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ESE 499 – Feedback Control Systems
SECTION 7: STEADY-STATE ERROR
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K. Webb ESE 499
Introduction2
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Steady-State Error – Introduction
Consider a simple unity-feedback system
The error is the difference between the reference and the output𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝑌𝑌 𝑠𝑠
The input to the controller, 𝐷𝐷 𝑠𝑠
Consider a case where: Reference input is a step Plant has no poles at the origin – finite DC gain Controller is a simple gain block
In steady state, the forward path reduces to a constant gain:
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Steady-State Error – Introduction
In steady state, we’d like: Output to be equal to the input: 𝑦𝑦𝑠𝑠𝑠𝑠 = 𝑟𝑟𝑠𝑠𝑠𝑠 Zero steady-state error: 𝑒𝑒𝑠𝑠𝑠𝑠 = 0
Is that the case here?𝑒𝑒𝑠𝑠𝑠𝑠 = 𝑟𝑟𝑠𝑠𝑠𝑠 − 𝑦𝑦𝑠𝑠𝑠𝑠 = 𝑟𝑟𝑠𝑠𝑠𝑠 − 𝑒𝑒𝑠𝑠𝑠𝑠𝐾𝐾
𝑒𝑒𝑠𝑠𝑠𝑠 = 𝑟𝑟𝑠𝑠𝑠𝑠1
1 + 𝐾𝐾
No, if 𝑟𝑟𝑠𝑠𝑠𝑠 ≠ 0, then 𝑒𝑒𝑠𝑠𝑠𝑠 ≠ 0
Non-zero steady-state error to a step input for finite steady-state forward-path gain Finite DC gain implies no poles at the origin in 𝐷𝐷 𝑠𝑠 or 𝐺𝐺 𝑠𝑠
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Steady-State Error – Introduction
Now, allow a single pole at the origin An integrator in the forward path
Now the error is𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝐸𝐸 𝑠𝑠 ⋅
𝐾𝐾𝑠𝑠
𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑆𝑆𝑠𝑠
𝑠𝑠 + 𝐾𝐾 For a step input
𝐸𝐸 𝑠𝑠 =1𝑠𝑠
𝑠𝑠𝑠𝑠 + 𝐾𝐾
=1
𝑠𝑠 + 𝐾𝐾
Applying the final value theorem gives the steady-state error
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑠𝑠 + 𝐾𝐾
= 0
Zero steady-state error to a step input when there is an integrator in the forward path
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Steady-State Error – Introduction
Next, consider a ramp input to the same system
𝑟𝑟 𝑡𝑡 = 𝑡𝑡 ⋅ 1 𝑡𝑡 and 𝑅𝑅 𝑠𝑠 = 1𝑠𝑠2
Now the error is
𝐸𝐸 𝑠𝑠 =1𝑠𝑠2
𝑠𝑠𝑠𝑠 + 𝐾𝐾
=1
𝑠𝑠 𝑠𝑠 + 𝐾𝐾
The steady-state error is
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑠𝑠 𝑠𝑠 + 𝐾𝐾
=1𝐾𝐾
Non-zero, but finite, steady-state error to a ramp input when there is an integrator in the forward path
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Steady-State Error – Introduction
Two key observations from the preceding example involving unity-feedback systems:
Steady-state error is related to the number of integrators in the open-loop transfer function
Steady-state error is related to the type of input
We’ll now explore both of these observations more thoroughly
First, we’ll introduce the concept of system type
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System Type and Steady-State Error8
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System Type
System TypeThe degree of the input polynomial for which the steady-state error is a finite, non-zero constant
Type 0: finite, non-zero error to a step input
Type 1: finite, non-zero error to a ramp input
Type 2: finite, non-zero error to a parabolic input
For the remainder of this sub-section, and the one that follows, we’ll consider only the special case of unity-feedback systems
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System Type – Unity-Feedback Systems
For unity-feedback systems, system type is determined by the number of integrators in the forward path
Type 0: no integrators in the open-loop TF, e.g.:
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 4
𝑠𝑠 + 6 𝑠𝑠2 + 4𝑠𝑠 + 8
Type 1: one integrator in the open-loop TF, e.g.:
𝐺𝐺 𝑠𝑠 =15
𝑠𝑠 𝑠𝑠2 + 3𝑠𝑠 + 12
Type 2: two integrators in the open-loop TF, e.g.:
