Transient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n


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1 Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2 n Figure above; The time response of the second order underdamped system 1
2 The percentage overshoot, %OS, is given by %OS = c max c final c final Note that %OS is a function only of the damping ratio, ζ. %OS = e (ζπ/ 1 ζ 2) 100 The inverse is given by ζ = ln(%os/100) π 2 + ln 2 (%OS/100) There is also a relationship between damping ratio and phase margin. The phase margin is obtained by solving G(jω) = 1 to obtain the frequency as ω 1 = ω n 2ζ 2 + The phase margin is Φ M = arctan 1 + 4ζ 4 2ζ 2ζ ζ 4 2
3 Thus if we can vary the phase margin, we can vary the percent overshoot, via a simple gain adjustment. Figure above; Bode plots showing gain adjustment for a desired phase margin. 3
4 Problem: For the position control system shown below, find the value of preamplifier gain, K, to yield a 9.5% overshoot in the transient response for a step input. Use only frequency response methods. Figure above; Bode plots for the example above. 4
5 Solution: 1. G(s) = 100K s(s+36)(s+100). Choose K = 3.6 to start the magnitude plot at 0dB. 2. Use ζ = ln(%os/100) π 2 + ln 2 (%OS/100) to find ζ = 0.6 for %OS/100 = 0.095, and then use 2ζ Φ M = arctan 2ζ ζ 4 to find Φ M = 59.2 for ζ = Locate on the phase plot the frequency that yields a 59.2 margin. This means phase angle at frequency of 14.8 rad/s. 4. At 14.8 rad/s on the magnitude plot, the gain is found to be 44.2dB. This magnitude has to be raise to 0dB to yield the required phase margin. 44.2dB increase is in gain, Thus K = =
6 Lag compensation The steady error constants are: position constant K p = lim s 0 G(s), velocity constant K v = lim s 0 sg(s), and acceleration constant K a = lim s 0 s 2 G(s). The value of the steadystate error decreases as the steady error constants increases. The function of the lag compensator is to (i) improve the steady error constant by increasing only the lowfrequency gain without any resulting instability, and (2) increase the phase margin of the system to yield the desired transient response. 6
7 Visualizing lag compensation The transfer function of the lag compensator is where α > 1. G c (s) = s + 1 T s + 1 αt Figure above; Frequency response of a lag compensator s+0.1 s
8 In the figure below, the uncompensated system is unstable since the gain at 180 is greater than 0dB. The lag compensator, while not changing the lowfrequency gain, does reduce the high frequency gain. The magnitude curve can be shaped to go through 0dB at the desired phase margin to obtain the desired transient response. Figure above; visualizing lag compensation 8
9 Design procedure 1. Set the gain,k, to the value that satisfies the steadystate error specification and plot the Bode plots based on this gain. 2. Find the frequency where the phase margin is 5 to 12 greater than the phase margin that yields the desired transient response. 3. Select a lag compensator whose magnitude response yields a composite Bode diagram that goes 0dB at the frequency found in step 2; (For detail see the next example) 4. Reset the system gain, K, to compensate for any attenuation in the lag network in order to keep the static error constant the same as that found in step 1. 9
10 Problem: For the same previous position control system, use Bode Diagrams to design a lag compensator to yield a a tenfold improvement in steadystate error over the gaincompensated system while keeping the percent overshoot at 9.5%. Solution: 1. The gaincompensated system is G(s) = s(s + 36)(s + 100) Thus K v = 16.22, a tenfold improvement changes K v = Thus K needs to be set at K = 5839, and the open loop transfer function G(s) = s(s + 36)(s + 100) 10
11 Figure above; Bode plots showing lag compensator design. 2. The phase margin required for %OS/100 = (ζ = 0.6) is Φ M = We increase this value by 10 to Φ M = 69.2 (i.e. phase angle ). The frequency is 9.8rad/s. At this frequency the magitude plot must go through 0dB. The magnitude at 9.8rad/s is 24dB. Thus the lag compensator must provide 24dB attenuation at 9.8rad/s. 11
12 3.&4. We now design the compensator. First draw the high frequency asymptote at 24dB. Arbitrarily select the higher break frequency to be about one decade below the phase margin frequency, i.e. at 0.98rad/s, to start drawing a 20dB/decade until 0dB is reached. The lower break frequency is found to be 0.062rad/s. Hence the lag compensator s transfer function is 0.063(s ) G c (s) = (s ) where the gain of the compensator is set to achieve a dc gain of unity. The compensated system s forward transfer function is thus G c (s)g(s) = 36786(s ) s(s + 36)(s + 100)(s ) 12
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