[PPT] - Plan of the Lecture Review: Nyquist stability criterion Todays PowerPoint Presentation

SLIDE 1

Plan of the Lecture

◮ Review: Nyquist stability criterion ◮ Today’s topic: Nyquist stability criterion (more examples);

phase and gain margins from Nyquist plots.

SLIDE 2

Plan of the Lecture

◮ Review: Nyquist stability criterion ◮ Today’s topic: Nyquist stability criterion (more examples);

phase and gain margins from Nyquist plots. Goal: explore more examples of the Nyquist criterion in action.

SLIDE 3

Plan of the Lecture

◮ Review: Nyquist stability criterion ◮ Today’s topic: Nyquist stability criterion (more examples);

phase and gain margins from Nyquist plots. Goal: explore more examples of the Nyquist criterion in action. Reading: FPE, Chapter 6

SLIDE 4

Review: Nyquist Plot

Consider an arbitrary transfer function H. Nyquist plot: Im H(jω) vs. Re H(jω) as ω varies from −∞ to ∞

Im H(jω) Re H(jω)

SLIDE 5

Review: Nyquist Stability Criterion

G(s) Y

+ −

R K Goal: count the number of RHP poles (if any) of the closed-loop transfer function KG(s) 1 + KG(s) based on frequency-domain characteristics of the plant transfer function G(s)

SLIDE 6

The Nyquist Theorem

G(s) Y

+ −

R K

Nyquist Theorem (1928) Assume that G(s) has no poles on the imaginary axis∗, and that its Nyquist plot does not pass through the point −1/K. Then N = Z − P #( of −1/K by Nyquist plot of G(s)) = #(RHP closed-loop poles) − #(RHP open-loop poles)

∗ Easy to fix: draw an infinitesimally small circular path that goes around

the pole and stays in RHP

SLIDE 7

The Nyquist Stability Criterion

G(s) Y

+ −

R K

N

#( of −1/K)

= Z

#(unstable CL poles)

− P

#(unstable OL poles)

Z = N + P Z = 0 ⇐ ⇒ N = −P

SLIDE 8

The Nyquist Stability Criterion

G(s) Y

+ −

R K

N

#( of −1/K)

= Z

#(unstable CL poles)

− P

#(unstable OL poles)

Z = N + P Z = 0 ⇐ ⇒ N = −P Nyquist Stability Criterion. Under the assumptions of the Nyquist theorem, the closed-loop system (at a given gain K) is stable if and only if the Nyquist plot of G(s) encircles the point −1/K P times counterclockwise, where P is the number

f unstable (RHP) open-loop poles of G(s).

SLIDE 9

Applying the Nyquist Criterion

Workflow: Bode M and φ-plots − → Nyquist plot

SLIDE 10

Applying the Nyquist Criterion

Workflow: Bode M and φ-plots − → Nyquist plot Advantages of Nyquist over Routh–Hurwitz

SLIDE 11

Applying the Nyquist Criterion

Workflow: Bode M and φ-plots − → Nyquist plot Advantages of Nyquist over Routh–Hurwitz

◮ can work directly with experimental frequency response

data (e.g., if we have the Bode plot based on measurements, but do not know the transfer function)

SLIDE 12

Applying the Nyquist Criterion

Workflow: Bode M and φ-plots − → Nyquist plot Advantages of Nyquist over Routh–Hurwitz

◮ can work directly with experimental frequency response

data (e.g., if we have the Bode plot based on measurements, but do not know the transfer function)

◮ less computational, more geometric (came 55 years after

Routh)

SLIDE 13

Example 1 (From Last Lecture)

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles)

SLIDE 14

Example 1 (From Last Lecture)

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Characteristic equation: (s + 1)(s + 2) + K = 0 ⇐ ⇒ s2 + 3s + K + 2 = 0

SLIDE 15

Example 1 (From Last Lecture)

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Characteristic equation: (s + 1)(s + 2) + K = 0 ⇐ ⇒ s2 + 3s + K + 2 = 0 From Routh, we already know that the closed-loop system is stable for K > −2.

SLIDE 16

Example 1 (From Last Lecture)

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Characteristic equation: (s + 1)(s + 2) + K = 0 ⇐ ⇒ s2 + 3s + K + 2 = 0 From Routh, we already know that the closed-loop system is stable for K > −2. We will now reproduce this answer using the Nyquist criterion.

