Re: Hilbert transform

[Przemek Klosowski]

> octave:37> foo = center(linspace(0,1,100));
> octave:38> max(abs(foo + imag(hilbert(imag(hilbert(foo))))))
> ans =  0.0050505   - the expected answer is 0.
>
> And, even more interesting:
>
> octave:39> min(abs(foo + imag(hilbert(imag(hilbert(foo))))))
> ans =  0.0050505
>
> i.e. the error is constant.

Curiously, as the vector length increases from two to 20, odd vector 
lengths satisfy Sergei's identity u = -Hilb(Hilb(u)), but odd ones have 
quite large errors:

 0 0.5 0 0.16667 0 0.10000 0 0.07143 0 0.05556 0 0.04545 0 0.03846 0 
0.03333 0 0.02941 0 0.02632

In fact, min(abs(foo(N) + imag(hilbert(imag(hilbert(foo(N))))))) seems 
to be equal to 1/(2*N-2)

as the following code shows:

for i=2:33
  foo = center(linspace(0,1,i));
  printf("%d\t%f\t%f\n", i,min(abs(foo + 
imag(hilbert(imag(hilbert(foo)))))),1/(2*i-2));
end;

Could someone check that on Matlab?