Lecture 11/12 Analysis and design in the time domain using root locus North China Electric Power University Sun Hairong

Topics of this class  Magnitude and phase equations (Reading Module 9)  Rules for plotting the root locus (Reading Module 10)  System design in the complex plane using the root locus. (Reading Module 11)

1. Magnitude and phase equations We begin by writing the characteristic equation for the generalized feedback control system as The value of s that satisfy the above equation may be real or complex, so rewrite the equation in the form It may be inferred that both the magnitude and phase of each are equal and write

Given an second-order feedback system The characteristic equation 2. Root-locus The roots of the equation are Now we plot the closed-loop poles on the complex plane for K=0, 0.5, 1, 2, 10, ∞.

Firstly, for each value of K calculate s 1,2 ： k ……∞ s1s1 0….. s2s2 -2….. Then plot the closed-loop poles on the complex plane…..

3.Rules for plotting the root locus Suppose we are required to draw the root locus for the system with open –loop transfer function We can plot the closed-loop poles on the complex plane for each value of K calculate s 1,2,3 The roots locus can also be drawn using the following rules.

Rule #1. Draw the complex plane and mark the n open-loop poles and m zeros. The locus starts at a poles for K=0, and finishes at a zero or infinity when n>m, the number of segments going to infinity is therefore n-m. Rule #2. Segments of the real axis to the left of an odd number of poles and zeros are segments of the root locus, and the complex poles and zeros have no effect.

Rule #3. The loci are symmetrical about the real axis. The angle between adjacent asymptote is and to obey the symmetry rule, the negative real axis is one asymptote when n-m is odd. Rule #4. The asymptotes intersect the real axis at

Rule #5. The angle of emergence from complex poles is given by (angles of all the vectors from all other open-loop poles to the pole in question)+ (angles of all the vectors from all other open-loop zeros to the pole in question) Rule #6. The point where the locus crosses the imaginary axis may be obtained by substituting s=jω into the characteristic equation and solving for ω.

Rule #7. The point at which the locus leaves a real-axis segment is found by determining a local maximum value of K, while the point at which the locus enters a real-axis segment is found by determining a local minimum value of K. Rule #8. The angle between the directions of emergence (or entry) of q coincident poles (or zeros) on the real axis is given by

Rule #9. The gain at a selected point on the locus is given by Rule #10. If there are at least two more open-loop poles than open-loop zeros, the sum of the real parts of the closed-loop poles is constant, independent of K, and equal to the sum of the real parts of the open-loop poles.

1 ． Sketch the root locus based on the open-loop pole-zero map. Practice

Solution

4. Performance requirement as complex- plane constraints It has to satisfy the performance requirements respectively 1. overshoot less than 10%, 2. steady-state error to a unit ramp less than 10%, 3. dominant time constant less than 0.1s. 4. the settling time to a unit step less than 1s Determine the K value for each requirements. Solve the problems using root locus. Example, suppose a system described by

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