This document comprises the bulk of the text describing the root locus, and is formatted slightly differently to facilitate printing.
To understand what root locus plots are, and why they are important, let's examine the behavior of a system when it is in a control system. Assume that the system is defined by the transfer function:
We'll control this system with a very simple proportional controller in which the input to the system to be controlled is proportional (with gain, K) to the difference between the input, R(s), and the output, C(s).
The loop gain is K·G(s), so the closed loop gain is given by
We want to examine how the behavior of the system varies as K changes, so let's try several values of K. Let's arbitrarily try K=1, 10 and 100 so that we have a wide range of K values.
| K | Xfer Function | Step Response |
|---|---|---|
| K=1 |
 |
 |
| K=10 |
 |
 |
| K=100 |
 |
 |
The response with K=1 was too slow, the response with K=100 was too oscillatory, and the response with K=10 is almost just right, though we may want to adjust K to get a little bit less overshoot. Clearly this method is rather "hit-or-miss" and it may take us a long time to find a suitable value for K.
A more analytical method might involve finding the poles of the closed loop transfer function. Since the transfer function is second order, we can factor the denominator using the quadratic equation. The roots of the denominator are at:
This will give overdamped response for 9>4K, underdamped response for 9<4K, and critically damped response for 9=4K. If we specify that we want an underdamped response with ζ=1/√2 we must set the magnitudes of the real and imaginary parts of the roots equal to each other. So starting from
we know that the real and imaginary parts must have equal magnitude
and we can solve for K.
Let's try this value. If we use K=4.5 we get the transfer function
and the step response is shown below. As you can see, it is superior to our other attempts; it is reasonably fast with fairly small overshoot.
While this technique is suitable for simple problems, like the one given, it quickly becomes untenable for more complicated systems. For example what would we do if
We will need another approach to deal with such systems. This is introduced in the next section (though in the context of the simple system we have been dealing with).
For the very simple problem described above, it was possible to calculate the precise location of the roots, and choose a value of K that gave us a good response. For more complicated systems it is not so straightforward so we need a more general method for approaching the problem. This more general method is called the "root locus" method. With this technique we make a plot of the path of the roots as a parameter (usually the gain, as above) varies. (Note).
Consider the simple example from above,
If we want to plot the path of the roots as K varies we can calculate the roots of the equation
for many values of K by using the quadratic equation.
| K | Roots |
|---|---|
| 1 | -2.62, -0.38 |
| 2 | -2, -1 |
| 4 | -1.5 ± j1.32 |
| 10 | -1.5 ± j2.78 |
| 20 | -1.5 ± j4.21 |
| 40 | -1.5 ± j6.14 |
| 100 | -1.5 ± j9.89 |
These values are plotted below (for complex conjugate roots, the value of K is only shown for the root with a positive imaginary part).
There is a lot of information in this diagram. It tells us that the system starts out overdamped for small values of K, and becomes underdamped as K increases, and becomes increasingly underdamped as K continues to increase.
A root locus plot is a variation on this kind of plot. It shows the path of the roots as K is varied, but does not show the actual values of K. This kind of plot is sufficiently important that Matlab has a command specifically for generating these plots. A root locus plot for this system is shown below, along with the Matlab used to create it.
>> G=tf(1,[1 3 0]) %Define gain of system in the loop
Transfer function:
1
---------
s^2 + 3 s
>> rlocus(G)
>> axis([-4 0 -10 10])
The starting points for the roots, when K=0 (note), are shown by the two small "x" marks at s=0 and s=-3. As K increases, the two roots move horizontally towards each other, meet at s=-1.5, and then move vertically away from each other.
Though the value of K isn't plotted on the graph it is easily found. For example, if we want to know the value of K at s=-1.5 we can use the fact that the characteristic equation of
is given by the denominator,
.
This equation is true for any point on the root locus (more on this on the next page), and in particular it is true at s=-1.5. Since at s=-1.5
We can use this information to calculate the desired value of K:
.
For the very simple system of this problem, there were many ways to find how the roots varied as we varied the gain of the system. For a more complicated system this is not easy. The root locus plot gives us a graphical way to observe how the roots move as the gain, K, is varied. The next page gives a description of techniques for sketching the location of the closed loop poles of a system for systems that are much more complicated.
The root locus plot indicates how the closed loop poles of a system vary with a system parameter (typically a gain, K).
Derive Rules for
Sketching the Root Locus
© Copyright 2005-2009
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
This document will discuss the derivation of rules for
sketching the root locus. It is not necessary to understand all of these in
order to do the sketches, but it can be helpful to understand whence come the various
rules. Instead of presenting examples in this document, there are links
to files that contain five separate examples. After each rule, you can select
the link for each of the five examples, and the application of that specific rule
to the selected rule is displayed (along with a brief discussion). Each of
the five
examples can also be examined in its entirety by clicking on the link below.
Note: the example files are edited versions of web pages generated automatically
with a MatLab script, RLocusGui.
To sketch a root locus there are several techniques that can be used as a guide. Not all of these are applicable to all loci. The steps used to sketch a root locus plot are enumerated below:
The closed loop transfer function of the system shown is
So the characteristic equation is
We can write the loop gain as a ratio of polynomials, (we will assume K>0, a0>0, b0>0; generally a0=1). N(s), the numerator polynomial, is defined to be mth order; D(s) is nth order. N(s) has zeros at zi (i=1..m); D(s) has zeros at pi (i=1..n). Note the zeros of D(s) are the poles of the loop gain. The difference between the orders of the numerator and denominator polynomial, n and m, is q, so q=n-m. We assume here that the transfer function is proper - in other words q≥0.
,
or we can write the loop gain in its factored form:
As K changes, so do locations of closed loop poles (i.e., zeros of characteristic equation).
It is convenient for the derivation of many of the rules that follow to rewrite the characteristic equation as follows.
Many of the rules discussed below come from two conditions imposed by the characteristic equation. Since this equation involves a complex quantity both the magnitude and phase of the two sides of the equation must be equal.
The magnitude condition is expressed as
Since K≥0, we can rewrite this as
The phase angle is expressed as
Since K≥0, it has a phase of 0° and can be ignored. The angle of -1 is any odd multiple of 180º.
| Magnitude Condition | Phase Condition |
|
|
Note: an alternate set of rules, for K<0 can be derived; this is referred to as the complementary root locus.
The loop gain is KG(s)H(s) which can be rewritten as KN(s)/D(s).
N(s), the numerator polynomial, is mth order; D(s)
is nth order:
K>0, a0>0, b0>0.
N(s) has zeros at zi (i=1..m); D(s) has zeros
at pi (i=1..n):
.
