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LECTURE 27: FEEDBACK CONTROL. Objectives: Typical Feedback System Feedback Example Feedback as Compensation Proportional Feedback Applications

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LECTURE 27: FEEDBACK CONTROL

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Lecture 27 feedback control

LECTURE 27: FEEDBACK CONTROL

  • Objectives:Typical Feedback SystemFeedback ExampleFeedback as CompensationProportional FeedbackApplications

  • Resources:MIT 6.003: Lecture 20MIT 6.003: Lecture 21Wiki: Control SystemsBrit: Feedback ControlJC: Crash CourseWiki: Root LocusWiki: Inverted Pendulum CJC: Inverted Pendulum

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Lecture 27 feedback control

System Function For A Closed-Loop System

  • The transfer function of thissystem can be derived usingprinciples we learned inChapter 6:

  • Black’s Formula: Closed-loop transfer function is given by:

  • Forward Gain: total gain of the forward path from the inputto the output, where the gain of a summer is 1.

  • Loop Gain: total gain along the closed loop shared by all systems.

Loop


Lecture 27 feedback control

The Use Of Feedback As Compensation

  • Assume the open loopgain is very large(e.g., op amp):

 Independent of P(s)

  • The closed-loop gain depends only on the passive components (R1 and R2) and is independent of the open-loop gain of the op amp.


Lecture 27 feedback control

Stabilization of an Unstable System

  • If P(s) is unstable, can westabilize the system byinserting controllers?

  • Design C(s) and G(s) so thatthe poles of Q(s) are in the LHP:

  • Example: Proportional Feedback (C(s) = K)

  • The overall system gain is:

  • The transfer function is stable for K > 2.

  • Hence, we can adjust K until the system is stable.


Lecture 27 feedback control

Second-Order Unstable System

  • Try proportional feedback:

  • One of the poles is at

  • Unstable for all values of K.

  • Try damping, a term proportional to :

  • This system is stable as long as:

  • K2 > 0: sufficient damping force

  • K1 > 4: sufficient gain

  • Using damping and feedback, we have stabilized a second-order unstable system.


Lecture 27 feedback control

The Concept of a Root Locus

  • Recall our simple control systemwith transfer function:

  • The controllers C(s) and G(s) can bedesigned to stabilize the system, but that could involve a multidimensional optimization. Instead, we would like a simpler, more intuitive approach to understand the behavior of this system.

  • Recall the stability of the system depends on the poles of 1 + C(s)G(s)P(s).

  • A root locus, in its most general form, is simply a plot of how the poles of our transfer function vary as the parameters of C(s) and G(s) are varied.

  • The classic root locus problem involves a simplified system:

Closed-loop poles are the same.


Lecture 27 feedback control

Example: First-Order System

  • Consider a simple first-order system:

  • The pole is at s0 = -(2+K). Vary Kfrom 0 to :

  • Observation: improper adjustment of the gain can cause the overall system to become unstable.

Becomes less stable

Becomes more stable


Lecture 27 feedback control

Example: Second-Order System With Proportional Control

  • Using Black’s Formula:

  • How does the step responsevary as a function of the gain, K?

  • Note that as K increases, thesystem goes from too little gainto too much gain.


Lecture 27 feedback control

How Do The Poles Move?

Desired Response

  • Can we generalize this analysis to systems of arbitrary complexity?

  • Fortunately, MATLAB has support for generation of the root locus:

  • num = [1];

  • den = [1 101 101]; (assuming K = 1)

  • P = tf(num, den);

  • rlocus(P);


Lecture 27 feedback control

Example


Lecture 27 feedback control

Feedback System – Implementation


Lecture 27 feedback control

Summary

  • Introduced the concept of system control using feedback.

  • Demonstrated how we can stabilize first-order systems using simple proportional feedback, and second-order systems using damping (derivative proportional feedback).

  • Why did we not simply cancel the poles?

  • In real systems we never know the exact locations of the poles. Slight errors in predicting these values can be fatal.

  • Disturbances between the two systems can cause instability.

  • There are many ways we can use feedback to control systems including feedback that adapts over time to changes in the system or environment.

  • Discussed an application of feedback control involving stabilization of an inverted pendulum.


Lecture 27 feedback control

More General Case

  • Assume no pole/zero cancellation in G(s)H(s):

  • Closed-loop poles are the roots of:

  • It is much easier to plot the root locus for high-order polynomials because we can usually determine critical points of the plot from limiting cases(e.g., K= 0, ), and then connect the critical points using some simple rules.

  • The root locus is defined as traces of s for unity gain:

  • Some general rules:

  • At K= 0, G(s0)H(s0) =   s0are the poles of G(s)H(s).

  • At K= , G(s0)H(s0) = 0 s0are the zeroes of G(s)H(s).

  • Rule #1: start at a pole at K= 0 and end at a zero at K= .

  • Rule #2: (K  0) number of zeroes and poles to the right of the locus point must be odd.


Lecture 27 feedback control

Inverted Pendulum

  • Pendulum which has its mass above its pivot point.

  • It is often implemented with the pivot point mounted on a cart that can move horizontally.

  • A normal pendulum is stable when hanging downwards, an inverted pendulum is inherently unstable.

  • Must be actively balanced in order to remain upright, either by applying a torque at the pivot point or by moving the pivot point horizontally (Wiki).


Lecture 27 feedback control

Feedback System – Use Proportional Derivative Control

  • Equations describing the physics:

  • The poles of the system are inherentlyunstable.

  • Feedback control can be used to stabilize both the angle and position.

  • Other approaches involve oscillatingthe support up and down.


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