtain the steady-state part of the total response simply by letting
t → ∞ and determining e(∞). Hopefully it will tend to zero, or at
least some small number.
Important: A prerequisite to our analysis is that the system is
stable. For an unstable system e(∞) = ∞.
Steady-State Error
Steady-State Error for Closed-Loop Systems
Steady-State Error for Unity Feedback Systems
Steady-State Error for Disturbances
Steady-State Error for Nonunity Feedback Systems
We will be interested in the steady-state error in response to
different inputs: step functions, ramps, and parabolas. These
particular inputs are important because they represent common
real-world control inputs for control of position, velocity, and
acceleration respectively.
ENGI 5821
Unit 6: Steady-State Error
Consider the total responses shown below for step and ramp
inputs.
Steady-State Error
Steady-State Error for Closed-Loop Systems
Steady-State Error for Unity Feedback Systems
Steady-State Error for Disturbances
Steady-State Error for Nonunity Feedback Systems
Steady-state error can occur because of nonlinearities in our
system, but this is beyond our scope. We consider error that arises
because of the system itself and its input.
For example, consider a system with a step input and pure gain, K ,
The steady-state error cannot be reduced to zero for such a
system. If e(t) is zero then so to must c(t). The best we can do is
increase the gain. This will tend to minimize the steady-state error
esteady −state for a particular steady-state output, csteady −state .
esteady −state =
ENGI 5821
1
csteady −state
K
Unit 6: Steady-State Error
Steady-State Error
Steady-State Error for Closed-Loop Systems
Steady-State Error for Unity Feedback Systems
Steady-State Error for Disturbances
Steady-State Error for Nonunity Feedback Systems
If the forward-path gain is replaced by an integrator zero
steady-state error becomes possible.
The difference is that an integrator can have a constant output
without any input.
A motor can be considered a form of integrator. Consider a
position-control application (e.g. Lab 2). With a non-zero voltage
input, the shaft will turn. When the desired position is reached
e(t) will be reduced to zero. Still the motor maintains position
without any input.
ENGI 5821
Unit 6: Steady-State Error
Steady-State Error for Closed-Loop Systems
Consider a general closed-loop system. The error
E (s) = R(s) − C (s) can be shown as an output:
We have the following two facts:
E (s) = R(s) − C (s)
C (s) = R(s)T (s)
We eliminate C (s) to obtain,
E (s) = R(s)(1 − T (s))
Given T (s) and a particular input R(s) we could take the ILT and
let t → ∞ to obtain the steady-state error e(∞). However, there
is a more direct way that also yields more insight...
The Final Value Theorem (item 11 in table of LT theorems)
f (∞) = lim sF (s)
s→0
Derivation: Start with the differentiation theorem,
Z ∞
d
f (t) e −st dt = sF (s) − f (0)
dt
0
Z ∞
d
f (t) e −st dt = lim [sF (s) − f (0)]
lim
s→0
s→0 0
dt
Z ∞
d
f (t) dt =
dt
0
∞
f (t)
=
0
f (∞) − f (0) =
f (∞) − f (0) =
f (∞) =
lim sF (s) − f (0)
s→0
lim sF (s)
s→0
Continuing with our analysis of steady-state error for a closed-loop
system...
E (s) = R(s)(1 − T (s))
We apply the Final Value Theorem to obtain e(∞).
e(∞) = lim sE (s) = lim sR(s)(1 − T (s))
s→0
s→0
e.g. What is the steady-state error for R(s) = 1/s and
T (s) = 5/(s 2 + 7s + 10). First we obtain E (s),
E (s) =
s 2 + 7s + 5
s(s 2 + 7s + 10)
Applying the Final Value Theorem yields e(∞) = 1/2.
Steady-State Error for Unity Feedback Systems
Given a unity feedback, or any other kind of linear system, we can
apply the previous method to determine the steady-state error. We
consider here unity feedback systems in particular so that we can
draw a direct relationship between the transfer functions for such
systems and their steady-state error. The existence of this
relationship aids in analysis and design.
Start with a general unity feedback system:
We will establish E (s) in terms of the system and its input.
E (s) = R(s) − C (s)
C (s) = E (s)G (s)
R(s)
E (s) =
1 + G (s)
E (s) =
R(s)
1 + G (s)
Apply the Final Value Theorem:
e(∞) = lim
s→0
sR(s)
1 + G (s)
We now determine e(∞) for step, ramp, and parabolic inputs.
Step input: R(s) = 1/s:
s(1/s)
s→0 1 + G (s)
1
=
1 + lims→0 G (s)
estep (∞) =
lim
The quantity lims→0 G (s) is the DC gain of G (s). It is known as
the position constant Kp .
To achieve estep (∞) = 0 the position constant must equal ∞.
If lims→0 G (s) = ∞ what does this imply about G (s)? G (s) must
have at least one pole at s = 0. That is, G (s) must have the
following form,
G (s) =
(s + z1 )(s + z2 ) · · ·
+ p1 )(s + p2 ) · · ·
s n (s
where n ≥ 1. This means that this system must have at least one
pure integration in the forward path. Otherwise, we will be stuck
with some finite value for estep (∞).
Ramp input: R(s) = 1/s 2 :
s(1/s 2 )
s→0 1 + G (s)
1
=
lims→0 sG (s)
eramp (∞) =
lim
The quantity lims→0 sG (s) is known as the velocity constant Kv .
To achieve eramp (∞) = 0 it must equal ∞.
