10.7 Performance of Second-Order System (Unit Step Response)

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1 Lecture Notes on Control Systems/D. Ghose/ Performance of Second-Order System (Unit Step Response) Consider the second order system a ÿ + a ẏ + a 0 y = b 0 r So, Y (s) R(s) = b 0 a s + a s + a 0 where, ω n s +ω n s + ωn Apply a unit step input r(t) =u(t) = s Then, ω n = a 0 a, K = b 0 a 0, = a a 0 a This can be written as, ω n Y (s) =K s (s +ω n s + ωn) Y (s) = k s + k s + k 3 s +ω n s + ω n Let us find the values of k, k,andk 3 by equating the coefficients of the numerator polynomials. ω n Y (s) =K s(s +ω n s + ωn) = (k + k )s +(ω n k + k 3 )s + ωnk s(s +ω n s + ωn) Comparing the coefficients, k + k = 0 ω n k + k 3 = 0 ωnk ωn which can be solved to yield k k = K k 3 = ω n K

2 Lecture Notes on Control Systems/D. Ghose/0 58 Then, Y (s) s s +ω n s +ω n s + ωn s s +ω n s +ω n s + ωn ωn + ωn s s +ω n (s + ω n ) + ωn( ) s s + ω n (s + ω n ) + ωn( ) ω n (s + ω n ) + ωn( ) { } s s + ω n (s + ω n ) + ωn( ) ω n ω n ω n (s + ω n ) + ωn( ) { } s s + ω n (s + ω n ) + ωn( ) ω n (s + ω n ) + ωn( ) Taking inverse Laplace transforms (assuming 0 <<, i.e., an underdamped system), y(t) e ωnt cos ω n t e ωnt sin ω n t { } e ωnt cos ω n t + sin ω n t Let =cosφ and =sinφ, then { } y(t) e ωnt cos φ cos ω n t +sinφsin ω n t e ωnt cos(ω n t φ) e ωnt cos(ω d t φ) Let us normalize y(t) by defining, so that ȳ(t) ast. ȳ(t) = y(t) K Then, the step response of the underdamped system is, ȳ(t) = e ωnt cos(ω d t φ)

9 of the final value y( ). Rise Time (T r ): This is defined for second-order underdamped systems. This is the time for y(t) to go from 0 to of the final value y( ).

3 Lecture Notes on Control Systems/D. Ghose/0 59 Figure 0.: Unit step response and location of the poles of a second order system Rise Time (T r ): This is the same as in the first order system and is the time for y(t) to go from 0. to 0.9 of the final value y( ). Rise Time (T r ): This is defined for second-order underdamped systems. This is the time for y(t) to go from 0 to of the final value y( ). Settling Time (T s ): This is the same as in first order system and is defined as the time after which y(t) remains within % (or 5%) of the final value y( ). Maximum Overshoot (M p ): This is the maximum overshoot value of y(t) over the final value y( ) and is usually expressed as a percentage. Peak Time (T p ): Time at which the maximum overshoot occurs. For second-order underdamped system, these quantities can be related to ω n and τ in an approximate sense. Settling Time (T s ): The envelope of ȳ(t) from below is governed by the equation, ȳ e (t) = e ωnt So, if we take the % criterion for settling time, ȳ e (T s )= e ωnts =0.98

4 Lecture Notes on Control Systems/D. Ghose/0 60 For e ωnts =0.0 e ωnts =0.0 ω n T s =ln 0.0 T s = ln 0.0 ω n 3.9 ln So. approximately, we can say that, So, mainly T s depends on ω n =/τ T s = 4 ω n = 4τ Design Rule : To decrease T s Increase ω n decrease τ Peak Time (T p ): The time at which ȳ(t) reaches its peak, the slope is zero. ȳ(t) = e ωnt (cos ω d t + ) sin ω dt So, ȳ(t) = ω n e ωnt (cos ω d t + ) sin ω dt ( e ωnt ω d sin ω d t + ω ) d cos ω dt ω ne ωnt sin ω d t + ω n e ωnt sin ω d t = = ω n e ωnt sin ω d t = ω n e ωnt sin ω d t + = ω n e ωnt sin ω d t Slope is zero implies that ȳ(t) = 0, which happens when, sin ω d t =0 ω d t = nπ, n =0, ±, ±,... t = nπ, ω d n =0, ±, ±,...

