﻿ Root Locus: Example 4

# 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 ## 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 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 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 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 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 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 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.