Contents
Given a single loop feedback system
we would like to be able to determine whether or not the closed loop system, T(s), is stable. This is equivalent to asking whether the denominator of the transfer function (which is the characteristic equation of the system)
has any zeros in the right half of the s-plane (recall that the natural response of a transfer function with poles in the right half plane grows exponentially with time).
If we perform a mapping (as explained on the previous page) of the function "1+L(s)" with a path in "s" that encircle the entire right half plane and we count the encirclements of the origin in the "1+L(s)" domain in the clockwise direction we get the number N=Z-P (where Z is the number of zeros, and P is the number of poles). What we want, though, is Z, the number of zeros in the right half plane. But since we know L(s), we can easily find P, the numbers of poles of "1+L(s)." This is because any pole of L(s) is also a pole of "1+L(s)." So now we know N (from the mapping) and we know P (from L(s)), so we can easily determine Z.
Before continuing we make one small change. Instead of mapping from "s" to "1+L(s)" and counting encirclements of the origin, we map from "s" to "L(s)" and count encirclements of the point -1+j0 in the complex plane. This is because the origin in "1+L(s)" corresponds to the "-1+j0" point in "L(s)" (if L(s)=-1, then 1+L(s)=0).
To determine the stability of a system we:
In the previous section, we specified that the path of s should enclose the entire right half plane. To start, we assume that the function L(s) has no poles on the jω axis. We define the path as starting at the origin, moving up the imaginary axis to j∞, following a semicircle (in the clockwise direction), and then moving up the -jω axis and ending at the origin. This is shown below.
Let's examine this procedure with a couple of examples (followed by a video of the two examples). A variety of examples follow on the next page (Examples).
Consider a system with plant G(s), and unity gain feedback (H(s)=1)
If we map this function from "s" to "L(s)" with the variable s following the Nyquist path we get the following image (note: the image on the left is the "Nyquist path" the image on the right is called the "Nyquist plot")
If we zoom in on the graph in "L(s)"
the first thing we notice is the multiple arrowheads at the origin. This is because as the path in "s" traverses a semicircle at ∞ the path in "L(s)" remains at the origin, but the angle of L(s) changes. More importantly, we can see that it does not encircle the -1+j0, so N=0. We also know that P=0, and since N=Z-P, Z is also equal to zero. This tells us that the system is stable. And, if we close the loop, we find that the characteristic equation of the closed loop transfer function is
which has roots at -2±17.2j so the system is indeed stable. Recall that the roots of the characteristic equation are the poles of the transfer function.
A very large segment of the path in "s" occurs when |s|→∞, shown as the large semicircle in the s-domain plot (i.e., the left plot). However, during this segment of the plot, the path in L(s) will not move, assuming L(s) is a proper transfer function. The order of the numerator polynomial of a proper transfer function is less than or equal to that of the denominator. Let's consider, first, the case when the order of both polynomials is equal to n:
If |s|→∞, then the highest order term of the polynomial dominates and we get
So, when |s|→∞ the path in L(s) is at a single point. (Note: In the Nyquist diagrams, since there are several arrowheads on the path in "s" as it makes it excursion at infinity, there are also several arrowheads at this single point in "L(s).")
If the loop gain, L(s), is strictly proper (i.e., the order of the numerator is less than that of the denominator)
then
and the path in L(s) is at the origin while |s|→∞.
Consider the previous system, with a sensor in the feedback loop
If we map this function from "s" to "L(s)" with the variable s following the Nyquist path we get the following image
We can see that this graph encircles the -1+j0 twice in the clockwise direction so N=2. We also know that P=0, and since N=Z-P, we can calculate that Z=2. This tells us that the system is unstable, because the characteristic equation of the closed loop transfer function has two zeros in the right half plane (or, equivalently, the transfer function has two poles there). We can check this by closing the loop to get the characteristic equation of the closed loop transfer function:
which has roots at s=-7.5 and s=1.26±8.45j so the system is indeed unstable with two poles in the right half plane.
