# Chapter10 - CHAPTER 10 COMPENSATOR DESIGN USING BODE PLOTS...

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SOLUTION MANUAL 99 10.4 Lead compensator D ( s ) = 10 0 5 1 01 1 (. ) . s s + + meets the phase margin and steady- state accuracy requirements. These requirements are also met by the lag compensator D ( s ) = 10 10 1 100 1 () s s + + In this particularly simple example, specifications could be met by either compensation. In more realistic situations, there are additional performance specifications such as bandwidth and there are constraints on loop gain. Had there been additional specifications and constraints, it would have influenced the choice of compensator (lead or lag). 105. The settling time and peak overshoot requirements on performance may be translated to the following equivalent specifications: z = 0.45 ; w n = 2.22 z is related to the phase margin by the relation F M » z 001 . = 45º w n is related to the bandwidth w b by the relation w b = w n 12 24 4 22 4 -+ zz z / For a closed-loop system with z = 0.45, we estimate from this relation w b = 1.33 w n Therefore, we require a closed-loop bandwidth w b ~ - 3. The gain K of the compensator D ( s )= Ks s t at + + 1 1 may be set a value given by K = w n 2 . This gives K ~ - 5. To provide a suitable margin for settling time, we select K = 10. The phase margin of the uncompensated system is 0º because the double integration results in a constant 180º phase lag. Therefore, we must add a 45º phase lead at the gain crossover frequency of the compensated magnitude curve. Evaluating the value of a , we have a = 14 5 5 - + sin º sin º = 0.172 To provide a margin of safety, we select a = 1/6.
100 CONTROL SYSTEMS: PRINCIPLES AND DESIGN The frequency at which the uncompensated system has a magnitude of – 20 log (1/ a ) = – 7.78 dB is 4.95 rad/sec. Selecting this frequency as

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Chapter10 - CHAPTER 10 COMPENSATOR DESIGN USING BODE PLOTS...

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