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 5
𝑠𝑠2 𝑠𝑠 + 3 𝑠𝑠 + 7
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Types of Inputs
When characterizing a control system’s error performance we focus on three main inputs: Step Ramp Parabola
We will derive expressions for the steady-state error due to each
Step:
𝑟𝑟 𝑡𝑡 = 1 𝑡𝑡 ⟷ 𝑅𝑅 𝑠𝑠 = 1𝑠𝑠
For a positioning system, this represents a constant position
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Types of Inputs
Parabola:
𝑟𝑟 𝑡𝑡 = 12𝑡𝑡2 ⋅ 1 𝑡𝑡 ⟷ 𝑅𝑅 𝑠𝑠 = 1
𝑠𝑠3
For a positioning system, this represents a constant acceleration
Ramp:
𝑟𝑟 𝑡𝑡 = 𝑡𝑡 ⋅ 1 𝑡𝑡 ⟷ 𝑅𝑅 𝑠𝑠 = 1𝑠𝑠2
For a positioning system, this represents a constant velocity
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Steady-State Error – Unity-Feedback
For unity-feedback systems steady-state error can be expressed in terms of the open-loop transfer function, 𝐺𝐺 𝑠𝑠
𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝑌𝑌 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝐸𝐸 𝑠𝑠 𝐺𝐺 𝑠𝑠
𝐸𝐸 𝑠𝑠 =𝑅𝑅 𝑠𝑠
1 + 𝐺𝐺 𝑠𝑠
Steady-state error is found by applying the final value theorem
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑅𝑅 𝑠𝑠1 + 𝐺𝐺 𝑠𝑠
We’ll now consider this expression for each of the three inputs of interest
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Steady-State Error – Step Input
For a step input
𝑟𝑟 𝑡𝑡 = 1 𝑡𝑡 ⟷ 𝑅𝑅 𝑠𝑠 =1𝑠𝑠
Steady-state error to a step input is
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠 1𝑠𝑠
1 + 𝐺𝐺 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
11 + 𝐺𝐺 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 =1
1 + lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠
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Steady-State Error – Step Input
𝑒𝑒𝑠𝑠𝑠𝑠 =1
1 + lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠
In order to have 𝑒𝑒𝑠𝑠𝑠𝑠 = 0, as we’d like, we must have
lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠 = ∞
That is, the DC gain of the open-loop system must be infinite
If 𝐺𝐺 𝑠𝑠 has the following form
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
then
lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠 =𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2⋯
≠ ∞
and we’ll have non-zero steady-state error
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Steady-State Error – Step Input
However, consider 𝐺𝐺 𝑠𝑠 of the following form
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯
𝑠𝑠𝑛𝑛 𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
where 𝑛𝑛 ≥ 1 That is, 𝐺𝐺 𝑠𝑠 includes 𝑛𝑛 integrators
It is a type 𝒏𝒏 system
lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠 = ∞ and 𝑒𝑒𝑠𝑠𝑠𝑠 = 0
A type 1 or greater system will exhibit zero steady-state error to a step input
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Steady-State Error – Ramp Input
For a ramp input
𝑟𝑟 𝑡𝑡 = 𝑡𝑡 ⋅ 1 𝑡𝑡 ⟷ 𝑅𝑅 𝑠𝑠 =1𝑠𝑠2
Steady-state error to a ramp input is
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠 1𝑠𝑠2
1 + 𝐺𝐺 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
1𝑠𝑠 + 𝑠𝑠𝐺𝐺 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 =1
lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠
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Steady-State Error – Ramp Input
𝑒𝑒𝑠𝑠𝑠𝑠 =1
lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠
In order to have 𝑒𝑒𝑠𝑠𝑠𝑠 = 0, the following must be truelim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠 = ∞
If there are no integrators in the forward path, then
lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
= 0
and𝑒𝑒𝑠𝑠𝑠𝑠 = ∞