SLIDE 17

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles)

SLIDE 18

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Strategy:

SLIDE 19

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Strategy:

◮ Start with the Bode plot of G

SLIDE 20

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Strategy:

◮ Start with the Bode plot of G ◮ Use the Bode plot to graph Im G(jω) vs. Re G(jω) for

0 ≤ ω < ∞

SLIDE 21

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Strategy:

◮ Start with the Bode plot of G ◮ Use the Bode plot to graph Im G(jω) vs. Re G(jω) for

0 ≤ ω < ∞

◮ This gives only a portion of the entire Nyquist plot

(Re G(jω), Im G(jω)) , −∞ < ω < ∞

SLIDE 22

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Strategy:

◮ Start with the Bode plot of G ◮ Use the Bode plot to graph Im G(jω) vs. Re G(jω) for

0 ≤ ω < ∞

◮ This gives only a portion of the entire Nyquist plot

(Re G(jω), Im G(jω)) , −∞ < ω < ∞

◮ Symmetry:

G(−jω) = G(jω)

SLIDE 23

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Strategy:

◮ Start with the Bode plot of G ◮ Use the Bode plot to graph Im G(jω) vs. Re G(jω) for

0 ≤ ω < ∞

◮ This gives only a portion of the entire Nyquist plot

(Re G(jω), Im G(jω)) , −∞ < ω < ∞

◮ Symmetry:

G(−jω) = G(jω) — Nyquist plots are always symmetric w.r.t. the real axis!!

SLIDE 24

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles)

SLIDE 25

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0. 0.1 1 10

175.
150.
125.
100.
75.
50.
25.

0. 1/2 0◦ −180◦

SLIDE 26

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0. 0.1 1 10

175.
150.
125.
100.
75.
50.
25.

0. 1/2 0◦ −180◦ −90◦ A

SLIDE 27

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0. 0.1 1 10

175.
150.
125.
100.
75.
50.
25.

0. 1/2 0◦ −180◦ −90◦ A

Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A

SLIDE 28

Example 1

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0. 0.1 1 10

175.
150.
125.
100.
75.
50.
25.

0. 1/2 0◦ −180◦ −90◦ A

Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

SLIDE 29

Example 1: Applying the Nyquist Criterion

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

SLIDE 30

Example 1: Applying the Nyquist Criterion

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=0

SLIDE 31

Example 1: Applying the Nyquist Criterion

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=0

= ⇒ K ∈ R is stabilizing if and only if #( of −1/K) = 0

SLIDE 32

Example 1: Applying the Nyquist Criterion

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=0

= ⇒ K ∈ R is stabilizing if and only if #( of −1/K) = 0

◮ If K > 0, #( of −1/K) = 0

SLIDE 33

Example 1: Applying the Nyquist Criterion

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=0

= ⇒ K ∈ R is stabilizing if and only if #( of −1/K) = 0

◮ If K > 0, #( of −1/K) = 0 ◮ If 0 < −1/K < 1/2,

#( of −1/K) > 0

SLIDE 34

Example 1: Applying the Nyquist Criterion

G(s) = 1 (s + 1)(s + 2) (no open-loop RHP poles) Nyquist plot:

0.1 0.2 0.3 0.4 0.5

0.3
0.2
0.1

0.1 0.2 0.3

−A G(∞) = 0

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=0

= ⇒ K ∈ R is stabilizing if and only if #( of −1/K) = 0

◮ If K > 0, #( of −1/K) = 0 ◮ If 0 < −1/K < 1/2,

#( of −1/K) > 0 = ⇒ closed-loop stable for K > −2

SLIDE 35

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) = 1 s3 + s2 + s − 3

SLIDE 36

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) = 1 s3 + s2 + s − 3 #(RHP open-loop poles) = 1 at s = 1

SLIDE 37

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) = 1 s3 + s2 + s − 3 #(RHP open-loop poles) = 1 at s = 1 Routh: the characteristic polynomial is s3 + s2 + s + K − 3 — 3rd degree

SLIDE 38

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) = 1 s3 + s2 + s − 3 #(RHP open-loop poles) = 1 at s = 1 Routh: the characteristic polynomial is s3 + s2 + s + K − 3 — 3rd degree — stable if and only if K − 3 > 0 and 1 > K − 3.