The difference between n and m is q, so q=n-m.
Since the characteristic equation has real coefficients, any zeros must occur in complex conjugate pairs (which are symmetric about the real axis). Since the root locus is just a diagram of the roots of the characteristic equation as K varies, it must also be symmetric about the real axis.
The Root locus is symmetric about the real axis.
Since the root locus is just a diagram of the roots of the characteristic equation as K varies, and the order of the characteristic equation is the same as that of the denominator of the loop gain, the number of branches is n, the order of the denominator polynomial.
The number of branches of the root locus is equal to the order of characteristic equation.
Start from the magnitude condition:
It is apparent that if K→0, the only way the left hand side of the equation can be equal to 1 is if the quantity in the absolute value goes to infinity. This happens when D(s)→0. So the poles of the loop gain (D(s)=0) are the starting points for the loci (when K=0).
It is also apparent that if K→∞, the only way the left hand side of the equation can be equal to 1 is if the quantity in the absolute value goes to zero. This happens when N(s)→0, and it also happens as s→∞ if the order of the denominator is greater than the order of the numerator. So the zeros of the loop gain (which occur at N(s)=0, and perhaps as s→∞) are the ending points for the loci (when K→∞).
The locus starts (when K=0) at poles of the loop gain, and ends (when K→∞ ) at the zeros. Note: there are q zeros of the loop gain as s→∞ .
Start from the angle condition:
Write out the ratio of polynomials:
We will take a0>0, b0>0 (generally a0=1), so:
So the angle criterion can be restated as:
Now consider the angle between a point "s" (the red vector) on the real axis, and a point "z" (the blue vector) that is also on the real axis. The diagrams below show the vector "s-z" f(the green vector) or the case when "s" is to the left of "z," and when s is to the right of z. (Review: How to subtract vectors).
 |
 |
In both figures, "s" is shown by a red vector, and "z" is shown by a blue vector. The difference can be found by drawing a vector from the point "z" to the point "s," which is shown by a green vector. When "s" is to the left of "z" (left diagram), the angle of the vector "s-z" is 180º (or any odd multiple of 180º). When "s" is to the right of "z" (right diagram), the angle of the vector "s-z" is 0º (or any even multiple of 180º).
However, we still need to consider complex conjugate poles and zeros. To see their contribution, consider the diagram below.
In this diagram the vector "s" is red, "z" and its conjugate "z*" are blue and "s-z" and "s-z*" are green. Clearly the angle contributions from "z" and its conjugate "z*" (shown in dotted green) are equal and opposite, and so cancel each other out. Therefore we need not consider the contribution of complex conjugate zeros, or poles; we need only consider the contribution of zeros and poles that are on the real axis.
Now reconsider our statement of the angle criterion:
This equation indicates that any zeros to the left of a quantity "s" on the real axis contributes 180º, a pole to the left will contribute -180º, but a pole or zero to the right of "s" on the real axis contributes 0º. Since the sum of angle contributions from zeros on the real axis minus the sum of contributions from poles on the real axis is an odd multiple of 180º, this indicates that if a point "s" on the real axis will only be on the locus if it is to the left of an odd number of zeros and poles that are on the axis.
The locus exists on real axis to the left of an odd number of poles and zeros on the axis.
If q>0 (in other words, the denominator polynomial of the loop gain is of higher order than is the numerator), then the loop gain has q zeros as |s|→∞ . We will show that we can make some simple approximations that will describe the behavior of the closed loop poles as |s|→∞ . If q=0, you need not do this step.
First we will do a very simple approximation to study behavior of the root locus as |s|→∞ . Consider the characteristic equation.
We can rewrite this, and then if we let |s|→∞ , all but the highest order terms of the polynomials go become insignificant.
Now we can apply the angle criterion
Now we can write "s" as
and complete the derivation
K, a0, b0 and M are all positive so they don't contribute to the angle, so
To summarize: as |s|→∞ , the loci are asymptotic to a set of lines that that radiate outward from the origin with angles of θ=±r180°/q where r=1, 3, 5...
If this isn't obvious, consider a positive number, β, with
Now use the polar form of "s,"
,
and we get
Since the number is complex, both the magnitude and phase must be equal.
Consider first the magnitude.
This tells us that as β goes from 0 to infinity, so does the magnitude of "s."
Now consider the phase.
This tells us that the angle of "s" is given by ±r180°/q, where r=1, 3, 5...
Taking both the magnitude and phase into consideration this shows that
is represents lines emanating from the origin at equally spaced angles as β goes from 0 to ∞ .
In the approximation above, we kept only the highest order term of the numerator and denominator polynomial as |s|→∞ . We can get a better approximation if we keep the two highest order terms. Let's start with the factored form of the loop gain, and multiply it out:
I have only written out the two highest order terms of the polynomial. If the second term in the polynomial is not obvious, then examining the case of a third order polynomial should make it obvious.
(The equation on the second line agrees with our result.)
Now let's do some manipulations to get this into a more useful form. First, we let |s|→∞ and only keep the two highest order terms of the polynomials.
We can multiply the numerator and denominator by the same term. This simplifies the numerator.
Note that the second term in the numerator polynomial is small, so we can use the binomial approximation.
Keep only the highest order terms from the denominator polynomial (since |s|→∞ ):
Now we note that a polynomial with repeated roots is given by
Again, if this isn't obvious, consider the third order case (which generalizes to higher order):
As |s|→∞ we can, again, simplify this by keeping only the highest order term
This is expression has the same form as the one in the denominator of the expression we just derived (i.e., loop gain as |s|→∞). We can make the substitution:
our expression for the loop gain as |s|→∞ becomes
Putting this back into our expression for the characteristic equation we get (as |s|→∞ ):
This represents a set of vectors that intersect the real axis at s=-σ, that radiate outward with angles of θ= ±r180°/q where r=1, 3, 5...
as s→∞ .Since
represents lines emanating from the origin at equally spaced angles as β goes from 0 to ∞ , then the expression
is just shifted by σ. In other words, it represents lines emanating from σ at equally space angles.
This tells us that the locus is asymptotic to these lines because both the locus and (s-σ)q have the same form as |s|→∞ .
If q>0 there are asymptotes of the root locus, and...
asymptotes intersect real axis
at 
,
asymptotes radiate out with
angles 
, where
r=1, 3, 5…
To find where the locus breaks away from the axis (or converges on the axis), we note that this always occurs when two (or more) roots intersect. It is a well know fact, that when a polynomial has multiple roots, not only is the value of the polynomial zero, but its derivative is zero also (Background).