Video: Mapping with no poles on the jω axis
1 minute video created with the Matlab script NyquistGui
If there are poles on the jω axis, we can no longer use the Nyquist Path as specified above because the path will go through poles of L(s) where its value is undefined. To handle that situation we make small "detours" around the poles that are on the axis. These detours are small semicircles in the counterclockwise direction around these poles (so that the path stays in the right half plane), However, we make the radius of the detours infinitesimally small, so they don't exclude any part of the right half plane. For instance if we have
The path now makes small semicircular detour of infinitesimally small radius around the poles at s=±j4. However, because the path is so close to the pole, the magnitude of the path in "L(s)" is at infinity. And because the path is going around the pole in the counterclockwise direction in "s" the path in "L(s)" is in the clockwise direction. This is shown below (in this diagram the radius of the detours is exaggerated so they can be seen on the graph).
Note that:
Again, we can double check this by finding the zeros of the characteristic equation.
This has roots at s=-0.25±j2.63, so the system is stable.
Video: Mapping with poles on the jω axis
1½ minute video created with the Matlab script NyquistGui
If the path in "s" makes a 180° counterclockwise detour around a pole on the jω axis, then the corresponding path in "L(s)" has infinite radius and goes through 180° in the clockwise direction. If the detour in "s" is around two coincident poles (e.g., two poles at the origin) the angle in "L(s) will be 360°.
Counting the number of encirclements of the -1+j0 point is obviously of critical importance to determining the stability of the system (the number of encirclements in the clockwise direction is the "N" in the equation "N=Z-P"). There are two ways to do this. The first is easier conceptually, the second is easier practically.
Imagine you are presented with the Nyquist diagram below
How many encirclements of the -1+j0 point are there. To visualize the answer, assume you have a pointer whose base is at the -1+j0 point and whose head is anywhere on the Nyquist plot. For example we can start at the point near the origin and call this angle zero (Note: since we will be traversing the entire path the place where we start on the graph is arbitrary.)
We follow the path with the tip of our arrow to the first crossing of the jω axis (180° counterclockwise) and continue following the path until we get back to the starting point. As we do so we keep track of the total angle that has been swept out by the arrow, as shown below.
1) 180° counterclockwise | 2) 0° (total) | 3) 180° clockwise (total) | 4) 0° (total) |
This diagram show 180° movement, ccw |
Here we have 180°, cw, for a total of 0° |
Another 180° cw, for a total of 180°, cw |
A final 180° movement ccw yields 0° total. |
When we are finished we have traversed a net of 0°, so the -1+j0 point is not encircled.
However, with a slightly different Nyquist diagram we get a different result. Start as before,
and then follow the path with our arrow
1) 180° counterclockwise | 2) 360° ccw | 3) 360° ccw |
4) 360° ccw | 5) 540° ccw | 6) 720° ccw |
In this case we have two encirclements in the counter clockwise direction.
An easier way to determine the number of encirclements of the -1+j0 point is to simply draw a line out from the point, in any directions. Consider the first example from above, shown below with a vertical line drawn from -1+j0.
If you count the number of times that the Nyquist path crosses the line in the clockwise direction (i.e., left to right in the image, and denoted by a red circle) and subtract the number of times it crosses in the counterclockwise direction (the blue dot), you get the number of clockwise encirclements of the -1+j0 point. A negative number indicates counterclockwise encirclements. In the image above, there is one crossing in each direction, and therefore zero encirclements (as determined previously).
The direction of the line draw is arbitrary. The image below shows the same Nyquist path, but a different line. In this case there are two clockwise crossings (red) and two counterclockwise crossings, for a total of zero encirclements, as expected.
We now apply this technique to another example (the second example from above), in which we have the same path shifted to the right.
Here we see two counterclockwise crossings of the line and hence two counterclockwise encirclements of -1+j0.
If we choose a different line we get one clockwise crossing of the line, and three counterclockwise crossings. Hence we have two counterclockwise encirclements of -1+j0, as expected.
