16.06 Principles of Automatic Control
Recitation 11
Sketch a Nichols plot of:
Gpsq “
1
ps ` 1q3
s “ ´ 3 tan ´1 pωq “ ´180˝
ω “1.73
´ 1 ¯3
1
1
“ ?
“ “K
6 K“ ?
3
8
1`3
p 1 ` 3q
Magnitude, dB
Start with Bode plot of the system:
0
10
−3
−2
10
−3
10
−2
10
−1
0
10
10
1
10
2
10
0
Phase (deg)
−50
−100
−150
−200
−250
−300 −3
10
−2
10
−1
0
10
10
Frequency, ω (rad/sec)
1
1
10
2
10
Nichols:
|G(jω)|
100
10
G(jω)
-360
-270
-90
0
0.1
0.01
But, similar to Nyquist, we need to reflect the plot, so what we really have is:
|G(jω)|
100
k>0
k<0
10
G(jω)
-360
-270
-90
0.1
0
0.125
0.01
Now the question is how do we count the number of encirclements? Z “ N ` P still applies!
A clockwise encirclement in Nyquist is equivalent to leftward crossing in Nichols. In Nyquist,
when counting encirclements, always on imaginary axis. In Nichols, this corresponds to 0˝
vertical line. Along 0˝ line there is one leftward crossing, which makes N “ 1. Along ´180˝
2
line for magnitudes greater than 0.125, N “ 0. For magnitudes lower than 0.125, N “ 2.
What gain corresponds to boundary point?
1
1
ą
K
8
0 ăK ă 8
´1
0ă
ă 1
K
K ą ´ 1
so stable for ´ 1 ăK ă 8
8ą
Check with Nyquist and get the same result!
3
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16.06 Principles of Automatic Control
Fall 2012
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