On 14 Okt, 13:46, "Peter123" <p...@[EMAIL PROTECTED]
> wrote:
> >On 13 Okt, 00:22, "Peter123" <p...@[EMAIL PROTECTED]
> wrote:
> >> I have implemented Jens Joergen Nielsen FFT code in my windows
program
> >> that converts V(t) to V(f). =3DA0I am using 2500 Hz sampling with
> N=3D3D8192 =3D
> >point
> >> FFT.
> >> I am plotting the FFT amplitude as sqrt(Re^2 + Im^2)/N. Im getting
the
> >> correct peak frequency after FFT, however, I am getting only about
1/2
> of
> >> the peak =3DA0amplitude (of the input sine wave input amplitude) with
> >> rectangular window. (If my input wave is 20.0*sin(omega*t) the FFT
> gives
> >> ~10 for peak height at omega).
> >> Using various windows (Hamming, Bartlett, etc) the V(f) peak
amplitude
> >> becomes even smaller.
>
> >> Any suggestion why do I get 1/2 peak heights?
>
> >It's because of Euler's equations:
>
> >cos(x) =3D3D 1/2 =A0(exp(jx)+exp(-jx))
> >sin(x) =3D3D 1/j2 (exp(jx)-exp(-jx))
>
> >> What correction factors should I use for the spectrum height for
> various
> >> window types?
>
> >Don't bother. You will not see the exact numbers you use
> >for the amplitudes unless the frequency of the sinusoidal
> >is an integer fraction of the sampling frequency,
>
> >f =3D3D k/N
>
> >wher k and N integers and k < N/2.
>
> >Rune
>
> Thanks Rune.
> I assume 3D is the amplitude.. =A0(why 3D?)
No 3D anywhere. It seems there are character coding
issues when the posts are read with different readers.
This is an equal sign: =3D
In some readers it appears as '3D<equal sign>' or
something like that.
Rune
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