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3.5.5 Ramp response of first-order systems Example 3.6 Find an expression for the response of a first-order system to a ramp function of slope Q. Solution From Figure 3.14 X o ( s ) QK s 2 (1 Ts ) QK = T s 2 ( s 1 = T ) A s B s 2 C ( s 1 = T ) (3 : 32) (See partial fraction expansion equation (3.13)). Multiplying both sides by s 2 ( s 1/ T ), we get QK T As s 1 T ± ² B s 1 T ± ² Cs 2 i : e : QK T As 2 A T s Bs B T Cs 2 (3 : 33) Equating coefficients on both sides of equation (3.33) ( s 2 ) : 0 A C (3 : 34) ( s 1 ) : 0 A T B (3 : 35) ( s 0 ) : QK T B T (3 : 36) From (3.34) C ± A From (3.36) B QK Substituting into (3.35) A ± QKT Hence from (3.34) C QKT K 1+ Ts X s Q s i ( )= / 2 X s o ( ) Fig. 3.14 Ramp response of a first-order system (see also Figure A1.1). Time domain analysis 47

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Inserting values of A , B and C into (3.32) X o ( s ) ± QKT s QK s 2 QKT ( s 1 = T ) (3 : 37) Inverse transform, and factor out KQ x o ( t ) KQ t ± T T e ± t = T ³ ´ (3 : 38) If Q 1 (unit ramp) and K 1 (unity gain) then x o ( t ) t ± T T e ± t = T (3 : 39) The first term in equation (3.39) represents the input quantity, the second is the steady-state error and the third is the transient component. When time t is expressed as a ratio of time constant T , then Table 3.3 and Figure 3.15 can be constructed. In Figure 3.15 the distance along the time axis between the input and output, in the steady-state, is the time constant. Table 3.3 Unit ramp response of a first-order system t / T 0 1 2 3 4 5 6 7 x i ( t )/ T 0 1 2 3 4 5 6 7 x o ( t )/ T 0 0.368 1.135 2.05 3.018 4.007 5 6 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 x t T o ( ) × (1/ ) Number of Time Constants 6 7 Fig. 3.15 Unit ramp response of a first-order system. 48 Advanced Control Engineering
3.6 Time domain response of second-order systems 3.6.1 Standard form Consider a second-order differential equation a d 2 x o d t 2 b d x o d t cx o ex i ( t ) (3 : 40) Take Laplace transforms, zero initial conditions as 2 X o ( s ) bsX o ( s ) cX o ( s ) eX i ( s ) ( as 2 bs c ) X o ( s ) eX i ( s ) (3 : 41) The transfer function is G ( s ) X o X i ( s ) e as 2 bs c To obtain the standard form, divide by c G ( s ) e c a c s 2 b c s 1 which is written as G ( s ) K 1 ! 2 n s 2 2 ± ! n s 1 (3 : 42) This can also be normalized to make the s 2 coefficient unity, i.e. G ( s ) K ! 2 n s 2 2 ±! n s ! 2 n (3 : 43) Equations (3.42) and (3.43) are the standard forms of transfer functions for a second- order system, where K steady-state gain constant, ! n undamped natural frequency (rad/s) and ± damping ratio. The meaning of the parameters ! n and ± are explained in sections 3.6.4 and 3.6.3. 3.6.2 Roots of the characteristic equation and their relationship to damping in second-order systems As discussed in Section 3.1, the transient response of a system is independent of the input. Thus for transient response analysis, the system input can be considered to be zero, and equation (3.41) can be written as ( as 2 bs c ) X o ( s ) 0 If X o ( s ) 6 0, then as 2 bs c 0 (3 : 44) Time domain analysis 49

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Table 3.4 Transient behaviour of a second-order system Discriminant Roots Transient response type b 2 > 4 ac s 1 and s 2 real and unequal ( ± ve) Overdamped Transient Response b 2 4 ac s 1 and s 2 real and equal ( ± ve)
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