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MAE 4421 – Control of Aerospace & Mechanical Systems
SECTION 3: STABILITY
K. Webb MAE 4421
Introduction2
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Stability
Consider the following 2nd‐order systems
and
has two real poles:and
has a complex‐conjugate pair of poles:
,
The step response of each system is:
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Stability
Both step responses are a superposition of: Natural response (transient) Driven or forced response (steady‐state)
1.5 2.5 1
cos 2 sin 2 1
In both cases, the natural response decays to zero as
Natural Response Driven Response
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Stability
Both step responses are characteristic of stable systems
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Stability
Now, consider the following similar‐looking systems:
and
has two real polesand
has a complex‐conjugate pair of poles
,
The step responses of these systems are:
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Stability
Again, step responses consist of a natural response component and a driven component
1.5 2.5 1
cos 2 sin 2 1
Now, as , the natural responses do not decay to zero They blow up – why? Exponential terms are positive
Natural Response Driven Response
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Stability
Step responses characteristic of unstable systems
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Stability
Why are the exponential terms positive? Determined by the system poles
For the over‐damped system, the poles areand
And, the step response is
For the under‐damped system, the poles are
,
The step response is
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Stability and System Poles
Sign of the exponentials determined by , the real part of the system poles
If Pole is in the left half‐plane (LHP) Natural response as System is stable
If Pole is in the right half‐plane (RHP) Natural response as System is unstable
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Purely‐Imaginary Poles
LHP poles correspond to stable systems RHP poles correspond to unstable systems It seems that the imaginary axis is the boundary for stability
What if poles are on the imaginary axis? Consider the following system
Two purely‐imaginary poles
,
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Marginal Stability
Step response for this undamped system is
Natural response neither decays to zero, nor grows without bound Oscillates indefinitely System is marginally stable
Natural Response Driven Response
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Marginal Stability
Step response is characteristic of a marginally‐stablesystem
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Repeated Imaginary Poles
We’ll look at one more interesting case before presenting a formal definition for stability
Consider the following system168 16
164
Repeated poles on the imaginary axis
, 2 and , 2 The step response for this system is
cos 2 ⋅ sin 2 1
Natural Response Driven Response
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Repeated Imaginary Poles
Multiplying time factor causes the natural response to grow without bound An unstable system Results from repeated poles
Multiple identical poles on the imaginary axis implies an unstable system
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Repeated Imaginary Poles
Step response shows that the system is unstable
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Definitions of Stability17
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Definitions of Stability – Natural Response
We know that system response is the sum of a natural response and a driven response
Can define the categories of stability based on the natural response:
Stable A system is stable if its natural response → 0 as → ∞
Unstable A system is unstable if its natural response → ∞ as → ∞
Marginally Stable A system is marginally stable if its natural response neither decays nor grows, but remains constant or oscillates
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BIBO Stability
Alternatively, we can define stability based on the total response
Bounded‐input, bounded‐output (BIBO) stability
Stable A system is stable if every bounded input yields a bounded output
Unstable A system is unstable if any bounded input yields an unbounded output
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Closed‐Loop Poles and Stability
Stable A stable system has all of its closed‐loop poles in the left‐half plane
Unstable An unstable system has at least one pole in the right half‐plane and/or repeated poles on the imaginary axis
Marginally Stable A marginally‐stable system has non‐repeated poles on the imaginary axis and (possibly) poles in the left half‐plane
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Determining System Stability21
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Determining Stability
Stability determined by pole locations Poles determined by the characteristic polynomial, Δ
Factoring the characteristic polynomial will always tell us if a system is stable or not Easily done with a computer or calculator
Would like to be able to detect RHP poles without a computer Form of Δ may indicate RHP poles directly, or Routh‐Hurwitz Criterion
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Stability from Coefficients
A stable system has all poles in the LHP
⋯ Poles: For all LHP poles, 0, ∀ Result is that all coefficients of Δ are positive
If any coefficient of Δ is negative, there is at least one RHP pole, and the system is unstable
If any coefficient of Δ is zero, the system is unstable or, at best, marginally stable
If all coefficients of Δ are positive, the system may be stable or may be unstable
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Routh‐Hurwitz Criterion
Need a method to detect RHP poles if all coefficients of are positive: Routh‐Hurwitz criterion
General procedure:1. Generate a Routh table using the characteristic
polynomial of the closed‐loop system
2. Apply the Routh‐Hurwitz criterion to interpret the table and determine the number (not locations) of RHP poles