Start by considering the characteristic equation
At the break-away (and -in) points, the derivative of the characteristic equation is also zero.
If we simplify this we get:
I prefer to switch the order of the subtraction (though it really makes no difference),
.
There are break-away or –in points of the locus on the real
axis wherever .
Note: Many times, especially for simple root loci, there are no break-away or break-in points. In these cases, this step is not necessary.
If the loop gain, G(s)H(s), has a simple pole on the real axis, we know that the locus will leave from the pole, as K→0, along the axis. However, if the pole is complex it can leave at any angle. To find the angle at which the locus leaves a complex pole, we start from the re-stated angle criterion (from the "Locus on Real Axis" rule):
To find the angle at which the locus leaves from the pole pj, we can rewrite the angle criterion by isolating the angle between the locus and pj.
or
In this equation we have taken r=1 since the solutions are the same for all values of r. Now if we consider a point "s" on the locus that is very close to pj, then all the terms on the right hand side can be approximated by the angle between the pole or zero and pj. In other words, if "s" is very close to pj, then we can approximate the angle criterion as:
This is demonstrated by an example, below which shows a Root Locus plot of a function G(s)H(s) that has one zero at s=-1, and three poles at s=-2, and s= -1±j. :
To find the angle of departure from the pole at s=-1+j (which we will call p2), we choose a point on the locus very near p2 and then find the angles from the zero and the other poles.
The angle of departure is shown in grey on the diagram.
The angle of departure from a complex pole, pj, is 180 degrees + (sum of angles between pj and all zeros) - (sum of angles between pj and all other poles).
Note: Many times, especially for simple root loci, there are no complex poles in loop gain. In these cases, this step is not necessary.
This follows closely the derivation for the previous rule ("Angle of Departure") and will be brief.
If the loop gain, G(s)H(s), has a simple zero on the real axis, we know that the locus will arrive at the zero, as K→∞ , along the axis. However, if the zero is complex it can arrive at any angle. To find the angle at which the locus arrives at a complex zero, we start from the re-stated angle criterion (from the "Locus on Real Axis" rule):
To find the angle at which the locus arrives from the pole zj, we can rewrite the angle criterion as
or
In this equation we have taken r=1 since the solutions are the same for all values of r. Now if we consider a point "s" on the locus that is very close to zj, then all the terms on the right hand side can be approximated by the angle between the pole or zero and zj. In other words, if "s" is very close to zj, then we can approximate the angle criterion as:
The angle of arrival at a complex pole, zj, is 180 degrees + (sum of angles between zj and all other zeros) - (sum of angles between zj and all poles).
Note: Many times, especially for simple root loci, there are no complex zeros in loop gain. In these cases, this step is not necessary.
If it becomes apparent that the root locus crosses the imaginary axis (i.e., it is unstable for some values of K), use a technique such as Routh-Horwitz to find where the locus crosses the imaginary axis (i.e., the frequency of oscillation when it becomes unstable).
Use Routh-Horwitz to determine where the locus crosses the imaginary axis.
Note: Many times, especially for simple root loci, the root locus does not cross the imaginary axis, or does so along the real axis. In these cases, this step is not necessary.
Recall that the characteristic equation can be written in the form:
or
The values a0...an and b0...bm are all known. So given a value of K we can determine the resulting polynomial and factor it to find the roots of the characteristic equation (this may require a computer).
Rewrite characteristic equation as D(s)+KN(s)=0. Put value of K into equation, and find roots of characteristic equation.
Note: Many times this step is not necessary.
Recall that the characteristic equation can be written in the form:
or
So, given a value of "s" that is on the locus, it is possible to solve for the corresponding value of K.
Note that if the value of "s" is obtained by inspection of a root locus plot, it is only approximate. If the chosen value does not actually lie on the locus, the resulting value of K may be complex. If this happens, the imaginary part will be small, so just take the imaginary part of K. You should then use this value of K (see above) to find the exact value of the root location.
Rewrite characteristic equation as
, replace “s” by the desired pole location
and solve for K.
Note: Many times this step is not necessary, especially when the task is simply to draw the root locus.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
For the open loop transfer function, G(s)H(s):
We have n=2 poles at s = 0, -3. We have m=0 finite zeros. So there exists
q=2 zeros as s goes to infinity (q = n-m = 2-0 = 2).
We can rewrite the open loop transfer function as G(s)H(s)=N(s)/D(s) where N(s)
is the numerator polynomial, and D(s) is the denominator polynomial.
N(s)=
1, and D(s)= s2 + 3 s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s2 + 3 s+ K( 1 ) = 0
As you can see, the locus is symmetric about the real axis
The open loop transfer function, G(s)H(s), has 2 poles, therefore the locus
has 2 branches. Each branch is displayed in a different color.
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s).
These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the
open loop transfer function, G(s)H(s). Don't forget we have we also have q=n-m=2
zeros at infinity. (We have n=2 finite poles, and m=0 finite zeros).
The root locus exists on real axis to left of an odd number of poles and zeros of open loop transfer function, G(s)H(s), that are on the real axis. These real pole and zero locations are highlighted on diagram, along with the portion of the locus that exists on the real axis.
Root locus exists on real axis between:
0 and -3
... because on the real axis, we have 2 poles at s = -3, 0, and we have no
zeros.
In the open loop transfer function, G(s)H(s), we have n=2 finite poles, and m=0
finite zeros, therefore we have q=n-m=2 zeros at infinity.
Angle of asymptotes at odd multiples of ±180°/q, (i.e., ±90°)
There exists 2 poles at s = 0, -3, ...so sum of poles=-3.
There exists 0 zeros, ...so sum of zeros=0.
(Any imaginary components of poles and zeros cancel when summed because they appear
as complex conjugate pairs.)
Intersect of asymptotes is at ((sum of poles)-(sum of zeros))/q = -1.5.
Intersect is at ((-3)-(0))/2 = -3/2 = -1.5 (highlighted by five pointed star).
Break Out (or Break In) points occur where N(s)D'(s)-N'(s)D(s)=0, or 2 s + 3
= 0. (details below*)
This polynomial has 1 root at s = -1.5.
From these 1 root, there exists 1 real root at s = -1.5. These are highlighted
on the diagram above (with squares or diamonds.)
These roots are all on the locus (i.e., K>0), and are highlighted with squares.
* N(s) and D(s) are numerator and denominator polynomials of G(s)H(s), and the tick
mark, ', denotes differentiation.
N(s) = 1
N'(s) = 0
D(s)= s2 + 3 s
D'(s)= 2 s + 3
N(s)D'(s)= 2 s + 3
N'(s)D(s)= 0
N(s)D'(s)-N'(s)D(s)= 2 s + 3
Here we used N(s)D'(s)-N'(s)D(s)=0, but we could multiply by -1 and use N'(s)D(s)-N(s)D'(s)=0.
