196 System Stability
4. The breakaway points can be found from
dF(z) = 0, d z
or
0.368(z − 1)(z − 0.368) − 0.368(z + 0.717)(2z − 1.368) = 0,
which gives
z2 + 1.434z − 1.348 = 0
and the roots are at
z = −2.08 and z = 0.648.
5. The value of k at the breakaway points can be calculated from
F (z) z=−2.08,0.648
which gives k = 15 and k = 0.196.
The root locus of the system is shown in Figure 8.3. The locus is a circle starting from the poles, breaking away at z = 0.648 on the real axis, and then joining the real axis at z = −2.08.
Figure 8.3 Root locus for Example 8.8
Figure 8.4 Root locus with unit circle
At this point one part of the locus moves towards the zero at z = −0.717 and the other moves towards the zero at −∞.
Figure 8.4 shows the root locus with the unit circle drawn on the same axis. The system will become marginally stable when the locus is on the unit circle. The value of k at these points can be found either from Jury’s test or by using the Routh–Hurwitz criterion.
Using Jury’s test, the characteristic equation is
1 |
+ |
K |
0.368(z + 0.717) |
= |
0, |
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(z |
− |
1)(z |
− |
0.368) |
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or |
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z2 − z(1.368 − 0.368K ) + 0.368 + 0.263K = 0. |
Applying Jury’s test |
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F (1) = 0.631 |
for K > 0 . |
Also, |
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|0.263K + 0.368| < 1
which gives K = 2.39 for marginal stability of the system.
198 System Stability
Example 8.9
For Example 8.8, calculate the value of k for which the damping factor is ζ = 0.7.
Solution
In Figure 8.5 the root locus of the system is redrawn with the lines of constant damping factor and constant natural frequency.
From the figure, the roots when ζ = 0.7 are read as s1,2 = 0.61 ± j 0.25 (see Figure 8.6). The value of k can now be calculated as
F (z) z=0.61± j 0.25
which gives k = 0.324.
Example 8.10
A closed-loop system has the characteristic equation
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1 |
+ |
GH(z) |
= |
1 |
+ |
K |
(z − 0.2) |
= |
0. |
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z2 − 1.5z + 0.5 |
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Draw the root locus and hence determine the stability of the system. What will be the value of K for a damping factor ζ > 0.6 and a natural frequency of ωn > 0.6 rad/s?
Figure 8.5 Root locus with lines of constant damping factor and natural frequency
Figure 8.6 Reading the roots when ζ = 0.7
Solution
The above equation is in the form 1 + kF(z) = 0, where
F (z) |
= z2 |
z − 0.2 |
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− 1.5z + 0.5 |
The system has two poles at z = 1 and at z = 0.5. There are two zeros, one at z = −0.2 and the other at infinity. The locus will start from the two poles and terminate at the two zeros.
1.The section on the real axis between z = 0.5 and z = 1 is on the locus. Similarly, the section on the real axis between z = −∞ and z = 0.2 is on the locus.
2. Since n p − nz = 1, there is one asymptote and the angle of this asymptote is
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180r |
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180◦ for r |
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= n p − nz |
= ± |
= ± |
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Note that since the angle of the asymptotes are ±180◦ it meaningless to find the real axis intersection point of the asymptotes.
3. The breakaway points can be found from
dF(z) = 0