8/8     Is there a technical reason why there exists a tradeoff between
        resolution in frequency and resolution in time? When I took
        EE120 with Kahn, he attributed it to Heisenberg Uncertainty,
        and although that involves a tradeoff too (velocity/location)
        I don't see how that explains the frequency/time tradeoff.
        Thanks.
        \_ Isn't it the idea that by "zooming in" on a unit of time,
           you're missing the "big picture" frequency?  Eg, mistaking the
           high frequency carrier wave in AM radio for the actual lower
           frequency signal?
        \_ Here is the crux of the problem, how can you resolve a 10 Hz
           signal if you observe for a mere 1 us?
        \_ The posters above give a good intuitive explanation of the
           time-frequency trade-off.  For a more technical explanation
           imagine that you want to measure some properties of a
           signal, s(t), through a finite time-slice.  You can get a
           finite time-slice via multiplying by a "boxcar", b(at), which
           is 1 for -.5 < t < .5 and 0 elsewhere.  In frequency, multiplying
           by b(at) is convolving with (1/a)B(f/a) where B(f) is a sinc.
           Choosing a to be large gets you very fine time-resolution, but
           makes the sinc sidelobes wide smearing out X(f) and giving poor
           frequency resolution.  Choosing a to be small makes the sinc narrow
           and reduces the smearing in frequency, but gives you poor
           time-resolution since you are including a long time-interval
           of s(t).  There are many ways to do time-frequency analysis where
           instead of using a boxcar you use other functions, but you can't
           escape the tradeoff.  See wavelets for more info.  -emin
                \_ Thanks, but that's more of a way of just
                   demonstrating the tradoff. Is there a natural or
                   technical reason why it exists?
                   \_ I'm not sure what you mean by "a natural or technical
                      reason why it exists".  If you could be more
                      specific about what you find unsatisfying in the
                      previous three explanations, I could do better. -emin
                        \_ I'm just wondering why the tradeoff exists. Is
                           it simply because time and frequency are
                           are inverses? (i.e. t = 1/f)
                    \_ I tried to understand this in particular and EPR in
                       general at one time.  I went to a friend of mine
                       who now is a physics instructor at CAL (part time)
                       and he stepped me through a bunch of math (including
                       Lornez transformation) and I realized I was over my
                       head and gave up.
        \_ I found this confusing too. The equations that describe
           a particle position/momentum and the equations that describe
           the time/frequency resolution are the same equations. So,
           the phenomena of the time/frequency trade-off is called
           Heisenberg Uncertainty, not because there is some physical
           connection between particles and Fourier analysis, but because
           the math works out exactly the same.