# Frequency Response 5

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```Definitions - Unity Feedback
R(s) +
C(s)
Num( s)
Den(s)
K
-
Stability in Frequency Domain
Open-loop
transfer function
Num( s)
= 1 = 0 dB
Magnitude criterion: K
Den (s)
 Num ( s )
angle 
 Den ( s )
Phase criterion:
• Gain Margin, GM - How much open-loop
gain K can we add such that the 0 db
crossover occurs at -180&deg; phase?
• Phase Margin, Φ M - How much phase
angle could we subtract at 0 dB crossover (unity gain) to reach -180&deg; phase?
Both gain and phase margin computed from the
open-loop transfer function’s frequency response

 = − 180 o

Phase Margin and ζ
Estimations
• We can estimate some unity feedback,
closed-loop time responses from the
open loop frequency response:
–damping ratio from phase margin
–static error constants from low frequency
slopes
• From p. 587 of text, for an ideal 2nd
order system,


Φ M = tan − 1 


2ς
− 2ς 2 +



4 
1 + 4ς 
Eqn. 10.73
A plot of phase margin vs. damping ratio is
shown in Fig. 10.48 and on next slide
Plot of Phase Margin vs. ζ
Example #1
100
90
R(s) +
True Phase Margin
Phase Margin, degrees
80
Phase~100*Damping
Ratio
70
-
K
C(s)
2
s( s + 12 s + 20 )
60
50
Find phase margin, ΦM, for K = 30
40
30
&quot;Good&quot; Approximation
Use
Matlab™
What are the closed loop roots at K = 30?
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Damping Ratio, z
1
Example #2
• Figure 10.51 of your text shows how to
determine the static error constants
from the low frequency response of the
open-loop system
C(s)
K
2
s( s + 4 s + 8 )
-
Find phase margin, ΦM, for K = 10
Use
Matlab™
What are the closed loop roots at K = 10?
– Kp (position error constant) for Type 0 systems
– Kv (velocity error constant) for Type 1 systems
– Ka (acceleration error constant) for Type 2 systems
Example #4
90
20
10
60
10
0
30
0
0
-20
-30
Type 0 system, so
20log10(Kp)=14dB
-30
-40
-60
-90
-50
-120
-60
-70
-80
Magnitude
-90
Phase
K p = 10
(+14 20 ) ≈ 5
-100
1
10
100
1000
-150
-180
180
150
Type 1 system,
so find Kv
-10
Magnitude, dB
-10
Phase Angle, degrees
Magnitude, dB
Example #3
20
-20
120
90
60
-30
30
-40
0
Magnitude = 0 dB
at ω=2, so Kv=2 rad/sec
-50
-60
-30
-60
-70
-90
-210
-80
Magnitude
-120
-240
-90
Phase
-150
-270
-100
10000
-180
1
Definition - Bandwidth
– “zero” frequency would be approximated by a very
small frequency
– if you roughly approximated your closed loop system
as first order (i.e., a low pass filter), this would be the
“break” frequency
10
100
1000
10000
Frequency, rad/sec
Frequency, rad/sec
• Bandwidth (ω BW), is the frequency at
which the closed-loop frequency
response is 3 dB below its “zero”
frequency value (p. 579 of text)
Phase Angle, degrees
R(s) +
Static Error Constants
Example #5
R(s) +
-
K
C(s)
2
s( s + 12 s + 20 )
Find bandwidth, ω MW, for K = 30
Use
Matlab™
2
```