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 1 (see also Appendix 1, sec_ord.m). 3.7
3.7.1 Step response analysis and performance
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 1) second-order underdamped system responding to
an input of the form
xi (t) 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
%Overshoot Â 100
a1 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 BeÀ!n (=2)
B (3:66) Since the frequency of transient oscillation is !d , then,
!n 1 À 2 (3:67) Substituting (3.67) into (3.66) p
%Overshoot eÀ2!n =2!n 1À Â 100
%Overshoot eÀ= 1À (3:68) Method (b): Logarithmic decrement. Consider the ratio of successive peaks a1 and a2
a1 BeÀ!n (=2) (3:69) a2 BeÀ!n (3=2) (3:70) a1
À! (3=2) efÀ!n (=2)!n (3=2)g
a2 e n
e!n 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
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
(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 BeÀ!n ts
50 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 !n ts
!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.
From Figure 3.22
Xo (s) 25
s(1 2s)(s2 s 25) (3:74) 12:5
s(s 0:5)(s2 s 25) (3:75) A
s (s 0:5) (s 0:5)2 (4:97)2 (3:76) Resistance Thermometer Xi (s )=1/s 1
1 + 2s Valve
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.
- Winter '12
- Sali mon