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.
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.