Lecture 18 Nyquist Stability Criterion - BioMechaTronics · 1/38 MECE 3350U Control Systems Lecture...
Transcript of Lecture 18 Nyquist Stability Criterion - BioMechaTronics · 1/38 MECE 3350U Control Systems Lecture...
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MECE 3350UControl Systems
Lecture 18Nyquist Stability Criterion
MECE 3350 - C. Rossa 1 / 38 Lecture 18
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Outline of Lecture 18
By the end of today’s lecture you should be able to
• Extend the concept of gain and phase
• Understand the Nyquist stability criterion
• Determine the stability based on open loop transfer function
MECE 3350 - C. Rossa 2 / 38 Lecture 18
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Applications
Knowing the open-loop transfer function of the system below, how can weevaluate its stability without computing the closed-loop transfer function?
MECE 3350 - C. Rossa 3 / 38 Lecture 18
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Gain and phase - review
Any system can be characterized by its frequency response to a sinusoidalexcitation.
The ratio B/A is called the gain of G(s) for given frequency.
The phase shift φ is the phase of G(s) for a given frequency.
Data can be obtained experimentally is G(s) is unknown.
MECE 3350 - C. Rossa 4 / 38 Lecture 18
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Gain and phase - reviewFor a generic transfer function G(s)
G(s) = k∏n
i=1(s + zi )∏mk=1(s + pk )
we can evaluate the gain at a frequency ω by letting s = jω.
The gain is
G(jω) = |k|∏n
i=1 |jω + zi |∏mk=1 |jω + pk |
where |jω ± a| =√ω2 + (±a)2
MECE 3350 - C. Rossa 5 / 38 Lecture 18
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Gain and phase - reviewFor a generic transfer function G(s)
G(s) = k∏n
i=1(s + zi )∏mk=1(s + pk )
we can evaluate the phase at a frequency ω by letting s = jω.
The phase is
∠G(jω) = ∠|k|+n∑
i=1
∠(jω + zi )−m∑
k=1
∠(jω + pk )
where ∠(jω + a) = tan−1 ω/a
MECE 3350 - C. Rossa 6 / 38 Lecture 18
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Open loop vs closed loop stability
Generally the process and controller transfer functions are known
The open loop transfer function is L(s) = C(s)H(s).
Closing the loop changes the transfer function to
T (s) = C(s)G(s)1 + C(s)G(s) = L(s)
1 + L(s)
MECE 3350 - C. Rossa 7 / 38 Lecture 18
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Open loop vs closed loop stabilityOpen-loop stability
T (s) = C(s)G(s) (1)
→ Evaluate the location of the poles of C(s)G(s)
Closed-loop stability
T (s) = C(s)G(s)1 + C(s)G(s)
→ Evaluate the location of the zeros of 1 + C(s)G(s)
Example: If C(s)G(s) = s+as+b
→ Open-loop stable if C(s)G(s) has real negative poles: i.e., b > 0
→ Closed-loop stable if 1 + C(s)G(s) has real negative zeros:
T (s) =s+as+b
1 + s+as+b
=s+as+b
(s+b)+(s+a)s+b
MECE 3350 - C. Rossa 8 / 38 Lecture 18
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Open loop vs closed loop stability
The open loop transfer function
C(s)G(s) = s + as + b (2)
has a zero at −a and pole at −b.
The characteristic equation of the closed-loop transfer function in a unitfeedback system becomes
1 + C(s)G(s) = 1 + s + as + b = s + a + s + b
s + b (3)
and has a pole at −b.
The pole is (3) the same as in (2) !
For close-loop stability, the zeros of the characteristic equation, i.e. the zerosof 1 + C(s)G(s), must have negative real parts.
MECE 3350 - C. Rossa 9 / 38 Lecture 18
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Function mappingConsider the hypothetical function
H(s) = s − 2
How can we determine the location of the zeros of H(s) graphically?
We can map any point from the s-plane into the "w" plane:→ s = 2 + 2j becomes H(2 + 2j) = 2j→ s = 1 + j becomes H(1 + j) = −1 + j→ s = −j becomes H(−j) = −2− j
MECE 3350 - C. Rossa 10 / 38 Lecture 18
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Cauchy’s argument principle
H(s) =|H(s)|∠H(s)
=|k|∏n
i=1 |s + zi |∏mk=1 |s + pk |
(n∑
i=1
∠(s + zi )−m∑
k=1
∠(s + pk )
)=|H(s)|
(∑φi −
∑φk
)
MECE 3350 - C. Rossa 11 / 38 Lecture 18
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Cauchy’s argument principle
1 - Select a point P in the s-plane
2 - Draw the vectors from P to each zero and pole
3 - Calculate the magnitude of each vector
4 - The magnitude is the product of magnitude of zeros divided by the productof the magnitude of poles
5 - The angle isφ = φ1 + φ2 − φ3 − φ4
MECE 3350 - C. Rossa 12 / 38 Lecture 18
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Cauchy’s argument principle
As s traverses Γ2, the net angle change of φ is
As s traverses Γ1, the net angle change of |φ| is
MECE 3350 - C. Rossa 13 / 38 Lecture 18
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Cauchy’s argument principle
As s traverses Γ1, the net angle change of φ is
MECE 3350 - C. Rossa 14 / 38 Lecture 18
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Cauchy’s argument principle
As s traverses Γ1, the net angle change of φ is
MECE 3350 - C. Rossa 15 / 38 Lecture 18
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Cauchy’s argument principleAssume that the characteristic equation of 1 + C(s)G(s) has:
→ A number P of poles in the right-half plane.→ A number N of zeros in the right-half plane.
