INC 342 Lecture 3: Design with Root Locus Dr. Benjamas - - PowerPoint PPT Presentation

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INC 342 Lecture 3: Design with Root Locus Dr. Benjamas - - PowerPoint PPT Presentation

INC 342 Lecture 3: Design with Root Locus Dr. Benjamas Panomruttanarug Benjamas.pan@kmutt.ac.th BP INC342 1 Review Gain adjustment: Calculate Max. K that still makes the system stable Design K to get a desired damping ratio BP


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INC 342

Lecture 3: Design with Root Locus

  • Dr. Benjamas Panomruttanarug

Benjamas.pan@kmutt.ac.th

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Review

  • Gain adjustment:

– Calculate Max. K that still makes the system stable – Design K to get a desired damping ratio

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3

Compensator

  • Allows us to meet transient and steady

state error.

  • Composed of poles and zeros.
  • Increased an order of the system.
  • The system can be approx. to 2nd order

using some techniques.

BP INC342

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Types of compensator

  • 1. Active compensator

– PI, PD, PID use of active components, i.e., OP‐AMP – Require power source – ss error converge to zero – Expensive

  • 2. Passive compensator

– Lag, Lead use of passive components, i.e., R L C – No need of power source – ss error nearly reaches zero – Less expensive

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Improving steady‐state error using compensator

Placing a pole at the origin to increase system order; decreasing ss error as a result!!

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The pole at origin affects the transeint response  adds a zero close to the pole to get an ideal integral compensator

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Example

Damping ratio = 0.174 in both uncompensated and PI cases Choose zero at ‐0.1

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  • Draw root locus

without compensator

  • Draw a straight line of

damping ratio

  • Evaluate K from the

intersection point

  • From K, find the last

pole (at ‐11.61)

  • Calculate steady‐state

error

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  • Draw root locus with compensator

(system order is up by 1‐‐from 3rd to 4th)

  • Needs complex poles corresponding

to damping ratio of 0.174 (K=158.2)

  • From K, find the 3rd and 4th poles (at

‐11.55 and ‐0.0902)

  • Pole at ‐0.0902 can do phase

cacellation with zero at ‐1 (3th order approx.)

  • Compensated system and

uncompensated system have similar transient response (closed loop poles and K are aprrox. The same)

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Comparison of step response

  • f the 2 systems

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PI Controller

s K K s K s K K s Gc            

2 1 1 2 1

) (

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Example

K = 164.6 in P controller and K = 158.2 in PI controller

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SS response improvement conclusions

  • Can be done either by PI controller (pole at
  • rigin) or lag compensator (pole closed to
  • rigin).
  • Improving ss error without affecting the

transient response.

  • Next step is to improve the transient response

itself.

BP INC342