The Nyquist plot (one is shown in the video above) is a very useful tool for determining the stability of a
system. It has advantages over the root locus and Routh-Horwitz because it
easily handles time delays. However, it is most useful because it gives us a way to use the
Bode plot to determine stability. We can not only use the Bode plot to
determine stability, but it also gives us insight on how to improve the
stability of a system. To fully understand the Bode plot based methods for
determining stability it is important to first understand the Nyquist plot.
This presents some difficulties because the Nyquist plot uses some rather
abstract mathematical techniques. These web pages try to demystify the
process by starting with some simple examples (with animations) and then
proceeding to more complicated examples. The treatment given here is not
meant to be mathematically rigorous (it does not even use contour integration in
the complex plane), but rather to give engineering students an
intuition for what the Nyquist plot represents, and how it is used to determine
the stability of a system.
The next several documents describe the Nyquist plot.
Mapping in the complex plane.
This gives mathematical background needed to understand the Nyquist plot
Stability as determined by the
Nyquist plot. How does the Nyquist plot determine stability.
Examples. Several examples of
determination of stability of systems from Nyquist plots, this can be used
as a tutorial for the interpretation of Nyquist plots.
Bode plots. How we can go from
Nyquist plots to Bode plots, and how can we determine stability from Bode
plots.
Bode Examples. Several examples of
determination of stability of systems from Bode plots.
NyquistGui. A Matlab program that
can be helpful for understanding Nyquist plots.
Printable. The documents above combined
into one file (for easier printing).
The mapping of functions in the complex plane is conceptually simple, but will
lead us to a very powerful technique for determining system stability. In
addition it will give us insight into how to avoid instability. To introduce
the concept we will start with some simple examples. There are several videos
on this page - they merely support the written material, but are not absolutely
vital.
The simplest functions to map are those with a single zero. Several examples
follow.
Example: Mapping with circle in "s", zero at the origin in "L(s)"
Consider the trivial function L(s)=s (we will deal with more complicated
functions later, this simple function allows us to introduce concepts associated
with mapping). We let the variable s traverse a circular path centered at the
origin with a radius of 5 moving in the clockwise direction (the left
side graph), In other words
We then plot L(s) on the right hand graph. This is
shown in the image below, followed soon thereafter by a
video that better demonstrates the mapping.
Notes about image above
There are several aspects about the image above that are important (many
of these come up later):
The path in the "s" plane is shown at the left. It is a circle
of radius 5, centered at the origin and moving in a clockwise direction,
as indicated by the arrows.
The arrows show the direction of motion, but the spacing of the arrows
is arbitrary.
The colors of the two plots match, so that the portion of "s" that is
yellow is also yellow on "L(s)". This is obvious here, but becomes
a useful distinction in more complex plots to come.
There is a pink circle centered on the zero in the "s" plane.
This shows the angle subtended between the zero and the path in s.
You can see the growth of this circle (in the clockwise direction)
in the video below.
There is also a pink circle centered on the origin in the "L(s)" plane.
This shows the angle subtended between the origin and the path in s.
The color indicates the direction; pink indicates the clockwise direction.
If the direction had been counterclockwise it would be blue.
Note that as the path in "s" encircles the zero one time in the counterclockwise
direction, the path in "L(s)" encircles the origin once in the same direction.
2½ minute video created with the Matlab script
NyquistGui)
Example: Mapping with circle in "s", zero at s=-4
If we now place a zero at s=-4 so that L(s)=s+4, the mapping is still very
straightforward. Every location in "s" simply maps to a location that
is 4 units to the right in "L(s)". The path in s remains as before.
(Note: this example is also considered as part of the
video above.)
Notes about image above
The path in "s" is still a circle, as is the path in
"L(s)", but the path
is simply offset by 4 (since L(s)=s+4.
Note that as the path in "s" encircles the zero one time in the counterclockwise
direction, the path in "L(s)" encircles the origin once in the same direction.
As the path in "s" gets near the pole in "L(s)" (where the path has
a light blue color), the path in "L(s)" becomes the smalles (i.e., the closest
to the origin). This is because the term in the numerator, in this
case (s+4), is minimal when "s" is near the pole at -4. Also, when
the distance from the zero to the path in "s" is maximal (red), so is the
distance from L(s) to the origin.
Example: Mapping with circle in "s", zero at s=-6
If we now place a zero at s=-4 so that L(s)=s+6, the mapping is still very
straightforward. Every location in "s" simply maps to a location that
is 4 units to the right in "L(s)." The path in s remains as before.
(Note: this example is also considered as part of the
video above.)
