Chapter 6 Examples

# Chapter 6 Examples - EEL4657 Dr Haniph Latchman Chapter 6...

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Chapter 6: Lead and Lag Compensator Design Examples: ω - Do- main Analysis I. Design a lead compensator for the system: G ( s ) = 1 ( s )(1 + 0 . 3 s ) where the compensator is given by the form: C ( s ) = k (1 + ατs ) 1 + τs Which meets the following requirements: i. Steady-state error < 0.30% ii. Percent Overshoot < 10% Solution: We will approach the problem similar to the process shown in the example in section 6.3.2.1 1) The gain, k , can be chosen from the steady-state error condition e ss = lim s 0 sR ( s ) 1 + kG ( s ) = s 1 s 1 + k 1 s (1+0 . 3 s ) = 1 k < 0 . 003 = k > 333 . 33 k can be chosen to be: k = 334 2) In a similar manner, the damping ratio can be found through the percent overshoot requirement P.O. = 10% = 0 . 10 = e - ζπ 1+ ζ 2 = ζ = 0 . 59 And from the damping ratio, we can ﬁnd the phase margin φ m = 100 · ζ = 100 · 0 . 59 = 59 If the gain is set equal to unity, we can solve for the crossover frequency: | kG

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Chapter 6 Examples - EEL4657 Dr Haniph Latchman Chapter 6...

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