the program:
В
Fig.4
4) determine settling time and overshoot of the uncompensated system;
Theoretical information:
Settling time is the period of time from the beginning of transients to the moment of time, after which the inequality
В
takes place, where is given small constant value, which is usually equal to of the steady state value. Or in other words, it is the time for the system output to settle down to within a tolerance band of the final value, normally between В± 2 or 5%. p> Overshoot is maximal deflection of the output value in the transient process of the transient process from the steady state value. It normally expressed as percentage determined as:
,
where is maximal deflection of the transient process.the most cases a system would has sufficient stability margins if overshoot is less or equal to.
Results of the program:
From the Fig.4: of closed loop compensated system is 81.8%. time of closed loop uncompensated system is 6.76 sec.
5) check whether the system is stable or not using Hurwitz criterion;
Theoretical information:
Hurwitz stability criterion
The characteristic equation of order system is
.
For this system to be stable it is necessary and sufficient that the determinant of Hurwitz matrix and the determinants of all its diagonal minors are positive.matrix is ​​
.
The main diagonal of Hurwitz matrix contains coefficients of characteristic equation beginning with to. Elements which are located above the main diagonal have increasing indices; elements which are located under the main diagonal have decreasing indices. So odd-numbered lines contain coefficients of characteristic equation with odd indexes, and even-numbered lines contain coefficients of characteristic equation with even indexes. Places, in which coefficients are absent, are filled by zeros. p align="justify"> Program code:
[n0, d0] = tfdata (Wclun, 'v') d0 (1)> 0 & d0 (2)> 0 & d0 (3)> 0 & d0 (4) > 0 & (d0 (2) * d0 (3) - d0 (1) * d0 (4))> 0 ('System is stable by Hyrwitz criterion') ('because all coeficients in Hurwitz matrix are positive') ('and particular case for 3rd order sm is true') disp ('system is unstable')% task 1.5
Results of the program:
n0 =
0 6.3000 63.0000 =
.1000 1.6000 6.0000 63.0000is stable by Hyrwitz criterion because all coeficients in Hurwitz matrix are positive and particular case for 3rd order sm is true
6) check whether the system is stable or not using Nyquist criterion;
