MIT 2 004 - Second-Order System Response Characteristics

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MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.ty = 1y ( t )s t e pz < 1d a m p e d o s c i l l a t o r y r e s p o n s e0y ( t ) = 1 -s t e pe1 - z2-z w tnc o s (w t - f)dyp e a ks sTpo v e r s h o o t� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 211 Reading: • Nise: Secs. 4.6 – 4.8 (pp. 168 - 186) 1 Second-Order System Response Characteristics (contd.) 1.1 Percent Overshoot The height of the first peak of the response, expressed as a percentage of the steady-state response. %OS = ypeak − yss × 100 yss At the time of the peak y(Tp) ypeak = y(Tp) = 1 + e−(ζπ/√1−ζ2) and since yss = 1 %OS = e−(ζπ/√1−ζ2) × 100. Note that the percent overshoot depends only on ζ. Conversely we can find ζ to give a specific percent overshoot from the above: − ln (%OS/100)ζ = �π2 + ln2 (%OS/100) 1copyright cD.Rowell 2008 21–1Example 1 Find the damping ratio ζ that will generate a 5% overshoot in the step response of a second-order system. Using the above formula − ln (%OS/100) − ln(0.05)ζ = = = 0.69�π2 + ln2 (%OS/100) �π2 + ln2(0.05) Example 2 Find the location of the poles of a second-order system with a damping ratio ζ = 0.707, and find the corresponding overshoot. The complex conjugate poles line on a pair of radial lines at an angle θ = cos−1 0.707 = 45◦ from the negative real axis. The percentage overshoot is e−(ζπ/√1−ζ2)%OS = × 100 e−(0.707π/√1−0.5)= × 100 = 4.3% (≈ 5%) The value ζ = .707 = √2/2 is a commonly used specification for system design and represents a compromise between overshoot and rise time. 1.2 Settling Time The most common definition for the settling time Ts is the time for the step response ystep(t) to reach and stay within 2% of the steady-state value yss. A conservative estimate can be found from the decay envelop e, that is by finding the time for the envelope to decay to less than 2% of its initial value, e−ζωnt �1 − ζ2 < 0.02 giving ln(0.02�1 − ζ2)Ts = − ζωn or 4 Ts ≈ ζωn for ζ2 � 1. 21–2jwss - p l a n exx4 5oj 6 . 2 8- j 6 . 2 8j 8 . 8 8- j 8 . 8 88 . 8 8- 6 . 2 8d e s i r e d p o l e p o s i t i o n sExample 3 Find (i) the pole locations for a system under feedback control that has a peak time Tp = 0.5 sec, and a 5% overshoot. Find the settling time Ts for this system. From Example 2 we take the desired damping ratio ζ = 0.707. Then π Tp = = 0.5 s ωn�1 − ζ2 so that π π ωn = = = 8.88rad/s. Tp�1 − ζ2 0.5√1 − 0.5 The pole locations are shown below: Then p1, p2 = −8.88 cos �π 4 � ± j8.88 sin �π 4 � = −6.28 ± j6.28 The indicated settling time Ts from the approximate formula is 4 4 Ts = = 0.64 s. ≈ ζωn 0.707 × 8.88 Note that in this case ζ does not meet the criterion ζ2 � 1 and the full expression ln(.02�1 − ζ2) ln(.02√1 − 0.5)Ts = − ζωn = − 0.5 × 8.88 = 0.68 s gives a slightly larger value. 21–354 ( s + 1 ) 54 ( s + 2 s + 5 )+-y ( t )u ( t )2 Higher Order Systems For systems with three or more poles, the system be analyzed as a parallel combination of first- and second-order blocks, where complex conjugate poles are combined into a single second-order block with real coefficients, using partial fractions. The total system output is then the superposition of the individual blocks. Example 4 Express the system 5 G(s) = (s + 1)(s2 + 2s + 5) as a parallel combination of first- and second-order blocks. 5 A Bs + C G(s) = = + (s + 1)(s2 + 2s + 5) s + 1 s2 + 2s + 5 5 1 5 s + 1 = 4 s + 1 − 4 s2 + 2s + 5 using partial fractions. The system is described by the following block diagram and the response to an input u(t) may be found as the (signed) sum of the responses of the two blocks. 3 Some Fundamental Properties of Linear Systems 3.1 The Principle of Superposition For a linear system at rest at time t = 0, if the response to an input u(t) = f(t) is yf (t), and the response to a second input u(t) = g(t) is yg(t), then the response to an input that is a linear combination of f (t) and g(t), that is u(t) = af(t) + bg(t) where a and b are constants is y(t) = ayf (t) + byg(t). 21–4G ( s )f ( t )y ( t )i m p l i e sG ( s )d fd td yd tG ( s )f ( t )y ( t )i m p l i e sG ( s )òf d t0tòy d t0ti n t e g r a t i o nd i f f e r e n t i a t i o n0t i m e1 . 0u ( t )ts0t i m e0d( t )tu ( t )1 . 0001 . 0t i m eti n t e g r a t i o nd i f f e r e n t i a t i o nr� � � 3.2 The Derivative Property For a linear system at rest at time t = 0, if the response to an input u(t) = f(t) is yf (t), then the response to an input that is the derivative of f (t), that is df u(t) = dt is dyf y(t) = . dt 3.3 The Integral Property For a linear system at rest at time t = 0, if the response to an input u(t) = f(t) is yf (t), then the response to an input that is the integral of f(t), that is t u(t) = f(t)dt 0 is t y(t) = yf (t)dt. 0 Example 5 We can use the derivative and integral properties to find the impulse and ramp responses from the step response. We have seen therefore d yδ(t) = ystep(t)dtt yr(t) = ystep(t)dt 0 21–5K N ( s )u ( t )y ( t )D ( s )1x ( t )K N ( s )y ( t )D ( s )1v ( t )u ( t )For example, consider b G(s) = s + a with step response b �1 − e−at� .ystep = a The impulse response is d yδ(t) = ystep(t) = …


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