How to choose Spectrogram parameter ?
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I have a signal with 3Hz frequency and runs from 0:10 second. The signal is zeroed everywhere except from 2 to 4 and 7 to 8 second in the signal as per the image below. I tried to to get the spectrogram but it didn't give correct representation and accurate one. When i surf the spectrogram i can see the 3 signal but the time is shifted. How to choose the correct parameters for spectrogram ? My code:
t=0:1/50:10;
x=sin(2*pi*t*3);
x[1:100]=0;
x[200:350]=0;
x[400:501]=0;
[s,ff,tt,p]=spectrogram(x,50,25,2048,50);
surf(tt,ff,(p),'edgecolor','none'); axis tight; view(0,90);
Thank you
1 个评论
omar thamer
2013-10-28
another question: Below is the spectrogram for my signal. Ohnestly I can't do the interpretation. This signal is different and it runs from (1:1:515 Sec) and fs=1 and max freq is 0.5. the signal is triangular and have a mean of near zero.
采纳的回答
Where you set it to zero, you should be using parentheses and not brackets. not really sure what you were plotting since that syntax should just throw an error.
This should create what you would expect to see, two pulses at around 3 Hz
t=0:1/50:10;
x=sin(2*pi*t*3);
x(1:100)=0;
x(200:350)=0;
x(400:501)=0;
[s,ff,tt,p]=spectrogram(x,50,25,2048,50);
surf(tt,ff,(p),'edgecolor','none'); axis tight; view(0,90);
更多回答（1 个）
Wayne King
2013-10-29
0 个投票
NFFT is based on the length of the segment, not the length of the signal. Choosing the segment length is the most important parameter in the spectrogram because that determines and fixes your frequency resolution. Picking a value of NFFT greater than the segment length only provides an interpolation of the DFT estimates at the fundamental (Fourier) frequencies, it does not improve your frequency resolution.
4 个评论
omar thamer
2013-10-29
Jeremy
2013-10-29
the nfft is not a function of the sampling rate, the results of any Fourier transform will always be over a frequency range from 0 to the 1/2 the sampling rate (nyquist frequency). The result of the spectrogram in my example will always run from 0 to 25 Hz. the nfft determines the bin width, or the number of steps between 0 and the nyquist frequency. the bin width is equal to the sampling rate/nfft. In the case of a spectrogram you will need to balance the need for time vs. frequency resolution for the nfft you use. an nfft equal to the length of your time history will give the greatest frequency resolution, but only one time step...
This example makes the inputs a little more clear. you can increase the last nfft to smooth out the display, but it just zero padding so it is not really increasing you frequency resolution.
fs=50; t=1/fs:1/fs:10; x=sin(2*pi*t*3); x(1:100)=0; x(200:350)=0; x(400:501)=0; nfft=100; [s,ff,tt,p]=spectrogram(x,hanning(nfft),nfft/2,nfft*2,fs); surf(tt,ff,(p),'edgecolor','none'); axis tight;view(0,90)
omar thamer
2013-10-29
Jeremy
2013-10-29
your spectrogram is very odd. it would appear there is something like a burst of a square wave with a consistent duration starting at random intervals. I'm not sure about the blue stripes, they are are consistent over time when there is no signal, they might be side lobes due to zero padding. There is also a DC component to the signal; it is not zero when there is no square wave.
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