Brian Mace Root Locus

Question 1

What is the effect of changing the gain of the system?

Answer 1

It will change the locations of the poles.

Question 2

What is the root locus?

Answer 2

It will tell us how the poles change along with the changing gain, from 0 -> infinity. It is in the complex s plane.

Question 3

All values on the settling time line will have the same settling time. Is this line horizontal or vertical?

Answer 3

It is vertical.

Question 4

All values on the peak time line will have the same peak time. Is this line horizontal or vertical?

Answer 4

It is horizontal.

Question 5

All values on the damping line will have the same damping factor? What is the orientation of this line?

Answer 5

It will come diagonally from the origin at a specific angle.

Question 6

How do we multiply two complex numbers together that are in polar form?

Answer 6

We multiply the magnitudes and add the angles.

Question 7

How do we divide two complex numbers that are in polar form?

Answer 7

Divide the magnitudes and subtract the angles.

Question 8

How can we represent a variable of s?

Answer 8

s = jw + o, where w will lie on the imaginary axis and o is along the real axis.

Question 9

What is the magnitude of a complex function?

Answer 9

Zero distances / Pole Distances

Question 10

What is the angle of a complex function?

Answer 10

SUM(zero angles) - SUM(pole angles)

Question 11

If both our Open Loop function G and our Feedback function H have a numerator and denominator, what are the zeros of T(s)?

Answer 11

The zeros will be the zeros of the OL function and the poles of the Feedback.

Question 12

If we get complex values for the poles, what does this mean for the root locus?

Answer 12

It means that the root locus will begin to diverge away from the real axis.

Question 13

What is the angle criteria that a point must satisfy if it is too lie on the root locus?

Answer 13

ANGLE(G(s) * H(s)) = 180 +- 360n

Question 14

What is the value of K?

Answer 14

Zero distances / Pole Distances

Question 15

What is the first rule of sketching the root locus?

Answer 15

Any value k of the characteristic equation will have n solutions so n closed loop poles.

Question 16

What is the second rule of sketching the root locus?

Answer 16

The coefficients of the characteristic equation are real therefore any solutions are either real or come in complex conjugate pairs.

Question 17

Rule 3 for sketching the root locus?

Answer 17

At any point p on the real axis:
Any real pole or zero to the left of it will contribute zero to the overall angle.
Any pole or zero to the right will contribute 180
Complex conjugate will contribute zero to the angle.

Question 18

What happens if we have more poles than zeros?

Answer 18

n > m therefore T(s) = 1 / s^n-m

Question 19

What happens if we more zeros than poles?

Answer 19

n < m therefore T(s) = s^m-n

Question 20

Rule 4 for sketching the root locus?

Answer 20

The root locus will always start at the open loop poles and finish at the open loop zeros. This simulates starting at zero gain and finishing at infinite gain

Question 21

Rule 5 for sketching the root locus?

Answer 21

As K increases, the number of m - n branches of the root locus diverge away from a point on the real axis and approach n - m asymptotes

Question 22

What is the formula for the break away point?

Answer 22

SUM(finite poles) - SUM(finite zeros) / n - m

Question 23

What is the angle of break away?

Answer 23

(2K + 1) * 180 / n-m

Question 24

What is a breakaway point

Answer 24

It is the point on the root locus where the poles converge and diverge away from the real axis, becoming complex conjugates.

Question 25

What is the value of K at the break away and break in points?

Answer 25

At the break away points the value of K will be max and at the break in points the value of K will be min.

Question 26

Since the break in break away points occur at the min max values of K, how can we determine the break away point?

Answer 26

By taking the derivative when it equals zero.
dK / dO = ND’ - N’D = 0
SOLVE this to find O
N and D correspond to the numerator and denominator of our transfer function.

Question 27

What does it mean if our root locus crosses the imaginary axis

Answer 27

It means that we are crossing from stable to unstable.

Question 28

How can we find the position w where we cross the imaginary axis?

Answer 28

Either with a Routh array or by letting s = jw in the characteristic equation and solving for w.

Question 29

How can we find the angles of arrival and departure?

Answer 29

Using the angle criterion. We know that the sum of all angles must add up to 180.