Crc Press Mechatronics Handbook 2002 By Laxxuss Episode 3 Part 6 pps

The output signal, y[t] = y h [t] + y p [t], is shown in Fig. 24.6, for different values of the eigenvalue h ᐉ . The transient is given by y h [t] = − , and the steady state response by y p [t] = 1. We observed in Eq. (24.114) that the system eigenvalues define the damping of its transient response, but also determine its frequency of oscillation (when the eigenvalues have a nonzero imaginary part). The potential problem when resonant modes exist is the same problem we found in the context of continuous-time systems, i.e., the system input contains a sine wave or another kind of signal, with energy at a frequency close to one of the natural frequencies of the system. The system output still remains bounded, although it grows to undesirable amplitudes. Example 24.8 Consider the discrete-time system described by the state space model (24.120) (24.121) The eigenvalues of the system are obtained from A d : (24.122) And the associated natural modes, present in the transient response, are (24.123) The natural modes are slightly damped, because |h 1,2 | is close to 1, and they show an oscillation of frequency p/4. In the plots shown in Fig. 24.7 we appreciate a strongly resonant output. The upper plot corresponds to an input u[t] = sin( t), i.e., the input frequency coincides with the frequency of the natural modes. In the lower plot the input is a square wave of frequency input signal p/12. In this case, the input third harmonic has a frequency equal to the frequency of the natural modes. Effect of Different Sampling Periods We observe in Eq. (24.95) that A d and B d depend on the choice of the sampling period ∆. This choice determines the position of the eigenvalues of the system too. If we look at the Eq. (24.96), assuming that A has been diagonalized, we have that (24.124) FIGURE 24.6 Step response of the system for different eigenvalues. 0 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1 discrete time t y [ t ] η = 0.2 η = 0.6 η = 0.8 h ᐉ t x t 1+[] 1.2796 0.81873– 10 x t[] 1 0 ut[]+= yt[] 0 0.5391 x t[]= h 1,2 0.6398 j0.6398± 0.9048 e jp/4 ()== h 1,2 t 0.9048 t e j p 4 t 0.9048 t p 4 t   j p 4 t   sin±cos== p 4 A d e diag l 1 ,…,l n {}∆ diag e l 1 ∆ ,…, e l n ∆ {}== 0066_Frame_C24 Page 20 Thursday, January 10, 2002 3:44 PM ©2002 CRC Press LLC The output signal, y[t] = y h [t] + y p [t], is shown in Fig. 24.6, for different values of the eigenvalue h ᐉ . The transient is given by y h [t] = − , and the steady state response by y p [t] = 1. We observed in Eq. (24.114) that the system eigenvalues define the damping of its transient response, but also determine its frequency of oscillation (when the eigenvalues have a nonzero imaginary part). The potential problem when resonant modes exist is the same problem we found in the context of continuous-time systems, i.e., the system input contains a sine wave or another kind of signal, with energy at a frequency close to one of the natural frequencies of the system. The system output still remains bounded, although it grows to undesirable amplitudes. Example 24.8 Consider the discrete-time system described by the state space model (24.120) (24.121) The eigenvalues of the system are obtained from A d : (24.122) And the associated natural modes, present in the transient response, are (24.123) The natural modes are slightly damped, because |h 1,2 | is close to 1, and they show an oscillation of frequency p/4. In the plots shown in Fig. 24.7 we appreciate a strongly resonant output. The upper plot corresponds to an input u[t] = sin( t), i.e., the input frequency coincides with the frequency of the natural modes. In the lower plot the input is a square wave of frequency input signal p/12. In this case, the input third harmonic has a frequency equal to the frequency of the natural modes. Effect of Different Sampling Periods We observe in Eq. (24.95) that A d and B d depend on the choice of the sampling period ∆. This choice determines the position of the eigenvalues of the system too. If we look at the Eq. (24.96), assuming that A has been diagonalized, we have that (24.124) FIGURE 24.6 Step response of the system for different eigenvalues. 0 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1 discrete time t y [ t ] η = 0.2 η = 0.6 η = 0.8 h ᐉ t x t 1+[] 1.2796 0.81873– 10 x t[] 1 0 ut[]+= yt[] 0 0.5391 x t[]= h 1,2 0.6398 j0.6398± 0.9048 e jp/4 ()== h 1,2 t 0.9048 t e j p 4 t 0.9048 t p 4 t   j p 4 t   sin±cos== p 4 A d e diag l 1 ,…,l n {}∆ diag e l 1 ∆ ,…, e l n ∆ {}== 0066_Frame_C24 Page 20 Thursday, January 10, 2002 3:44 PM ©2002 CRC Press LLC . 10 0 0.2 0.4 0 .6 0.8 1 discrete time t y [ t ] η = 0.2 η = 0 .6 η = 0.8 h ᐉ t x t 1+[] 1.27 96 0.818 73 10 x t[] 1 0 ut[]+= yt[] 0 0. 539 1 x t[]= h 1,2 0. 63 9 8 j0. 63 9 8± 0.9048 e jp/4 ()== h 1,2 t 0.9048 t e j p 4 . e l 1 ∆ ,…, e l n ∆ {}== 0 066 _Frame_C24 Page 20 Thursday, January 10, 2002 3: 44 PM 2002 CRC Press LLC The output signal, y[t] = y h [t] + y p [t], is shown in Fig. 24 .6, for different values of. 10 0 0.2 0.4 0 .6 0.8 1 discrete time t y [ t ] η = 0.2 η = 0 .6 η = 0.8 h ᐉ t x t 1+[] 1.27 96 0.818 73 10 x t[] 1 0 ut[]+= yt[] 0 0. 539 1 x t[]= h 1,2 0. 63 9 8 j0. 63 9 8± 0.9048 e jp/4 ()== h 1,2 t 0.9048 t e j p 4