61_adv_control_eng

66 p 2 overshoot e2n 2n 1 100 p 2 overshoot e 1 368

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Unformatted text preview: 14 14.5 ωnt (rad) Fig. 3.19 Unit step response of a second-order system. The generalized second-order system response to a unit step input is shown in Figure 3.19 for the condition K ˆ 1 (see also Appendix 1, sec_ord.m). 3.7 3.7.1 Step response analysis and performance specification Step response analysis It is possible to identify the mathematical model of an underdamped second-order system from its step response function. Consider a unity-gain (K ˆ 1) second-order underdamped system responding to an input of the form xi (t) ˆ B (3:64) The resulting output xo (t) would be as shown in Figure 3.20. There are two methods for calculating the damping ratio. Method (a): Percentage Overshoot of first peak a1 %Overshoot ˆ  100 B Now a1 ˆ BeÀ!n (=2) (3:65) //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 56 ± [35±62/28] 9.8.2001 2:27PM 56 Advanced Control Engineering xo ( t ) –ζω t Be n (with reference to final value) B a1 a2 B τ τ/2 t Fig. 3.20 Step response analysis. Thus, %Overshoot ˆ BeÀ!n (=2)  100 B (3:66) Since the frequency of transient oscillation is !d , then, ˆ ˆ 2 !d 2 p !n 1 À  2 (3:67) Substituting (3.67) into (3.66) p 2 %Overshoot ˆ eÀ2!n =2!n 1À  100 p 2 %Overshoot ˆ eÀ= 1À (3:68) Method (b): Logarithmic decrement. Consider the ratio of successive peaks a1 and a2 a1 ˆ BeÀ!n (=2) (3:69) a2 ˆ BeÀ!n (3=2) (3:70) a1 eÀ!n (=2) ˆ À! (3=2) ˆ efÀ!n (=2)‡!n (3=2)g a2 e n p 2 ˆ e!n  ˆ e2= 1À (3:71) Hence //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 57 ± [35±62/28] 9.8.2001 2:27PM Time domain analysis 57 Equation (3.71) can only be used if the damping is light and there is more than one overshoot. Equation (3.67) can now be employed to calculate the undamped natural frequency 2 !n ˆ p  1 À 2 3.7.2 (3:72) Step response performance specification The three parameters shown in Figure 3.21 are used to specify performance in the time domain. (a) Rise time tr : The shortest time to achieve the final or steady-state value, for the first time. This can be 100% rise time as shown, or the time taken for example from 10% to 90% of the final value, thus allowing for non-overshoot response. (b) Overshoot: The relationship between the percentage overshoot and damping ratio is given in equation (3.68). For a control system an overshoot of between 0 and 10% (1 <  > 0:6) is generally acceptable. (c) Settling time ts : This is the time for the system output to settle down to within a tolerance band of the final value, normally between Æ2 or 5%. Using 2% value, from Figure 3.21 0:02B ˆ BeÀ!n ts Invert 50 ˆ e!n ts –ζω t xo(t ) Be n (with reference to final value) Overshoot + – 2 or 5% of B B Rise t Time tr Settling Time ts Fig. 3.21 Step response performance specification. //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 58 ± [35±62/28] 9.8.2001 2:27PM 58 Advanced Control Engineering Take natural logs ln 50 ˆ !n ts giving  ts ˆ  1 ln 50 !n (3:73) The term (1/!n ) is sometimes called the equivalent time constant Tc for a secondorder system. Note that ln 50 (2% tolerance) is 3.9, and ln 20 (5% tolerance) is 3.0. Thus the transient period for both first and second-order systems is three times the time constant to within a 5% tolerance band, or four times the time constant to within a 2% tolerance band, a useful rule-of-thumb. 3.8 Response of higher-order systems Transfer function techniques can be used to calculate the time response of higherorder systems. Example 3.8 (See also Appendix 1, examp38.m) Figure 3.22 shows, in block diagram form, the transfer functions for a resistance thermometer and a valve connected together. The input xi (t) is temperature and the output xo (t) is valve position. Find an expression for the unit step response function when there are zero initial conditions. Solution From Figure 3.22 Xo (s) ˆ 25 s(1 ‡ 2s)(s2 ‡ s ‡ 25) (3:74) ˆ 12:5 s(s ‡ 0:5)(s2 ‡ s ‡ 25) (3:75) ˆ A B Cs ‡ D ‡ ‡ s (s ‡ 0:5) (s ‡ 0:5)2 ‡ (4:97)2 (3:76) Resistance Thermometer Xi (s )=1/s 1 1 + 2s Valve 25 s2 + s + 25 Fig. 3.22 Block diagram representation of a resistance thermometer and valve. Xo (s )...
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This note was uploaded on 01/18/2014 for the course MECH 107 taught by Professor Sali mon during the Winter '12 term at Bingham University.

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