Lecture 10 Nyquist Plot
16
Control Systems I Emam Fathy email: [email protected] http:// www.aast.edu/cv.php?disp_unit=346&ser=68525 Lecture 10 Nyquist Plot 1
Transcript of Lecture 10 Nyquist Plot
Control Systems I
Emam Fathyemail: [email protected]
http://www.aast.edu/cv.php?disp_unit=346&ser=68525
Lecture 10
Nyquist Plot
1
Nyquist Plot (Polar Plot)
• The polar plot of a sinusoidaltransfer function G(jω) is a plot ofthe magnitude of G(jω) versus thephase angle of G(jω) on polarcoordinates as ω is varied fromzero to infinity.
• Thus, the polar plot is the locus ofvectors as ω isvaried from zero to infinity.
Nyquist Plot (Polar Plot)
• Each point on the polar plot ofG(jω) represents the terminalpoint of a vector at aparticular value of ω.
• The projections of G(jω) onthe real and imaginary axesare its real and imaginarycomponents.
Nyquist Plot (Polar Plot)
• An advantage in using a polar plot isthat it depicts the frequencyresponse characteristics of a systemover the entire frequency range in asingle plot.
• One disadvantage is that the plotdoes not clearly indicate thecontributions of each individualfactor of the open-loop transferfunction.
Nyquist Plot of Integral and Derivative Factors
• The polar plot of G(jω)=1/jω is the negative imaginaryaxis, since
jjG
1)(
11j
j
j
jjG
)(
901
)( jG formpolar In
Im
Re-90o
ω=0
ω=∞
Nyquist Plot of Integral and Derivative Factors
• The polar plot of G(jω)=jω is the positive imaginary axis,since
jjG )(
90 )( jG formpolar In
Im
Re
90o
ω=0
ω=∞
Nyquist Plot of First Order Factors
• The polar plot of first order factor in numerator is
1 jjG )(
Im
Re
ω Re Im
0 1 0
1 1 1
2 1 2
∞ 1 ∞
ω=0
1
1 ω=1
ω=22
ω= ∞
Nyquist Plot of First Order Factors
• The polar plot of first order factor in denominator is
1
1
jjG )(
j
j
jjG
1
1
1
1)(
21
1
jjG )(
22 11
1
jjG )(
ω Re Im
0 1 0
0.5 0.8 0.4
1 1/2 -1/2
2 1/5 -2/5
∞ 0 undefined
Nyquist Plot of First Order Factors
• The polar plot of first order factor in denominator is
Im
Reω=0
1
ω=1
ω=2
-0.5
ω= ∞
ω Re Im
0 1 0
0.5 0.8 -0.4
1 0.5 -0.5
2 0.2 -0.4
∞ 0 undefined
0.50.2
-0.4
0.8
ω=0.5
Nyquist Plot of First Order Factors
• The polar plot of first order factor in denominator is
Im
Re
ω=2
ω= ∞
ω Re Im
0 1 0 1 0o
0.5 0.8 -0.4 0.9 -26o
1 0.5 -0.5 0.7 -45o
2 0.2 -0.4 0.4 -63o
∞ 0 0 0 -90
)( jG )( jG
ω=0
ω=0.5
ω=1
Example#1
• Draw the polar plot of following open loop transfer function.
)()(
1
1
jjjG
)()(
1
1
sssG
Solution
jsPut
jjG
2
1)(
j
j
jjG
2
2
2
1)(
24
2
jjG )(
Example#1
24
2
jjG )(
2424
2
jjG )(
)()(
1
1
1
122
jjG
ω Re Im
0 -1 ∞
0.1 -1 -10
0.5 -0.8 -1.6
1 -0.5 -0.5
2 -0.2 -0.1
3 -0.1 -0.03
∞ 0 0
Example#1
ω Re Im
0 -1 ∞
0.1 -1 -10
0.5 -0.8 -1.6
1 -0.5 -0.5
2 -0.2 -0.1
3 -0.1 -0.03
∞ 0 0
Im
Re
ω=0
-10
-1
ω=0.1
ω=1
ω=2 ω=3
ω=0.5
ω=∞
Nyquist Stability Criterion
• The Nyquist stabilitycriterion determines thestability of a closed-loopsystem from its open-loopfrequency response andopen-loop poles.
• A minimum phase closedloop system will be stable ifthe Nyquist plot of openloop transfer function doesnot encircle (-1, j0) point.
Im
Re(-1, j0)
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Gain cross-over point
Phase cross-over point
Gain Margin
Phase Margin
END OF LEC