Notes about image above
The path in "s" is still a circle, as is the path in
"L(s)", but the path
is simply offset by 4 (since L(s)=s+4.
Note that as the path in "s" no longer encircles the zero, and the path
in "L(s)" no longer encircles the origin.
Comments about mapping with a pole in "L(s)"
There are three important, and general, statements we can now make about mapping
from "s" to "L(s)" when there is a zero in "L(s)":
If the path in "s" is in the clockwise direction, then the path in "L(s)"
is in the clockwise direction.
As the path in "s" gets close to the zero in "L(s)" the path in "L(s)" goes
to its smallest value.
If the path in "s" encircles the pole of "L(s)," then the path
in "L(s)" encircles
the origin once in the same direction.
Mapping of functions with a single pole is not much more difficult than mapping
with a single zero. There are two important facts about encirclements of
poles that can be shown by considering L(s) with a single pole at the origin,
and the path in "s" being a clockwise circle of radius 'r' around the origin:
The first characteristic to be realized is that as the path in "s" comes close to a pole, the path in
"L(s)" gets large Clearly as the radius of the
encirclement, r, becomes small, the magnitude of L(s) becomes large.
The second characteristic is that the path in "L(s)" is in the opposite
direction of the path in "s." In this example, the path in "s" is
clockwise, so the path in "L(s)" is counterclockwise.
The mapping around various functions, L(s), with a single
poles are shown in the diagrams in the examples below, followed by a video that
shows several functions. Each example in the video is also included in the
examples that follow. It is useful to read the examples before viewing the
video.
Example: Mapping with circle in "s", pole at origin
If we choose L(s) such that it has a pole at the origin,
(note: the constant multiplier makes the plots looks nicer,
but isn't necessary for the mathematics to work)
and we let s follow the same path as before
We get
Because the ejθ was originally in the
denominator, its sign is changed when it moves to the numerator. In other
words, the path in "L(s)" is a circle of radius 2 that encircles the origin once
in the clockwise direction.
Notes about image above
The path in the "s" plane is shown at the left. It is a circle
of radius 5, centered at the origin and moving in a clockwise direction.
There is a blue circle centered on the pole in the "s" plane (a pink
circle will be used for zeros, as in the previous angles). This shows
the angle subtended between the pole and the path in s. You can see
the growth of this circle in the video below.
There is a grey circle centered on the origin in the "L(s)" plane.
This shows the angle subtended between the origin and the path in s.
Note that as the path in "s" encircles the pole one time in the clockwise
direction, the path in "L(s)" encircles the origin once in the opposite
(counterclockwise) direction.
Example: Mapping with circle in "s", pole at s=-4
If we now place a pole at s=-4 so that
Note: this example is also considered as part of the video
above.)
The path in s remains as before, but the path in "L(s)" has
changed. The center and extent of the path in "L(s)" have both changed.
(note: if you can show that the path in "L(s)" is also
a circle and derive equations for the radius and center, I'll include it here,
with an acknowledgement for the first person who sent it to me)
Notes about image above
The path in "s" still encircles the pole in a clockwise direction, and
the path in "L(s)" still encircles the origin in a counterclockwise (opposite)
direction.
As the path in "s" gets near the pole in "L(s)" (where the path has
a light blue color), the path in "L(s)" becomes the largest (i.e., the farthest
from the origin). This is because the term in the denominator, in
this case (s+4), is minimal when "s" is near the pole at -4. Also,
when the distance from the pole to the path in "s" is maximal (red), the
distance from L(s) to the origin is minimal
Example: Mapping with circle in "s", pole at s=-6
If we now place a pole at s=-6 so that
Note: this example is also considered as part of the video
above.)
The path in s remains as before, but the path in "L(s)" has
changed. The center and extent of the path in "L(s)" have both changed.
Notes about image above
Note that as the path in "s" no longer encircles the pole in a clockwise
direction, and the path in "L(s)" no longer encircles the origin.
Example: Mapping with circle in "s", pole at s=-4.8
If we now place a pole at s=-4.8 so that
Note: this example is also considered as part of the video
above.)
The path in s remains as before, but the path in "L(s)" has
changed. The center and extent of the path in "L(s)" have both changed.
The shape
(note: if you can show that the path in "L(s)" is also
a circle and derive equations for the radius and center, I'll include it here,
with an acknowledgement for the first one to send it to me)
Notes about image above
Note that as the path in "s" encircles the pole in a clockwise direction,
and the path in "L(s)" still encircles the origin in a counterclockwise direction.
