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MIT16_30F10_rec02

# MIT16_30F10_rec02 - 16.30/31 Fall 2010 Recitation 2 In this...

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16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R G ( s ) C G c ( s ) E - + Figure 1: The standard block diagram for a unit-feedback loop. Problem 1 (VDV 8.14) If Figure 1 models a radar tracking system with G = 1 / [(0 . 1 s + 1) s ], design series compen- sation to meet the following specifications: 1. The steady-state error following ramp inputs must not exceed 2%. 2. The error in response to sinusoidal inputs up to 5 rad/sec should not exceed about 5%. 3. The crossover frequency should be about 50 rad/sec to meet bandwidth requirements while limiting the response to high-frequency noise. 4. The ratio of the break frequencies of G c should not exceed 5 to limit noise effects. 5. The phase margin should be about 50 . Solution The very first thing we should do is check closed-loop stability. A quick root locus sketch will reveal that the closed-loop system is stable for all positive gains, i.e., it is stable for G c ( s ) = K , for K > 0. For simplicity, we can choose K = 1 to get a baseline stabilizing controller. Re( s ) Im( s ) Figure 2: Root locus sketch

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Just for practice, let us figure out where the closed-loop poles would be for K = 1, us- ing root-locus techniques. First of all, write the transfer function in root-locus form, i.e., 1 G ( s ) = 10 s ( s +10) . Applying the magnitude condition, for G c = K = 1, we know that the closed-loop poles will be at locations that satisfy | s |·| s + 10 | = 10. Let us check whether the closed-loop poles would be on the part of the root locus lying on the real axis. In such a case, s would be a negative real number, and the magnitude condition would read s ( s +10) = 10 K , i.e., s 2 + 10 s + 10 = 0, which has the two roots s = 5 ± 25 10 = {− 8 . 8730 , 1 . 1270 } . Let us sketch the Bode plot of such a system (recall we chose G c = 1 for the time being, for convenience). The straight-line approximation is shown in Figure 3 . Since we have an integrator, the magnitude Bode plot starts with a negative “unit” slope (i.e., -20dB/decade). It will cross the line ω = 1 at K = 1. The effect of the pole at s = 10 is to decrease the slope to -40 dB/decade, starting at the break frequency ω = 10. The phase will start at -90 degrees for low frequencies (because of the integrator), be equal to -135 degrees at the pole break frequency ω = 10, and will ultimately approach -180 degrees at high frequencies (because of the additional pole). Since the phase never drops below -180 degrees, the phase margin will always be positive—but will be small if
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MIT16_30F10_rec02 - 16.30/31 Fall 2010 Recitation 2 In this...

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