A system with feedback becomes stable when equations describing that system possess roots that follow certain patterns.
Otherwise, the system will become unstable. Example of such an unstable system is when microphones create screeches. Part of the loudspeaker voice feedbacks to microphone and becomes amplified by amplifiers and then goes into the loudspeakers and again feds into the microphone and loops again and again until it saturates amplifiers into creating a high pitched noise.
Feedback sometimes keeps the system just in the margin of instability and starts to make the system oscillating. This might be useful in electronics and elsewhere to have a steady oscillation; in a device such as a clock. But if the margin has not been carefully calculated a small change might devastate the system into the destruction. This is seen when some bridges have collapsed due to become oscillating and then into the instability runaway when people or cars or trains are passing over them. A newly constructed London bridge opened for pedestrians for millennium was near to this runaway in the first day of its inauguration but as it was still under careful observation of the constructors was shut down and disaster did not happen. Root locus helps engineers to predict specification of their system to meet stability criteria. Although all academia are full of a plethora of software for drawing the "Root Locus" still it is fascinating for all learners of engineering to know the conceptual sketch of this method.
Root Locus: Example
Transfer function
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
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
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).
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
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.
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
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).
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
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.
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
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.
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
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
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
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).
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.
Reference: http://www.wikihow.com/Draw-Root-Locus-of-a-System
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