resolution in the fft

[Matthias Brennwald]

On Jun 9, 2009, at 8:46 PM, help-octave-request at octave.org wrote:

> Message: 5
> Date: Tue, 09 Jun 2009 20:34:49 +0200
> From: Markus Feldmann <feldmann_markus at gmx.de>
> Subject: resolution in the fft
> To: help at octave.org
> Message-ID: <h0ma09$mh7$1 at ger.gmane.org>
> Content-Type: text/plain; charset=ISO-8859-15; format=flowed
>
> Hi All,
>
> i made a spectrum on a charge/discharge function, but i am not sure  
> how
> to raise the xscale resoultion in my spectrum.
>
> Any ideas ?
>
> Here comes my function:
> %Konstanten
> FS=10e6
> T1=1e-6;
> T2=2e-6;
> t1=0:1/FS:5e-6;
> t2=0:0.1e-6:10e-6;
> t=[t1,t2+t1(end)];
> U0=10;
> %charge
> u1=U0*(1-exp(-t1/T1));
> %discharge
> u2=U0*exp(-t2/T2);
> u=[u1,u2];
> subplot(2,1,1)
> plot(t,u,'-+o')
> xlabel('Time/s');
> ylabel('Voltage/V');
> grid('on')
> %FFT
> nfft= 2^(nextpow2(length(u)))
> fftu = fft(u,nfft);
> NumUniquePts = ceil((nfft+1)/2)
> fftu = fftu(1:NumUniquePts);
> mu = abs(fftu)/length(u);
> mu = mu.^2;
> if rem(nfft, 2)
> 	mu(2:end) = mu(2:end)*2;
> else
> 	mu(2:end -1) = mu(2:end -1)*2;
> end
> f = (0:NumUniquePts-1)*FS/nfft;
> subplot(2,1,2)
> plot(f,mu,'-+o')
> xlabel('Frequency/Hz');
> ylabel('Power');
> grid('on')
> axis([0,1e6])
> print -deps >exponential.eps

Dear Markus

1. The short answer is: you can't increase the resolution of the fft,  
because the resolution is given by the number of samples in your signal.
2. The somewhat longer answer is: you can increase the resolution of  
the fft by adding more samples to your signal. You can do this either  
by using a higher sampling rate (I'd recommend to do this if possible)  
or by padding zeroes at the end of the signal (I'd classify this as  
cheating, though).

Apart from that I believe you do not need the line "nfft=  
2^(nextpow2(length(u)))". I modified your code accordingly (see below).

Matthias



---------------------------------------
FS=10e6
T1=1e-6;
T2=2e-6;
t1=0:1/FS:5e-6;
t2=0:0.1e-6:10e-6;
t=[t1,t2+t1(end)];
U0=10;
%charge
u1=U0*(1-exp(-t1/T1));
%discharge
u2=U0*exp(-t2/T2);
u=[u1,u2];
subplot(2,1,1)
plot(t,u,'-+o')
xlabel('Time/s');
ylabel('Voltage/V');
grid('on')
%FFT
% nfft= 2^(nextpow2(length(u)))
fftu = fft(u); % the fft function is smart enough to handle any non- 
power-of-2 length
NumUniquePts = ceil((length(u)+1)/2)
fftu = fftu(1:NumUniquePts);
mu = abs(fftu)/length(u);
mu = mu.^2;
if rem(length(u), 2)
	mu(2:end) = mu(2:end)*2;
else
	mu(2:end -1) = mu(2:end -1)*2;
end
f = (0:NumUniquePts-1)*FS/nfft;
subplot(2,1,2)
plot(f,mu,'-+o')
xlabel('Frequency/Hz');
ylabel('Power');
grid('on')
axis([0,1e6])
print -deps >exponential.eps
---------------------------------------