Try this instead:
# just as you had before:
r=2*2*pi
Pole1=r*exp(I*2*pi/3)
Pole2=-r
Pole3=r*exp(-I*2*pi/3)
Ts=0.1
# one change here:
var('f')
z=exp(I*2*pi*Ts*f)
p=(2/Ts)*(z-1)/(z+1)
H = (p^3)/((p-Pole1)*(p-Pole2)*(p-Pole3))
Then H (or show(H)) will print out a complicated expression in terms
of f, and
H.subs(f=3)
will plug in f=3.
H.subs(f=3).n(4)
will gives a decimal approximation to H.subs(f=3) to 4 bits of
precision.
H.subs(f=3).n(digits=12)
will give an approximation to 12 decimal digits of precision.
Or, as Tim Lahey hints, you could define a function:
HH(x) = (x^3)/((x-Pole1)*(x-Pole2)*(x-Pole3))
Then 'HH(p)' will return the same thing as 'H' does.
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