61_adv_control_eng

4 the transient response of a second order system is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ero or negative which will make the roots real and unequal, real and equal or complex. This gives rise to the three different types of transient response described in Table 3.4. The transient response of a second-order system is given by the general solution xo (t) ˆ Aes1 t ‡ B es2 t (3:46) This gives a step response function of the form shown in Figure 3.16. xo (t ) Underdamping (s1 and s2 complex) Critical damping (s1 and s2 real and equal) Overdamping (s1 and s2 real and unequal) t Fig. 3.16 Effect that roots of the characteristic equation have on the damping of a second-order system. //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 51 ± [35±62/28] 9.8.2001 2:26PM Time domain analysis 51 3.6.3 Critical damping and damping ratio Critical damping When the damping coefficient C of a second-order system has its critical value Cc , the system, when disturbed, will reach its steady-state value in the minimum time without overshoot. As indicated in Table 3.4, this is when the roots of the Characteristic Equation have equal negative real roots. Damping ratio z The ratio of the damping coefficient C in a second-order system compared with the value of the damping coefficient Cc required for critical damping is called the Damping Ratio  (Zeta). Hence ˆ C Cc (3:47) Thus  ˆ 0 No damping  < 1 Underdamping  ˆ 1 Critical damping  > 1 Overdamping Example 3.7 Find the value of the critical damping coefficient Cc in terms of K and m for the spring±mass±damper system shown in Figure 3.17. C Cxo Kxo K xo(t ) F(t ) m xo(t ) 1o(t ) m F(t ) xo(t ) Lumped Parameter Diagram (a) Fig. 3.17 Spring^ mass ^ damper system. Free-Body Diagram (b) +ve //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC03.3D ± 52 ± [35±62/28] 9.8.2001 2:26PM 52 Advanced Control Engineering Solution From Newton's second law ˆ  Fx ˆ mxo From the free-body diagram •  F (t) À Kxo (t) À C xo (t) ˆ mxo (t) (3:48) Taking Laplace transforms, zero initial conditions F (s) À KXo (s) À CsXo (s) ˆ ms2 Xo (s) or (ms2 ‡ Cs ‡ K )Xo (s) ˆ F (s) (3:49) Characteristic Equation is ms2 ‡ Cs ‡ K ˆ 0 CK ‡ ˆ0 mm i:e...
View Full Document

Ask a homework question - tutors are online