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### Filtering view of the Fourier transform

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```Hello Forum,

I found in a book the Fourier transform interpreted as a filtering
operation:

Given a signal f(t), The complex coefficient at frequency w0 given by the
Fourier transform is X(w0).
Say that the signal spectrum is X(w+w0) as if it was the result of a
modulation. X(w+w0) is centered at w0.

The Fourier integral has {f(t)*exp(-j*w0*t)} as integrand. This step can be
seen as a demodulation which shifts the spectrum of f(t) back to the
origin.

The {integral of f(t)*exp(-j*w0*t)} can be seen as a convolution integral
of f(t)*exp(-j*w0*t) with the signal g(t)=1. This is equivalent to
multiplication in the frequency domain by a delta(w) at the origin, which
sifts the value of the spectrum of X(w+w0) at w0, i.e. X(w0) which is a
constant, and therefore does not vary with time ,and exists for all
time....

Does this make any sense?

If we put limits other than infinity in the integral the convolution would
not be with the constant 1 but with rectangular function.....
That would make X(w0) not a constant anymore, but a function that changes
with time.....

thanks for any correction, or better explanation....
I think this is an interesting way to view the Fourier integral...

thanks
fisico32

```
 0
Reply fisico32 4/30/2010 3:42:04 PM

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```On Apr 30, 11:42=A0am, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com>
wrote:
> Hello Forum,
>
> I found in a book the Fourier transform interpreted as a filtering
> operation:
>
> Given a signal f(t), The complex coefficient at frequency w0 given by the
> Fourier transform is X(w0).
> Say that the signal spectrum is X(w+w0) as if it was the result of a
> modulation. X(w+w0) is centered at w0.
>
> The Fourier integral has {f(t)*exp(-j*w0*t)} as integrand. This step can =
be
> seen as a demodulation which shifts the spectrum of f(t) back to the
> origin.
>
> The {integral of f(t)*exp(-j*w0*t)} can be seen as a convolution integral
> of f(t)*exp(-j*w0*t) with the signal g(t)=3D1. This is equivalent to
> multiplication in the frequency domain by a delta(w) at the origin, which
> sifts the value of the spectrum of X(w+w0) at w0, i.e. X(w0) which is a
> constant, and therefore does not vary with time ,and exists for all
> time....
>
> Does this make any sense?
>
> If we put limits other than infinity in the integral the convolution woul=
d
> not be with the constant 1 but with rectangular function.....
> That would make X(w0) not a constant anymore, but a function that changes
> with time.....
>
> thanks for any correction, or better explanation....
> I think this is an interesting way to view the Fourier integral...
>
> thanks
> fisico32

Well not only can the Fourier transform be viewed as a "filter." But
you will find convolution is a property common to many integral
transforms and not just the Fourier transform. Also since the idea of
finite limits seems to intrigue you - you may also wish to look at
Green's functions. With these you replace a variable limit on the
integral with infinite limits and multiply the integrand with a
"gating function -  an adjustable window." An advantage of doing such
is the delta function can be expanded into a sum of basis functions
which in some cases simplfies the integration.

Clay
```
 0
Reply Clay 4/30/2010 4:22:24 PM

```>On Apr 30, 11:42=A0am, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com>
>wrote:
>> Hello Forum,
>>
>> I found in a book the Fourier transform interpreted as a filtering
>> operation:
>>
>> Given a signal f(t), The complex coefficient at frequency w0 given by
the
>> Fourier transform is X(w0).
>> Say that the signal spectrum is X(w+w0) as if it was the result of a
>> modulation. X(w+w0) is centered at w0.
>>
>> The Fourier integral has {f(t)*exp(-j*w0*t)} as integrand. This step can
=
>be
>> seen as a demodulation which shifts the spectrum of f(t) back to the
>> origin.
>>
>> The {integral of f(t)*exp(-j*w0*t)} can be seen as a convolution
integral
>> of f(t)*exp(-j*w0*t) with the signal g(t)=3D1. This is equivalent to
>> multiplication in the frequency domain by a delta(w) at the origin,
which
>> sifts the value of the spectrum of X(w+w0) at w0, i.e. X(w0) which is a
>> constant, and therefore does not vary with time ,and exists for all
>> time....
>>
>> Does this make any sense?
>>
>> If we put limits other than infinity in the integral the convolution
woul=
>d
>> not be with the constant 1 but with rectangular function.....
>> That would make X(w0) not a constant anymore, but a function that
changes
>> with time.....
>>
>> thanks for any correction, or better explanation....
>> I think this is an interesting way to view the Fourier integral...
>>
>> thanks
>> fisico32
>
>Well not only can the Fourier transform be viewed as a "filter." But
>you will find convolution is a property common to many integral
>transforms and not just the Fourier transform. Also since the idea of
>finite limits seems to intrigue you - you may also wish to look at
>Green's functions. With these you replace a variable limit on the
>integral with infinite limits and multiply the integrand with a
>"gating function -  an adjustable window." An advantage of doing such
>is the delta function can be expanded into a sum of basis functions
>which in some cases simplfies the integration.
>
>Clay
>
Hi Clay,
thanks for the message. Please bear with me and lets see if I truly
understand.
The way the Fourier transform is viewed as a filter, is that given a signal
x(t) and its spectrum X(w), the FT will filter such a signal and only allow
the single frequency w0 to go through, as if it was an infinitely sharp
passband filter with gain 1.
This is done for all different frequencies w.....
I am still, conceptually, not sure why the amplitudes X(w) of such filter
cannot be time-dependent. I know its due to the infinity limits of
integration.

Can you give me a better explanation of why the Fourier transform works as
a filter and exactly what type of filter it is?
If the limits of integration were finite we would get the STFT (short time
Fourier transform)
Wy would the FT work well only for stationary signals and not for
nonstationary?

Two different signals can have the same power spectra but different phase
spectra. the phase spectra seem to characterize the signals at every
instant uniquely......that seems like the FT works well for stationary and
nonstationary signals....
What am i missing?

Thanks!
fisico32
```
 0
Reply fisico32 4/30/2010 8:34:14 PM

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