Examples (Click on Transfer Function) | |||||
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A Weakness of the root locus |
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)= s^{2} + 3 s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) =
s^{2}
+ 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)= s^{2}
+ 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)
= s^{2} + 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) = s^{2}
+ 3 s + 2.2522( 1 ) = 0, or
s^{2} + 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) = -( s^{2} + 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.
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)= s^{3} + 5 s^{2} + 6 s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s)
= s^{3} + 5 s^{2} + 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 s^{2}
+ 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)= s^{3} + 5 s^{2} + 6 s
D'(s)= 3 s^{2}
+ 10 s + 6
N(s)D'(s)= 3 s^{2} + 10 s + 6
N'(s)D(s)= 0
N(s)D'(s)-N'(s)D(s)=
3 s^{2} + 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)
= s^{3} + 5 s^{2} + 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)
= s^{3} + 5 s^{2} + 6 s + 4.0019( 1 ) = 0, or
s^{3} +
5 s^{2} + 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)
= -( s^{3} + 5 s^{2} + 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.
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)= s^{2} - 1 s -
2.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s)
= s^{2} - 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 Kââ 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
s^{2}
+ 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)=
s^{2} - 1 s - 2
D'(s)= 2 s - 1
N(s)D'(s)= 2 s^{2} + 5 s -
3
N'(s)D(s)= s^{2} - 1 s - 2
N(s)D'(s)-N'(s)D(s)= s^{2} +
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)
= s^{2} - 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) = s^{2}
- 1 s - 2 + 7.1593( s + 3 ) = 0, or
s^{2} + 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)
= -( s^{2} - 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.
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)= s^{3} + 4 s^{2}
+ 6 s + 4.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s^{3} + 4 s^{2} + 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 s^{3}
+ 7 s^{2} + 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)= s^{3} + 4 s^{2}
+ 6 s + 4
D'(s)= 3 s^{2} + 8 s + 6
N(s)D'(s)= 3 s^{3} + 11
s^{2} + 14 s + 6
N'(s)D(s)= s^{3} + 4 s^{2} + 6 s + 4
N(s)D'(s)-N'(s)D(s)= 2 s^{3} + 7 s^{2} + 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
θ_{z1} =angle((Departing
pole)- (zero at -1) ).
θ_{z1} =angle((-1+1j) - (-1)) = angle(0+1j) =
90°
θ_{p1} =angle((Departing pole)- (pole at -2) ).
θ_{p1}
=angle((-1+1j) - (-2)) = angle(1+1j) = 45°
θ_{p3} =angle((-1+1j) - (-1-1j))
= angle(0+2j) = 90°
Angle of Departure is equal to:
θ_{depart}
= 180° + sum(angle to zeros) - sum(angle to poles).
θ_{depart}
= 180° + 90 - 135.
θ_{depart} = 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)
= s^{3} + 4 s^{2} + 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)
= s^{3} + 4 s^{2} + 6 s + 4 + 2.6026( s + 1 ) = 0, or
s^{3}
+ 4 s^{2} + 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)
= -( s^{3} + 4 s^{2} + 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.
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)= s^{2} + 2 s + 2, and
D(s)= s^{5} + 9 s^{4} + 33 s^{3} + 51 s^{2} + 26
s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s)
= s^{5} + 9 s^{4} + 33 s^{3} + 51 s^{2} + 26 s+
K( s^{2} + 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 s^{6}
+ 26 s^{5} + 97 s^{4} + 204 s^{3} + 274 s^{2} +
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) = s^{2} + 2 s + 2
N'(s) = 2 s + 2
D(s)= s^{5}
+ 9 s^{4} + 33 s^{3} + 51 s^{2} + 26 s
D'(s)= 5 s^{4}
+ 36 s^{3} + 99 s^{2} + 102 s + 26
N(s)D'(s)= 5 s^{6}
+ 46 s^{5} + 181 s^{4} + 372 s^{3} + 428 s^{2} +
256 s + 52
N'(s)D(s)= 2 s^{6} + 20 s^{5} + 84 s^{4} +
168 s^{3} + 154 s^{2} + 52 s
N(s)D'(s)-N'(s)D(s)= 3 s^{6}
+ 26 s^{5} + 97 s^{4} + 204 s^{3} + 274 s^{2} +
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
θ_{z1} =angle((Departing
pole)- (zero at -1+1j) ).
θ_{z1} =angle((-3+2j) - (-1+1j)) = angle(-2+1j)
= 153.4349°
θ_{z2} =angle((-3+2j) - (-1-1j)) = angle(-2+3j) = 123.6901°
θ_{p1} =angle((Departing pole)- (pole at 0) ).
θ_{p1} =angle((-3+2j)
- (0)) = angle(-3+2j) = 146.3099°
θ_{p3} =angle((-3+2j) - (-3-2j)) =
angle(0+4j) = 90°
θ_{p4} =angle((-3+2j) - (-2)) = angle(-1+2j) = 116.5651°
θ_{p5} =angle((-3+2j) - (-1)) = angle(-2+2j) = 135°
Angle of Departure
is equal to:
θ_{depart} = 180° + sum(angle to zeros) - sum(angle to poles).
θ_{depart} = 180° + 277.125-487.875.
θ_{depart} = -30.7°.
This angle is shown in gray. It may be hard to see if it is near 0°.
Find angle of arrival to zero at -1+1j
θ_{z2} =angle( (Arriving
zero) - (zero at -3+2j) ).
θ_{z2} =angle((-1+1j) - (-1-1j)) = angle(0+2j)
= 90°
θ_{p1} =angle( (Arriving zero) - (pole at 0) ).
θ_{p1}
=angle((-1+1j) - (0)) = angle(-1+1j) = 135°
θ_{p2} =angle((-1+1j) - (-3+2j))
= angle(2-1j) = -26.5651°
θ_{p3} =angle((-1+1j) - (-3-2j)) = angle(2+3j)
= 56.3099°
θ_{p4} =angle((-1+1j) - (-2)) = angle(1+1j) = 45°
θ_{p5}
=angle((-1+1j) - (-1)) = angle(0+1j) = 90°
Angle of arrival is equal to:
θ_{arrive} = 180° - sum(angle to zeros) + sum(angle to poles).
θ_{arrive }= 180° - 90 + 299.7449.
θ_{arrive }= 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)
= s^{5} + 9 s^{4} + 33 s^{3} + 51 s^{2} + 26 s+
K( s^{2} + 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) = s^{5} + 9 s^{4}
+ 33 s^{3} + 51 s^{2} + 26 s + 20.3683( s^{2} + 2 s + 2
) = 0, or
s^{5} + 9 s^{4} + 33 s^{3} + 71.3683 s^{2}
+ 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)
= -( s^{5} + 9 s^{4} + 33 s^{3} + 51 s^{2} + 26
s ) / ( s^{2} + 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.
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 s^{3}+6s^{2}+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 (ω_{n}=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.
© Copyright 2005-2013 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College