No complex poles in loop gain, so no angles of departure.
No complex zeros in loop gain, so no angles of arrival.
Locus crosses imaginary axis at 1 value of K. These values are normally
determined by using Routh's method. This program does it numerically, and
so is only an estimate.
Locus crosses where K = 0, corresponding to crossing imaginary axis at s=0.
These crossings are shown on plot.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s2 + 3 s+ K( 1 ) = 0
So, by choosing K we determine the characteristic equation whose roots are the closed
loop poles.
For example with K=2.25225, then the characteristic equation is
D(s)+KN(s) = s2 + 3 s + 2.2522( 1 ) = 0, or
s2 + 3 s + 2.2522= 0
This equation has 2 roots at s = -1.5±0.047j. These are shown by the large
dots on the root locus plot
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0, or
K = -D(s)/N(s) = -( s2 + 3 s ) / ( 1 )
We can pick a value of s on the locus, and find K=-D(s)/N(s).
For example if we choose s= -1.6 + 1.6j (marked by asterisk),
then D(s)=-4.87 + -0.243j, N(s)= 1 + 0j,
and K=-D(s)/N(s)= 4.87 + 0.243j.
This s value is not exactly on the locus, so K is complex, (see note below), pick
real part of K ( 4.87)
For this K there exist 2 closed loop poles at s = -1.5± 1.6j. These poles
are highlighted on the diagram with dots, the value of "s" that was originally specified
is shown by an asterisk.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on the
locus, then the calculated value of K will be complex instead of real. Just ignore
the the imaginary part of K (which will be small).
Note also that only one pole location was chosen and this determines the value
of K. If the system has more than one closed loop pole, the location of the other
poles are determine solely by K, and may be in undesirable locations.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
For the open loop transfer function, G(s)H(s):
We have n=3 poles at s = 0, -3, -2. We have m=0 finite zeros. So there
exists q=3 zeros as s goes to infinity (q = n-m = 3-0 = 3).
We can rewrite the open loop transfer function as G(s)H(s)=N(s)/D(s) where N(s)
is the numerator polynomial, and D(s) is the denominator polynomial.
N(s)= 1, and D(s)= s3 + 5 s2 + 6 s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s3 + 5 s2 + 6 s+ K( 1 ) = 0
As you can see, the locus is symmetric about the real axis.
The open loop transfer function, G(s)H(s), has 3 poles, therefore the locus
has 3 branches. Each branch is displayed in a different color.
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s).
These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the
open loop transfer function, G(s)H(s). Don't forget we have we also
have q=n-m=3 zeros at infinity. (We have n=3 finite poles, and m=0 finite
zeros).
The root locus exists on real axis to left of an odd number of poles and zeros
of open loop transfer function, G(s)H(s), that are on the real axis.
These real pole and zero locations are highlighted on diagram, along with the portion
of the locus that exists on the real axis.
Root locus exists on real axis between:
0 and -2
-3 and negative infinity
... because on the real axis, we have 3 poles at s = -2, -3, 0, and we have no
zeros.
In the open loop transfer function, G(s)H(s), we have n=3 finite poles, and m=0
finite zeros, therefore we have q=n-m=3 zeros at infinity.
Angle of asymptotes at odd multiples of ±180°/q, (i.e., ±60°
, ±180°)
There exists 3 poles at s = 0, -3, -2, ...so sum of poles=-5.
There exists 0 zeros, ...so sum of zeros=0.
(Any imaginary components of poles and zeros cancel when summed because they
appear as complex conjugate pairs.)
Intersect of asymptotes is at ((sum of poles)-(sum of zeros))/q = -1.67.
Intersect is at ((-5)-(0))/3 = -5/3 = -1.67 (highlighted by five pointed star).
Break Out (or Break In) points occur where N(s)D'(s)-N'(s)D(s)=0, or
3 s2 + 10 s + 6 = 0. (details below*)
This polynomial has 2 roots at s = -2.5, -0.78.
From these 2 roots, there exists 2 real roots at s = -2.5, -0.78. These
are highlighted on the diagram above (with squares or diamonds.)
Not all of these roots are on the locus. Of these 2 real roots, there exists 1 root
at s = -0.78 on the locus (i.e., K>0). Break-away (or break-in) points
on the locus are shown by squares.
(Real break-away (or break-in) with K less than 0 are shown with diamonds).
* N(s) and D(s) are numerator and denominator polylnomials of G(s)H(s), and the
tick mark, ', denotes differentiation.
N(s) = 1
N'(s) = 0
D(s)= s3 + 5 s2 + 6 s
D'(s)= 3 s2 + 10 s + 6
N(s)D'(s)= 3 s2 + 10 s + 6
N'(s)D(s)= 0
N(s)D'(s)-N'(s)D(s)= 3 s2 + 10 s + 6
Here we used N(s)D'(s)-N'(s)D(s)=0, but we could multiply by -1 and use N'(s)D(s)-N(s)D'(s)=0.
No complex poles in loop gain, so no angles of departure.
No complex zeros in loop gain, so no angles of arrival.
Locus crosses imaginary axis at 2 values of K. These values are normally
determined by using Routh's method. This program does it numerically,
and so is only an estimate.
Locus crosses where K = 0, 30.2, corresponding to crossing imaginary axis at s=0,
±2.45j, respectively.
These crossings are shown on plot.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s3 + 5 s2 + 6 s+ K( 1 ) = 0
So, by choosing K we determine the characteristic equation whose roots are the closed
loop poles.
For example with K=4.00188, then the characteristic equation is
D(s)+KN(s) = s3 + 5 s2 + 6 s + 4.0019( 1 ) = 0, or
s3 + 5 s2 + 6 s + 4.0019= 0
This equation has 3 roots at s = -3.7, -0.67± 0.8j. These are shown
by the large dots on the root locus plot
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0, or
K = -D(s)/N(s) = -( s3 + 5 s2 + 6 s ) / ( 1 )
We can pick a value of s on the locus, and find K=-D(s)/N(s).
For example if we choose s= -0.7 + 0.84j (marked by asterisk),
then D(s)=-4.15 + -0.222j, N(s)= 1 + 0j,
and K=-D(s)/N(s)= 4.15 + 0.222j.