5 Lecture Notes on Control Systems/D. Ghose/0 6 When does this happen for the first time after t = 0? When n =. So, T p = π ω d = π ω n Design Rule : To decrease T p Increase ω d Increase ω n. Maximum Overshoot (M p ): We have seen earlier, Where, ȳ(t) = e ωntp (cos ω d T p + = e ωntp (cos π + = +e ωntp = +e ω n ωn = +e π = +M p π M p = e π ) sin ω dt p ) sin π For, 0 0.6, Figure 0.: Maximum overshoot vs. damping factor M p = 0.6 Design Rule 3: Overshoot decreases with increasing (almost linear decrease for lower ranges of ). So, To decrease M p Increase.

6 Lecture Notes on Control Systems/D. Ghose/0 6 Rise Time (T r ): For underdamped system we use the 0% to 00% rise time criterion. We will denote this as T r for convenience. ȳ(t r ) = e ωntr (cos ω d T r + cos ω d T r + sin ω dt r =0 tan ω d T r = T r = ) tan ( ω d sin ω dt r ) = Let, cos φ = sin φ = tan φ = ( tan φ + π ) = ) tan ( = π + φ = π +sin π = + (for small ) So, the rise time is, T r = ) tan ( ω d = ( ) π ω d +sin = π/+sin (exact expression) ω n π/+ = ω = π + π n ω n = (for small ) ω n Design Rule 4: To decrease T r Increase ω n. Actually T r can also be decreased by decreasing, but this causes larger overshoot. Let us summarize all the above design rules in the following table.

7 Lecture Notes on Control Systems/D. Ghose/0 63 Performance Exact Approximate Design measure formula Formula rule T s : Settling time ln ω n ω n To decrease T s Increase ω n T p :Peak = π ω d To decrease T p time Increase ω d = ω n π ω n M p : Maximum e π overshoot To decrease M 0.6 p Increase T r : Rise time ( π ω n +sin ) π ω n To decrease T r (0% to 00%) Increase ω n Now, consider the s-plane interpretation of second order systems as given earlier. Remember how we got these loci. p, = ω n ± jω n = τ ± jω d

8 Lecture Notes on Control Systems/D. Ghose/0 64 Figure 0.3: Constant parameter loci of poles

9 Lecture Notes on Control Systems/D. Ghose/0 65 Figure 0.4: Regions in the s-plane

10 Lecture Notes on Control Systems/D. Ghose/0 66 PROBLEM SET 4. Find the poles, zeros, time constant, bandwidth, natural frequency, damped frequency, DC gain, damping ratio and whether the system is overdamped or underdamped, settling time, rise time, peak time, and maximum overshoot (whichever is relevant) of the following systems: (a) G(s) = (c) G(s) = s +3 s +4 s +4s + (b) G(s) = (d) G(s) = 0 s +6s +0 (s +)(s +3). For all the above systems determine if the final value theorem can be used to find the steady state value of the time response under unit step input. If yes, then find the final value. 3. For all the above systems, determine the equation for the time response to a unit step input. What kind of response are these. Sketch qualitatively. 4. Consider the second order system ω n G(s) =K s +ω n s + ωn (a) Suppose ω n =0and varies between 0. and, draw the locus of the poles. (b) Suppose =0. andω n varies between and 0, draw the locus of the poles. (c) Assume that the system is underdamped and τ =0.6, then sketch the locus of the poles as both and ω vary. (d) Assume that the system is underdamped and ω d =0.8 then sketch the locus of the poles as both and ω vary. 5. (MATLAB assignment) (a) Plot the time response of the following systems, driven by step input signal. K s + G(s) =K s + K s +K 3 Assume three different combinations of (K, K,K,K 3 ). All of them should be positive.

11 Lecture Notes on Control Systems/D. Ghose/0 67 (b) Determine and plot the poles and zeros of the above system for the same three combinations. (c) Consider a combination of (K, K,K,K 3 ) such that the resulting system is underdamped and has all the poles on the left hand s-plane. Then, select another such combination so that the system remains underdamped but has better performance in terms of rise time, maximum overshoot, peak time, and settling time. Plot both the time responses.

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