To eliminate steady-state error lims→0 G (s) = ∞. What does this
imply about G (s)? Again, it must have the following form,
G (s) =
(s + z1 )(s + z2 ) · · ·
+ p1 )(s + p2 ) · · ·
s n (s
only now n ≥ 2. The system must have at least two integrations in
the forward path in order to eliminate steady-state error. With one
integration Kv would be finite, and therefore so would eramp . With
no integrations Kv = 0 and eramp = ∞.
Parabolic input: R(s) = 1/s 3 :
s(1/s 3 )
s→0 1 + G (s)
1
=
lims→0 s 2 G (s)
eramp (∞) =
lim
The quantity lims→0 s 2 G (s) is known as the acceleration constant
Ka . To achieve eramp (∞) = 0 we must have Ka = ∞. Again this
requires the form,
G (s) =
(s + z1 )(s + z2 ) · · ·
+ p1 )(s + p2 ) · · ·
s n (s
with n ≥ 3. Otherwise the system will exhibit finite steady-state
error (n = 2) or infinite steady-state error (n = 0 or n = 1).
Examples
e.g. Determine the steady-state error for the system below and the
following inputs: 5u(t), 5tu(t), 5t 2 u(t). Firstly, we should assess
for stability (not shown).
sR(s)
5
5
=
=
1 + G (s)
1 + lims→0 G (s)
21
s(5/s 2 )
5
= lim
=
=∞
s→0 1 + G (s)
lims→0 sG (s)
s(5 · s23 )
10
= lim
=
=∞
s→0 1 + G (s)
lims→0 s 2 G (s)
estep =
eramp
eparabola
lim
s→0
e.g. As above, but now the system has one integration.
sR(s)
5
=
=0
1 + G (s)
1 + lims→0 G (s)
5
5
1
s(5/s 2 )
=
=
=
= lim
s→0 1 + G (s)
lims→0 sG (s)
100
20
2
s(5 · s 3 )
10
= lim
=
=∞
s→0 1 + G (s)
lims→0 s 2 G (s)
estep =
eramp
eparabola
lim
s→0
e.g. Find the value of K so that the system below exhibits 10%
steady-state error for an appropriate input.
What is an ‘appropriate input’ ? A system with one integration will
exhibit 0 error step response, finite error ramp response, and
infinite parabolic response. Since we are expecting finite error, we
assume a ramp input.
1
s(1/s 2 )
=
= 0.1
s→0 1 + G (s)
lims→0 sG (s)
eramp = lim
K ×5
= 10
6×7×8
Therefore K = 672 satisfies the specification. One should also
ensure that the system is stable for this value of K (it is).
=⇒ lim sG (s) =
s→0
System Type
We characterize unity feedback systems by the number of
integrations in their forward path. The most common types are
Type 0, 1, and 2. The type is given by the number of integrations:
G (s) =
(s + z1 )(s + z2 ) · · ·
+ p1 )(s + p2 ) · · ·
s n (s
where the type is given by n. The following table summarizes the
steady-state error obtained by inputting different inputs into
different types of systems:
Input
Step, u(t)
Ramp, tu(t)
Parabola, t 2 u(t)
Type 0: e(∞)
1
1+Kp
∞
∞
Type 1: e(∞)
0
1
Kv
∞
Type 2: e(∞)
0
0
1
Ka
Integral Control
The simplest form of feedback control systems have a pure gain
component acting as the controller:
This is known as a Proportional Control. The system type is
completely governed by G (s).
We often wish to increase the system type to achieve zero
steady-state error for step inputs (as well as improving the
response to other inputs). An integrator can be added to serve as
a compensator and increase the system type by 1.
This is known as Integral Control. The combination of
proportional and integral control is known as PI Control.
Steady-State Error for Disturbances
Feeback systems can compensate for disturbances.
C (s) = E (s)G1 (s)G2 (s) + D(s)G2 (s)
C (s) = R(s) − E (s)
We eliminate C (s) and solve for E (s),
E (s) =
1
G2 (s)
R(s) −
D(s)
1 + G1 (s)G2 (s)
1 + G1 (s)G2 (s)
We can think of part of this error as being attributable to the input
and the rest as being from the disturbance.
Apply the Final Value Theorem to obtain e(∞),
lim sE (s) = lim
s→0
s→0
sG2 (s)
s
R(s) − lim
D(s)
s→0 1 + G1 (s)G2 (s)
1 + G1 (s)G2 (s)
Call the first term eR (∞), the steady-state error due to R(s). This
is what we have considered already. The second term is eD (∞),
the steady-state error due to the disturbance.
To analyze eD (∞) we need to know D(s). Assume the disturbance
is the unit step. We obtain,
eD (∞) = −
lims→0 G21(s)
1
+ lims→0 G1 (s)
To minimize eD (∞) we can increase the dc gain of G1 (s) or
decrease the dc gain of G2 (s).
ED (s) = −
G2 (s)
D(s)
1 + G1 (s)G2 (s)
Another way of looking at this situation is to draw the block
diagram relating D(s) as an input to E (s) as an output.
If D(s) and G1 (s)E (s) are close then the error will be small. This
can occur by increasing G1 (s) (all for s → 0). Similarly, we can
reduce E (s) by decreasing G2 (s) (for s → 0).
Steady-State Error for Nonunity Feedback Systems
Often the feedback path has something other than unity gain.
There may be an output transducer and/or a compensating
component in the feedback path. Our approach is simply to
transform such a system into an equivalent unity feedback
representation...
e.g. Determine the steady-state error for the system below for a
step input.
Ge (s) =
G (s)
100(s + 5)
= 3
1 + G (s)H(s) − G (s)
s + 15s 2 − 50s − 400
e(∞) = lim
s→0
s(1/s)
= −4
1 + Ge (s)