Yaskawa SMC–4000 - Manual (Page 38)

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SMC–4000 User Manual
Now, determine the magnitude and phase of L(s) at the frequency ω
c
= 500:
L(j500) = 1.27*10 7 /[(j500) 2 (j500+2000)]
This function has a magnitude of:
|L(j500)| = 0.025
and a phase:
Arg[L(j500)] = -180 ° - tan -1
(500/2000) = -194 °
G(s) is selected so that A(s) has a crossover frequency of 500 rad/s and a phase margin of 45 degrees. This
requires that:
|A(j500)| = 1
Arg [A(j500)] = -135 °
However, since:
A(s) = L(s) G(s)
then it follows that G(s) must have magnitude of:
|G(j500)| = |A(j500)/L(j500)| = 40
and a phase:
arg [G(j500)] = arg [A(j500)] - arg [L(j500)] = -135 ° + 194° = 59°
In other words, we need to select a filter function G(s) of the form:
G(s) = P + sD
so that at the frequency ω
c
=500, the function would have a magnitude of 40 and a phase lead of 59
degrees.
These requirements may be expressed as:
|G(j500)| = |P + (j500D)| = 40
and:
arg [G(j500)] = tan -1 [500D/P] = 59 °
The solution of these equations leads to:
P = 40cos 59 ° = 20.6
500D = 40sin 59 ° = 34.3
Therefore:
D = 0.0686
and:
G = 20.6 + 0.0686s
The function G is equivalent to a digital filter of the form:
D(z) = KP + KD(1-z
-1
)
where:
KP = P
and:
KD = D/T