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.
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
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Jeremy
2013-10-28
编辑:Jeremy
2013-10-28
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
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
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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