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Interpretation of sinc interpolation in Fourier domain

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Would someone be kind enough to check the argument below. I'm a bi
uncomfortable with the conclusion. The equations are in LaTeX format.

Some signal, f(t), is bandlimited such that we can represent it wit
samples every T seconds.
f(t) = \sum_{n=-\infty}^{\infty} f(nT) sinc(\frac{\pi t}{T} - nT)

The linearity property of the Fourier transform gives that

F(w) = \sum_{n=-\infty}^{\infty} \mathcal{F}\{f(nT) sinc(\frac{\pi t}{T}
nT)\}(w)

where \mathcal{F}\{g(t)\}(w) is the Fourier transform of g(t).

As f(nT) is a single complex number for any n, this becomes

F(w) = \sum_{n=-\infty}^{\infty} f(nT) \mathcal{F}\{sinc(\frac{\pi t}{T}
nT)\}(y)

This means that the Fourier transform of a sampled signal can be perfectl
reconstructed by a linear combination of rect functions.

I feel this conclusion is wrong. Is it? And if so, why?


 0
Reply johnjhealy (7) 1/28/2008 1:46:38 PM

On 28 Jan, 14:46, "jhealy" <johnjhe...@gmail.com> wrote:
> Would someone be kind enough to check the argument below. I'm a bit
> uncomfortable with the conclusion. The equations are in LaTeX format.
>
> Some signal, f(t), is bandlimited such that we can represent it with
> samples every T seconds.
> f(t) = \sum_{n=-\infty}^{\infty} f(nT) sinc(\frac{\pi t}{T} - nT)
>
> The linearity property of the Fourier transform gives that
>
> F(w) = \sum_{n=-\infty}^{\infty} \mathcal{F}\{f(nT) sinc(\frac{\pi t}{T} -
> nT)\}(w)
>
> where \mathcal{F}\{g(t)\}(w) is the Fourier transform of g(t).

Don't know how well the arguments fit till here...

> As f(nT) is a single complex number for any n, this becomes
>
> F(w) = \sum_{n=-\infty}^{\infty} f(nT) \mathcal{F}\{sinc(\frac{\pi t}{T} -
> nT)\}(y)
>
> This means that the Fourier transform of a sampled signal can be perfectly
> reconstructed by a linear combination of rect functions.
>
> I feel this conclusion is wrong. Is it? And if so, why?

The argument is wrong. The trick is to see that for each time
t =nT there is exactly one sinc function which is nonzero.
The sincs are lined up such that the top of one occurs at
a time where *all* the others are zero. at those times there
is only one non-zero term in the reconstruction sum.

Rune

 0
Reply allnor (8474) 1/28/2008 2:31:31 PM

On 28 Jan., 14:46, "jhealy" <johnjhe...@gmail.com> wrote:
> Would someone be kind enough to check the argument below. I'm a bit
> uncomfortable with the conclusion. The equations are in LaTeX format.
>
> Some signal, f(t), is bandlimited such that we can represent it with
> samples every T seconds.
> f(t) = \sum_{n=-\infty}^{\infty} f(nT) sinc(\frac{\pi t}{T} - nT)
>
> The linearity property of the Fourier transform gives that
>
> F(w) = \sum_{n=-\infty}^{\infty} \mathcal{F}\{f(nT) sinc(\frac{\pi t}{T} -
> nT)\}(w)
>
> where \mathcal{F}\{g(t)\}(w) is the Fourier transform of g(t).
>
> As f(nT) is a single complex number for any n, this becomes
>
> F(w) = \sum_{n=-\infty}^{\infty} f(nT) \mathcal{F}\{sinc(\frac{\pi t}{T} -
> nT)\}(y)
>
> This means that the Fourier transform of a sampled signal can be perfectly
> reconstructed by a linear combination of rect functions.
>
> I feel this conclusion is wrong. Is it? And if so, why?

The argument is correct but the conclusion is wrong. You forgot the
linear phase-shift terms exp(-i w n/T) that you get from time-shifting
the sinc-kernel. So it's a linear combination of those terms exp(-i w
n/T), which is just a complex Fourier series.

You have just prooved that the time-domain samples are the Fourier
coeffcients of the series expansion of the (periodic) spectrum. Using
the injectivity of the Fourier transform, you can now argue that the
samples are enough to characterize the signal - this is Whittaker's
(or was it Shannon's?) original proof that the samples are enough to
fully describe a bandlimited signal.

Regards,
Andor


 0
Reply andor.bariska (1307) 1/28/2008 3:23:01 PM

>On 28 Jan., 14:46, "jhealy" <johnjhe...@gmail.com> wrote:
>> Would someone be kind enough to check the argument below. I'm a bit
>> uncomfortable with the conclusion. The equations are in LaTeX format.
>>
>> Some signal, f(t), is bandlimited such that we can represent it with
>> samples every T seconds.
>> f(t) = \sum_{n=-\infty}^{\infty} f(nT) sinc(\frac{\pi t}{T} - nT)
>>
>> The linearity property of the Fourier transform gives that
>>
>> F(w) = \sum_{n=-\infty}^{\infty} \mathcal{F}\{f(nT) sinc(\frac{\p
t}{T} -
>> nT)\}(w)
>>
>> where \mathcal{F}\{g(t)\}(w) is the Fourier transform of g(t).
>>
>> As f(nT) is a single complex number for any n, this becomes
>>
>> F(w) = \sum_{n=-\infty}^{\infty} f(nT) \mathcal{F}\{sinc(\frac{\p
t}{T} -
>> nT)\}(y)
>>
>> This means that the Fourier transform of a sampled signal can b
perfectly
>> reconstructed by a linear combination of rect functions.
>>
>> I feel this conclusion is wrong. Is it? And if so, why?
>
>The argument is correct but the conclusion is wrong. You forgot the
>linear phase-shift terms exp(-i w n/T) that you get from time-shifting
>the sinc-kernel. So it's a linear combination of those terms exp(-i w
>n/T), which is just a complex Fourier series.
>
>You have just prooved that the time-domain samples are the Fourier
>coeffcients of the series expansion of the (periodic) spectrum. Using
>the injectivity of the Fourier transform, you can now argue that the
>samples are enough to characterize the signal - this is Whittaker's
>(or was it Shannon's?) original proof that the samples are enough to
>fully describe a bandlimited signal.
>
>Regards,
>Andor
>
>
>

neglected them as 'just phase'!

 0
Reply johnjhealy (7) 1/28/2008 3:44:48 PM

On Jan 28, 10:44 am, "jhealy" <johnjhe...@gmail.com> wrote:
> neglected them as 'just phase'!

yeah, "famous last words".

:-)

r b-j


 0
Reply rbj (3909) 1/28/2008 7:07:47 PM

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