As the path in "s" comes very close to the pole in "L(s)" (where the
paths are light blue), the path in "L(s)" becomes very large. In these
diagrams this is shown as going towards infinity. In these plots,
anything with a radius of more than twelve is truncated and shown at infinity.
If you understand the concept of mapping of functions with individual poles and
zeros, it is not much harder to understand mapping of functions with multiple poles
and zeros. A few examples will illustrate this. You should read
through the first example carefully, it has a lot of important information.
Example: L(s) has two poles, one zero; all are encircled
Consider mapping of the transfer function
where s follows a clockwise circular path of radius 5 around
the origin , as before.
To get an idea of what the mapping will look like, let's express the function
in polar notation.
At this point we are mostly interested in the angle of L(s),
so lets examine it more closely.
Recall that, in general, the angle
is
determined by drawing a line from s0 to s, and finding the angle
between that line and the horizontal (described
here). So if we let s0=-2, then the angle
would be determined by drawing a line from s=-2 to s and
finding the angle to the horizontal. This is shown in the diagram
below on the left for angles between the location s=-5j to the zero at -2,
and the poles at -1 and -3.
Since the angle of L(s) is given by
then this is the angle shown in the image above on the
right. Since we know the first term in the angle of L(s) goes from 0→-2*π,
and we subtract the other two terms, then the angle of L(s) must
go from 0→2*π, that is it encircles the origin once in the counterclockwise
direction. This can be seen in the image below, and in the
video that follows.
Notes about image above
The path in "s" encircles the poles and zero in the clockwise direction.
The angle between the path and the zero is shown in red, and the angle
between the path and the pole is shown in blue. The path in "L(s)",
which is shown in gray, is the sum of the angles from the zeros (red) minus
the angles from the poles (blue). This is more obvious in
the video.
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. In this case Z=1, P=2 so N=-1,
and we have one encirclement of the origin in "L(s)" in the
counterclockwise direction.
Aside: Magnitude of L(s)
The magnitude of L(s) is given by
This means that if s is very close to a zero (i.e.,
near s=-2) that the magnitude of L(s) becomes very small, and if s is
very close to a pole (i.e., near s=-1 or s=-3) that the magnitude of
L(s) becomes very large.
Key Concept: Magnitude and phase of L(s)
The phase of L(s) is simply the sum of the angles from the zeros of L(s)
to s, minus the angles from the poles of L(s) to s.
The magnitude of L(s) is small near zeros of L(s) and large near poles of
L(s).
Example: L(s) has two poles, one zero, one pole; one zero is encircled
Now if we change the transfer function to
then the path in s encircles the zero, but only one of the
poles.
Notes about image above
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. In this case Z=1, P=1 so N=0,
and we have no encirclements of the origin in "L(s)".
Example: L(s) has two poles, one zero; one zero is encircled
Now if
then the path in "s" encircles only the zero
Notes about image above
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. In this case Z=1, P=0 so N=1,
and we have one encirclement of the origin in "L(s)" in the clockwise
direction.
Example: L(s) has two poles, one zero; two poles are encircled
Now if
then the path in "s" encircles only the zero
Notes about image above
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. In this case Z=0, P=2 so N=-2,
and we have two encirclements of the origin in "L(s)" in the clockwise
direction.
You can tell there are two encirclements because the circle in
"L(s)"
is twice as dark. This is more clear in
the video.
Example: L(s) has two complex conjugate poles encircled by s
Consider
Poles at -2±4j which are inside a circle with radius 5.
Notes about image above
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. In this case Z=0, P=2 so N=-2,
and we have two encirclements of the origin in "L(s)" in the clockwise
direction.
Example: L(s) has two complex conjugate poles not encircled by s
Consider
Poles at -4±4j, which are outside a circle of radius 5.
Notes about image above
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. In this case Z=0, P=0 so N=0 and
we have no encirclements of the origin in "L(s)".
Video: Mapping with a multiple
poles and zeros (3½ minute video created with the Matlab script
NyquistGui)
Nothing that we have done so far depends on the fact that the path in "s" be
circular, which is important to the development of the Nyquist stability criterion
on the next web page. Two examples below (and a video) demonstrate this.
In the first example, immediately below, the path in "s" encircles a zero in the clockwise direction,
and the path in "L(s)" encircles the origin in the same direction.
In this first example L(s)=s, i.e., there is a zero at the origin.
In the second example, below, the path in "s" encircles a pole in the
clockwise direction, and the path in "L(s)" encircles the origin in the opposite
direction, Though the path is a very different shape. Note also that where
the distance from the path in "s" to the pole is minimal (i.e., where the path is
light blue), then the distance of path in "L(s)" to the origin is maximal
(and where the distance in "s" is maximal (where the path is red), the distance in "L(s)" is minimal).
