Today’s goals
•
•
So far
– Modeling the dynamics of electro-mechanical systems
– Controling the dynamics of electro-mechanical systems
• s-domain (root locus)
• State space
• Frequency domain (Bode plots)
– Types of compensators
• Proportional
• Proportional-Derivative
• Proportional-Integral (aka Ideal Integral)
• Proportional-Integral-Derivative
Today
– Passive compensators
• Lag
• Lead
• Lead-Lag
– Time delays
2.004 Fall ’07
Lecture 37 – Friday, Dec. 7
Lag and lead compensators
Compensator transfer function
Gc (s) =
z > p : Lag
M (ω) [dB]
1
T , α > 1.
Gc (s) =
1
s+
αT
s+
20log10 (z/p)
z < p : Lead
M (ω) [dB]
1
αT , α > 1.
Gc (s) =
1
s+
T
s+
0
20log10 (z/p)
0
p
0
s+z
.
s+p
z
ω
Φ(ω) deg.
p
z
ω
p
z
ω
Φ(ω) deg.
+90
−90
0
p
2.004 Fall ’07
z
ω
Lecture 37 – Friday, Dec. 7
Lag compensation
•
•
Gc (s) =
Improve steady-state error
Increase phase margin
s+z
,
s+p
z>p
Steady—state error without compensation:
e(∞) =
1
, Kp = Gp (0).
1 + Kp
Steady—state error with lag compensation:
e(∞) =
1
.
1 + zKp /p
Images removed due to copyright restrictions.
Please see: Fig. 11.4 and 9.10 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 37 – Friday, Dec. 7
Lead compensation
•
•
Gc (s) =
s+z
,
s+p
z<p
Increase bandwidth (faster response)
Increase phase margin
Images removed due to copyright restrictions.
Please see: Fig. 11.7, 9.24, and 9.25 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 37 – Friday, Dec. 7
Lead-lag compensation
1
1
s+
T1
T2
Gc (s) =
×
,
α
1
s+
s+
T1
αT2
s+
Essentially equivalent to PID compensation
– Lead component
0
-5
fixes transient
-10
– Lag component
-15
-20
fixes steady-state error
-25
-30
Three degrees of freedom:
-35
0.001
0.01
T1, T2, α
•
Phase (degrees)
20 logM
•
α > 1.
80
60
40
20
0
-20
40
-40
50
-60
-80
0.001
γ = 10
20
30
40
50
0.1
1
Frequency (rad/s)
10
100
40
50
30
γ = 10
20
γ = 10
20
30
0.01
Figure by MIT OpenCourseWare.
2.004 Fall ’07
Asymptotes for γ = 10
Lecture 37 – Friday, Dec. 7
0.1
1
Frequency (rad/s)
10
100
Nise Figure 11.11
Systems with time delay
c(t)
Gd (s) = e
c(t − T )
G(jω)
−sT
=
6
|G(jω)| =
G(jω)
=
e−ωT
1
−ωT
Delay element (or “delay line”)
M (ω) [dB]
Time delay reduces
the phase margin;
therefore time delay
results in reduced stability
0
ω
0
Φ(ω) deg.
−90
ω
2.004 Fall ’07
Lecture 37 – Friday, Dec. 7
Cascading phase delay to a plant
r(t)
Gd (s) = e
−sT
r(t − T )
Gp (s) (plant)
Image removed due to copyright restrictions.
Please see: Fig. 10.54 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 37 – Friday, Dec. 7
c(t)
Example: time delay in feedback configuration
r(t)
+
−
Gd (s) = e
K
Gp (s) =
s(s + 1)(s + 10)
−sT
c(t)
Time delay T=1sec
(K = 1)
Images removed due to copyright restrictions.
Please see: Fig. 10.55 in Nise, Norman S. Control Systems Engineering.
4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
1. Range of gain for stability
Without phase delay, the phase angle is −180◦
at frequency 3.16rad/sec so the gain margin is 40.84dB.
The closed—loop system is stable for 0 < K ≤ 1040.84/20 = 110.2.
With phase delay, the phase angle is −180◦
at frequency 0.81rad/sec so the gain margin is 20.39dB.
The closed—loop system is stable for 0 < K ≤ 1020.39/20 = 10.46.
2. Percent overshoot for K = 5
Since K = 5, the magnitude curve is raised by 20log10 5 = 13.98dB.
The zero dB crossing occurs at frequency 0.47rad/sec.
Without phase delay, the phase at the zero dB crossing
is −118◦ , so the phase margin is −118◦ − (−180◦ ) = 62◦ .
From the phase margin vs damping curve (2nd—order approx.)
we find ζ = 0.64 ⇒ %OS = 7.3%.
With phase delay, the phase at the zero dB crossing
where the phase is −145◦ , so the phase margin is 35◦ .
From the phase margin vs damping curve (2nd—order approx.)
we find ζ = 0.33 ⇒ %OS = 33%.
Lecture 37 – Friday, Dec. 7
Example: time delay in feedback configuration
r(t)
Gd (s) = e
+
−
With phase delay T=1sec
Without phase delay
1.4
1.6
1.2
1.4
1.0
1.2
1.0
c(t)
0.8
0.6
0.8
0.6
0.4
0.4
0.2
0.2
0
c(t)
K
Gp (s) =
s(s + 1)(s + 10)
−sT
0
0
2
4
6
Time (seconds)
8
10
0
5
15
20
Time (seconds)
25
30
Figure by MIT OpenCourseWare.
Figure 10.56
2.004 Fall ’07
10
Lecture 37 – Friday, Dec. 7