A type 0 system has infinite steady-state error to a ramp input
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Steady-State Error – Ramp Input
If there is a single integrator in the forward path, i.e. a type 1 system
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯
𝑠𝑠 𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
then lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠 =𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2⋯
and 𝑒𝑒𝑠𝑠𝑠𝑠 =
𝑝𝑝1𝑝𝑝2 ⋯𝑧𝑧1𝑧𝑧2 ⋯
A type 1 system has non-zero, but finite, steady-state error to a ramp input
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Steady-State Error – Ramp Input
If there are two or more integrators in the forward path, i.e. a type 2 or greater system
𝐺𝐺 𝑠𝑠 = 𝑠𝑠+𝑧𝑧1 𝑠𝑠+𝑧𝑧2 ⋯𝑠𝑠𝑛𝑛 𝑠𝑠+𝑝𝑝1 𝑠𝑠+𝑝𝑝2 ⋯
, 𝑛𝑛 ≥ 2
then lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯
𝑠𝑠𝑛𝑛 𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯= ∞
and 𝑒𝑒𝑠𝑠𝑠𝑠 = 0
A type 2 or greater system has zero steady-state error to a ramp input
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Steady-State Error – Parabolic Input
For a Parabolic input
𝑟𝑟 𝑡𝑡 =𝑡𝑡2
2 ⋅ 1 𝑡𝑡 ⟷ 𝑅𝑅 𝑠𝑠 =1𝑠𝑠3
Steady-state error to a parabolic input is
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠 1𝑠𝑠3
1 + 𝐺𝐺 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
1𝑠𝑠2 + 𝑠𝑠2𝐺𝐺 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 =1
lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠
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Steady-State Error – Parabolic Input
𝑒𝑒𝑠𝑠𝑠𝑠 =1
lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠
In order to have 𝑒𝑒𝑠𝑠𝑠𝑠 = 0, the following must be truelim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠 = ∞
If there are no integrators in the forward path, then
lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠2𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
= 0
and𝑒𝑒𝑠𝑠𝑠𝑠 = ∞
A type 0 system has infinite steady-state error to a parabolic input
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Steady-State Error – Parabolic Input
If there is a single integrator in the forward path, i.e. a type 1 system
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯
𝑠𝑠 𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
then lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2 ⋯
= 0
and 𝑒𝑒𝑠𝑠𝑠𝑠 = ∞
A type 1 system has infinite steady-state error to a parabolic input
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Steady-State Error – Parabolic Input
If there are two integrators in the forward path, i.e. a type 2 system
𝐺𝐺 𝑠𝑠 =𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯
𝑠𝑠2 𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
then lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯
=𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2 ⋯
and 𝑒𝑒𝑠𝑠𝑠𝑠 =
𝑝𝑝1𝑝𝑝2 ⋯
𝑧𝑧1𝑧𝑧2 ⋯
A type 2 system has non-zero, but finite, steady-state error to a parabolic input
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Steady-State Error – Parabolic Input
If there are three or more integrators in the forward path, i.e. a type 3 or greater system
𝐺𝐺 𝑠𝑠 = 𝑠𝑠+𝑧𝑧1 𝑠𝑠+𝑧𝑧2 ⋯𝑠𝑠𝑛𝑛 𝑠𝑠+𝑝𝑝1 𝑠𝑠+𝑝𝑝2 ⋯
, 𝑛𝑛 ≥ 3
then lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠2𝑠𝑠 + 𝑧𝑧1 𝑠𝑠 + 𝑧𝑧2 ⋯
𝑠𝑠𝑛𝑛 𝑠𝑠 + 𝑝𝑝1 𝑠𝑠 + 𝑝𝑝2 ⋯= ∞
and 𝑒𝑒𝑠𝑠𝑠𝑠 = 0
A type 3 or greater system has zero steady-state error to a parabolic input
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Static Error Constants26
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Static Error Constants – Unity-Feedback
We’ve seen that the steady-state error to each of the inputs considered is Step: 𝑒𝑒𝑠𝑠𝑠𝑠 = 1
