What follows are several examples of Nyquist plots. In general each example has five sections: 1) A definition of the loop gain, 2) A Nyquist plot made by the NyquistGui program, 3) a Nyquist plot made by Matlab, 4) A discussion of the plots and system stability, and 5) a video of the output of the NyquistGui program. In some cases, one or more of these sections will be omitted because they are unnecessary to the understanding of the example.
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The -1+j0 point is not encircled so N=1. There are no poles of L(s) in the right half plane so P=0. Since N=Z-P, Z=0. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The system is stable.
We can check this by finding the location of the zeros of the characteristic equation:
This has roots at s=-4.5±9.4j so the system is stable as expected.
Note: This plot is zoomed in (the original Matlab plot has
axes with much higher limits, making it hard to decipher).
This Nyquist Diagram is a little hard to decipher because the branches go off towards infinity. However, because there is a pole at the origin, we can infer that the counterclockwise 180° detour around the origin in "s" yields a clockwise 180° detour in "L(s)" that is not shown in the Matlab plot. Hence the -1+j0 point is not encircled and the system is stable.
This system has a pole at the origin (i.e., on the jω axis) so we must take a detour around it. This is clear on the NyquistGui plot, but is not shown on the Matlab plot. Whenever a detour around a pole is required, this is not shown on the Matlab plot, and the user must understand what happens going around the detour. In this case, since there is a single pole at the origin and the detour in "s" has radius approaching zero and moves in the counterclockwise direction, you know that the part of the Nyquist plot that is not shown must be a semicircle at infinity in the clockwise direction.
The -1+j0 point is not encircled, so N=0. L(s) has no poles in the right half plane, so P=0. Since N=Z-P, Z=0. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The system is stable.
We can check this by finding the location of the zeros of the characteristic equation:
This has roots at s=-4.64, -0.18±2.07 so the system is stable as expected.
Since the gain margin is 3.52 dB (=1.5), this tells us that we could increase the gain by up to a factor of 1.5 before the system goes unstable. Let's test this. If we multiply L(s) by 5.
we get the Nyquist plot shown, which has negative gain and phase margins, so the system is indeed unstable.
The Matlab plot is initially quite hard to decipher,
But it becomes clear if we zoom in (and display the stability margins, which are both negative, indicating instability).
We can check this by finding the location of the zeros of the characteristic equation:
This has roots at s=-5.36, 1.18±4.15j so the system is unstable as expected.
The -1+j0 point is encircled one in the counterclockwise direction so N=-1. There is one pole of L(s) in the right half plane so P=-1. Since N=Z-P, Z=0. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The system is stable.
We can check this by finding the location of the zeros of the characteristic equation:
This has roots at s=-5±1j, so the system is stable as expected.
The -1+j0 point is encircled one in the clockwise direction so N=1. There is one pole of L(s) in the right half plane so P=-1. Since N=Z-P, Z=2. This means that the characteristic equation of the closed loop transfer function has two zeros in the right half plane (the closed loop transfer function has two poles there). The system is unstable.
We can check this by finding the location of the zeros of the characteristic equation:
This has roots at s=-1.53, 1.26±7.95j so the system is unstable as expected.
This plot is hard to understand because the two branches go off to infinity. However because there is a double pole at the origin, we can infer that the 180° counterclockwise detour around the origin in "s" yields a 360° clockwise detour in "L(s)." Hence the because of the large scale involved.
The -1+j0 point is not encircled so N=0. There are no poles of L(s) in the right half plane so P=0. Since N=Z-P, Z=0. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The system is stable.
We can check this by finding the location of the zeros of the characteristic equation:
This has roots at s=-2.35, -0.321±2.90j, so the system is stable as expected.
If we move the zero so that it is to the left of the poles
then we get a slightly, but very significantly, different Nyquist plot:
The NyquistGui plot (above) makes it clear that the origin is encircled twice in the clockwise direction (N=2, P=0 so Z=2), so the closed loop system is unstable with two poles in the right half plane. If we were using just the Matlab plot (below) we have to remember that L(s) has a double pole at the origin which gives a 360° clockwise path with infinite radius in the "L(s)" domain.
Note: this is the
same transfer function that was used in example 1, with the
addition of a time delay.
We can see some useful detail if we zoom in on the plot in "L(s)"
Original | Significantly magnified | |
The shape of this plot is significantly different from those that preceded it, as is evident from the spiral as the path in L(s) approaches the origin. The cause of this spiral is the delay term
Along the jω axis the delay evaluates to
(Note: magnitude and phase of a time delay are discussed
here)
This term has a magnitude of 1, but a phase of -0.05·ω . So as we move up the imaginary axis, the magnitude of the Nyquist plot doesn't change but the phase is constantly decreasing; this causes the spiral. (Note: It may be helpful to look at the Nyquist plots above alongside those of example 1).
The -1+j0 point is not encircled so N=0. There are no poles of L(s) in the right half plane so P=0. Since N=Z-P, Z=0. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The system is stable.
Because of the time delay, it is impossible to find the roots of the characteristic equation, but simulation (using Simulink or some other program) will show that the system is stable.
If we increase the time delay so that
then we get a slightly, but very significantly, different Nyquist plot:
The NyquistGui plot (above) clearly shows the plot in "L(s)" spiraling towards the origin because of the negative phase added by the time delay. This is even more clear in the Matlab plot.
Original | Significantly magnified | |
The -1+j0 point is encircled, so the system is unstable. Because of the time delay, it is impossible to find the roots of the characteristic equation, but simulation (using Simulink or some other program) will show that the system is unstable.
© Copyright 2005-2013 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College