| Examples (Click on Transfer Function) | ||||
|---|---|---|---|---|
1 |
2 |
3 |
4 |
5 |
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
Back