Fourier Transform of a frequency linearly modulated signal 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.
 A: 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.
A: 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.
