Hi,

What is the meaning of a negative frequency? (in the context of the
spectrum of a stochastic process)

I have seen time to time that we extend integration to negative frequencies
to benefit from Fourier transform calculus, and that it was virtually the
same from a physical point of view because the contributions of negative
(unphysical) frequencies were negligible. However if you compute the power
spectrum S(omega) of the random telegraph signal, you get for S a
Lorentzian centered about 0, i.e., fluctuations of frequencies -omega_0 are
as important as +omega_0, so no question here to say that unphysical
negative frequencies are negligibles.

For this special case of telegraph signal, how do you understand its
spectrum? Why is it centered about 0? What is the meaning of this zero
frequency (no oscillations; I'd think the main frequency, i.e., the peak of
the spectrum, should be centered on the inverse rate of transition)? Why
are there negative frequencies on an equal footing with positive ones?

Thanks.

Student wrote in message ...
Hi,

What is the meaning of a negative frequency? (in the context of the
spectrum of a stochastic process)


When looking at the definition of the Fourier transform, you see that
your signal is decomposed in a superposition of functions of the form
exp(-j w t), where w goes from -infinity to + infinity. When you look
at a definite frequency, you should in fact take into account that
frequency and its negative:
exp(-j w t) and exp (+ j w t). The sum of both gives you a real
function, namely Cos(w t). The difference of both gives you another
real function, namely Sin(w t). So if you combine exp(-j w t) and
exp(+j w t) with equal weight, but opposite complex phases, you get a
real function Cos(w t + phi). The only way to get REAL combinations
is that the coefficient of the "negative frequency" and the one of the
"positive frequency" are of equal magnitude and opposite complex
phase, meaning they are conjugate.
So the Fourier transform of a real signal is always a function that
has
the property: H(w) = H(-w)*
(so what you write that the negative frequencies in real signals are
neglegible is not true: they are just as present as the positive
ones).
Of course there is no extra information in the negative frequency part
of the Fourier transform: if you know H(w) for w0, then you know it
for w0, exactly by the relation given above.
Concerning spectral power density and so on, it's just a matter of
convention. Given the fact that |H(w)| = |H(-w)|, you can chose to
just integrate over the positive frequencies and put a factor of 2 in
front, or just integrate from -INfinity to +Infinity.

cheers,
Patrick.

In article ,
Student wrote:

For this special case of telegraph signal, how do you understand its
spectrum? Why is it centered about 0? What is the meaning of this
zero frequency (no oscillations; I'd think the main frequency, i.e.,
the peak of the spectrum, should be centered on the inverse rate of
transition)? Why are there negative frequencies on an equal footing
with positive ones?

Negative frequencies appear because you have a real-valued signal and
you are taking a complex Fourier transform. If you did sine and cosine
transforms you would find no conceptual problems, I presume.

In general you expect the power of the positive and negative
frequencies to be the same, so the spectrum will be symmetric about
zero. I still expect you to find peaks away from zero, associated to
physical frequencies or inverses of characteristic times.

Regards,

Miguel Carrion

The zero frequency is the average dc value of the signal. The negative
frequencies are a result of how the Fourier transform breaks down the
signal.

"Student" wrote in message
...

What is the meaning of a negative frequency? (in the context of the
spectrum of a stochastic process)