SLIDE 39

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) = 1 s3 + s2 + s − 3 #(RHP open-loop poles) = 1 at s = 1 Routh: the characteristic polynomial is s3 + s2 + s + K − 3 — 3rd degree — stable if and only if K − 3 > 0 and 1 > K − 3. Stability range: 3 < K < 4

SLIDE 40

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) = 1 s3 + s2 + s − 3 #(RHP open-loop poles) = 1 at s = 1 Routh: the characteristic polynomial is s3 + s2 + s + K − 3 — 3rd degree — stable if and only if K − 3 > 0 and 1 > K − 3. Stability range: 3 < K < 4 Let’s see how to spot this using the Nyquist criterion ...

SLIDE 41

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole)

SLIDE 42

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Bode plot:

0.1 1 10

80.
60.
40.
20.

0.1 1 10

260.
240.
220.
200.
180.

−180◦ −270◦

SLIDE 43

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Bode plot:

0.1 1 10

80.
60.
40.
20.

0.1 1 10

260.
240.
220.
200.
180.

−180◦ −270◦

Nyquist plot: ω = 0 M = 1/3, φ = −180◦ ω = 1 M = 1/4, φ = −180◦

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

SLIDE 44

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Bode plot:

0.1 1 10

80.
60.
40.
20.

0.1 1 10

260.
240.
220.
200.
180.

−180◦ −270◦

Nyquist plot: ω = 0 M = 1/3, φ = −180◦ ω = 1 M = 1/4, φ = −180◦ ω → ∞ M → 0, φ → −270◦

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

SLIDE 45

Example 2

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Bode plot:

0.1 1 10

80.
60.
40.
20.

0.1 1 10

260.
240.
220.
200.
180.

−180◦ −270◦

Nyquist plot: ω = 0 M = 1/3, φ = −180◦ ω = 1 M = 1/4, φ = −180◦ ω → ∞ M → 0, φ → −270◦

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

SLIDE 46

Example 2: Applying the Nyqiust Criterion

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole)

SLIDE 47

Example 2: Applying the Nyqiust Criterion

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Nyquist plot:

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=1

SLIDE 48

Example 2: Applying the Nyqiust Criterion

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Nyquist plot:

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=1

K ∈ R is stabilizing if and only if #( of −1/K) = −1

SLIDE 49

Example 2: Applying the Nyqiust Criterion

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Nyquist plot:

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=1

K ∈ R is stabilizing if and only if #( of −1/K) = −1 Which points −1/K are encircled once by this Nyquist plot?

SLIDE 50

Example 2: Applying the Nyqiust Criterion

G(s) = 1 (s − 1)(s2 + 2s + 3) (1 open-loop RHP pole) Nyquist plot:

0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.10
0.05

0.05 0.10

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=1

K ∈ R is stabilizing if and only if #( of −1/K) = −1 Which points −1/K are encircled once by this Nyquist plot?

nly − 1/3 < −1/K < −1/4

= ⇒ 3 < K < 4

SLIDE 51

Example 2: Nyquist Criterion and Phase Margin

Closed-loop stability range for G(s) = 1 (s − 1)(s2 + 2s + 3) is 3 < K < 4 (using either Routh or Nyquist).

SLIDE 52

Example 2: Nyquist Criterion and Phase Margin

Closed-loop stability range for G(s) = 1 (s − 1)(s2 + 2s + 3) is 3 < K < 4 (using either Routh or Nyquist). We can interpret this in terms of phase margin:

0.1 1 10

260.
240.
220.
200.
180.

−180◦ −270◦ for 3 < K < 4, ωc is here

SLIDE 53

Example 2: Nyquist Criterion and Phase Margin

Closed-loop stability range for G(s) = 1 (s − 1)(s2 + 2s + 3) is 3 < K < 4 (using either Routh or Nyquist). We can interpret this in terms of phase margin:

0.1 1 10

260.
240.
220.
200.
180.

−180◦ −270◦ for 3 < K < 4, ωc is here

So, in this case, stability ⇐ ⇒ PM > 0 (typical case).

