How can I use the time-frequency uncertainty principle? I have a signal composed of the summation of a set of sine waves of different frequencies. The amplitude of these sub-signals can change so many times a second.
I have been told that, if I want to retain the ability to distinguish each of the frequencies, the time-frequency uncertainty principal means there will be a limiting relationship between the duration of the time window between amplitude changes and the smallest interval between frequencies.
I found a website which seems to deal with the problem, but as a non-mathematician, I'm not sure how to utilise the formulas it shows. To be honest, I'm not even sure if it's relevant.
My question then: What is the relationship between the time interval and the minimum frequency interval?
 A: The insight and engineering rule of thumb is that
we need to observe at least half of a sine wave to determine its amplitude and frequency.
Of course, as a rule of thumb it is a drastic cutoff, but it is quite intuitive and close to "real".
Now take the sum of two sinusoids with near frequencies
$$
\sin \left( {2\pi \left( {f - \Delta f/2} \right)t} \right) + \sin \left( {2\pi \left( {f + \Delta f/2} \right)t} \right) = 2\cos \left( {2\pi \,\Delta f/2\,t} \right)\sin \left( {2\pi \,f\,t} \right)
$$
we obtain a sine of frequency $f$ amplitude-modulated by a (co)sine of frequency $\Delta f/2$.
(a well known phenomenon of which I do not remember the name in english).
According to the given rule, we need to "see" at least half of the modulating wave to be able
and reconstruct the signal, and thus the difference in frequency. If the duration , i.e. "persistance" of the
signals, is much less we will only see one sinusoid with the average frequency and double amplitude.
Calling $T$ the duration of the observation window we shall have
$$
{1 \over 2}{1 \over {\Delta f/2}} < T\quad  \Rightarrow \quad 1 < T\Delta f
$$
Passing to a more rigorous mathematical analysis, consider that
the Box function (a window of duration $T$)
$$
R(t,T) = \left\{ {\matrix{
   1 & { - T/2 \le t \le T/2}  \cr 
   0 & {otherwise}  \cr 
 } } \right. = U(t + T/2) - U(t - T/2)
$$
where $U$ is step function, has a frequency spectrum (bilateral Fourier Transform) given by
$$
G(f,T) = T{\rm sinc}(f\,T) = {{{\rm sin}(\pi f\,T)} \over {\pi f}}
$$

A sinusoid of duration $T$ is the product of the Box function for
a sine function.
The spectrum will be the convolution of a Dirac at frequency $f$ (apart that at $-f$)
with the $G(f,T) =T{\rm sinc}(f\,T) $ function , i.e. the same centered at $f$.
Then, approximating the $G(f,T)$ to a box function with a cutoff at $f \pm 1/(2T)$
leads to that we can distinguish between frequencies with a separation
of at least $1/T$, i.e. to the formula given above.
A: for a energy signal i.e physically realizable, exist a uncertainty principle between duration of signal and her bandwidth. If Bq is her quadratic band and Dq is her quadratic duration then BqDq >= (8*pi)^-1 . Therefore a signal can't have both duration and band zero. But if duration tend to zero then band tend to infinity and vice versa. To notice that the particular constant (8*pi)^-1 to derive from particular systems of mesure that is used.