At this point you may well ask, "Why go through this whole procedure if we can just find the closed loop transfer function and see if it is stable?" The reason we use the Nyquist Stability Criterion is that it gives use information about the relative stability of a system and gives us clues as to how to make a system more stable. In particular, there are two quantities, the gain margin and the phase margin, that can be used to quantify the stability of a system.
Consider a system with
Let us draw the Nyquist plot:
If we zoom in, we can see that the plot in "L(s)" does not encircle the -1+j0, so the system is stable.
We can verify this by finding the roots of the characteristic equation
The roots are at s=-5.5 and s=-0.24±2.88j so the system is stable, as expected.
This can be seen even more clearly using Matlab's "nyquist" command. In fact, the "nyquist" command is generally more useful for examining the details of a Nyquist plot (called here the Nyquist diagram) than the plots we have been using, but the plots we have been using may be better for learning about the plots.
Now lets draw a unit circle around the origin (using Matlab's "ltiview"
command).
There are two spots on the Nyquist plot that are emphasized. The first is where the Nyquist plot crosses the real axis in the left half plane. If we zoom in and put the cursor over this point we get the following image.
As you can see, the plot crosses the real axis at about -2/3, or -0.67. This tells us that if we multiply L(s) by a number greater than 3/2 that the path would encircle the -1+j0 and the systems would be unstable. So the "gain margin" is 3/2 or 20log_{10}(3/2)=3.5dB. The greater the gain margin, the more stable the system. If the gain margin is zero, the system is marginally stable. (Note: the text also shows that the Nyquist plot crosses the real axis when the Nyquist path is going through the point s=j3.32 (this is the "frequency" shown).)
The second point shows where the Nyquist plot crosses the unit circle as displayed on the images below.
The angle between between the point at which the plot crosses the unit circle (when the Nyquist path is at s=j2.73) has an angle of 14° to the horizontal axis. This tells us that if we decrease (where a decrease moves the plot in the clockwise direction) the phase of L(s) by more than 14° that the -1+j0 point becomes encircled and the system becomes unstable. We say that the system has a phase margin of 14°. A higher phase margin yields a more stable system. A phase margin of 0° indicates a marginally stable system. Note: if you know about the frequency response time delays, recall that a time delay corresponds to a change in phase - for this system we could have a delay of 0.089 seconds (corresponding to 14° at 2.73 rad/sec). If you don't know about time delays, you can skip this.
If we multiply L(s) by 3/2 we get
and we see that the system gain and phase margins go to zero so we expect the system to be marginally stable.
Nyquist Plot | Nyquist plot zoomed with stability margins=0 | |
We can verify this by finding the roots of the characteristic equation
The roots are at s=-6 and s=±3.32j so the system is marginally stable, as expected.
If we multiply the original L(s) by 4 we get
and we see that the system gain and phase margins become negative so we expect the system to be unstable.
Nyquist Plot | Zoomed with negative gain margin shown | Zoomed with negative phase margin shown | ||
The margins tell us that we'd have to decrease the gain by -8.52dB (multiply by 0.375) or change the phase by -25.9° to make this system stable. (Note: we can readily verify the gain margin because we know that multiplication of L(s) by 3/2 made the system marginally stable, and 4·0.375=3/2 (the current gain (4) multiplied by the gain margin (0.375) yields the gain that creates marginal stability (3/2).) We can verify that the system is unstable by finding the roots of the characteristic equation
The roots are at s=-7.5 and s=+0.75±4.64j so the system is unstable, as expected.
The gain margin of a system will be infinite if the phase of the loop gain never reaches -180° (i.e., if the Nyquist plot never crosses the real axis in the left half plane). If then the Nyquist path is as shown
The gain margin is infinite because the path never crosses the real axis in the left half plane (the path goes to the origin as |s|→∞.
The phase margin will be infinite if the magnitude of gain of L(s) is never greater than one. If then the Nyquist path is as shown
The phase margin is infinite because the gain is always less than one, so no matter how much the phase changes, the -1+j0 point will never be encircled.