1+lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠
Ramp: 𝑒𝑒𝑠𝑠𝑠𝑠 = 1lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠
Parabola: 𝑒𝑒𝑠𝑠𝑠𝑠 = 1lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠
The limit term in each expression is the static error constantassociated with that particular input:
Position constant: 𝐾𝐾𝑝𝑝 = lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠
Velocity constant: 𝐾𝐾𝑣𝑣 = lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠
Acceleration constant: 𝐾𝐾𝑎𝑎 = lim𝑠𝑠→0
𝑠𝑠2𝐺𝐺 𝑠𝑠
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Steady-State Error vs. System Type
Steady-state error vs. input and system type
System Type
Input
Step Ramp Parabola
0 11 + 𝐾𝐾𝑝𝑝
∞ ∞
1 0 1𝐾𝐾𝐾𝐾
∞
2 0 0 1𝐾𝐾𝐾𝐾
3 0 0 0
Note that the given steady-state error is for inputs of unit magnitude Actual error is scaled by the magnitude of the reference
input
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Non-Unity-Feedback Systems29
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Non-Unity-Feedback Systems
So far, we’ve focused on the special case of unity-feedback systems
System type determined by # of integrators in the forward path – i.e., # of open-loop poles at the origin
Steady-state error determined using static error constants Static error constants determined from the open-loop
transfer function
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Non-Unity-Feedback Systems
More general approach to determining steady-state error is to use the closed-loop transfer function Applicable to non-unity-feedback systems, e.g.:
The error is
𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝑌𝑌 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝑅𝑅 𝑠𝑠 𝑇𝑇 𝑠𝑠
𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 1 − 𝑇𝑇 𝑠𝑠
𝑇𝑇 𝑠𝑠 =𝐺𝐺1 𝑠𝑠 𝐺𝐺2 𝑠𝑠
1 + 𝐺𝐺2 𝑠𝑠 𝐻𝐻 𝑠𝑠
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Non-Unity-Feedback Systems
Apply the final value theorem to determine the steady-state error:
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐸𝐸 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑅𝑅 𝑠𝑠 1 − 𝑇𝑇 𝑠𝑠
Here, system type is determined by using the more general definition:System type is the degree of the input polynomial for which the steady-state error is a finite, non-zero constant
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Non-Unity-Feedback Systems
Alternatively, find steady-state error by converting to a unity-feedback configuration, e.g.:
Add and subtract unity-feedback paths:
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Non-Unity-Feedback Systems
Combine the two upper parallel feedback paths:
Collapsing the inner feedback form leaves a unity-feedback system Can now apply unity-
feedback error analysis techniques
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Steady-State Error – Examples 35
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Steady-State Error – Example 1
What is the steady-state error to a constant reference input, 𝑟𝑟 𝑡𝑡 =3 𝑐𝑐𝑐𝑐 ⋅ 1 𝑡𝑡 , for the following feedback positioning system?
A type 0 system Non-zero error to a constant reference
Position constant:𝐾𝐾𝑝𝑝 = lim
𝑠𝑠→0𝐺𝐺 𝑠𝑠 = 10
Steady-state error:
𝑒𝑒𝑠𝑠𝑠𝑠 = 𝑟𝑟𝑠𝑠𝑠𝑠1
1 + 𝐾𝐾𝑝𝑝= 3 𝑐𝑐𝑐𝑐
11 + 10
𝑒𝑒𝑠𝑠𝑠𝑠 = 0.27 𝑐𝑐𝑐𝑐
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Steady-State Error – Example 1
𝑒𝑒𝑠𝑠𝑠𝑠 = 0.27 𝑐𝑐𝑐𝑐
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Steady-State Error – Example 1
What is the same system’s steady-state error to a unit ramp input, 𝑟𝑟 𝑡𝑡 = 𝑡𝑡 ⋅ 1 𝑡𝑡 ? A type 0 system, so error to a ramp reference will be infinite
Verify using closed-loop transfer function
