Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on an oscillating signal whose trend can be modelled as a frequency linearly varying function. An example may be as follows: $$ \Gamma(t)=\sin(2\pi\nu(t)t) $$ with $$ \nu(t)=\nu_0 + at $$ My signal is defined in a time interval as the following: $$ t=[0,t_\mathrm{end}] $$

When I Fourier Transform $\Gamma(t)$ getting $\Phi(\nu)$ ($\Phi(\nu)=FT[\Gamma(t)]$), I expect in the frequency domain a large peak extending from $\nu_0$ to $\nu_0 + at_\mathrm{end}$. Instead, what I obtain is a large peak extending from $\nu_0$ to $\nu_0 + 2at_\mathrm{end}$, centered at $\nu_0 + at_\mathrm{end}$.

Is this a feature of the Fourier Transform? I cannot understand what's going on.

Thank you very much.

share|cite|improve this question
Engineers call this kind of signal a chirp signal. You might want to try your question on dsp.SE where a fair number of participants are quite familiar with these signals. – Dilip Sarwate May 15 '12 at 0:08
Yes, a sinuosoid with an "intantaneous frequency" that varies continuosly between $f_1$ and $f_2$ (FM) would seem to have a Fourier transform with support (non zero values) in the $f_1 f_2$. But things are more complicated, FM introduces harmonics and hence to have larger frencuencies is to be expected. – leonbloy May 15 '12 at 0:51
I expected it to be much more trivial than it was.. thanks a lot for the link! – Francesco Pochetti May 15 '12 at 7:41

The instantaneous frequency in hertz is $f=\frac{d}{dt}(\nu(t)t)=\nu_0+2at$, so basically that's why the FT extends from $\nu_0$ to $\nu_0+2at_{end}$.

You are thinking of $\nu(t)$ as the frequency, which is incorrect. That's the source of the confusion.

share|cite|improve this answer

Yes, this seems to be a real effect. The Fourier transform of $$ \Gamma(t) = \cases{\exp(2 \pi i (\nu_0 + at) t), & $0 \le t \le T$\cr 0 & otherwise\cr}$$ is, according to Maple, $$ \eqalign{\widehat{\Gamma}(s) &= \int_0^T e^{-2 \pi i st} \Gamma(t)\ dt \cr &= \frac{1+i}{4\sqrt{a}} {{\rm e}^{{\dfrac {-i\pi \, \left( s- \nu_{{0}} \right) ^{2}}{2a}}}} \left( {{\rm erf}\left({\frac { \left( 1-i \right) \sqrt {\pi } \left( s-\nu_{{0}} \right) }{2\sqrt {a}}}\right)} - {{\rm erf}\left({\frac { \left( 1-i \right) \sqrt {\pi } \left( s-2aT-\nu_{{0}} \right) }{2\sqrt {a}}}\right)} \right) \cr}$$

The difference of the two erf terms is near $2$ for approximately $\nu_0 < s < \nu_0 + 2aT$, and near $0$ outside that interval.

share|cite|improve this answer
Dear @Robert ,this is quite weird... Anyway you are right! Thanks a lot! – Francesco Pochetti May 15 '12 at 6:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.