EEET2197 TUTORIAL EIGHT

EEET2197 CONTROL SYSTEMS

TUTORIAL 8 - SOLUTION.
Question 1.
Apply the Zeigler-Nichols oscillation method to tune a PI controller for the following plant.

G (s ) =
1
s + 3s + 3s + 1
3 2

[Answer]
The closed loop transfer function under proportional control only takes the form:
K p G (s ) Kp
T (s ) = =
1 + K p G (s ) s + 3s + 3s + 1 + K p
3 2

To determine the oscillation point we require a pair of complex conjugate poles on the jω axis, or:
( )
j − ω 3 + 3ω − 3ω 2 + 1 + K p = 0
This yields two equations involving the real and imaginary terms as:
(
−ω ω2 − 3 = 0 )
K p = 3ω 2 − 1
Thus we have:
ωc = 3 rad/s
Kc = 8
Applying the ZN table to define the PI parameters we have:
K p = 3.6, Tr = 3.023s

Question 2.
Apply the Zeigler-Nichols oscillation method to tune a PID controller for the following first order
plant incorporating a simple transport delay.
e−s
G (s ) =
s +1
[Answer]
The closed loop transfer function under proportional control only takes the form:
K p G (s ) K pe−s
T (s ) = =
1 + K p G (s ) s + 1 + K pe−s

To determine the oscillation point we require a pair of complex conjugate poles on the jω axis, or:
j ω + 1 + K p e − jω = 0

[
1 + K p cos ω + j ω − K p sin ω = 0 ]
This requires that:

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K p cos ω = −1
K p sin ω = ω

Dividing the second equation by the first equation gives:

tan ω = −ω
This is a non-linear algebraic equation of the form:
g (ω ) = f (ω )
Plotting these functions we find:

Apart from the trivial solution at w = 0, we find the next solution at:
ω = 2.0288 rad/s
Substituting this into K p cos ω = −1 , we find a value for the critical gain as:

K p = 2.2616
This means that the critical gain and the critical period are given by:
K c = 2.2616, Pc = 3.097 s
Substituting these values into the ZN tuning table we obtain:
K p = 1.3751, Tr = 1.548s, Td = 0.3871s
Then choose:
τ D = 0.1Td = 0.03871s

Question 3.
A system with a transfer function G(s) below is to be controlled using an integral controller. Sketch
the root locus of the system as a function of the controller gain KI.

G (s ) =
1
s + 2 s + 10
2

[Answer]
When an integral controller is used, the open loop gain takes the form:

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G (s )H (s ) =
KI
(
s s 2 + 2 s + 10 )
The poles of the forward path gain function G(s)H(s) are found to be:
p1 = 0, p2 = −1 + 3i, p3 = −1 − 3i
Plotting these values in the S-plane:

Next determine the real axis branch of the root locus. This must exist to the left of an odd number of
real axis poles or zeros. There is only one real axis pole located at the origin, so the locus branch is
placed to the left of this pole.

Next determine the number and positions of the locus asymptotes. There are no zeros, and three
poles so:
n−m =3
Therefore there are three asymptotes, which radiate from the point:

α=∑
p−∑z 2
=−
n−m 3
These asymptotes depart from the point α, at the angles:
180° + 360°(l − 1)
φ= , l = 1,2,3
n−m
φ = 60°, 180°, − 60°
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Drawing the asymptotes:

Next determine the locus departure angles for the two remaining branches. The departure angles are
given by:
φdep 2 = ∑ψ − ∑ φ − 180 ° = −90° − 108° − 180° + 360° = −18°

φdep 3 = ∑ψ − ∑ φ − 180 ° = −270° − 252° − 180° + (2)360° = 18°

Plotting the final root locus:

Question 4.
A system with a transfer function G(s) below is to be controlled using an integral controller. Sketch
the root locus of the system as a function of the controller gain KI.
s 2 + 8s + 32
G (s ) =
s 2 + 4 s + 20
[Answer]
When an integral controller is used, the open loop gain takes the form:

G (s )H (s ) =
(
K I s 2 + 8s + 32)
s (s + 4 s + 20 )
2

The poles and zeros of the forward path gain function G(s)H(s) are found to be:

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z1 = −4 + 4i, z 2 = −4 − 4i
p1 = 0, p2 = −2 + 4i, p3 = −2 − 4i
Plotting these values in the S-plane:

Next determine the real axis branch of the root locus. This must exist to the left of an odd number of
real axis poles or zeros. There is only one real axis pole located at the origin, and no real axis zero,
so the locus branch is placed to the left of this pole at the origin.

