Compensation Filter
G(s) = P + sD
The next step is to combine all the system elements, with the exception of G(s), into one function, L(s).
L(s) = M(s) K
a
K
d
K
f
H(s) =3.17∗10
6
/[s
2
(s+2000)]
Then the open loop transfer function, A(s), is
A(s) = L(s) G(s)
Now, determine the magnitude and phase of L(s) at the frequency ω
c
= 500.
L(j500) = 3.17∗10
6
/[(j500)
2
(j500+2000)]
This function has a magnitude of
|L(j500)| = 0.00625
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)| = 160
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 160 and a phase lead of 59 degrees.
These requirements may be expressed as:
|G(j500)| = |P + (j500D)| = 160
and
arg [G(j500)] = tan
-1
[500D/P] = 59°
The solution of these equations leads to:
P = 160cos 59° = 82.4
500D = 160sin 59° = 137
Therefore,
D = 0.274
and
G = 82.4 + 0.2744s
The function G is equivalent to a digital filter of the form:
D(z) = KP + KD(1-z
-1
)
DMC-40x0 User Manual Chapter 10 Theory of Operation • 194
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