[R] Is R's fast fourier transform function different from "fft2" in Matlab?
Rolf Turner
r.turner at auckland.ac.nz
Fri Aug 1 05:30:27 CEST 2008
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
######################################################################
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
More information about the R-help
mailing list