[R] Is R's fast fourier transform function different from "fft2" in Matlab?

[stephen sefick]

 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.  you have to take the inverse of
the fft to get the original series.  I don't know anything about
matlab but it looks like the imaginary part is hidden.

On Thu, Jul 31, 2008 at 5:12 PM, Li Li <lilycai2007 at gmail.com> wrote:
> Thanks for both replies.
> Then I found the "ifft2" from Matlab gives different result from "fft( ,
> inverse=T)" from R.
> An example:
> in R:
>> temp  <- matrix(c(1,4,2, 20), nrow=2)
>> fft(temp)
>       [,1]   [,2]
> [1,]  27+0i -17+0i
> [2,] -21+0i  15+0i
>> fft(temp,inverse=T)
>       [,1]   [,2]
> [1,]  27+0i -17+0i
> [2,] -21+0i  15+0i
>
> In Matlab:
>> A = [1,2;4,20];
>> fft2(A)
> Ans =
>   27    -17
>  -21     15
>>ifft2(A)
> Ans=
>   6.7500    -4.2500
>  -5.2500      3.7500
>
> I also tried mvfft with inverse but can't get same result with "ifft2". Does
> any function work?
> Thanks,
>
> Li
>
> On 5/2/07, Sundar Dorai-Raj < sundar.dorai-raj at pdf.com> wrote:
>>
>>
>> I don't know Matlab or any of its functions, but the following produces
>> the same output.
>>
>> z <- matrix(c(1, 4, 2, 20), nrow = 2)
>> Re(fft(z))
>>
>> And from ?fft:
>>
>> When 'z' contains an array, 'fft' computes and returns the multivariate
>> (spatial) transform.
>>
>> HTH,
>>
>> --sundar
>>
>
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>
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