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EE471 –
CONTROLS 1
- Dr. Weisong Tian
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Stability of Control Systems
• When considering the design and analysis of feedback control system,
stability is of utmost importance.
Bounded
input
Bounded
output
T(s)
• Definition of stability:
• For any bounded input, if the output is bounded (BIBO), then the system is
said to be stable.
• Definition of bounded: for any time t, if there exists a positive constant c,
such that 𝑓 𝑡 ≤ 𝑐, then we consider the function 𝑓 𝑡 is bounded.
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Stability of Control Systems
• The stability status of a control system can be categorized into:
1. Stable
2. Neutral (called marginally stable, M.S.)
3. Unstable
Stable
M.S.
System
Not Stable
Unstable
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Stability
• Stable: BIBO
• M.S.: only certain bounded inputs (sinusoids of the frequency of the poles) will
drive the output unbounded.
• Unstable: system output is unbounded for bounded inputs.
• Theorem: The stability of a system is determined by the location of the poles
of the closed-loop transfer function.
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Determine Stability
imaginary
• Assume 𝑃𝐾 are the poles of the closed-loop
transfer function.
Stable
Unstable
• Stable: All the poles are in the left half plane,
i.e., 𝑅𝑒 𝑃𝐾 < 0
• Marginally stable (M.S.): Poles are either in
the left half plane or are simple roots on the
imaginary axis
• Unstable: Either one pole in the right half
plane, or repeated poles on the imaginary
axis.
real
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Example: determine the stability of the following systems
1.
2.
3.
4.
5.
6.
7.
1
𝑠+1 𝑠+2
1
𝑠−1 𝑠+2
1
𝑠+1 𝑠
1
𝑠+1 𝑠2
1
𝑠 𝑠+1 𝑠2 +4
1
𝑠+2 𝑠2 +4 2
1
𝑠2 +16𝑠+64
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Example
• For the system 𝑌 𝑠 =
1. 𝑟 𝑡 = 1
2. 𝑟 𝑡 = sin 𝑡
3. 𝑟 𝑡 = sin 2𝑡
1
𝑅
𝑠 2 +1
𝑠 , Find y(t) for the following reference signals
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Design control system
• Try to avoid unstable systems
• For M.S. systems, the steady state output will be sustained oscillation for a
bounded input, unless the input is a sinusoid signal whose frequency is equal
to that of the imaginary-axis root.
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Challenges:
• Suppose
𝑇 𝑠 =
1
𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + ⋯ + 𝑎1 𝑠 + 𝑎0
• Determine the stability of T(s) if we cannot compute the roots.
• Motivations:
1.
2.
It is difficult to compute the roots of a high-order polynomial
Some of the constants 𝑎𝑖 ’s may unknown or TBD.
• Motivating example: design K such that the closed loop system is stable
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A quick way to spot unstable system
• Theorem: If the parameters of the characteristic equation are not with the
same sign, then the system must be unstable.
• Example:
Different sign: unstable.
1. 𝑃1 𝑠 = 𝑠 2 − 𝑠 + 2
2. 𝑃2 𝑠 = −𝑠 4 + 𝑠 3 + 𝑠 2 + 𝑠 + 1
• The “sign” method is only a necessary condition. It can only be used to spot
some unstable systems.
Different
sign
Unstable
Same sign
Stable
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Routh-Hurwitz Criterion
• A sufficient and necessary condition for stability
• A method to determine stability without computing the roots
• A useful method for controller design
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Method: Routh Array
• The Routh-Hurwitz criterion is based on ordering the coefficients of the
•
•
•
•
characteristic equation
𝑃 𝑠 = 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + ⋯ + 𝑎1 𝑠 + 𝑎0
into a Routh array as:
Further rows of the schedule are then completed as
where
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Routh-Hurwitz Criterion
• We stop at 𝑠 0 .
• Then examine the first column of the schedule
• The number of roots with positive real part is
equal to the number of changes in sign of the
first column of the Routh array.
• If there is no change of sign in the first
column, then the system is stable.
• This is a necessary and sufficient condition.
• In addition, the number of sign changes
represents the number of unstable poles.
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Example
• 𝑃 𝑠 = 𝑠 3 + 𝑠 2 + 2𝑠 + 24
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Example
• 𝑃 𝑠 = 𝑠 6 + 4𝑠 5 + 3𝑠 4 + 2𝑠 3 + 𝑠 2 + 4𝑠 + 4
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Four possible scenarios
• Four distinct cases or configurations of the first column of the Routh array
must be considered, and each must be treated separately and requires
suitable modifications of the array calculation procedure:
1. No element in the first column is zero
2. There is a zero in the first column, but some other elements of the row
containing the zero in the first column are nonzero
3. There is a zero in the first column, and the other elements of the row
containing the zero are also zero (an entire zero row)
4. Multiple zero rows
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Case 2
• Case 2: There is a zero in the first column, but some other elements of the row
containing the zero in the first column are nonzero
• Solution: replace this zero in the first column by a small constant 𝜀 > 0.
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Example
• 𝑠 5 + 2𝑠 4 + 2𝑠 3 + 4𝑠 2 + 11𝑠 + 10 = 0
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Case 3
• An entire row is zero
• Solution: Use auxiliary polynomial
1. Step 1: Go the row above the zero row to get the auxiliary polynomial 𝑝(𝑠)
2. Step 2: Take derivative 𝑝′ 𝑠
3. Step 3: Use the coefficients in 𝑝′ 𝑠 to replace the zero row
• Note: The order of the auxiliary polynomial is always even and indicates the
number of symmetrical root pairs.
• Result 1: Sign change: unstable
• Result 2: No sign change
Stable
• Need to solve auxiliary polynomial 𝑝 𝑠 = 0 see if zeros on the imaginary axis
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Example
• 𝑠 4 + 3𝑠 3 + 6𝑠 2 + 12𝑠 + 8 = 0
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Case 4
• Multiple rows of zeros => repeated imaginary poles => unstable
• Example: 𝑃 𝑠 = 𝑠 5 + 𝑠 4 + 2𝑠 3 + 2𝑠 2 + 𝑠 + 1
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Example
• Determine the range of K, such that the following polynomial is stable
• 𝑠 4 + 𝑠 3 + 3𝑠 2 + 2𝑠 + 𝐾
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Example
• Design the control gain K, so
that the system is stable
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Use Routh Array to Design a Controller
In the control system, 𝑎 > 0 is a
constant. Design the gain K, so
that the system is stable.