# MIT 2 004 - Determining Stability Bounds in Closed-Loop Systems (10 pages)

Previewing pages*1, 2, 3*of 10 page document

**View the full content.**## Determining Stability Bounds in Closed-Loop Systems

Previewing pages
*1, 2, 3*
of
actual document.

**View the full content.**View Full Document

## Determining Stability Bounds in Closed-Loop Systems

0 0 50 views

- Pages:
- 10
- School:
- Massachusetts Institute of Technology
- Course:
- 2 004 - Dynamics and Control II

**Unformatted text preview:**

MIT OpenCourseWare http ocw mit edu 2 004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use visit http ocw mit edu terms Massachusetts Institute of Technology Department of Mechanical Engineering 2 004 Dynamics and Control II Spring Term 2008 Lecture 261 Reading Nise Chapter 6 Nise Chapter 8 1 Determining Stability Bounds in Closed Loop Systems Consider the closed loop third order system with proportional controller gain K with openloop transfer function K Gf s 3 s 3s2 5s 2 shown below R s P c o n tr o lle r p la n t K s 3 3 s 2 1 C s 5 s 2 The closed loop transfer function is Gcl s N s K 3 2 D s N s s 3s 5s 2 K Let s examine the closed loop stability by using the pzmap function in MATLAB sys tf 1 1 3 5 2 pzmap sys hold on for K 2 2 30 sys tf K 1 3 5 2 K pzmap sys end which superimposes the closed loop pole zero plots for K 0 30 on a single plot 1 c D Rowell 2008 copyright 26 1 Pole Zero Map 3 K 30 2 K increasing K 0 Imaginary Axis 1 K increasing K 30 K 0 0 1 K 0 K increasing 2 3 4 K 30 3 5 3 2 5 2 1 5 Real Axis 1 0 5 0 0 5 1 From the plot we note the following This system always has two complex conjugate poles and a single real pole When K 0 the poles are the open loop poles As K increases the real pole moves deeper into the l h plane and the complex con jugate poles approach and cross the imaginary j axis and the system becomes unstable Close examination of the plot shows that the system becomes unstable at a value of K between K 12 and K 14 We now look at three methods for determining the stability limit of the proportional gain K for this system Example 1 Use the Routh Hurwitz method to nd the range of proportional controller gain K for which the above system will be stable The rst two rows of the Routh array are taken directly from D s s3 s2 1 5 3 2 K 0 0 26 2 and the next two rows are computed as above 1 1 1 1 an an 2 5 b1 K 13 3 2 K a a an 1 3 3 n 1 n 3 1 an 1 1 0 an 4 b2 0 an 1 an 1 an 5 3 3 0 Similarly the s0 row is 1 an 1 c1 b1 b1 1 a c2 n 1 b1 b1 computed 3 an 3 3 24 2 K b2 K 13 K 13 3 2 K 3 an 3 3 0 0 b3 K 13 K 13 3 0 and the complete Routh array is s3 s2 s1 s0 1 5 0 3 2 K 0 K 13 3 0 2 K 0 We now examine the rst column to determine the range of proportional gain for which this system will be stable In order for there to be no sign changes we require 2 K 13 We conclude that if K 2 there will be one therefore real unstable pole while if K 13 there will be two unstable poles When K 13 the denominator is D s s3 3s2 5s 15 s 3 s j2 236 s j2 236 so that the closed loop system has a pair of poles on the imaginary axis The system will be marginally stable a pure oscillator at a frequency of 2 236 r s Example 2 Use the stability criterion for third order systems developed in Example 3 of Lecture 25 to determine the stability bounds for the above system In Lecture 25 we showed that for a third order system with characteristic equa tion D s a3 s3 a2 s2 a1 s a0 0 the system is stable only if a1 a2 a0 a3 26 3 In this case D s s3 3s2 5s 2 K and therefore for stability we require 15 2 K or K 13 Example 3 Use the characteristic equation directly to nd the closed loop stability limits for the above system There are three closed poles We conjecture that at the stabil ity boundary marginal stability there will be a pair of poles on the imaginary axis at s j and a single real pole at s a The closed loop characteristic polynomial will therefore be D s s a s2 2 s3 as2 2 s a 2 Comparing coe cients with the actual closed loop characteristic polynomial D s s3 3s2 5s 2 K we determine a 3 2 5 2 a K 2 2 5 K 13 Root Locus Methods We have seen that the closed loop poles change as controller parameters vary A root locus is is an s plane plot of the paths that the closed loop poles take as a controller parameter varies Let s start with some simple examples Example 4 Consider the rst order plant under proportional control as shown below 26 4 P c o n tr o lle r R s p la n t C s 1 K s a The closed loop transfer function is K s a K Gcl s with a single pole pc a K The root locus is simply the path of this pole as K varies from K 0 to K Clearly as K 0 the closed loop pole approaches the open loop pole s a and as K the closed loop pole p This is all the information we need to construct the root locus for this system jw s p la n e K in c r e a s in g K 0 x a th e ro in d ic a o f th e v a r ie s s o t lo c u s te s th e p a th p o le a s K Example 5 Construct the root locus plot for the rst order system under P D control with Gc s Kp Kd s R s P D c o n tr o lle r K p K d s 26 5 p la n t 1 s a C s If we write Gc s Kd Kp s Kd we have a open loop pole at s a and an open loop zero at s Kp Kd b The closed loop transfer function is Gcl s Kd s b Kd 1 s a Kd b with a single pole a bKd 1 Kd We now construct the root locus as Kd varies from 0 to Clearly as Kd 0 the closed loop pole pcl a approaches the open loop pole at s a and as Kd the closed loop pole pcl b in other words the closed loop pole approaches the open loop zero There are two possibilities for the root locus based on …

View Full Document