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ECE 8443 – Pattern Recognition EE 3512 – Signals: Continuous and Discrete Objectives: Typical Feedback System Feedback Example Feedback as Compensation.

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Presentation on theme: "ECE 8443 – Pattern Recognition EE 3512 – Signals: Continuous and Discrete Objectives: Typical Feedback System Feedback Example Feedback as Compensation."— Presentation transcript:

1 ECE 8443 – Pattern Recognition EE 3512 – Signals: Continuous and Discrete Objectives: Typical Feedback System Feedback Example Feedback as Compensation Proportional Feedback Applications Resources: MIT 6.003: Lecture 20 MIT 6.003: Lecture 21 Wiki: Control Systems Brit: Feedback Control JC: Crash Course Wiki: Root Locus Wiki: Inverted Pendulum CJC: Inverted Pendulum MIT 6.003: Lecture 20 MIT 6.003: Lecture 21 Wiki: Control Systems Brit: Feedback Control JC: Crash Course Wiki: Root Locus Wiki: Inverted Pendulum CJC: Inverted Pendulum LECTURE 27: FEEDBACK CONTROL Audio: URL:

2 EE 3512: Lecture 27, Slide 1 System Function For A Closed-Loop System The transfer function of this system can be derived using principles we learned in Chapter 6: Blacks Formula: Closed-loop transfer function is given by: Forward Gain: total gain of the forward path from the input to the output, where the gain of a summer is 1. Loop Gain: total gain along the closed loop shared by all systems. Loop

3 EE 3512: Lecture 27, Slide 2 The Use Of Feedback As Compensation Assume the open loop gain is very large (e.g., op amp): The closed-loop gain depends only on the passive components (R 1 and R 2 ) and is independent of the open-loop gain of the op amp. Independent of P(s)

4 EE 3512: Lecture 27, Slide 3 Stabilization of an Unstable System If P(s) is unstable, can we stabilize the system by inserting controllers? Design C(s) and G(s) so that the 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.

5 EE 3512: Lecture 27, Slide 4 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: K 2 > 0: sufficient damping force K 1 > 4: sufficient gain Using damping and feedback, we have stabilized a second-order unstable system.

6 EE 3512: Lecture 27, Slide 5 The Concept of a Root Locus Recall our simple control system with transfer function: The controllers C(s) and G(s) can be designed 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.

7 EE 3512: Lecture 27, Slide 6 Example: First-Order System Consider a simple first-order system: The pole is at s 0 = -(2+K). Vary K from 0 to : Observation: improper adjustment of the gain can cause the overall system to become unstable. Becomes more stable Becomes less stable

8 EE 3512: Lecture 27, Slide 7 Example: Second-Order System With Proportional Control Using Blacks Formula: How does the step response vary as a function of the gain, K? Note that as K increases, the system goes from too little gain to too much gain.

9 EE 3512: Lecture 27, Slide 8 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 = [ ]; (assuming K = 1) P = tf(num, den); rlocus(P);

10 EE 3512: Lecture 27, Slide 9 Example

11 EE 3512: Lecture 27, Slide 10 Feedback System – Implementation

12 EE 3512: Lecture 27, Slide 11 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.

13 EE 3512: Lecture 27, Slide 12 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(s 0 )H(s 0 ) = s 0 are the poles of G(s)H(s). At K =, G(s 0 )H(s 0 ) = 0 s 0 are 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.

14 EE 3512: Lecture 27, Slide 13 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).Wiki

15 EE 3512: Lecture 27, Slide 14 Feedback System – Use Proportional Derivative Control Equations describing the physics: The poles of the system are inherently unstable. Feedback control can be used to stabilize both the angle and position. Other approaches involve oscillating the support up and down.


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