On 14 Okt, 13:46, "Peter123" <p...@cornell.edu> wrote:
> >On 13 Okt, 00:22, "Peter123" <p...@cornell.edu> wrote:
> >> I have implemented Jens Joergen Nielsen FFT code in my windows program
> >> that converts V(t) to V(f). =A0I am using 2500 Hz sampling with
> N=3D8192 =
> >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 =A0amplitude (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) =3D 1/2 �(exp(jx)+exp(-jx))
> >sin(x) =3D 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 =3D k/N
>
> >wher k and N integers and k < N/2.
>
> >Rune
>
> Thanks Rune.
> I assume 3D is the amplitude.. �(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: =
In some readers it appears as '3D<equal sign>' or
something like that.
Rune
Reply by Peter123●October 14, 20082008-10-14
>On 13 Okt, 00:22, "Peter123" <p...@cornell.edu> wrote:
>> I have implemented Jens Joergen Nielsen FFT code in my windows program
>> that converts V(t) to V(f). =A0I am using 2500 Hz sampling with
N=3D8192 =
>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 =A0amplitude (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) =3D 1/2 (exp(jx)+exp(-jx))
>sin(x) =3D 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 =3D k/N
>
>wher k and N integers and k < N/2.
>
>Rune
>
Thanks Rune.
I assume 3D is the amplitude.. (why 3D?)
Peter
Reply by Rune Allnor●October 12, 20082008-10-12
On 13 Okt, 00:22, "Peter123" <p...@cornell.edu> wrote:
> I have implemented Jens Joergen Nielsen FFT code in my windows program
> that converts V(t) to V(f). �I am using 2500 Hz sampling with N=8192 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 �amplitude (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) = 1/2 (exp(jx)+exp(-jx))
sin(x) = 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 = k/N
wher k and N integers and k < N/2.
Rune
Reply by Peter123●October 12, 20082008-10-12
I have implemented Jens Joergen Nielsen FFT code in my windows program
that converts V(t) to V(f). I am using 2500 Hz sampling with N=8192 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 amplitude (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?
What correction factors should I use for the spectrum height for various
window types?
Appreciate all help.
Peter