This s value is not exactly on the locus, so K is complex,
(see note below), pick real part of K ( 4.15)
For this K there exist 3 closed loop poles at s = -3.7, -0.66±0.83j.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on
the locus, then the calculated value of K will be complex instead of real. Just
ignore the imaginary part. These poles are highlighted on the diagram
with dots, the value of "s" that was originally specified is shown by an asterisk.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on
the locus, then the calculated value of K will be complex instead of real. Just
ignore the the imaginary part of K (which will be small).
Note also that only one pole location was chosen and this determines the value of
K. If the system has more than one closed loop pole, the location of the other poles
are determine solely by K, and may be in undesirable locations.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
For the open loop transfer function, G(s)H(s):
We have n=2 poles at s = 2, -1.
We have m=1 finite zero at s = -3.
So there exists q=1 zero as s goes to infinity (q = n-m = 2-1 = 1).
We can rewrite the open loop transfer function as
G(s)H(s)=N(s)/D(s) where N(s) is the numerator polynomial, and
D(s) is the denominator polynomial.
N(s)= s + 3, and
D(s)= s2 - 1 s - 2.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s2 - 1 s - 2+ K( s + 3 ) = 0
As you can see, the locus is symmetric about the real axis
The open loop transfer function, G(s)H(s), has 2 poles, therefore the locus
has 2 branches. Each branch is displayed in a different color.
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s).
These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the open loop
transfer function, G(s)H(s). Finite zeros are shown by a "o" on the diagram
above. Don't forget we have we also have q=n-m=1 zero at infinity. (We have n=2 finite poles, and m=1 finite zero).
The root locus exists on real axis to left of an odd number of poles and zeros
of open loop transfer function, G(s)H(s), that are on the real axis.
These real pole and zero locations are highlighted on diagram, along with the portion
of the locus that exists on the real axis.
Root locus exists on real axis between:
2 and -1
-3 and negative infinity
... because on the real axis, we have 2 poles at s = -1, 2, and we have 1 zero at s = -3.
In the open loop transfer function, G(s)H(s), we have n=2 finite poles, and
m=1 finite zero, therefore we have q=n-m=1 zero at infinity.
Angle of asymptotes at odd multiples of ±180°/q, (i.e., ±180°)
There exists 2 poles at s = 2, -1, ...so sum of poles=1.
There exists 1 zero at s = -3, ...so sum of zeros=-3.
(Any imaginary components of poles and zeros cancel when summed because they
appear as complex conjugate pairs.)
Intersect of asymptotes is at ((sum of poles)-(sum of zeros))/q = 4.
Intersect is at ((1)-(-3))/1 = 4/1 = 4 (highlighted by five pointed star).
Since q=1, there is a single asymptote at ±180°
(on negative real axis), so intersect of this asymptote
on the axis s not important (but it is shown anyway).
Break Out (or Break In) points occur where N(s)D'(s)-N'(s)D(s)=0, or
s2 + 6 s - 1 = 0. (details below*)
This polynomial has 2 roots at s = -6.2, 0.16.
From these 2 roots, there exists 2 real roots at s = -6.2, 0.16. These
are highlighted on the diagram above (with squares or diamonds.)
These roots are all on the locus (i.e., K>0), and are highlighted with squares.
* N(s) and D(s) are numerator and denominator polylnomials of G(s)H(s), and the
tick mark, ', denotes differentiation.
N(s) = s + 3
N'(s) = 1
D(s)= s2 - 1 s - 2
D'(s)= 2 s - 1
N(s)D'(s)= 2 s2 + 5 s - 3
N'(s)D(s)= s2 - 1 s - 2
N(s)D'(s)-N'(s)D(s)= s2 + 6 s - 1
Here we used N(s)D'(s)-N'(s)D(s)=0, but we could multiply by -1 and use N'(s)D(s)-N(s)D'(s)=0.
No complex poles in loop gain, so no angles of departure.
No complex zeros in loop gain, so no angles of arrival.
Locus crosses imaginary axis at 2 values of K. These values are normally
determined by using Routh's method. This program does it numerically,
and so is only an estimate.
Locus crosses where K = 0.646, 1, corresponding to crossing imaginary axis at s=0,
±0.994j, respectively.
These crossings are shown on plot.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s2 - 1 s - 2+ K( s + 3 ) = 0
So, by choosing K we determine the characteristic equation whose roots are the closed
loop poles.
For example with K=7.15931, then the characteristic equation is
D(s)+KN(s) = s2 - 1 s - 2 + 7.1593( s + 3 ) = 0, or
s2 + 6.1593 s + 19.4779= 0
This equation has 2 roots at s = -3.1± 3.2j. These are shown by the
large dots on the root locus plot
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0, or
K = -D(s)/N(s) = -( s2 - 1 s - 2 ) / ( s + 3 )
We can pick a value of s on the locus, and find K=-D(s)/N(s).
For example if we choose s= -3.2 + 3.3j (marked by asterisk),
then D(s)=0.672 + -24.8j, N(s)=-0.234 + 3.32j,
and K=-D(s)/N(s)= 7.44 + -0.322j.
This s value is not exactly on the locus, so K is complex,
(see note below), pick real part of K ( 7.44)
For this K there exist 2 closed loop poles at s = -3.2± 3.2j.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on
the locus, then the calculated value of K will be complex instead of real. Just
ignore the imaginary part. These poles are highlighted on the diagram
with dots, the value of "s" that was originally specified is shown by an asterisk.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on
the locus, then the calculated value of K will be complex instead of real. Just
ignore the the imaginary part of K (which will be small).
Note also that only one pole location was chosen and this determines the value of
K. If the system has more than one closed loop pole, the location of the other poles
are determine solely by K, and may be in undesirable locations.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
For the open loop transfer function, G(s)H(s):
We have n=3 poles at s = -2, -1± 1j.
We have m=1 finite zero at s = -1.
So there exists q=2 zeros as s goes to infinity (q = n-m = 3-1 = 2).
We can rewrite the open loop transfer function as
G(s)H(s)=N(s)/D(s) where N(s) is the numerator polynomial, and
D(s) is the denominator polynomial.
N(s)= s + 1, and
D(s)= s3 + 4 s2 + 6 s + 4.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s3 + 4 s2 + 6 s + 4+ K( s + 1 ) = 0
As you can see, the locus is symmetric about the real axis
The open loop transfer function, G(s)H(s), has 3 poles, therefore the locus has
3 branches. Each branch is displayed in a different color.
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s).
These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the open loop transfer function, G(s)H(s). Finite zeros are shown by a "o" on the diagram above. Don't forget we have we also have q=n-m=2 zeros at infinity. (We have n=3 finite poles, and m=1 finite zero).
The root locus exists on real axis to left of an odd number of poles and zeros
of open loop transfer function, G(s)H(s), that are on the real axis.