In this example L(s)=10/(s+6), i.e., there is a pole at s=-6.
Video: Mapping with a multiple
poles and zeros (2 minute video created with the Matlab script
NyquistGui))
The key points to keep in mind as you move to the next page:
If a path in "s" encircles a zero of L(s) in the clockwise
direction, then this contributes 360° in the clockwise direction to
the path in "L(s)" as it moves around the origin.
If a path in "s" encircles a pole of L(s) in the clockwise
direction, then this contributes 360° in the counter clockwise
direction to the path in "L(s)" as it moves around the origin.
If a path in "s" encircles Z zeros and P poles of L(s) in the
clockwise direction, then the path in "L(s)" encircles the origin N=Z-P
times in the clockwise direction. If N is negative, this
corresponds to encirclements in the counterclockwise direction.
After reading through the material above, the question arises "So what?".
What we have done here is introduce a technique that gives use information about
the number of poles and zeros in a closed contour. To determine the
stability of a system, we want to determine if a system's transfer function has
any of poles in the right half plane. With just a little more work, we can
define our contour in "s" as the entire right half plane - then we can use this
to determine if there are any poles in the right half plane.
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).
Key Concept: Statement of the Problem
To determine the stability of a system we:
Start with a system whose characteristic equation is given by
"1+L(s)=0."
Make a mapping from the "s" domain to the "L(s)"
domain where the path of "s" encloses the
entire right half plane.
From the mapping we find the number N, which is the number of
encirclements of the -1+j0 point in "L(s)." Note: This is
equivalent to the number of encirclements of the origin in "1+L(s)."
We can factor L(s) to determine the number of poles that are in the
right half plane.
Since we know N and P, we can determine Z, the number of zeros of
"1+L(s)" in the right half plane (which is the same as the number
of poles of T(s)).
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).
Example: Nyquist path, no poles on jω axis, stable
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.
Behavior of the Nyquist path when |s|→∞
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|→∞.
Example: Nyquist path, no poles on jω axis, unstable
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.
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:
The counterclockwise detours around the poles at s=±j4 results in
a clockwise semicircle at L(s)=∞ in "L(s)" (see
video below)
The clockwise semicircle at infinity in "s" corresponds to a single
point in "L(s)"
If the counterclockwise detour was around a double pole on the axis (for example two
poles at the origin), the path in L(s) goes through an angle of 360° in
the clockwise direction.
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.
1½ minute video created with the Matlab script
NyquistGui
Key Concept: Detours around poles on the jω axis
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.
Counting encirclements by visualization
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.
Counting encirclements the easy way
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.
A stable 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 20log10(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.
A marginally stable system
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.
An unstable system
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.
Aside: Gain and Phase Margins may be Infinite
Infinite Gain Margin
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|→∞.
Infinite Phase Margin
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.
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.
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.
Discussion
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.
Example 6: L(s) has time delay
Loop gain:
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)"
Matlab plot
Original
Significantly magnified
Discussion
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.
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.
NyquistGui (Nyquist Graphical User Interface) is a Matlab program designed to help to understand Nyquist Plots.
It is not meant to be an accurate, flexible tool for creating Nyquist diagrams;
if that is what you are looking for, use Matlab's "nyquist" command. The gui has two basic "modes," one for learning about the process of mapping
from one domain to another (i.e., from the "s" domain to the "L(s)" domain).
Once you have a good understanding of the mapping process, you can choose to
learn about using the Nyquist plot to check for system stability. A brief (4 minute)video follows that shows how to start the program, and describes its
major controls.
You should put these files in a directory in the Matlab path so that Matlab
can find them. After that you just need to enter "NyquistGui" at the
command prompt.
There is no dynamic scaling of axes, so you must choose systems with
poles that fit within the axes. Again - this is a tool for learning,
not for design. If you need an accurate plot with scalable axes, use
Matlab's "nyquist" command.
System time delays are ignored.
In the case of a system that is marginally stable (i.e., it passes
through the -1+j0 point), the angles drawn on the "L(s)" plot may be
inaccurate, particularly if all poles of the closed loop system are on the
jω axis.
Any poles of L(s) that are very near the imaginary axis are placed on
the axis.
then the Nyquist path is as shown
then the Nyquist path is as shown
, straightforward case, no special conditions
, L(s) has pole at the origin
, L(s) has pole in RHP, and is stable
, L(s) has pole in the RHP and is unstable
, L(s) has double pole at the origin
, L(s) has time delay
but is otherwise equivalent to Example 1