SLIDE 54

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2

SLIDE 55

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Open-loop poles:

SLIDE 56

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Open-loop poles: s = −2 (LHP)

SLIDE 57

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Open-loop poles: s = −2 (LHP) s2 − s + 1 = 0

SLIDE 58

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Open-loop poles: s = −2 (LHP) s2 − s + 1 = 0

s − 1

2 2 + 3 4 = 0

SLIDE 59

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Open-loop poles: s = −2 (LHP) s2 − s + 1 = 0

s − 1

2 2 + 3 4 = 0 s = 1 2 ± j √ 3 2 (RHP)

SLIDE 60

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Open-loop poles: s = −2 (LHP) s2 − s + 1 = 0

s − 1

2 2 + 3 4 = 0 s = 1 2 ± j √ 3 2 (RHP) ∴ 2 RHP poles

SLIDE 61

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2

SLIDE 62

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Routh:

SLIDE 63

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Routh:

char. poly.

s3 + s2 − s + 2 + K(s − 1)

SLIDE 64

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Routh:

char. poly.

s3 + s2 − s + 2 + K(s − 1) s2 + s2 + (K − 1)s + 2 − K (3rd-order)

SLIDE 65

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Routh:

char. poly.

s3 + s2 − s + 2 + K(s − 1) s2 + s2 + (K − 1)s + 2 − K (3rd-order) — stable if and only if K − 1 > 0 2 − K > 0 K − 1 > 2 − K

SLIDE 66

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) = s − 1 s3 + s2 − s + 2 Routh:

char. poly.

s3 + s2 − s + 2 + K(s − 1) s2 + s2 + (K − 1)s + 2 − K (3rd-order) — stable if and only if K − 1 > 0 2 − K > 0 K − 1 > 2 − K — stability range is 3/2 < K < 2

SLIDE 67

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles)

SLIDE 68

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

SLIDE 69

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

SLIDE 70

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

◮ ω = 0 and ω → 0

SLIDE 71

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

◮ ω = 0 and ω → 0 ◮ ω = 1/

√ 2:

SLIDE 72

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

◮ ω = 0 and ω → 0 ◮ ω = 1/

√ 2: jω − 1 (jω − 1)((jω)2 − jω + 1)

ω=1/

√ 2

SLIDE 73

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

◮ ω = 0 and ω → 0 ◮ ω = 1/

√ 2: jω − 1 (jω − 1)((jω)2 − jω + 1)

ω=1/

√ 2

=

j √ 2 − 1

j

√ 2 + 2

− 1

2 − j √ 2 + 1

SLIDE 74

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

◮ ω = 0 and ω → 0 ◮ ω = 1/

√ 2: jω − 1 (jω − 1)((jω)2 − jω + 1)

ω=1/

√ 2

=

j √ 2 − 1

j

√ 2 + 2

− 1

2 − j √ 2 + 1

=

j √ 2 − 1

− 3

2

j

√ 2 − 1

= −2 3

SLIDE 75

Example 3

G(s) = s − 1 (s + 2)(s2 − s + 1) (2 open-loop RHP poles) Bode plot (tricky, RHP poles/zeros)

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

φ = 180◦ when:

◮ ω = 0 and ω → 0 ◮ ω = 1/

√ 2: jω − 1 (jω − 1)((jω)2 − jω + 1)

ω=1/

√ 2

=

j √ 2 − 1

j

√ 2 + 2

− 1

2 − j √ 2 + 1

=

j √ 2 − 1

− 3

2

j

√ 2 − 1

= −2 3 (need to guess this, e.g., by mouseclicking in Matlab)

SLIDE 76

Example 3

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles)

SLIDE 77

Example 3

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

SLIDE 78

Example 3

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

Nyquist plot: ω = 0 M = 1/2, φ = 180◦ ω = 1/ √ 2 M = 2/3, φ = 180◦

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

SLIDE 79

Example 3

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

Nyquist plot: ω = 0 M = 1/2, φ = 180◦ ω = 1/ √ 2 M = 2/3, φ = 180◦ ω → ∞ M → 0, φ → 180◦

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

SLIDE 80

Example 3

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Bode plot:

0.1 1 10

60.
50.
40.
30.
20.
10.

0.1 1 10 180. 190. 200. 210.

