EGGN307: Root Locus III: PD Design∗
Tyrone Vincent
Lecture 22
Contents
1
PD Design
1
2
Quiz Yourself
2.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
6
1
PD Design
Design Problem
• Choose parameters of a PD controller to achieve a desired closed loop step response.
R(s) +
E(s)
Kp
U (s)
+
−
G(s)
Y (s)
−
Kd s
To design the controller, we need to see how the control will affect the closed loop system response.
• Closed Loop Transfer Function:
Kp G(s)
Y (s)
=
R(s)
1 + G(s)(Kp + Kd s)
• Closed Loop Poles:
1 + G(s)(Kp + Kd s) = 0
Kp
1 + Kd s +
G(s) = 0
Kd
∗
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1
K
Note that the controller has a result similar to proportional control, except it adds an additional zero at − Kpd .
The control design has two parts: Place the zero, then choose the gain Kd . We will work through these steps using
the example open loop system
1
G(s) = 2
s + 3s + 2
• Step 1: Translate design specifications to desired closed loop pole locations
Example 1. Specifications:
– % OS ≤ 10 %
– 1% Settling time ≤ 2.3s
From these specifications, we find requirements on ζ and ωn :
ζ≥p
− ln(0.1)
π 2 + ln(0.1)2
≥ 0.59
4.6
ζωn
ζωn ≥ 2
2.3 ≥
This implies the following region of acceptable closed loop poles:
Im(s)
in this region,
%OS≤ 10%,
ts < 2.3s
cos−1 (.59) = 54◦
−2
Re(s)
• Step 2: Plot root locus of G(s) and check if design can be met using proportional control
Root locus with proportional control alone
» s = tf([1 0],[1]) % define s as the Laplace variable
» sys = 1/(sˆ2+3*s+2);
» rlocus(sys)
2
With the closed loop pole specifications overlaid on the root locus, we see that proportional control cannot meet
the specifications.
• Step 3: Choose corner point in specification and solve for zero to place root locus through this point
The point where the specifications intersect (and thus both specifications are just met) is given by
s̄ = −2 + 2 × tan(54◦ )j = −2 + 2.74j
The root locus criterion for the system plus zero needs to be satisfied at this point:
Kp
∠ s̄ +
G(z̄) = −180◦
Kd
For simplicity, let a =
Kp
Kd .
Then we want
∠
(s̄ + a)
= −180◦
(s̄ + 2)(s̄ + 1)
where we have factored the denominator s2 + 3s + 2 = (s + 2)(s + 1). This equation is solved for a,
∠(s̄ + a) − ∠(s̄ + 2) − ∠(s̄ + 1) = −180◦
∠ − 2 + j2.74 + a − ∠ − 2 + j2.7 + 2 − ∠(−2 + j2.7 + 1) = −180◦
∠a − 2 + j2.74 − ∠j2.7 − ∠ − 1 + j2.7 = −180◦
Paying attention to quadrants, we find that
∠j2.7 = 90◦
∠ − 1 + j2.7 = 110◦
Thus,
∠a − 2 + j2.74 = −180◦ + 90◦ + 110◦ = 20◦ .
So what does a need to be in order that the resulting angle is 20◦ ? First we note that 20◦ is in the first quadrant,
so the angle formula is simply the inverse tangent of the imaginary part over the real part
2.74
−1
∠a − 2 + j2.74 = tan
= 20◦
a−2
Take the tangent of both sides,
2.74
= tan(20◦ ) = .364
a−2
3
Solve for a
a = 2 + 2.74
1
= 9.53
.364
Finally, we have the ratio of Kp to Kd
Kp
= 9.53
Kd
• Step 4: The next step is to solve for Kd . To do this, we go back to the fundamental equation for closed loop
poles
Kp
G(s̄) = 0
1 + Kd s̄ +
Kd
which we want to be satisfied at s̄ = −2 + 2.74j. Moving the 1 to the other side and taking the magnitude of
both sides gives us
Kd s̄ + Kp G(s̄) = | − 1|
Kd
1
Kd = Kp
s̄ + Kd G(s̄)
Where we used our knowledge that Kd is positive and real. Plugging in what we know so far,
1
Kd = 1
(s̄ + 9.53) (s̄+2)(s̄+1) s̄=−2+2.74j
|(s̄ + 2)| |(s̄ + 1)|
=
|s̄ + 9.53|
s̄=−2+2.74j
|−2 + 2.74j + 2| |−2 + 2.74j + 1|
|−2 + 2.74j + 9.53|
|2.74j|| − 1 + 2.74j|
=
7.53 + 2.74j|
√
(2.74)( 1 + 2.742 )
= √
7.532 + 2.742
= 0.997
=
Finally, since Kd = 0.997, Kp = 9.53 × 0.997 = 9.5.
• Step 5: Check result with new root locus. The roots locus for the PD system is found using
s=tf([1 0],1);
sys=(s+9.53)*(1/(sˆ2+3*s+2));
rlocus(sys)
Root Locus with PD controller
4
Root Locus
10
System: sys
Gain: 1.02
Pole: −2.04 + 2.77i
Damping: 0.593
Overshoot (%): 9.9
Frequency (rad/sec): 3.44
8
Imaginary Axis
6
4
2
0
−2
−4
−6
−8
−10
−35
2
−30
−25
−20
−15
Real Axis
−10
−5
0
5
Quiz Yourself
2.1
Questions
1. An electro-magnet can be used to balance a ball made out of a ferromagnetic material. However, for a fixed
current, the force on the object becomes stronger when the ball moves closer, and weaker when the ball moves
farther away, and thus the system is unstable in open loop.
↑
iin (t)
y(t)
The transfer function for a magnetic levitation system has the form
Y (s)
K0
= 2
Iin (s)
(s − a2 )(s + τ )
You wish use a PD control system to stabilize the mag-lev system and achieve a 1% settling time less than 1.3
seconds and %OS less than 10%.
R(s)
+
Kp
−
Iin (s)
+
−
K0
(s2 −a2 )(s+τ )
Kd s
Suppose the parameters are K0 = 10, a = 1 and τ = 10.
5
Y (s)
(a) Sketch the root locus for Kd = 0, and indicate the allowable closed loop pole regions.
(b) Using the technique from class, choose Kd and Kp to place the closed loop poles on or inside the allowable
region
(c) Find the closed loop transfer function Y (s)/R(s). Using Matlab find the step response to verify that you
design specifications have been met.
2.2
Solutions
1.
• (a)
• (b)
6
• (c)
7
8