𝑇𝑇 𝑠𝑠 =𝐺𝐺 𝑠𝑠
1 + 𝐺𝐺 𝑠𝑠 =10
𝑠𝑠 + 11
Steady-state error is
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝑅𝑅 𝑠𝑠 1 − 𝑇𝑇 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠1𝑠𝑠2 1 −
10𝑠𝑠 + 11
𝑒𝑒𝑠𝑠𝑠𝑠 = lim𝑠𝑠→0
1𝑠𝑠
𝑠𝑠 + 1𝑠𝑠 + 11 = ∞
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Steady-State Error – Example 1
𝑒𝑒𝑠𝑠𝑠𝑠 = ∞
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Steady-State Error – Example 2
Design the controller, 𝐷𝐷 𝑠𝑠 , for error of 0.05 to a unit ramp input
Plant is type 0 Forward path must be type 1 for finite error to a ramp input 𝐷𝐷 𝑠𝑠 must be type 1, so one very simple option is:
𝐷𝐷 𝑠𝑠 =𝐾𝐾𝑠𝑠
Forward-path transfer function is
𝐷𝐷 𝑠𝑠 𝐺𝐺 𝑠𝑠 =𝐾𝐾 𝑠𝑠 + 2
𝑠𝑠 𝑠𝑠 + 1 𝑠𝑠 + 5
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Steady-State Error – Example 2
The velocity constant is
𝐾𝐾𝑣𝑣 = lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠 = lim𝑠𝑠→0
𝑠𝑠𝐾𝐾 𝑠𝑠 + 2
𝑠𝑠 𝑠𝑠 + 1 𝑠𝑠 + 5=2𝐾𝐾5
Steady-state error is
𝑒𝑒𝑠𝑠𝑠𝑠 =1𝐾𝐾𝑣𝑣
=52𝐾𝐾
For error of 0.05:𝑒𝑒𝑠𝑠𝑠𝑠 = 0.05 =
52𝐾𝐾
𝐾𝐾 = 50
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Steady-State Error – Example 2
𝑒𝑒𝑠𝑠𝑠𝑠 = 0.05
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Steady-State Error – Example 3
Next, consider a non-unity-feedback system:
Determine controller gain, 𝐾𝐾, to provide a 2% steady-state error to a constant reference input
First, convert to a unity-feedback system Combine forward-path blocks Simultaneously add and subtract
unity-feedback paths
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Steady-State Error – Example 3
Combine the top two parallel feedback paths
Simplifying the inner feedback form leaves a unity-feedback system
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Steady-State Error – Example 3
Steady-state error for this type 0 system is
𝑒𝑒𝑠𝑠𝑠𝑠 =1
1 + 𝐾𝐾𝑝𝑝where
𝐾𝐾𝑝𝑝 = lim𝑠𝑠→0
𝐺𝐺 𝑠𝑠 =20 ⋅ 𝐾𝐾
10= 2 ⋅ 𝐾𝐾
For 2% steady-state error
𝑒𝑒𝑠𝑠𝑠𝑠 = 0.02 =1
1 + 2 ⋅ 𝐾𝐾 The controller gain is
𝐾𝐾 = 24.5
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Steady-State Error – Example 3
Note that the controller gain has been set to satisfy a steady-state error requirement only Closed loop poles
are very lightly-damped
Dynamic response is likely unacceptable
𝑒𝑒𝑠𝑠𝑠𝑠 = 0.02
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Steady-State Error – Example 4
Now, consider a unity-feedback system with a disturbance input
where𝐷𝐷 𝑠𝑠 = 𝐾𝐾 and 𝐺𝐺 𝑠𝑠 = 1
𝑠𝑠+5
Determine the controller gain, 𝐾𝐾, such that error due to a constant disturbance is 1% of 𝑊𝑊 𝑠𝑠
For this value of 𝐾𝐾, what is the steady-state error to a constant reference input?
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Steady-State Error – Example 4
The total error is given by𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝑌𝑌 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 − 𝐸𝐸 𝑠𝑠 𝐷𝐷 𝑠𝑠 𝐺𝐺 𝑠𝑠 + 𝑊𝑊 𝑠𝑠 𝐺𝐺 𝑠𝑠
𝐸𝐸 𝑠𝑠 1 + 𝐷𝐷 𝑠𝑠 𝐺𝐺 𝑠𝑠 = 𝑅𝑅 𝑠𝑠 −𝑊𝑊 𝑠𝑠 𝐺𝐺 𝑠𝑠
𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠1
1 + 𝐷𝐷 𝑠𝑠 𝐺𝐺 𝑠𝑠−𝑊𝑊 𝑠𝑠
𝐺𝐺 𝑠𝑠1 + 𝐷𝐷 𝑠𝑠 𝐺𝐺 𝑠𝑠
Substituting in controller and plant transfer functions gives
𝐸𝐸 𝑠𝑠 = 𝑅𝑅 𝑠𝑠𝑠𝑠 + 5
𝑠𝑠 + 5 + 𝐾𝐾−𝑊𝑊 𝑠𝑠
1𝑠𝑠 + 5 + 𝐾𝐾
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K. Webb ESE 499
49
Steady-State Error – Example 4
Error due to a constant disturbance can be found by applying the final value theorem
𝑒𝑒𝑠𝑠𝑠𝑠,𝑤𝑤 = lim𝑠𝑠→0
𝑠𝑠 −𝑊𝑊 𝑠𝑠1
𝑠𝑠 + 5 + 𝐾𝐾
𝑒𝑒𝑠𝑠𝑠𝑠,𝑤𝑤 = lim𝑠𝑠→0
−𝑠𝑠1𝑠𝑠
1𝑠𝑠 + 5 + 𝐾𝐾
= −1
5 + 𝐾𝐾
We can calculate the required gain for 1% error
𝑒𝑒𝑠𝑠𝑠𝑠,𝑤𝑤 = 0.01 =1
5 + 𝐾𝐾⟶ 𝐾𝐾 = 95
At this gain value, the error due to a constant reference is
𝑒𝑒𝑠𝑠𝑠𝑠,𝑟𝑟 = lim𝑠𝑠→0
𝑠𝑠1𝑠𝑠
𝑠𝑠 + 5𝑠𝑠 + 5 + 𝐾𝐾
=5
100⟶ 5%