2.004 Dynamics and Control II Spring 2008

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1 MIT OpenCourseWare Dynamics and Control II Spring 008 For information about citing these materials or our Terms of Use, visit:

2 O A Reading: Massachusetts Institute of Technology Department of Mechanical Engineering Nise: Secs (pp ).004 Dynamics and Control II Spring Term 008 Lecture Second-Order System Response Characteristics (contd.). Percent Overshoot F A = O I A F L A H I D O I I 6 = F I? E = H O H A I F I A O I A F M? I M The height of the first peak of the response, expressed as a percentage of the steady-state response. At the time of the peak y(t p ) and since y ss = %OS = y peak y ss 00 y ss y peak = y(t p ) = + e (ζπ/ ζ ) %OS = e (ζπ/ ζ ) 00. 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/00) ζ = π + ln (%OS/00) copyright c D.Rowell 008

3 Example 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/00) ζ = ln(0.05) = = 0.69 π + ln (%OS/00) π + ln (0.05) Example 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 = 45 from the negative real axis. The percentage overshoot is e (ζπ/ ζ ) %OS = 00 e (0.707π/ 0.5) = 00 = 4.3% ( 5%) The value ζ =.707 = / is a commonly used specification for system design and represents a compromise between overshoot and rise time.. Settling Time The most common definition for the settling time T s is the time for the step response y step (t) to reach and stay within % of the steady-state value y ss. A conservative estimate can be found from the decay envelope, that is by finding the time for the envelope to decay to less than % of its initial value, e ζω nt ζ < 0.0 giving or ln(0.0 ζ ) T s = ζω n 4 T s ζωn for ζ.

4 N N I Example 3 Find (i) the pole locations for a system under feedback control that has a peak time T p = 0.5 sec, and a 5% overshoot. Find the settling time T s for this system. From Example we take the desired damping ratio ζ = Then π T p = = 0.5 s ω n ζ so that π π ω n = = = 8.88rad/s. T p ζ The pole locations are shown A I E H F A F I E E I & & & M I F = A & & & $ & $ & " # $ & & & & Then p, p ( π ) = 8.88 cos ± j8.88 sin 4 = 6.8 ± j6.8 ( π ) 4 The indicated settling time T s from the approximate formula is 4 4 T s = = 0.64 s. ζωn Note that in this case ζ does not meet the criterion ζ and the full expression ln(.0 ζ ) ln(.0 0.5) T s = = ζω n gives a slightly larger value. = 0.68 s 3

5 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 + )(s + s + 5) as a parallel combination of first- and second-order blocks. G(s) = 5 A Bs + C = + (s + )(s + s + 5) s + s + s + 5 = 5 5 s + 4 s + 4 s + s + 5 using partial fractions. The system is described by the following block diagram K # " I O # " I I # 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. 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 y f (t), and the response to a second input u(t) = g(t) is y g (t), then the response to an input that is a linear combination of f(t) and g(t), that is where a and b are constants is u(t) = af(t) + bg(t) y(t) = ay f (t) + by g (t). 4

6 A I H 3. The Derivative Property For a linear system at rest at time t = 0, if the response to an input u(t) = f(t) is y f (t), then the response to an input that is the derivative of f(t), that is df u(t) = is dy f y(t) =. B / I O E F E A B 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 y f (t), then the response to an input that is the integral of f(t), that is t u(t) = f(t) is y(t) = 0 0 t y f (t). B / I O E F E A I / I Example 5 We can use the derivative and integral properties to find the impulse and ramp responses from the step response. We have E A C H = E K E A C H = E E B B A H A E = E B B A H A E = E E E A E A therefore y δ (t) = d y step (t) t y r (t) = y step (t) 0 5

7 For example, consider with step response b G(s) = s + a b ( e at ). y step = a The impulse response is d y δ (t) = y step (t) = be at, and the ramp response is t b ( ( )) e at y r (t) = y step (t) = t + 0 a a 4 The Effect of Zeros on the System Response Consider a system with a transfer function: G(s) = K N(s) = K sm + b m s m + + b s + b 0. D(s) s n + a n s n + + a s + a 0 N(s), which defines the system zeros, is associated with the RHS of the differential equation, while D(s) is derived from the LHS of the differential equation. Therefore N(s) does not affect the homogeneous response of the system. We can draw G(s) as cascaded blocks in two forms: K I N, I O or K, I L I O In this case we consider the all-pole sys tem /D(s) to be excited by u(t) di rectly to generate x(t), and the out put is formed as a superposition of the derivatives of v(t) In this case we consider the all-pole system /D(s) to be excited by x(t), which is a superposition of the derivatives of u(t) m x(t) = K d k u b k k k=0 m y(t) = K b k d k v k k=0 6

8 N N I Example 6 Find the step response of s + 0 G(s) =. s + s + 5 Splitting up the transfer function N(s) = s + 0, D(s) = s + s + 5 and ω n = 5, and ζ = / 5. I F = A Method : For the case, K I N, I O if u(t) = u s (t), the unit-step function du x(t) = + 0u = δ(t) + 0u s (t) and for the all-pole system /D(s) y step (t) = ( ) e t cos(t) e t sin(t) 5 y δ (t) = e t sin(t). For the complete system y(t) = y δ (t) + 0y step (t) = e t cos(t) e t sin(t) Method : For the case, 7

9 K, I L I O from above the step-response to the all-pole system /D(s) is ( ) v(t) = e t cos(t) e t sin(t) 5 and the system output is dv y(t) = + 0v 0 ( ) = e t sin(t) + e t cos(t) sin(t) 5 = e t cos(t) e t sin(t) which is the same as found in Method. 8