����PRESENTATION ON �FEEDBACK SYSTEM

• BRANCH-E & TC ENGG
• SUBJECT- CONTROL SYSTEM & COMPONENT
• CHAPTER – 6 – FEEDBACK CHARACTRISTICS OF CONTROL SYSTEM
• TOPIC- METHODOLOGY
• SEM-6^TH
• FACULTY – Er. MANAS RANJAN MOHANTA (Sr. Lecturer E & TC ENGG DEPARTMENT)
• AY-2021-2022, SUMMER-2022

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A Typical Feedback System

• Why use feedback?
• Reducing Nonlinearities
• Reducing Sensitivity to Uncertainties and Variability
• Stabilizing Unstable Systems
• Reducing Effects of Disturbances
• Tracking
• Shaping System Response Characteristics (bandwidth/speed)

Feed Forward

Feedback

EE 3512: Lecture 30, Slide 1

Motivating Example

• Open loop system: aim and shoot.
• What happens if you miss?
• Can you automate the correction�process?

• Closed-loop system: automatically adjusts until the proper coordinates are achieved.
• Issues: speed of adjustment, inertia, momentum, stability, …

EE 3512: Lecture 30, Slide 2

System Function For A Closed-Loop System

• The transfer function of this�system can be derived using�principles we learned in�Chapter 6:
• Black’s 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

EE 3512: Lecture 30, Slide 3

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 (R₁ and R₂) and is independent of the open-loop gain of the op amp.

⮘ Independent of P(s)

EE 3512: Lecture 30, Slide 4

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.

EE 3512: Lecture 30, Slide 5

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.

EE 3512: Lecture 30, Slide 6

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:

EE 3512: Lecture 30, Slide 7

Example: First-Order System

• Consider a simple first-order system:
• The pole is at s₀ = -(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

EE 3512: Lecture 30, Slide 8

Example: Second-Order System With Proportional Control

• Using Black’s 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.

EE 3512: Lecture 30, Slide 9

How Do The Poles Move?

• 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);

EE 3512: Lecture 30, Slide 10

Example

EE 3512: Lecture 30, Slide 11

Feedback System – Implementation

EE 3512: Lecture 30, Slide 12

THANK YOU

EE 3512: Lecture 30, Slide 13