Introduction to Control 034040

Tutorial 10 – Nyquist stability criterion

Spring 2014

Department of Mechanical Engineering

Technion – Israel Institute of Technology

1

Short background

Consider the closed-loop control system

L(s)r y

n

−

and the Nyquist contours depicted in Fig. 1.

Re s

-R

R

Im s

Gs,2

Gs,1

Gs,3

R¥

(a) Nyquist contour

Re s

-R

R

Im s

´

Gs,0

Gs,4

Gs,2

Gs,1

Gs,3

R¥

(b) Nyquist contour for L(s) withpole(s) at the origin

Figure 1

• The Nyquist plot is the mapping ΓL of Γs by the loop transfer function, L(s).

• The graph of L(jω) in polar coordinates as ω runs from −∞ to +∞ is called the Nyquistplot of L(jω). The Nyquist plot can be constructed in two steps:

1. Construct the polar plot of L(jω).

2. Add the reflection of the polar plot about the real axis.

• When L(s) has poles on the imaginary axis, we add an infinite-radius arc connecting thepoint where ω → 0− and the point where ω → 0+ through the angle πl in the clockwisedirection.

2

• Nyquist stability criterion: If the Nyquist plot of L(jω) doesn’t cross the critical point(−1 + 0j), the number of closed-loop unstable poles, P+

clis

P+cl

= P+ol+ ν

where ν is the number of clockwise encirclements of the critical point by the Nyquist plot,and P+

olis the number of unstable open-loop poles. For stability, we require P+

cl= 0; hence,

P+ol

= −ν.

• If the Nyquist plot does intersect the critical point, say at the frequency ω = ω0, then χcl(s)has at least one root at s = jω.

3

Question 1

Use the Nyquist stability creterion to determine the stability of the closed-loop systems below (theplant P (s) = 150

(s+2)(s+5)(s+10)is controlled by the controller C(s) = 1):

1150

(s+2)(s+5)(s+10)r e u

dy

n

−ym

Question 1 Solution

Step 1: Construct the polar plot of L(jω)

Here, L(s) = P (s)C(s) = P (s) = 150(s+2)(s+5)(s+10)

. Compute the loop frequency response

L(jω) =150

(jω + 2) (jω + 5) (jω + 10)

⇓

|L(jω)| = 150√ω2 + 4 ·

√ω2 + 25 ·

√ω2 + 100

argL(jω) = − tan−1(ω

2

)

− tan−1(ω

5

)

− tan−1( ω

10

)

Compute the initial values,

|L(0)| = 1.5

⇓argL(0) = 0

and the final values,

|L(j∞)| = 0

⇓argL(jω) = −270

Since |L(jω)| and argL(jω) are monotonically decreasing functions of ω, there can be only oneintersection point of the polar plot with the negative part of the real axis. In order to find theintersection frequency, ωP , use the trigonometric formula

tan−1 (a)± tan−1 (b) = tan−1

(

a± b

1∓ ab

)

4

The intersection frequency, ωP , satisfies argL (jωP ) = −180.

argL(jω) = − tan−1(ω

2

)

− tan−1(ω

5

)

− tan−1( ω

10

)

= − tan−1

(

7ω

10− ω2

)

− tan−1( ω

10

)

= − tan−1

(

ω (80− ω2)

100− 17ω2

)

= −180

Hence,ω (80− ω2)

100− 17ω2= 0

The solution which satisfies the equation and drives the denominator to be negative is ωP =√80[

rad

sec

]

≈ 8.94[

rad

sec

]

(as ω0 = 0[

rad

sec

]

drives the denominator to be positive, which implies thatω0 = 0

[

rad

sec

]

is the intersection frequency with the positive part of the real axis). The magnitudeof L(jωP ) is

|L (jωP )| =150

√

ω2P+ 4 ·

√

ω2P+ 25 ·

√

ω2P+ 100

≈ 0.12

Hence, the polar plot intersects the negative part of the real axis at P = (−0.12, 0).

