t07 control system

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Curtin University of Technology
Department of Electrical and Computer Engineering
CS301/CS603
Tutorial 7
1. Consider the system shown in Figure 1, where
D(s) = K
s+ α
s2 + ω2
Σ
+
−
D(s) 1
s(s+1)
R Y
Figure 1. Block diagram for Problem 1.
Find the range of K and α such that the closed-loop system is stable. For the special case
where α = −1 is fixed, determine whether the closed-loop system would be stable for some
set of values for K.
Answer:
With the open-loop transfer function
G(s) =
K(s+ α)
s(s+ 1)(s2 + ω2)
the closed-loop transfer function is seen to be
Y (s)
R(s)
=
G(s)
1 +G(s)
=
K(s+ α)
s4 + s3 + ω2s2 + (ω2 +K)s+Kα
From the Routh array
1 ω2 Kα
1 ω2 +K 0
−K Kα 0
ω2 +K + α 0 0
Kα 0 0
it follows that the conditions for closed-loop stability are
K < 0 and − (ω2 +K) < α < 0
With α = −1, the above reduce to
1− ω2 < K < 0
Thus, there exists a value of K such that the system is stable when α = −1 is fixed provided
|ω| > 1.
2. A DC-motor speed control is described by the differential equation
y˙ + 60y = 600va − 1500w
where y is the motor speed, va is the armature voltage, and w is the load torque. Assume the
armature voltage is computed using the PI control law
va = −
(
K1y +Ki
∫ t
0
ydt
)
(a) Compute the transfer function from w to y.
(b) Compute values forK1 andKi so that the closed-loop system will have poles at −60±60j.
Answer:
Applying the Laplace transform to the differential equation yields
(s+ 60)Y (s) = 600Va(s)− 1500W (s)
As
Va(s) = −(K1 +Ki/s)Y (s)
there results
Y (s)
W (s)
=
−1500s
s2 + (60 + 600K1)s+ 600Ki
To have the closed-loop poles placed at −σ ± jωd = −60± 60j, there must hold
60 + 600K1 = 2σ = 120
600Ki = ω
2
n = σ
2 + ω2d = 7200
Solving these equations yields
K1 = 0.1 and Ki = 12
3. Consider the unity feedback system depicted in Figure 2. You are to design what is called a
proportional-integral (PI) controller so that the closed-loop poles lie within the shaded regions
shown in Figure 3.
R
Σ
+
−
K 1+KI
s
Kα
s+α
Y
Figure 2. The closed-loop system.
Re(s)
Im(s)
0-5 -4 -3 -2 -1 1 2 3
1
2
3
-3
-2
-1
Figure 3. The complex s plane.
(a) Let Kα = α = 2. Find values for K and KI so that the poles of the closed-loop system
lie within the shaded regions (edge inclusive).
(b) Prove that for any nonzero Kα and negative α, the proportional-integral controller can
provide enough flexibility to place the closed-loop poles anywhere in the open left half
s-plane.
Answer:
(a) The closed-loop transfer function of the system is
H(s) =
KKα(s+KI)
s2 + (α +KKα)s+KKαKI
=
2K(s+KI)
s2 + (2 + 2K)s+ 2KKI
and a pair of complex closed-loop poles can be found as
p = (−1−K)± j
√
2KKI − (1 +K)2
To ensure the complex closed-loop poles to lie within the shaded region, the following
two inequalities must be satisfied:
−4 ≤ −1−K ≤ −2
12 ≤ 2KKI − (1 +K)2 ≤ 32
which are equivalent to
1 ≤K ≤ 3
1 + (1 +K)2
2K
≤KI ≤ 9 + (1 +K)
2
2K
(b) We know that the characteristic equation of the given system is in the form
s2 + a1s+ a2 = 0
The closed-loop poles can be placed anywhere in the open left half plane if and only if
the characteristic equation can have arbitrary positive coefficients a1 and a2. Recalling
the transfer function derived in part (a) we have
a1 = α +KKα and a2 = KKαKI
Solving for K and KI yields
K =
a1 − α
Kα
and KI =
a2
a1 − α
Since a1 − α > 0, the PI controller parameters K and KI can always be determined
for any given a1 and a2. In other words, it is always possible to design the proper PI
controller to place the closed-loop poles at any desired positions in the open left half
plane.