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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.

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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.

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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.

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

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