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From: Gert Baars on 26 Sep 2005 19:17
With a little home-brew DFT program I am trying to some experiments.
When I draw one period of a sine with amplitude 1 no DC componenent and
analyse it with M #samples I get only 1 freq. component at Fs/M (n=1).
I think this is correct. However the amplitude of the DTF component is
on half of the sine's amplitude. When I add a virtual DC component
to the sine of 1 the DFT component for n =0 (DC) =1 but for the sine
still 1/2. I square sum and root the IM and RE components.

Why is the value of the for n=1 only 1/2 of the sine's amplitude?
From: Jerry Avins on 26 Sep 2005 19:50
Gert Baars wrote:
> With a little home-brew DFT program I am trying to some experiments.
> When I draw one period of a sine with amplitude 1 no DC componenent and
> analyse it with M #samples I get only 1 freq. component at Fs/M (n=1).
> I think this is correct. However the amplitude of the DTF component is
> on half of the sine's amplitude. When I add a virtual DC component
> to the sine of 1 the DFT component for n =0 (DC) =1 but for the sine
> still 1/2. I square sum and root the IM and RE components.
>
> Why is the value of the for n=1 only 1/2 of the sine's amplitude?

You should get both positive and negative frequencies if you use the
exponential form to represent your sinusoidal components, each at half
amplitude. So the question is not why is the amplitude 1/2, but rather
where is your missing line?

Jerry
--
Engineering is the art of making what you want from things you can get.
???????????????????????????????????????????????????????????????????????
From: Gert Baars on 26 Sep 2005 20:06
Jerry Avins wrote:
> Gert Baars wrote:
>
>> With a little home-brew DFT program I am trying to some experiments.
>> When I draw one period of a sine with amplitude 1 no DC componenent
>> and analyse it with M #samples I get only 1 freq. component at Fs/M
>> (n=1).
>> I think this is correct. However the amplitude of the DTF component is
>> on half of the sine's amplitude. When I add a virtual DC component
>> to the sine of 1 the DFT component for n =0 (DC) =1 but for the sine
>> still 1/2. I square sum and root the IM and RE components.
>>
>> Why is the value of the for n=1 only 1/2 of the sine's amplitude?
>
>
> You should get both positive and negative frequencies if you use the
> exponential form to represent your sinusoidal components, each at half
> amplitude. So the question is not why is the amplitude 1/2, but rather
> where is your missing line?
>
> Jerry

Thank you for the explaination. I guess it has to do with causiality.

From: Jerry Avins on 26 Sep 2005 20:26
Gert Baars wrote:

> ... I guess it has to do with causiality.

I don't see that. Would you care to explain?
--
Engineering is the art of making what you want from things you can get.
???????????????????????????????????????????????????????????????????????
From: John Monro on 27 Sep 2005 00:50
Gert Baars wrote:
> With a little home-brew DFT program I am trying to some experiments.
> When I draw one period of a sine with amplitude 1 no DC componenent and
> analyse it with M #samples I get only 1 freq. component at Fs/M (n=1).
> I think this is correct. However the amplitude of the DTF component is
> on half of the sine's amplitude. When I add a virtual DC component
> to the sine of 1 the DFT component for n =0 (DC) =1 but for the sine
> still 1/2. I square sum and root the IM and RE components.
>
> Why is the value of the for n=1 only 1/2 of the sine's amplitude?

Gert,
Your program is working well, but there should be one other component
present, which you have not mentioned.

If you have calculated the DFT for M frequency samples ( and not merely
M/2 samples), you will have found a component at n=(M-1).

This represents the amplitude of the (Fs/M) negative-frequency part of
the signal, and has an amplitude of -1/2.

The real signal you describe is made up of these two components, which
of course sum to 1.0.

Regards,
John
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