Steady State Errors ELEC304-Alper Erdogan
1–1
Lecture 1
Steady State Errors
• We define the error as the difference between the
input and output:
• For unity feedback systems error is the input of the
feedforward block:
And the steady state error
e(∞) = lim sE(s)
s→0
(1)
Steady State Errors ELEC304-Alper Erdogan
1–2
Steady State Error in Terms of T (s)
• Error can be written as
E(s) = R(s) − C(s)
Here
C(s) = R(s)T (s)
and therefore
E(s) = R(s) − R(s)T (s) = R(s)(1 − T (s))
As a result, the steady state error
e(∞) = lim sR(s)(1 − T (s))
s→0
Example: For T (s) =
R(s) = 1s
5
s2 +7s+10
and unit step input
1
s2 + 7s + 5
E(s) = (1 − T (s)) =
s
s(s2 + 7s + 10)
Steady State Errors ELEC304-Alper Erdogan
and the steady state error
s2 + 7s + 5
1
e(∞) = lim sE(s) = lim 2
=
s→0
s→0 s + 7s + 10
2
How about the steady state error for the ramp
input (R(s) = s12 )?
1
s2 + 7s + 5
E(s) = 2 (1 − T (s)) = 2 2
s
s (s + 7s + 10)
and the steady state error
s2 + 7s + 5
e(∞) = lim sE(s) = lim
=∞
s→0
s→0 s(s2 + 7s + 10)
1–3
Steady State Errors ELEC304-Alper Erdogan
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Steady State Error in Terms of G(s)
• Note that
E(s) = R(s) − C(s)
and
C(s) = E(s)G(s)
which yield
E(s) =
R(s)
1 + G(s)
Therefore the steady state error
sR(s)
e(∞) = lim sE(s) = lim
s→0
s→0 1 + G(s)
• For a step input R(s) =
1
s
s 1s
1
e(∞) = lim
=
s→0 1 + G(s)
1 + lims→0 G(s)
Steady State Errors ELEC304-Alper Erdogan
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To have zero steady state error
lim G(s) = ∞
s→0
which implies G(s) must take the form
(s + z1)(s + z2)...
G(s) = n
n≥1
s (s + p1)(s + p2)...
i.e. at least one pure integrator!
• For a ramp input R(s) =
1
s2
1
s s12
=
e(∞) = lim
s→0 1 + G(s)
lims→0 sG(s)
To have zero steady state error
lim sG(s) = ∞
s→0
which implies G(s) must take the form
(s + z1)(s + z2)...
n≥2
G(s) = n
s (s + p1)(s + p2)...
i.e. at least two pure integrators!
• For a parabolic input R(s) =
1
s3
s s13
1
e(∞) = lim
=
s→0 1 + G(s)
lims→0 s2G(s)
To have zero steady state error
lim s2G(s) = ∞
s→0
Steady State Errors ELEC304-Alper Erdogan
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which implies G(s) must take the form
G(s) =
(s + z1)(s + z2)...
sn(s + p1)(s + p2)...
i.e. at least three pure integrators!
n≥3
Steady State Errors ELEC304-Alper Erdogan
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Examples
Consider the System (with no integrator) in the
following figure
• The steady state error for r(t) = 5u(t) (R(s) = 5s )
s 5s
5
5
5
e(∞) = lim
=
=
=
s→0 1 + G(s)
1 + lims→0 G(s) 1 + 20 21
• The steady state error for r(t) = 5tu(t) (R(s) =
s s52
5
e(∞) = lim
=
=∞
s→0 1 + G(s)
lims→0 sG(s)
5
)
s2
Steady State Errors ELEC304-Alper Erdogan
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Examples
Consider the System (with one integrator) in the
following figure
• The steady state error for r(t) = 5u(t) (R(s) = 5s )
s 5s
5
5
e(∞) = lim
=
=
=0
s→0 1 + G(s)
1 + lims→0 G(s) ∞
• The steady state error for r(t) = 5tu(t) (R(s) =
5
)
s2
5
5
1
s s52
=
=
=
e(∞) = lim
s→0 1 + G(s)
lims→0 sG(s) 100 20
• The steady state error for r(t) = 5t2u(t) (R(s) =
10
)
s3
s 10
10
10
s3
e(∞) = lim
=
=
=∞
2
s→0 1 + G(s)
lims→0 s G(s)
0
Steady State Errors ELEC304-Alper Erdogan
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Type of the System
The type of the system (the number of integrators in
the open loop system) determines which type of
signals we can track:
Here n is the type of the system.
• If n = 0,
– It can track unit step with a nonzero error
1
1
e(∞) =
=
1 + lim G(s) 1 + Kp
s→0 {z
Kp
|
}
where Kp is the position constant.
– It can not track ramp or parabolic inputs.
• If n = 1,
– It can track unit step with a zero error.
– It can track ramp with a nonzero error.
1
1
e(∞) =
=
lim sG(s) Kv
s→0
|
{z
Kv
}
Steady State Errors ELEC304-Alper Erdogan
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where Kv is the velocity constant.
– It can not track parabolic inputs.
• If n = 2,
– It can track unit step and ramp input with zero
errors.
– It can track parabolic input with a nonzero error.
1
1
=
e(∞) =
lim s2G(s) Ka
s→0
|
{z
}
Ka
where Ka is the acceleration constant.
Steady State Errors ELEC304-Alper Erdogan
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Steady State Errors: Static Error
Constants
• Position constant
(1)
Kp = lim G(s)
s→0
Steady state error for unit step input is e(∞) =
1
1+Kp .
• Velocity constant
Kv = lim sG(s)
(2)
s→0
Steady state error for a ramp input is e(∞) =
1
KV .
• Acceleration constant
Ka = lim s2G(s)
s→0
Steady state error for a parabolic input is e(∞) =
(3)
1
Ka .
Steady State Errors ELEC304-Alper Erdogan
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Example
Given the control system
find K such that there is a 10 percent error in steady
state?
Solution: Since system type is 1 the finite steady
state error should be for a ramp input.
e(∞) =
So
1
= 0.1
Kv
5·K
Kv = 10 = lim sG(s) =
s→0
6·7·8
we obtain K = 672
Steady State Errors ELEC304-Alper Erdogan
How About Nonunity Feedback
Systems
• Easy! Use the following transformation:
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