Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) =
2 ωn . The closed loop transfer function is s(s+2ζωn)
2 ωn T (s) = 2 2 s + 2ζωs + ωn
Figure above; The time response of the second order underdamped system
1
The percentage overshoot, %OS, is given by %OS = cmax − cf inal cf inal
Note that %OS is a function only of the damping ratio, ζ. √ −(ζπ/ 1−ζ 2 ) × 100 %OS = e The inverse is given by ζ= π 2 + ln2(%OS/100) − ln(%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 ω1 = ωn −2ζ 2 + The phase margin is ΦM = arctan 2ζ −2ζ 2 + 1 + 4ζ 4
2
1 + 4ζ 4
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
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
100K Solution: 1. G(s) = s(s+36)(s+100) . Choose K = 3.6 to start the magnitude plot at 0dB.
2. Use ζ= π 2 + ln2(%OS/100) − ln(%OS/100)
to find ζ = 0.6 for %OS/100 = 0.095, and then use 2ζ ΦM = arctan −2ζ 2 + 1 + 4ζ 4 to find ΦM = 59.2◦ for ζ = 0.6.
3. Locate on the phase plot the frequency that yields a 59.2◦ margin. This means −120.8◦ 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 162.2 in gain, Thus K = 3.6 × 162.2 = 583.9.
5
Lag compensation The steady error constants are: position constant Kp = lim G(s),
s→0
velocity constant Kv = lim sG(s),
s→0
and acceleration constant Ka = lim s2G(s).
s→0
The value of...