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}}
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
