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Tutorial 10 e
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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
5
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
6
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
9
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.
11
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. .
13
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.
14