These real pole and zero locations are highlighted on diagram, along with the portion
of the locus that exists on the real axis.
Root locus exists on real axis between:
-1 and -2
... because on the real axis, we have 1 pole at s = -2, and we have 1 zero at s = -1.
In the open loop transfer function, G(s)H(s), we have n=3 finite poles, and m=1 finite zero, therefore we have q=n-m=2 zeros at infinity.
Angle of asymptotes at odd multiples of ±180°/q, (i.e., ±90°)
There exists 3 poles at s = -2, -1± 1j, ...so sum of poles=-4.
There exists 1 zero at s = -1, ...so sum of zeros=-1.
(Any imaginary components of poles and zeros cancel when summed because they appear as complex conjugate pairs.)
Intersect of asymptotes is at ((sum of poles)-(sum of zeros))/q = -1.5.
Intersect is at ((-4)-(-1))/2 = -3/2 = -1.5 (highlighted by five pointed star).
Break Out (or Break In) points occur where N(s)D'(s)-N'(s)D(s)=0, or
2 s3 + 7 s2 + 8 s + 2 = 0. (details below*)
This polynomial has 3 roots at s = -1.6±0.65j, -0.34.
From these 3 roots, there exists 1 real root at s = -0.34. These are
highlighted on the diagram above (with squares or diamonds.)
None of the roots are on the locus.
* N(s) and D(s) are numerator and denominator polylnomials of G(s)H(s), and the
tick mark, ', denotes differentiation.
N(s) = s + 1
N'(s) = 1
D(s)= s3 + 4 s2 + 6 s + 4
D'(s)= 3 s2 + 8 s + 6
N(s)D'(s)= 3 s3 + 11 s2 + 14 s + 6
N'(s)D(s)= s3 + 4 s2 + 6 s + 4
N(s)D'(s)-N'(s)D(s)= 2 s3 + 7 s2 + 8 s + 2
Here we used N(s)D'(s)-N'(s)D(s)=0, but we could multiply by -1 and use N'(s)D(s)-N(s)D'(s)=0.
Find angle of departure from pole at -1+1j
θz1
=angle((Departing pole)- (zero at -1) ).
θz1
=angle((-1+1j) - (-1)) = angle(0+1j) = 90°
θp1
=angle((Departing pole)- (pole at -2) ).
θp1
=angle((-1+1j) - (-2)) = angle(1+1j) = 45°
θp3
=angle((-1+1j) - (-1-1j)) = angle(0+2j) = 90°
Angle of Departure is equal to:
θdepart
= 180°
+ sum(angle to zeros) - sum(angle to poles).
θdepart
= 180° + 90 - 135.
θdepart
= 135°
This angle is shown in gray.
It may be hard to see if it is near 0°.
No complex zeros in loop gain, so no angles of arrival.
Locus does not cross imaginary axis.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s3 + 4 s2 + 6 s + 4+ K( s + 1 ) = 0
So, by choosing K we determine the characteristic equation whose roots are the closed
loop poles.
For example with K=2.60256, then the characteristic equation is
D(s)+KN(s) = s3 + 4 s2 + 6 s + 4 + 2.6026( s + 1 ) = 0, or
s3 + 4 s2 + 8.6026 s + 6.6026= 0
This equation has 3 roots at s = -1.4± 1.8j, -1.3. These are shown by
the large dots on the root locus plot
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0, or
K = -D(s)/N(s) = -( s3 + 4 s2 + 6 s + 4 ) / ( s + 1 )
We can pick a value of s on the locus, and find K=-D(s)/N(s).
For example if we choose s= -1.2 + 1.3j (marked by asterisk),
then D(s)=0.285 + -1.17j, N(s)=-0.225 + 1.27j,
and K=-D(s)/N(s)=0.929 + 0.0603j.
This s value is not exactly on the locus, so K is complex,
(see note below), pick real part of K (0.929)
For this K there exist 3 closed loop poles at s = -1.2± 1.3j, -1.6Note: it is often
difficult to choose a value of s that is precisely on the locus, but we can pick
a point that is close. If the value is not exactly on the locus, then
the calculated value of K will be complex instead of real. Just ignore the imaginary
part. These poles are highlighted on the diagram with dots, the value
of "s" that was originally specified is shown by an asterisk.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on
the locus, then the calculated value of K will be complex instead of real. Just
ignore the the imaginary part of K (which will be small).
Note also that only one pole location was chosen and this determines the value of
K. If the system has more than one closed loop pole, the location of the other poles
are determine solely by K, and may be in undesirable locations.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
For the open loop transfer function, G(s)H(s):
We have n=5 poles at s = 0, -3± 2j, -2, -1.
We have m=2 finite zeros at s = -1± 1j.
So there exists q=3 zeros as s goes to infinity (q = n-m = 5-2 = 3).
We can rewrite the open loop transfer function as G(s)H(s)=N(s)/D(s) where N(s)
is the numerator polynomial, and D(s) is the denominator polynomial.
N(s)= s2 + 2 s + 2, and
D(s)= s5 + 9 s4 + 33 s3 + 51 s2 + 26
s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s5 + 9 s4 + 33 s3 + 51 s2
+ 26 s+ K( s2 + 2 s + 2 ) = 0
As you can see, the locus is symmetric about the real axis
The open loop transfer function, G(s)H(s), has 5 poles, therefore the locus
has 5 branches. Each branch is displayed in a different color.
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s).
These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the
open loop transfer function, G(s)H(s). Finite zeros are shown by a "o" on
the diagram above. Don't forget we have we also have q=n-m=3 zeros at infinity.
(We have n=5 finite poles, and m=2 finite zeros).
The root locus exists on real axis to left of an odd number of poles and zeros of open loop transfer function, G(s)H(s), that are on the real axis. These real pole and zero locations are highlighted on diagram, along with the portion of the locus that exists on the real axis.
Root locus exists on real axis between:
0 and -1
-2 and negative infinity
... because on the real axis, we have 3 poles at s = -1, -2, 0, and we have no
zeros.
In the open loop transfer function, G(s)H(s), we have n=5 finite poles, and m=2
finite zeros, therefore we have q=n-m=3 zeros at infinity.
Angle of asymptotes at odd multiples of ±180°/q, (i.e., ±60°, ±180°)
There exists 5 poles at s = 0, -3± 2j, -2, -1, ...so sum of poles=-9.
There exists 2 zeros at s = -1± 1j, ...so sum of zeros=-2.
(Any imaginary components of poles and zeros cancel when summed because they appear
as complex conjugate pairs.)
Intersect of asymptotes is at ((sum of poles)-(sum of zeros))/q = -2.33.