180◦ ω = 1/ √ 2

Nyquist plot: ω = 0 M = 1/2, φ = 180◦ ω = 1/ √ 2 M = 2/3, φ = 180◦ ω → ∞ M → 0, φ → 180◦

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

SLIDE 81

Example 3: Applying the Nyqiust Criterion

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles)

SLIDE 82

Example 3: Applying the Nyqiust Criterion

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Nyquist plot:

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=2

SLIDE 83

Example 3: Applying the Nyqiust Criterion

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Nyquist plot:

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=2

K ∈ R is stabilizing if and only if #( of −1/K) = −2

SLIDE 84

Example 3: Applying the Nyqiust Criterion

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Nyquist plot:

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=2

K ∈ R is stabilizing if and only if #( of −1/K) = −2 Which points −1/K are encircled twice by this Nyquist plot?

SLIDE 85

Example 3: Applying the Nyqiust Criterion

G(s) = s − 1 s3 + s2 − s + 2 (2 open-loop RHP poles) Nyquist plot:

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.3
0.2
0.1

0.1 0.2 0.3

#( of −1/K) = #(RHP CL poles) − #(RHP OL poles)

=2

K ∈ R is stabilizing if and only if #( of −1/K) = −2 Which points −1/K are encircled twice by this Nyquist plot?

nly − 2/3 < −1/K < −1/2

= ⇒ 3 2 < K < 2

SLIDE 86

Example 2: Nyquist Criterion and Phase Margin

CL stability range for G(s) = s − 1 s3 + s2 − s + 2: K ∈ (3/2, 2)

SLIDE 87

Example 2: Nyquist Criterion and Phase Margin

CL stability range for G(s) = s − 1 s3 + s2 − s + 2: K ∈ (3/2, 2) We can interpret this in terms of phase margin:

0.1 1 10 180. 190. 200. 210.

for 3/2 < K < 2, ωc is here

SLIDE 88

Example 2: Nyquist Criterion and Phase Margin

CL stability range for G(s) = s − 1 s3 + s2 − s + 2: K ∈ (3/2, 2) We can interpret this in terms of phase margin:

0.1 1 10 180. 190. 200. 210.

for 3/2 < K < 2, ωc is here

So, in this case, stability ⇐ ⇒ PM < 0

SLIDE 89

Example 2: Nyquist Criterion and Phase Margin

CL stability range for G(s) = s − 1 s3 + s2 − s + 2: K ∈ (3/2, 2) We can interpret this in terms of phase margin:

0.1 1 10 180. 190. 200. 210.

for 3/2 < K < 2, ωc is here

So, in this case, stability ⇐ ⇒ PM < 0 (atypical case; Nyquist criterion is the only way to resolve this ambiguity of Bode plots).

SLIDE 90

Stability Margins

How do we determine stability margins (GM & PM) from the Nyquist plot? GM & PM are defined relative to a given K, so consider Nyquist plot of KG(s) (we only draw the ω > 0 portion):

−M180◦ ϕ

How do we spot GM & PM?

SLIDE 91

Stability Margins

How do we determine stability margins (GM & PM) from the Nyquist plot? GM & PM are defined relative to a given K, so consider Nyquist plot of KG(s) (we only draw the ω > 0 portion):

−M180◦ ϕ

How do we spot GM & PM?

◮ GM = 1/M180◦

SLIDE 92

Stability Margins

How do we determine stability margins (GM & PM) from the Nyquist plot? GM & PM are defined relative to a given K, so consider Nyquist plot of KG(s) (we only draw the ω > 0 portion):

−M180◦ ϕ

How do we spot GM & PM?

◮ GM = 1/M180◦

— if we divide K by M180◦, then the Nyquist plot will pass through (−1, 0), giving M = 1, φ = 180◦

SLIDE 93

Stability Margins

How do we determine stability margins (GM & PM) from the Nyquist plot? GM & PM are defined relative to a given K, so consider Nyquist plot of KG(s) (we only draw the ω > 0 portion):

−M180◦ ϕ

How do we spot GM & PM?

◮ GM = 1/M180◦

— if we divide K by M180◦, then the Nyquist plot will pass through (−1, 0), giving M = 1, φ = 180◦

◮ PM = ϕ

SLIDE 94

Stability Margins

How do we determine stability margins (GM & PM) from the Nyquist plot? GM & PM are defined relative to a given K, so consider Nyquist plot of KG(s) (we only draw the ω > 0 portion):

−M180◦ ϕ