Step 2: Draw the Nyquist plot and count the encirclements

Fig. 2 shows the Nyquist plot of L(jω).

-1 -0.12 1.5Re

Im

Figure 2: Nyquist plot of L(s)

Our loop transfer function is stable; namely, P+ol

= 0. Therefore, in order to have a stable closed-loop (P+

ol= 0), we need ν = 0. Namely, the Nyquist plot must not encircle the critical point. It

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is readily seen that there are no encirclements of the critical point; therefore, the closed-loop isstable.Q: What would change if the controller was C(s) = 8.333 or C(s) = 20 ?Hint: The Nyquist diagrams presented in Fig. 3 should help you out a little.

12.5Re

Im

(a) Nyquist diagram: C(s) = 8.333

30Re

Im

(b) Nyquist diagram: C(s) = 20

Figure 3

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Question 2

Use the Nyquist stability creterion to determine the stability of the closed-loop systems below (theplant P (s) = 1

s(s2+6s+25)is controlled by the controller C(s) = 10):

101

s(s2+6s+25)r e u

dy

n

−ym

Question 2 Solution

Here, the loop transfer function is

L(s) = P (s)C(s) =10

s (s2 + 6s+ 25)

Compute the frequency response

L(jω) =10

jω (25− ω2 + 6jω)

⇓

|L(jω)| = 10

ω

√

(25− ω2)2 + 36ω2

argL(jω) = −90 − tan−1

(

6ω

25− ω2

)

Find the initial values

|L(0)| = ∞argL(0) = −90

and the final values

|L(j∞)| = 0

argL(j∞) = −90 − 180 = −270

Namely, the polar plot goes from ∞ to 0. In order to understand it better, split L(jω) into its realand imaginary parts,

L(jω) = − 60

36ω2 + (25− ω2)2− 10 (25− ω2)

ω(

36ω2 + (25− ω2)2)j

When ω → 0, we have

L(0) = − 60

252−∞j = −0.096−∞j

7

Hence, we have an asymptote at −0.096. Now, compute the intersection of the polar plot with thenegative part of the real axis; namely, argL (jωP ) = −180

argL (jωP ) = −90 − tan−1

(

6ωP

25− ω2P

)

= −180

⇓

tan−1

(

6ωP

25− ω2P

)

= 90

⇓

ωP = 5

[

rad

sec

]

Compute the gain of L(jωP )

|L(jωP )| =10

ωP

√

(25− ω2P)2+ 36ω2

P

=10

5√

(25− 52)2 + 36 · 52=

1

15

Hence, the intersection point with the negative part of the real axis is P =(

− 115, 0)

. Now, con-struct the Nyquist plot (Fig. 4)

-1 -115

Re

Im

-0.096

Figure 4: Nyquist plot of L(s)

The open-loop is stable (the integrator does NOT count!); thus, P+ol

= 0. In order to obtain astable closed-loop, P+

cl= 0, we need ν = 0; namely, the polar plot must not encircle the critical

point. Indeed, there are no encirclements; therefore, the closed-loop is stable.Q: What would change if the controller was C(s) = 150 or C(s) = 300 ?Hint: The Nyquist diagrams presented in Fig. 5 should help you out a little.

8

-1Re

Im

(a) Nyquist diagram: C(s) = 150

-1Re

Im

(b) Nyquist diagram: C(s) = 300

Figure 5

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Question 3

Use the Nyquist stability creterion to determine the stability of the closed-loop systems below (aplant controlled by the integral controller C(s) = 1

s) for

1. P (s) =√3s+1

s(s+√3)

2. P (s) = s+√3

s(√3s+1)

1s

P (s)r e u

dy

n

−ym

Question 3 Solution

Item 1

The loop transfer function is

L(s) = P (s)C(s) =

√3s+ 1

s2(

s+√3)

Compute the frequency response

L(jω) =

√3jω + 1

−ω2(

jω +√3)

⇓

|L(jω)| =√3ω2 + 1

ω2√ω2 + 3

argL(jω) = tan−1

(√3ω

1

)

− tan−1

(

ω√3

)

− 180

The initial and the final values are

|L(0)| = ∞argL(0) = −180

|L(j∞)| = 0

argL(j∞) = −180

Since tan−1 α is a monotonically increasing function of α, tan−1(√

3ω1

)

> tan−1(

ω√3

)

, as√3ω1

>ω√3

∀ω.