Intersect is at ((-9)-(-2))/3 = -7/3 = -2.33 (highlighted by five pointed star).
Break Out (or Break In) points occur where N(s)D'(s)-N'(s)D(s)=0, or
33 s6 + 26 s5 + 97 s4 + 204 s3 + 274
s2 + 204 s + 52 = 0. (details below*)
This polynomial has 6 roots at s = -2.7± 1.1j, -0.65± 1.6j, -1.4, -0.46.
From these 6 roots, there exists 2 real roots at s = -1.4, -0.46. These are highlighted
on the diagram above (with squares or diamonds.)
Not all of these roots are on the locus. Of these 2 real roots, there exists 1 root
at s = -0.46 on the locus (i.e., K>0). Break-away (or break-in) points on
the locus are shown by squares.
(Real break-away (or break-in) with K less than 0 are shown with diamonds).
* N(s) and D(s) are numerator and denominator polylnomials of G(s)H(s), and the
tick mark, ', denotes differentiation.
N(s) = s2 + 2 s + 2
N'(s) = 2 s + 2
D(s)= s5 + 9 s4 + 33 s3 + 51 s2 + 26
s
D'(s)= 5 s4 + 36 s3 + 99 s2 + 102 s + 26
N(s)D'(s)= 5 s6 + 46 s5 + 181 s4 + 372 s3
+ 428 s2 + 256 s + 52
N'(s)D(s)= 2 s6 + 20 s5 + 84 s4 + 168 s3
+ 154 s2 + 52 s
N(s)D'(s)-N'(s)D(s)= 3 s6 + 26 s5 + 97 s4 + 204
s3 + 274 s2 + 204 s + 52
Here we used N(s)D'(s)-N'(s)D(s)=0, but we could multiply by -1 and use N'(s)D(s)-N(s)D'(s)=0.
Find angle of departure from pole at -3+2j
θz1
=angle((Departing pole)- (zero at -1+1j) ).
θz1
=angle((-3+2j) - (-1+1j)) = angle(-2+1j) = 153.4349°
θz2
=angle((-3+2j) - (-1-1j)) = angle(-2+3j) = 123.6901°
θp1
=angle((Departing pole)- (pole at 0) ).
θp1
=angle((-3+2j) - (0)) = angle(-3+2j) = 146.3099°
θp3
=angle((-3+2j) - (-3-2j)) = angle(0+4j) = 90°
θp4
=angle((-3+2j) - (-2)) = angle(-1+2j) = 116.5651°
θp5
=angle((-3+2j) - (-1)) = angle(-2+2j) = 135°
Angle of Departure is equal to:
θdepart
= 180° + sum(angle to zeros) - sum(angle to poles).
θdepart
= 180° + 277.125-487.875.
θdepart
= -30.7°.
This angle is shown in gray. It may be hard to see if it is near 0°.
Find angle of arrival to pole at -1+1j
θz2
=angle( (Arriving zero) - (zero at -3+2j) ).
θz2
=angle((-1+1j) - (-1-1j)) = angle(0+2j) = 90°
θp1
=angle( (Arriving zero) - (pole at 0) ).
θp1
=angle((-1+1j) - (0)) = angle(-1+1j) = 135°
θp2
=angle((-1+1j) - (-3+2j)) = angle(2-1j) = -26.5651°
θp3
=angle((-1+1j) - (-3-2j)) = angle(2+3j) = 56.3099°
θp4
=angle((-1+1j) - (-2)) = angle(1+1j) = 45°
θp5
=angle((-1+1j) - (-1)) = angle(0+1j) = 90°
Angle of arrival is equal to:
θarrive = 180° - sum(angle to zeros) + sum(angle to poles).
θarrive = 180° - 90 + 299.7449.
θarrive = 390°
This is equivalent to 30°.
This angle is shown in gray. It may be hard to see if it is near 0°.
Locus crosses imaginary axis at 2 values of K. These values are normally
determined by using Routh's method. This program does it numerically, and
so is only an estimate.
Locus crosses where K = 0, 123, corresponding to crossing imaginary axis at s=0,
±4.21j, respectively.
These crossings are shown on plot.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s5 + 9 s4 + 33 s3 + 51 s2
+ 26 s+ K( s2 + 2 s + 2 ) = 0
So, by choosing K we determine the characteristic equation whose roots are the closed
loop poles.
For example with K=20.3683, then the characteristic equation is
D(s)+KN(s) = s5 + 9 s4 + 33 s3 + 51 s2
+ 26 s + 20.3683( s2 + 2 s + 2 ) = 0, or
s5 + 9 s4 + 33 s3 + 71.3683 s2 + 66.7365
s + 40.7365= 0
This equation has 5 roots at s = -4.6, -1.6± 2.3j, -0.56±0.89j. These are
shown by the large dots on the root locus plot.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0, or
K = -D(s)/N(s) = -( s5 + 9 s4 + 33 s3 + 51 s2
+ 26 s ) / ( s2 + 2 s + 2 )
We can pick a value of s on the locus, and find K=-D(s)/N(s).
For example if we choose s= -2.1 + 2.2j (marked by asterisk),
then D(s)= 16.2 + 59.5j, N(s)=-2.46 + -4.91j,
and K=-D(s)/N(s)= 11 + 2.21j.
This s value is not exactly on the locus, so K is complex, (see note below), pick
real part of K ( 11)
For this K there exist 5 closed loop poles at s = -3.9, -2.1±2j, -0.48±0.66j.
These poles are highlighted on the diagram with dots, the value of "s" that was
originally specified is shown by an asterisk.
Note: it is often difficult to choose a value of s that is precisely on the locus,
but we can pick a point that is close. If the value is not exactly on
the locus, then the calculated value of K will be complex instead of real. Just
ignore the the imaginary part of K (which will be small).
Note also that only one pole location was chosen and this determines the value of
K. If the system has more than one closed loop pole, the location of the other poles
are determine solely by K, and may be in undesirable locations.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
© Copyright 2005-2009
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
Not yet complete.
Location of poles and zeros.
Root locus with respect to other terms.
The root locus is obviously a very powerful technique for design and analysis of control systems, but it must be used with some care, and results obtained with it should always be checked. To show potential pitfalls of this method, consider the two systems G1(s) and G2(s).
| G1(s) | G2(s) |
 |
 |
If we control these systems with a simple proportional controller, as shown,
we can examine the root locus of each of them.
| Root Locus with G1(s) | Root Locus with G2(s) |
 |
 |
The two root loci are clearly very different, but it turns out (because of the way that I chose the systems) that if we choose K=40, we get two closed loop systems with identical characteristic equations.
| Closed loop system with G1(s) | Closed loop system with G2(s) |
 |
 |
The roots of the characteristic equations are at s=-1 and s=-2.5±j5.8 (i.e., the roots of the characteristic equation s3+6s2+45s+40), so we might expect the behavior of the systems to be similar. Since the pole at s=-1 is closer to the origin, we would expect it to dominate somewhat, giving the system behavior similar to a first order system with a time constant of 1 second, and a settling time of 4 seconds. However, if we plot the two responses, we get something quite different.