Next determine the number and positions of the locus asymptotes. There are two zeros, and three
poles so:
n − m =1
Therefore there is one asymptote, which corresponds to the real axis branch.
The locus branches which originate at the open loop complex conjugate poles must clearly track to
the open loop zeros. We must therefore determine the departure and arrival angles, and also whether
the locus branches cross the jω axis. Making the substitution of the form :
s = jω
1 + G (s )H (s ) = 0
s 3 + 4 s 2 + 20 s + K I s 2 + 8K I s + 32 K I = 0
− jω 3 − 4ω 2 + j 20ω − K I ω 2 + j8K I ω + 32 K I = 0
[ ] [ ]
32 K I − ω 2 (4 + K I ) + j ω (20 + 8K I ) − ω 3 = 0

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Separating the real and imaginary terms, we require:

32 K I − ω 2 (4 + K I ) = 0
ω (20 + 8K I ) − ω 3 = 0
From the real component we have two possible solutions:
32 K I
ω=±
4 + KI
Substituting into the imaginary component we find:
ω (20 + 8K I ) − ω 3 = ω [(20 + 8K I ) − ω 2 ]
32 K I  32 K I 
=± (20 + 8 K I ) − 
4 + KI  4 + KI 
32 K I  8 K I + 20 K I + 80 
2
=±  
4 + KI  4 + KI 
32 K I  8 K I 2 + 20 K I + 80 
=±  
4 + KI  4 + KI 
The quadratic in the numerator of this expression does not have any real valued roots since:

− 20 ± 20 2 − 4 × 8 × 80 − 20 ± 400 − 2560
K I _1, 2 = =
2×8 16
This means that the root locus does not cross the jω axis, and is constrained to the LHP.

Next determine the locus departure angles for the two remaining branches. The departure angles are
given by:
φdep 2 = ∑ψ − ∑ φ − 180°
= 0° + 76° − 90° − 116° − 180° + 360°
= 50°

φdep 3 = ∑ψ − ∑ φ − 180°
= 0° + 284° − 270° − 243° − 180° + 360°
= −50°
Similarly the arrival angles are given by:
ψ arr1 = ∑ φ − ∑ψ + 180°
= 180° + 135° + 104° − 90° + 180° − 360°
= 149°

ψ arr 2 = ∑ φ − ∑ψ + 180 °
= 180° + 256° + 225° − 270° + 180° − 2 × 360°
= −149°
This enables us to draw the final root-locus, ensuring that the departure and arrival angles are
accurately sketched, and that no locus branch crosses the jω axis.

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Question 5.
A system with a transfer function G(s) below is to be controlled using an integral controller. Sketch
the root locus of the system as a function of the controller gain KI.
s+3
G (s ) =
(s + 1)(s 2 + 4s + 5)
[Answer]
When a proportional controller is used, the open loop gain takes the form:
K p (s + 3)
G (s )H (s ) =
(
s (s + 1) s 2 + 4 s + 5 )
The poles and zeros of the forward path gain function G(s)H(s) are found to be:
z1 = −3
p1 = 0, p2 = −1, p3 = −2 + i, p4 = −2 − i
Plotting these values in the S-plane:

Next determine the real axis branch of the root locus. This must exist to the left of an odd number of
real axis poles or zeros, as is shown below:
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EEET2197 TUTORIAL EIGHT

Next determine the number and positions of the locus asymptotes. There is one zero, and four poles
so:
n−m =3
Therefore there are three asymptotes, which corresponds to the real axis branch.
Therefore there are three asymptotes, which radiate from the point:

α=∑
p−∑z 2
=−
n−m 3
These asymptotes depart from the point α, at the angles:
180° + 360°(l − 1)
φ= , l = 1,2,3
n−m
φ = 60°, 180°, − 60°
Drawing the asymptotes:

To complete the root locus it is necessary to determine the departure and arrival angles as follows:
φdep1 = ∑ψ − ∑ φ − 180°
= 0° − 0° − 153.4° + 153.4° − 180°
= 180°

