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
Trang 1The output signal, y[t] = yh[t] + yp[t], is shown in Fig 24.6 , for different values of the eigenvalue h.
The transient is given by yh[t] = − , and the steady state response by yp[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 Ad:
(24.122) And the associated natural modes, present in the transient response, are
(24.123)
The natural modes are slightly damped, because |h1,2| 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 Ad and Bd 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 0.2 0.4 0.6 0.8 1
discrete time t
η = 0.2
η = 0.6
η = 0.8
ht
x t [ + 1 ] 1.2796 – 0.81873
0 u t [ ] +
=
y t [ ] = 0 0.5391 x t [ ]
h1,2 = 0.6398 ± j0.6398 = 0.9048 e ( jp/4)
h1,2t 0.9048te
j p
4
- t
0.9048t p
4
-t
4
-t
sin
± cos
p
4
-Ad ediag l{ 1 ,…,ln}∆
diag el1 ∆
,…, el n∆
0066_Frame_C24 Page 20 Thursday, January 10, 2002 3:44 PM
©2002 CRC Press LLC
Trang 2The output signal, y[t] = yh[t] + yp[t], is shown in Fig 24.6 , for different values of the eigenvalue h.
The transient is given by yh[t] = − , and the steady state response by yp[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 Ad:
(24.122) And the associated natural modes, present in the transient response, are
(24.123)
The natural modes are slightly damped, because |h1,2| 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 Ad and Bd 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 0.2 0.4 0.6 0.8 1
discrete time t
η = 0.2
η = 0.6
η = 0.8
ht
x t [ + 1 ] 1.2796 – 0.81873
0 u t [ ] +
=
y t [ ] = 0 0.5391 x t [ ]
h1,2 = 0.6398 ± j0.6398 = 0.9048 e ( jp/4)
h1,2t 0.9048te
j p
4
- t
0.9048t p
4
-t
4
-t
sin
± cos
p
4
-Ad ediag l{ 1 ,…,ln}∆
diag el1 ∆
,…, el n∆
0066_Frame_C24 Page 20 Thursday, January 10, 2002 3:44 PM
©2002 CRC Press LLC