Root Locus Examples


Examples  (Click on Transfer Function)  
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A Weakness of
the root locus

Root Locus: Example 1

Transfer function

Xfer Function Info

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

Completed Root Locus

RLTot

Root Locus Symmetry

As you can see, the locus is symmetric about the real axis

Number of Branches

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.

Start/End Points

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).

Locus on Real Axis

RLAx

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.

Asymptotes as |s| goes to infinity

RLAsym

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 and In Points on Real Axis

RLBOI

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.

Angle of Departure

No complex poles in loop gain, so no angles of departure.

Angle of Arrival

No complex zeros in loop gain, so no angles of arrival.

Cross Imag. Axis

RLImag

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.

Changing K Changes Closed Loop Poles

RLFR

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

Choose Pole Location and Find K

RLFG

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.


Root Locus: Example 2

Transfer function

Ex2

Xfer Function Info

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

Completed Root Locus

RLTot

Root Locus Symmetry

As you can see, the locus is symmetric about the real axis.

Number of Branches

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.

Start/End Points

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).

Locus on Real Axis

RLAx

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.

Asymptotes as |s| goes to infinity

RLAsym

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 and In Points on Real Axis

RLBOI

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.

Angle of Departure

No complex poles in loop gain, so no angles of departure.

Angle of Arrival

No complex zeros in loop gain, so no angles of arrival.

Cross Imag. Axis

RLImag

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.

Changing K Changes Closed Loop Poles

RLFR

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

Choose Pole Location and Find K

RLFG

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.

Root Locus: Example 3

Transfer function

Xfer Function Info

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

Completed Root Locus

RLTot

Root Locus Symmetry

As you can see, the locus is symmetric about the real axis

Number of Branches

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.

Start/End Points

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).

Locus on Real Axis

RLAx

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.

Asymptotes as |s| goes to infinity

RLAsym

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 and In Points on Real Axis

RLBOI

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.

Angle of Departure

No complex poles in loop gain, so no angles of departure.

Angle of Arrival

No complex zeros in loop gain, so no angles of arrival.

Cross Imag. Axis

RLImag

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.

Changing K Changes Closed Loop Poles

RLFR

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

Choose Pole Location and Find K

RLFG

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.

Root Locus: Example 4

Transfer function

Xfer Function Info

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

Completed Root Locus

RLTot

Root Locus Symmetry

As you can see, the locus is symmetric about the real axis

Number of Branches

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.

Start/End Points

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).

Locus on Real Axis

RLAx

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.

Asymptotes as |s| goes to infinity

RLAsym

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 and In Points on Real Axis

RLBOI

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.

Angle of Departure

RLDep

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°.

Angle of Arrival

No complex zeros in loop gain, so no angles of arrival.

Cross Imag. Axis

Locus does not cross imaginary axis.

Changing K Changes Closed Loop Poles

RLFR

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

Choose Pole Location and Find K

RLFG

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.

Root Locus: Example 5

Transfer function

Xfer Function Info

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

Completed Root Locus

RLTot

Root Locus Symmetry

As you can see, the locus is symmetric about the real axis

Number of Branches

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.

Start/End Points

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).

Locus on Real Axis

RLAx

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.

Asymptotes as |s| goes to infinity

RLAsym

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 and In Points on Real Axis

RLBOI

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.

Angle of Departure

RLDep

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°.

Angle of Arrival

RLArv

Find angle of arrival to zero 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°.

Cross Imag. Axis

RLImag

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.

Changing K Changes Closed Loop Poles

RLFR

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.

Choose Pole Location and Find K

RLFG

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.


A Weakness of the Root Locus

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)
G1(s) G2(s)

If we control these systems with a simple proportional controller, as shown,

Closed loop with G1 or G2

we can examine the root locus of each of them.

 Root Locus with G1(s) Root Locus with G2(s)
Root locus with G1  Root Locus with G2 

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)
Closed loop with G1  Closed loop with G2 

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.

Step Response

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)
Pole zero of T1  pole zero of T2 

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.

 

References

© Copyright 2005-2013 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College