<mrabcx@[EMAIL PROTECTED]
> wrote in message
news:f159935f-640f-44b0-8ea5-33ba525a985a@[EMAIL PROTECTED]
> Hi,
>
> I have a simple question regarding radars and Doppler ****fts.
> If the radars sends out a pulse p(t) then lets assume one
> moving targets reflects it, so that the receiving echo can
> be written as
> r(t)=a*p(t-d)*exp^(j*v),
> where a is the reflecting coefficient, d is some delay and
> exp^(j*v) coresponds to a Doppler ****ft due to moving target.
> I assume that * denotes standard multiplication.
>
> I think the model I have set up for r(t) should be quite reasable.
> Now what if I take a Fourier transform of it, can I then
> write it like this:
> R(w)=a*P(w)*exp(-jd) # F(exp^(j*v))
> where # denotes a convolution in Fourier domain and
> F(exp^(j*v)) is the Fourier transform of exp^(j*v) ?
>
> The reason I ask is that I have not seen this kind of notation
> being used in books but multiplication in time-domain corresponds
> to a convolution in Fourier so I hope writing it out this should
> be fine, or have I missed out something ?
> Or are there reasons why this notation should not be used or
> other simplified expressions are rather preferred ?
>
> Thanks for any help!
A true Doppler effect is really a compression or expansion of time and not
a
modulation. So, the relative bandwidth has to be somewhat low in order to
model it as a ****ft in frequency.
The Frequency ****fting Property says:
F[f(t)*e^jw0t] = F(w-w0)
But the Doppler effect doesn't modify F(w) in that manner.
Take time compression as an example (positive Doppler / closing range).
f(t) goes to f[(1+k)t] where k=v/V and v is the closing velocity and V is
the speed in the medium.
What I think you want to apply is the Scaling Property:
f(at) = (1/|a|)*F(w/a)
Compression in the time domain is equivalent to expansion in the frequency
domain.
{Lathi; Signals, Systems and Communication)
Fred
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