Yep you are totally right. I looked at the graphs to do the analysis
quickly, and sacrificed correctness.
z <- rnorm(5000)
z.ts <- ts(z)
f <- fft(z.ts)
d <- fft(f, inverse=T)
plot(z.ts, d/5000)
#this is how far off the algorithm was from recreating the series.
After it is divided by the signal length.
plot((z.ts)-(d/5000))
Does this hold for longer signals, too?
On Thu, Jul 31, 2008 at 11:30 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
>
> On 1/08/2008, at 2:56 PM, stephen sefick wrote:
>
>> z <- rnorm(5000)
>> f <- fft(z)
>> d <- fft(f, inverse=T)
>> plot(z, d)
>>
>> z <- rnorm(5000)
>> z.ts <- ts(z)
>> f <- fft(z.ts)
>> d <- fft(f, inverse=T)
>> plot(z.ts, d)
>>
>> temp <- matrix(c(1,4,2, 20), nrow=2)
>> d <- fft(temp)
>> f <- fft(d, inverse=T)
>> plot(temp, f)
>>
>> this, looks to me, to be the same.
>
> Then I think you'd better get your eyes checked, mate!
>
>> you have to take the inverse of the fft to get the original series.
>
> No you ***don't*** get the original series; you get n*(the original
> series)
> where n is the series length.
>
> I.e. the fft in R (and in S/Splus) does not apply any normalizing
> factor,
> so that the inverse transform only ``inverts'' up to a constant
> multiple.
>
> cheers,
>
> Rolf Turner
>
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