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A Scale-Invariant "Shannon" Sampling Theorem for the Wigner Distribution!

Started by Mark June 18, 2012
This is something I noticed a short while back which I've not seen
explicitly mentioned as such.

The Wigner distribution, when applied to the analytic signal
associated with a real signal, has a 1/2-continuous 1/2-discrete
sampling grid built into it that is scale invariant and hyperbolic,
much like the sampling grid for the discrete wavelet transform; and it
can be reproduced by its samples via a result similar to the Shannon
sampling theorem.

When doing the Wigner distribution W_x on a real signal x(t) = x(t)*,
in order to avoid aliasing artifacts arising from the negative
frequency part, one should first filter these out to get what is known
as the "analytic signal" z(t). In the frequency domain, up to factor,
denoting the Fourier transforms respectively by X(k) and Z(k), their
relations are given by
   X(k) = Z(k) if k > 0; 1/2 Z(k) if k = 0; 0 else.

The Wigner distribution W_f(q, p) for a complex function f(t) is given
by
   W_f(q, p) = integral f*(q - Q/2) 1^{-pQ} f(q + Q/2) dQ
where I'm using 1^{...} as shorthand for exp(2 pi i (...)) to denote
the "powers of unity".

In the frequency domain, denoting Fourier transforms by F and using
the convention
   F(k) = integral f(t) 1^{-kt} dt <==> f(t) = integral F(k) 1^{kt} dk
the Wigner transform becomes
   W_f(q, p) = W_F(q, p) = integral F*(p - P/2) 1^{Pq} F(p + P/2) dP.

Now apply this to the real signal x. After conversion to the analytic
signal z, one gets:
   W_z(q, p) = W_Z(q, p) = integral Z*(p - P/2) 1^{Pq} Z(p + P/2) dP.

But Z is just X, band-limited. The combination of the two factors in
the integral yields a band-limiting on P in the range [-2p, 2p]
resulting in:
   W_z(q, p) = integral X*(p - P/2) 1^{Pq} X(p + P/2) dP
the integral is now taken over the range P from -2p to 2p.

For each frequency p > 0 has an integrand is band-limited over all
frequencies with bandwidth 4p. The integral has non-zero values only
for p > 0.

Consequently, we can apply the Shannon sampling theorem. The function
to apply it to is
   F(P, p) = X*(p - P/2) X(p + P/2)
which is the Fourier transform from q to P of
   f(q, p) = W_Z(q, p).

This results in the reduction of the variable q to a discrete grid
with spacing delta(q) = 1/4p, when p > 0. Without going through the
details, the final result of the application of the theorem is the
following:

W_z(q, p) = sum_n W_z(n/4p, p) sinc(n - 4pq)
summed over all integers n
where
sinc(y) = sin(pi y)/(pi y).

The simplest way this can be applied is in discretizing the Wigner
distribution. Since the sampling takes place for each frequency p,
separately, one can compute W_z(q, p) over a discrete range of
"voices" (p = p_0, p_1, p_2, ...), in each case applying the above
sampling with sampling widths delta q_0 = 1/(4 p_0), delta q_1 = 1/(4
p_1), delta q_2 = 1/(4 p_2), ...

This grid is scale-invariant. If you rescale the function x(t) ->
root(s) x(st) for any scale multiplier s > 0, you end up getting the
same grid back. The sinc function only involves p and q in combination
as pq, while under rescaling p -> p/s, q -> sq.
On Mon, 18 Jun 2012 17:32:46 -0700 (PDT), Mark
<federation2005@netzero.com> wrote:

>This is something I noticed a short while back which I've not seen >explicitly mentioned as such. > >The Wigner distribution, when applied to the analytic signal >associated with a real signal, has a 1/2-continuous 1/2-discrete >sampling grid built into it that is scale invariant and hyperbolic, >much like the sampling grid for the discrete wavelet transform; and it >can be reproduced by its samples via a result similar to the Shannon >sampling theorem. > >When doing the Wigner distribution W_x on a real signal x(t) = x(t)*, >in order to avoid aliasing artifacts arising from the negative >frequency part, one should first filter these out to get what is known >as the "analytic signal" z(t). In the frequency domain, up to factor, >denoting the Fourier transforms respectively by X(k) and Z(k), their >relations are given by > X(k) = Z(k) if k > 0; 1/2 Z(k) if k = 0; 0 else. > >The Wigner distribution W_f(q, p) for a complex function f(t) is given >by > W_f(q, p) = integral f*(q - Q/2) 1^{-pQ} f(q + Q/2) dQ >where I'm using 1^{...} as shorthand for exp(2 pi i (...)) to denote >the "powers of unity". > >In the frequency domain, denoting Fourier transforms by F and using >the convention > F(k) = integral f(t) 1^{-kt} dt <==> f(t) = integral F(k) 1^{kt} dk >the Wigner transform becomes > W_f(q, p) = W_F(q, p) = integral F*(p - P/2) 1^{Pq} F(p + P/2) dP. > >Now apply this to the real signal x. After conversion to the analytic >signal z, one gets: > W_z(q, p) = W_Z(q, p) = integral Z*(p - P/2) 1^{Pq} Z(p + P/2) dP. > >But Z is just X, band-limited. The combination of the two factors in >the integral yields a band-limiting on P in the range [-2p, 2p] >resulting in: > W_z(q, p) = integral X*(p - P/2) 1^{Pq} X(p + P/2) dP >the integral is now taken over the range P from -2p to 2p. > >For each frequency p > 0 has an integrand is band-limited over all >frequencies with bandwidth 4p. The integral has non-zero values only >for p > 0. > >Consequently, we can apply the Shannon sampling theorem. The function >to apply it to is > F(P, p) = X*(p - P/2) X(p + P/2) >which is the Fourier transform from q to P of > f(q, p) = W_Z(q, p). > >This results in the reduction of the variable q to a discrete grid >with spacing delta(q) = 1/4p, when p > 0. Without going through the >details, the final result of the application of the theorem is the >following: > >W_z(q, p) = sum_n W_z(n/4p, p) sinc(n - 4pq) >summed over all integers n >where >sinc(y) = sin(pi y)/(pi y). > >The simplest way this can be applied is in discretizing the Wigner >distribution. Since the sampling takes place for each frequency p, >separately, one can compute W_z(q, p) over a discrete range of >"voices" (p = p_0, p_1, p_2, ...), in each case applying the above >sampling with sampling widths delta q_0 = 1/(4 p_0), delta q_1 = 1/(4 >p_1), delta q_2 = 1/(4 p_2), ... > >This grid is scale-invariant. If you rescale the function x(t) -> >root(s) x(st) for any scale multiplier s > 0, you end up getting the >same grid back. The sinc function only involves p and q in combination >as pq, while under rescaling p -> p/s, q -> sq.
Interesting. I'll have to look at this. Mark DeArman