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LECTURE 27: FEEDBACK CONTROLPowerPoint Presentation

LECTURE 27: FEEDBACK CONTROL

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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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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

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.

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.

- 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.

- 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.

- 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

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.

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);

- 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.

- 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.

- 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).

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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