For an contour that encircles the entire right-half plane:
The relation between P, Z , and the net number N of clockwise encirclementsof the origin is:
N =
MECE 3350 - C. Rossa 16 / 38 Lecture 18
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Cauchy’s argument principleA contour map of a complex function will encircle the origin
N = Z − P
times, where Z is the number of zeros and P is the number of poles of thefunction inside the contour.
The number of unstable poles are known: They are the same as in theopen-loop transfer function!
MECE 3350 - C. Rossa 17 / 38 Lecture 18
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Nyquist plot
Assuming that there are no poles in the right-half plane, are the followingsystems stable?
1 + C(s)G(s) = 0
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Nyquist plot
1 + C(s)G(s) = 0. (4)If there is a zero or pole of (4) in the right-half s-plane, the contour of (4)encircles the origin.
Subtracting 1 from the above equation shifts the contour to the left
Thus, if the open-loop equation
T (s) = C(s)G(s) (5)has a zero or pole in the right-half s-plane, the contour of (5) encircles -1.
MECE 3350 - C. Rossa 19 / 38 Lecture 18
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The Nyquist Stability Criterion
An open-loop transfer function L(s) has Z unstable closed-loop roots given by
Z = N + P
where
→ N is the number of clockwise encirclements of −1
→ P is the number of poles in the right-half s-plane
Note: If encirclements are in the counterclockwise direction, N is negative.
For stability, we wish to have Z = 0.
MECE 3350 - C. Rossa 20 / 38 Lecture 18
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The Nyquist Stability Criterion
A open-loop transfer function L(s) is closed-loop stable if and only if:
The number of counterclockwise encirclements of the −1 + 0j point is equal tothe number of poles of L(s) with positive real parts.
Z = N + P
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Example 1Is this closed-loop system stable?
The Nyquist plot of the open-loop transfer function L(s) = s+1s−0.5 is
-2 -1 0 1-1.5
1.5
Real Axis
Imag
inar
y A
xis
P = , N = , Z = N + P =
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Example 1 - continued
The closed-loop transfer function is
T (s) =s+1
s−0.5
1 + s+1s−0.5
= s + 12s + 0.5
Step response
0 40
0.5
2
Time (seconds)
Am
plitu
de
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Example 2Is this closed-loop system stable?
The Nyquist plot of the open-loop transfer function L(s) = s−1(s−0.25)2 is
-16 0 2
-10
10
Real Axis
Imag
inar
y A
xis
-1
P = , N = , Z = N + P =
MECE 3350 - C. Rossa 24 / 38 Lecture 18
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Example 2 - continued
The closed-loop transfer function is
T (s) =s−1
s2−0.5s+0.0625
1 + s−1s2−0.5s+0.0625
= s − 1s2 + 0.5s − 0.9375
The poles are: −1.25 and 0.75 and the step response is
0 10
0
Am
plitu
de
Time (seconds)
1 + L(s) has ONE unstable zero.
MECE 3350 - C. Rossa 25 / 38 Lecture 18
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Nyquist plot
How to create the Nyquist plot for a given function?
Next class!
MECE 3350 - C. Rossa 26 / 38 Lecture 18
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Exercise 105
The Nyquist plot of a conditionally stable open loop system is shown in thefigure.
(a) Determine whether the closed-loop system is stable
(b) Determine whether the closed-loop system is stable if the −1 point lies atthe dot on the axis.
MECE 3350 - C. Rossa 27 / 38 Lecture 18
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Exercise 105 - continued
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Exercise 106
A unit feedback system has a loop transfer function
L(s) = C(s)G(s) = kτs − 1
where k = 0.5 and τ = 1. Based on its Nyquist plot show below, determinewhether the system is stable.
-1 -0.8 -0.4 0 0.2
-0.2
0
0.2
Real Axis
Imag
inar
y A
xis
MECE 3350 - C. Rossa 29 / 38 Lecture 18
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Exercise 106 - continued
L(s) = C(s)G(s) = 0.5s − 1
MECE 3350 - C. Rossa 30 / 38 Lecture 18
-1 -0.8 -0.4 0 0.2
-0.2
0
0.2
Real Axis
Imag
inar
y A
xis
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Exercise 106 - continuedWhat value of k is required for stability?
L(s) = C(s)G(s) = ks − 1
MECE 3350 - C. Rossa 31 / 38 Lecture 18
-1 -0.8 -0.4 0 0.2
-0.2
0
0.2
Real Axis
Imag
inar
y A
xis
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Exercise 107
A loop transfer function is
L(s) = 1s + 2
Using the contour in the s-plane shown, determine the corresponding contourin the F (s) (or "w") plane.
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Exercise 107 - continued
L(s) = 1s + 2
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Exercise 108
Based on the Nyquist plot, evaluate the stability of
T (s) = s(s − 0.1)2
-5 -4 -3 -2 -1 0 1
-2
0
2
Real Axis
Imag
inar
y A
xis
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Exercise 109
Based on the Nyquist plot, evaluate the stability of
T (s) = 50(s + 5)(s − 9)
-1 -0.6 -0.2 0
-0.2
-0.1
0
0.1
0.2
Real Axis
Imag
inar
y A
xis
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Exercise 110
Based on the Nyquist plot, evaluate the stability of
T (s) = s + 0.5s3 + s2 + 1
-1.2 -1 0 0.6-1.5
0
1.5
Real Axis
Imag
inar
y A
xis
MECE 3350 - C. Rossa 36 / 38 Lecture 18
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Exercise 111
Based on the Nyquist plot, evaluate the stability of
T (s) = 10 s + 0.5s3 + s2 + 1
-10 -1 0 8-15
0
15
Real Axis
Imag
inar
y A
xis
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Next class...
• Nyquist plot
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