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Routh‐Hurwitz – Utility
Routh‐Hurwitz was very useful for determining stability in the days before computers Factoring polynomials by hand is difficult
Still useful for design, e.g.:
Stable for some range of gain, , but unstable beyond that range
Routh‐Hurwitz allows us to determine that range
6 8
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Routh Table
Consider a 4th‐order closed‐loop transfer function:
Routh table has one row for each power of in First row contains coefficients of even powers of (odd if the order of Δ is odd)
Second row contains coefficients of odd (even) powers of Fill in zeros if needed – if even order
0
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Routh Table
Remaining table entries calculated using entries from two preceding rows as follows:
0
000 0
00 0
00 0
000 0
00 0
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Routh Table – Example
Consider the following feedback system
The closed‐loop transfer function is5000
20 124 5240
The first two rows of the Routh table are
1 124201 5240262
Note that we can simplify by scaling an entire row by any factor
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Routh Table – Example
Calculate the remaining table entries:
1 124
201 5240262
1 1241 262
1 1381 01 01 0
1 262138 0
138 2621 0138 01 0
How do we interpret this table? Routh‐Hurwitz criterion
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Routh‐Hurwitz Criterion
Routh‐Hurwitz Criterion The number of poles in the RHP is equal to the number of sign changes in the first column of the Routh table
Apply this criterion to our example:1 1241 262138 0
262 0
Two sign changes in the first column indicate two RHP poles system is unstable
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Routh‐Hurwitz – Stability Requirements
Consider the same system, where controller gain is left as a parameter
Closed‐loop transfer function:100
20 124 240 100
Plant itself is stable Presumably there is some range of gain, , for which the closed‐loop system is also stable
Use Routh‐Hurwitz to determine this range
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Routh‐Hurwitz – Stability Requirements
10020 124 240 100
Create the Routh table
1 124
201 240 100 12 5
1 1241 12 5
1 112 51 01 01 0
1 12 5112 5 0
112 5 12 51 0138 01 0
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Routh‐Hurwitz – Stability Requirements
Since 0, only the third element in the first column can be negative
Stable for 112 5 0
22.4 Unstable (two RHP poles) for
112 5 022.4
1 1241 12 5112 5 012 5 0
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Routh Table – Special Cases
Two special cases can arise when creating a Routh table:1. A zero in only the first column of a row Divide‐by‐zero problem when forming the next row
2. An entire row of zeros Indicates the presence of pairs of poles that are mirrored
about the imaginary axis
We’ll next look at methods for dealing with each of these scenarios
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Routh Table – Zero in the First Column
If a zero appears in the first column1. Replace the zero with 2. Complete the Routh table as usual3. Take the limit as → 04. Evaluate the sign of the first‐column entries
For example:10
3 2 6 6 9
First two rows in the Routh table:
1 2 631 62 93
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First‐Column Zero – Example
Replace the first‐column zero with and proceed as usual
1 2 61 2 3
1 21 21 0
1 61 31 3
1 01 01 0
1 23 2 3
1 30 3
1 00 0
32 3 32 3 3
32 3
02 3 02 3 0
02 3 02 3 0
Continuing on the next page …
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First‐Column Zero – Example
1 2 61 2 3
3 02 3 3 0
33
2 30 0
2 3 3
3 32 3 0
3 32 3
3
2 3 0
3 32 3 0
3 32 3
0
2 3 0
3 32 3 0
3 32 3
0
Next, take the limit as → 0
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First‐Column Zero – Example
11
2 3
33
2 3
3
Taking the limit as → 0 and looking at the first column:
110
∞
0
3
Two sign changes Two RHP poles System is unstable
lim→
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Routh Table – Row of Zeros
A whole row of zeros indicates the presence of pairs of poles that are mirrored about the imaginary axis:
At best, the system is marginally stable Use a Routh table to determine if it is unstable
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Routh Table – Row of Zeros
If an entire row of zeros appears in a Routh table1. Create an auxiliary polynomial from the row above
the row of zeros, skipping every other power of
2. Differentiate the auxiliary polynomial w.r.t.
3. Replace the zero row with the coefficients of the resulting polynomial
4. Complete the Routh table as usual
5. Evaluate the sign of the first‐column entries
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Row of Zeros – Example
Consider the following system
15 11 23 28 12
The first few rows of the Routh table:
1 11 285 23 12
1 115 235 6.41
1 285 125 25.64
1 05 05 0
5 231 41 31
5 121 01 124
5 01 01 0
Continuing on the next page …
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Row of Zeros – Example
A row of zeros has appeared Create an auxiliary polynomial from the row
4 Differentiate
2
Replace the row with the / coefficients
1 11 285 23 121 4 01 4 0
1 41 41 0
1 41 41 0
5 01 01 0
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Row of Zeros – Example
2
Replacing the row with the coefficients of
1 11 285 23 121 4 01 4 002 0 0
1 42 02 4
1 02 02 0
1 02 02 0
No sign changes, so RHP poles, but Row of zeros indicates that system is marginally stable
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Stability Evaluation – Summary
If coefficients of have different signs System is unstable
If some coefficients of are zero System is, at best, marginally stable
If all coefficients have the same sign System may be stable or unstable Generate a Routh table and apply Routh‐Hurwitz criterion Replace any zero first‐column entries with and let take the limit as → 0
Replace a row of zeros with coefficients from the derivative of the auxiliary polynomial If no RHP poles are detected, the system is marginally stable