T1(s) resembles (somewhat) a first order system, and has no overshoot, and its settling time is almost exactly 4 seconds, as predicted. However, T2(s) behaves very differently, it is much faster and more oscillatory than expected. How can we explain this?
If we look more closely at T1(s) and T2(s), we can understand what happened. In particular, lets look at pole-zero plots of both closed loop transfer functions.
| Pole-Zero Plot of T1(s) | Pole-Zero Plot of T2(s) |
 |
 |
T1(s) has poles at s=-1 and s=-2.5±j5.8, and no zeros. T2(s) has poles at s=-1 and s=-2.5±j5.8 and zeros at -12.2 and -1.1. The zero at s=-1.1 is almost directly on top of the pole at s=-1, and so largely negates its effect. The closed loop system, T2(s), therefore behaves very much like a second order system with s=-2.5±j5.8 (ωn=6.3 rad/sec, and ζ=0.4).
The lesson here is that while the poles of a system (the roots of the denominator polynomial) are very important in determining the behavior of a system, the zeros of the system (the roots of the numerator polynomial) can also be important. After performing a root-locus design, it is critical to go back and test the closed loop system to ensure that it behaves as expected.
Stress - start with loop gain, plot closed loop poles.
Change title of this page.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
Complementary
Root Locus Rules
The table below summarizes how to sketch a root locus plot (K≥0). This is also available as a Word Document or PDF.
You can also find a page that includes the rules for the Complementary Root Locus (K≤0).
The closed loop transfer function of the system shown is
So the characteristic equation is
As K changes, so do locations of closed loop poles (i.e., zeros of characteristic equation). The table below gives rules for sketching the location of these poles as K varies from 0 to infinity (K>0).
| Rule Name | Description |
|---|---|
| Definitions |
|
| Symmetry | The locus is symmetric about real axis (i.e., complex poles appear as conjugate pairs). |
| Number of Branches | There are n branches of the locus, one for each pole of the loop gain. |
| Starting and Ending Points | The locus starts (when K=0) at poles of the loop gain, and ends (when K→∞ ) at the zeros. Note: there are q zeros of the loop gain as |s|→∞ . |
| Locus on Real Axis | The locus exists on real axis to the left of an odd number of poles and zeros. |
| Asymptotes as |s|→∞ | If q>0 there are asymptotes of the root
locus that intersect the real axis at
, and radiate out
with angles , where r=1, 3,
5… |
| Break-Away and -In Points on Real Axis | There are break-away or –in points of the
locus on the axis wherever
 . |
| Angle of Departure from Complex Pole | Angle of departure
from pole pj is
 |
| Angle of Arrival at Complex Zero | Angle of arrival at zero zj
is  |
| Locus Crosses Imaginary Axis | Use Routh-Horwitz to determine where the locus crosses the imaginary axis. |
| Determine Location of Poles, Given Gain "K" | Rewrite characteristic equation as D(s)+KN(s)=0. Put value of K into equation, and find roots of characteristic equation. (This may require a computer) |
| Determine Value of "K", Given Pole Locations | Rewrite characteristic equation as
, replace “s” by the desired
pole location and solve for K. Note: if
“s” is not exactly on locus, K may be complex, but the imaginary part
should be small. Take the real part of K for your answer.
|
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Complementary
Root Locus Rules
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
A Tool for Learning to Sketch Root Locus Plots
RLocusGui is a graphical user interface written in the Matlab® programming language. It takes a transfer function and applies the standard rules for sketching a root locus plot by hand. Of course, Matlab can do this more accurately, but it is important to know how pole and zero locations affect the final plot. It is hoped that the RLocusGui program will be a versatile program for teaching and learning the construction of root locus diagrams.
Files for the program are found here
Consider the transfer function:
This function consists of a zero at s=-3, and poles at s=0, -1, and -2±2j. We can define the system, and invoke the GUI.
Top>> mySys=tf([1 3],poly([0 -2+2j -2-2j -1])) %define Xfer function
Transfer function:
s + 3
----------------------------------------
s^4 + 5 s^3 + 12 s^2 + 8 s
>> RLocusGui(mySys) %Invoke GUI
The GUI generates a window as shown below.
Starting in the upper right and going counterclockwise, the window shows:
- The transfer function.
- The completed Root Locus plot (you could get the same root locus plot from Matlab directly:
>>rlocus(mySys)
- A set of radio buttons listing all of the root locus rules (see below).
- A text box that describes the application of the radio button selected in (3): "Description of Rule."
- A button, "Rule Detail (web)," that will open a web page describing the selected rule in detail.
- A button, "Make Web Page," that will generate a web page that describes the application of the root locus rules to the chosen transfer function (see below).
- A button, "Web Resources," that links to my files describing the root locus.
- A button, "Exit," that closes the GUI.
By selecting each of the radio buttons, you can change what is displayed. For example if you select the "Locus on Real Axis" radio button, the display changes as shown below.
There are two notable differences in the figure. First, the box in the lower right hand corner describes the application of the selected rule. Even more noticeable is the second graph (at the right). This graph shows the original root locus, and illustrates (graphically) the application of the selected rule.
Any of the other rules may be chosen in turn. The calculation of the "Angle of Departure" from the complex poles is illustrated below
If the rule doesn't apply (e.g., "Angle of Arrival"), the "Description of Rule" box states this, and there is no graphic (below)
The last two choices "Choose point of locus and determine gain" and "Choose gain and find point on locus" are not really rules of the locus, but are instead application (the former is more or less equivalent to the Matlab command "rlocfind", but describes how it would be done by hand). Both of them allow further interaction and exploration (through a check box that appears above "Description of Rule."
The program will also automatically generate a web page that displays all of the rules and the graphics illustrating the rule. For example, if you hit the "Make Web Page" button, each rule is invoked and a web page is created and opened. Note the web page and its associated files are created in the current Matlab directory (i.e., the directory you see when you use Matlab's "ls" command).
Please let me know (links are below) if you find any problems with this software, or have any suggestions.
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College
© Copyright 2005-2007
Erik Cheever This page may be freely used for educational
purposes.
Erik Cheever Department of Engineering Swarthmore College