hwsln-421-s06-13

hwsln-421-s06-13 - 1 ECE 421, Spring 2006 Solutions to HW...

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1 ECE 421, Spring 2006 Solutions to HW Assignment #13 Bode Compensator Design Problems 1) The first thing to check is the steady-state error requirement. For the given plant K v =2 , so with the specification that the steady-state error for a unit ramp input must be 0 . 02 , K c =(1 / 0 . 02) / 2=25 . This is from the relationship that K c = e ss plant e ss spec = K v required K v plant (1) and e ss =1 /K v . The Bode plots for the plant with and without K c are shown in Fig. 1. Since the phase margin speci- fication is not satisfied after K c is included, additional compensation is needed. The specification that the compensated settling time must be longer than the uncompensated value means that a lag compensator should be used since a lead compensator would certainly decrease the settling time by increasing the bandwidth. A lag-lead compensator could also be used if the gain crossover frequency was chosen correctly. The frequency where the phase shift is K c G p ( )= 180 +55 +10 = 115 is approximately. ω . 52 rad/sec. This is the frequency that will be made the gain crossover frequency ω xc of the compensated system. The magnitude is | K c G p ( ) | =28 . 3 db at that frequency, which corresponds to an absolute value of α g =26 . 1 . The general rule of thumb is that the maximum value of α g should be in the range of 10–20, so a two-stage compensator will be used for this problem since α g > 20 . Each stage has an equivalent parameter α g 1 that is the square root of the total α g . Therefore, α g 1 = 26 . 1=5 . 111 . The compensator zero and pole for each stage are located at z c = ω xc 10 × N stage =0 . 0757 ,p c = z c α q 1 = 0 . 0757 5 . 111 . 0148 (2) The Bode plots for the compensated system are given in Fig. 2. All the specifications are satisfied. The gain crossover frequency is ω xc . 52 rad/sec, the phase margin is 60 . 4 , and the steady-state error for a ramp input is 0 . 02 . The lag compensator is G c ( s K c μ s z c +1 2 μ s p c 2 = 25 ³ s 0 . 0757 ´ 2 ³ s 0 . 0148 ´ 2 (3) = μ K c α g ( s + z c ) 2 ( s + p c ) 2 = 0 . 957 ( s +0 . 0757) 2 ( s . 0148) 2 (4) where there are 2 poles and 2 zeros since a two-stage compensator was designed. 2) For this Type 2 system, K a , and with the specification that the steady-state error for a parabolic input must be 1 , the compensator gain can be K c . The Bode plots for the uncompensated system are shown in Fig. 3. Since the phase shift starts at 180 and becomes more negative, there is no frequency where the phase margin specification would be satisfied. Therefore, phase lead compensation is needed. In order to satisfy the gain crossover frequency specification, lag compensation is also needed. Thus, the final compensator will be lag-lead. The lead compensator will be designed first. Since a range is given for ω xc , a value in that range will be selected; both the lead and lag sections of the compensator will use that same value. I will choose the value ω xc . 4 rad/sec as the value for the compensated gain crossover frequency. At that frequency, the plant’s phase shift is G p ( 186 . 8 , so the lead compensator must provide a
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This note was uploaded on 10/25/2010 for the course ECE 421 taught by Professor Cook,g during the Spring '08 term at George Mason.

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hwsln-421-s06-13 - 1 ECE 421, Spring 2006 Solutions to HW...

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