10

tan-1H 3 ΩL tan-1HΩ 3 L

Ω

Therefore, argL(jω) is always greater than −180. Split L(jω) into its real and imaginary partsand obtain

L(jω) = −(ω2 + 1)√3

ω2 (ω2 + 3)− 2

ω (ω2 + 3)j

⇓

L(0) = limω→0

(

−√3

3ω2− 2

3ωj

)

Although there is no finite asymptote, the real part approaches zero faster than the imaginary. Atthis point we can draw the Nyquist diagram (Fig. 6). Note that we have two integrators in L(s);hence, we need to add a 360 arc from L (j0−) to L (j0+) in clockwise direction.

Ω=0-

Ω=0+

-1Re

Im

Figure 6: Nyquist plot of L(s)

Since the open-loop is stable (P+ol

= 0, as the integrators don’t count), for a stable closed-loop(P+

cl= 0) we need ν = 0 encirclements of the critical point. This actually happens here, so the

closed-loop system is stable.

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Item 2

The loop transfer function is

L(s) = P (s)C(s) =s+

√3

s2(√

3s+ 1)

Compute the frequency response

L(jω) =jω +

√3

−ω2(√

3jω + 1)

⇓

|L(jω)| =√ω2 + 3

ω2√3ω2 + 1

argL(jω) = tan−1

(

ω√3

)

− tan−1

(√3ω

1

)

− 180

The initial and the final values are

|L(0)| = ∞ |L(j∞)| = 0argL(0) = −180 argL(j∞) = −180

Since tan−1 α is a monotonically increasing function of α, tan−1(

ω√3

)

− tan−1(√

3ω1

)

< 0, as

ω√3<

√3ω1

∀ω.Therefore, argL(jω) is always smaller than −180. Construct the Nyquist diagram

(Fig. 7). Note that we have two integrators in L(s); hence, we need to add a 360 arc from L (j0−)to L (j0+) in clockwise direction.

Ω=0+

Ω=0-

-1Re

Im

Figure 7: Nyquist plot of L(s)

The open-loop is still stable, for a stable closed-loop (P+cl

= 0) we need ν = 0 encirclements of thecritical point. However, now we have two encirclements; therefore, P+

cl= P+

ol+ ν = 2. Namely, the

closed-loop is unstable (has two unstable poles).

12

Question 4

Use the Nyquist stability creterion to determine the stability of the closed-loop systems below (theplant P (s) = s+1

s2+1is controlled by the controller C(s) = 1):

1 s+1s2+1

r e ud

y

n

−ym

Determine the closed-loop stability.

Question 4 Solution

The loop transfer function is

L(s) = P (s)C(s) =s+ 1

s2 + 1

The frequency response is

L(jω) =jω + 1

1 − ω2

⇓

|L(jω)| =√ω2 + 1

|1− ω2|

argL(jω) =

tan−1 (ω) 0 ≤ ω < 1

tan−1 (ω)− 180 1 < ω < ∞

The initial and final values are

|L(0)| = 1

argL(0) = 0

|L(j∞)| = 0

argL(j∞) = −90

Also compute the values of L(jω) when ω → 1

|L(j1)| = ∞argL(j1−) = 45

argL(j1+) = −135

The Nyquist plot is presented in Fig. .

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Ω=1-

Ω=1+

Ω=-1 -

Ω=-1 +

Ω=

0+

Ω=0 -

Ω®+¥

Ω®-¥

-1Re

Im

Figure 8: Nyquist plot of L(s)

The open-loop is stable (the poles on jω axis don’t count); therefore, we need ν = 0 encirclementsof the critical point to have a stable closed-loop. Indeed, we have 0 encirclements, and the closed-loop is stable.

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