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EEET2197 TUTORIAL EIGHT

φdep 2 = ∑ψ − ∑ φ − 180°
= 0° − 180° − 135° + 135° − 180° + 360°
= 180°

φdep 3 = ∑ψ − ∑ φ − 180°
= 45° − 90° − 135° − 153.4° − 180° + 360°
= −153.4°

φdep 4 = ∑ψ − ∑ φ − 180°
= −45° + 90° + 135° + 153.4° − 180°
= 153.4°

ψ arr1 = ∑ φ − ∑ψ + 180°
= −135° + 135° + 180° + 180° − 180°
= 180°
Plotting the final root locus:

Question 6.
A system with a transfer function G(s) below is to be controlled using an integral controller. Sketch
the root locus of the system as a function of the controller gain KI.
s+2
G (s ) =
(s + 3)(s 2 + 2s + 2)
[Answer]
When an integral controller is used, the open loop gain takes the form:
K I (s + 2 )
G (s )H (s ) =
(
s (s + 3) s 2 + 2 s + 2 )
The poles and zeros of the forward path gain function G(s)H(s) are found to be:
z1 = −2
p1 = 0, p2 = −3, p3 = −1 + i, p4 = −1 − i
Plotting these values in the S-plane:

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EEET2197 TUTORIAL EIGHT

Next determine the real axis branch of the root locus. This must exist to the left of an odd number of
real axis poles or zeros, as is shown below:

Next determine the number and positions of the locus asymptotes. There is one zero, and four poles
so:
n−m =3
Therefore there are three asymptotes, which corresponds to the real axis branch.
Therefore there are three asymptotes, which radiate from the point:

α=∑
p−∑z
= −1
n−m
These asymptotes depart from the point α, at the angles:
180° + 360°(l − 1)
φ= , l = 1,2,3
n−m
φ = 60°, 180°, − 60°
Drawing the asymptotes:

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EEET2197 TUTORIAL EIGHT

To complete the root locus it is necessary to determine the departure and arrival angles as follows:
φdep1 = ∑ψ − ∑ φ − 180°
= 0° − 0° − 153.4° + 153.4° − 180°
= 180°

φdep 2 = ∑ψ − ∑ φ − 180°
= 180° − 180° + 153.4° − 153.4° − 180° + 360°
= 180°

φdep 3 = ∑ψ − ∑ φ − 180°
= 45° − 135° − 90° − 26.6° − 180° + 360°
= −26.6°

φdep 4 = ∑ψ − ∑ φ − 180°
= −45° + 135° + 90° + 26.6° − 180°
= 26.6°

ψ arr1 = ∑ φ − ∑ψ + 180°
= 0° + 180° + 135° − 135° − 180°
= 0°
Plotting the final root locus:

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EEET2197 TUTORIAL EIGHT

Question 7.
A system with a transfer function G(s) below is to be controlled using a proportional controller.
Sketch the root locus of the system as a function of the controller gain Kp.
s+2
G (s ) =
(
s s + 5s + 4
2
)
[Answer]
When a proportional controller is used, the open loop gain takes the form:
K p (s + 2 )
G (s )H (s ) =
(
s s 2 + 5s + 4 )
The poles and zeros of the forward path gain function G(s)H(s) are found to be:
z1 = −2
p1 = 0, p2 = −1, p3 = −4
Plotting these values in the S-plane:

Next determine the real axis branch of the root locus. This must exist to the left of an odd number of
real axis poles or zeros, as is shown below:

Next determine the number and positions of the locus asymptotes. There is one zero, and three poles
so:

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n−m = 2
Therefore there are three asymptotes, which corresponds to the real axis branch.
Therefore there are three asymptotes, which radiate from the point:

α=∑
p−∑z
= −1.5
n−m
These asymptotes depart from the point α, at the angles:
180° + 360°(l − 1)
φ= , l = 1,2
n−m
φ = 90°, − 90°
Drawing the asymptotes:

To complete the root locus it is necessary to determine the departure and arrival angles as follows:
φdep1 = ∑ψ − ∑ φ − 180°
= 0° − 0° − 0° − 180°
= 180°

φdep 2 = ∑ψ − ∑ φ − 180°
= 0° − 180° − 0° − 180°
= 0°

φdep 3 = ∑ψ − ∑ φ − 180°
= 180° − 180° − 180° − 180°
= 0°

ψ arr1 = ∑ φ − ∑ψ + 180°
= 0° + 180° + 180° − 180°
= 180°
Plotting the final root locus:

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EEET2197 TUTORIAL EIGHT

B. P. McGrath August 2016

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