I have some very short signals from oscilloscope (50k-200k samples) registered over about 2ms time length. Those are acoustic signals with registered signal of a spark of ESD (electrostatic discharge).
I'd like to get some frequency data of that signal, in near-acoustic frequency range (up to about 30kHz) with as high time resolution as possible.
I have tried ploting a spectrogram (specgram in Octave) to view the signal, but the output is not really usefull. Using specgram( x, N, fs );, where x is my signal of fs sampling rate, I receive plot starting at very high frequencies of about 500MHz for low values of N and I get better frequency resolution for big N values (like 2^12-13) but the window is too wide and I receive only 2 spectrum values over whole signal length.
I understand that it may be the limitation of Fourier transform which is probably used by the specgram function (actually, I don't know much about signal analysis).
Is there any other way to get some frequency (as a function of time) information of that kind of signal? I've read something about wavelets, but when I tried using dwt function of signal package, I received this error:
error: 'wfilters' undefined near line 51 column 14
error: called from
dwt at line 51 column 12
Even if this would work, I am not so sure if I'd know how to actually use the output of those wavelet functions ...
To get audio frequency information from such a high sample rate, you will need obtain a sample vector long enough to contain at least a few whole cycles at audio frequencies, e.g. many 10's of milliseconds of contiguous samples, which may or may not be more than your scope can gather. To reasonably process this amount of data, you might low pass filter the sample data to just contain audio frequencies, and then resample it to a lower sample rate, but above twice that filter cut-off frequency. Then you will end up with a much shorter sample vector to feed an FFT for your audio spectrum analysis.
Related
I've been given some digitized sound recordings and asked to plot the sound pressure level per Hz.
The signal is sampled at 40KHz and the units for the y axis are simply volts.
I've been asked to produce a graph of the SPL as dB/Hz vs Hz.
EDIT: The input units are voltage vs time.
Does this make sense? I though SPL was a time domain measure?
If it does make sense how would I go about producing this graph? Apply the dB formula (20 * log10(x) IIRC) and do an FFT on that or...?
What you're describing is a Power Spectral Density. Matlab, for example, has a pwelch function that does literally what you're asking for. To scale to dBSPL/Hz, simply apply 10*log10([psd]) where psd is the output of pwelch. Let me know if you need help with the function inputs.
If you're working with a different framework, let me know which, 100% sure they'll have a version of this function, possibly with a different output format in which case the scaling might be different.
A current source is exciting a Loud by an AC current of ±5mA. The voltage through the loud is measured using the NI data acquisition. The resistance of the loud changes with time, the peak-to-peak amplitude of the voltage signal changes accordingly. How to define the relationship between the loud's resistance and the voltage peak-to-peak amplitude?! in other words, how can I plot the graph of signal's peak-to-peak amplitude as a function of time in LabView?
measure with appropriate Sample-Rate, at least 10 times higher than the max. frequency of your signal.
use a DAQ, that has "synchronous sampling"
measure current and voltage (synchronously with high sample rate). You can use either a shunt or a current transducer for current measurement
Sample in "Blocks". This means: let the DAQ device store e.g. 10k Values (at a Sample rate of 100kHz) in it's internal memory, and read that buffer every 100 ms. Go to the Example finder (Help -> Find exampes) and look for "continous analog measurement" examples.
calculate the RMS-Value of both signals of each block and plot that in a graph. If you want it simple, feed both signals into a "Chart".
if the current is constant (which should be a strait line in the graph!), the voltage should rise over time, when the inner resistance of the loudspeaker rises ...
Note: be aware, that with the example numbers above (100kHz SR, Blocksizte 10k) calculating the RMS value will produce wrong results, when your signal main frequency is below 10 Hz!
According to what I have read on the internet, the normal range of fundamental frequency of female voice is 165 to 255 Hz .
I am using Praat and also python library called Parselmouth to get the fundamental frequency values of female voice in an audio file(.wav). however, I got some values that are over 255Hz(eg: 400+Hz, 500Hz).
Is it normal to get big values like this?
It is possible, but unlikely, if you are trying to capture the fundamental frequency (F0) of a speaking voice. It sounds likely that you are capturing a more easily resonating overtone (e.g. F1 or F2) instead.
My experiments with Praat give me the impression that the with good parameters it will reliably extract F0.
What you'll want to do is to verify that by comparing the pitch curve with a spectrogram. Here's an example of a fitting made by Praat (female speaker):
You can see from the image that
Most prominent frequency seems to be F2
Around 200 Hz seems likely to be F0, since there's only noise below that (compared to before/after the segment)
Praat has calculated a good estimate of F0 for the voiced speech segments
If, after a visual inspection, it seems that you are getting wrong results, you can try to tweak the parameters. Window length greatly affects the frequency resolution.
If you can't capture frequencies this low, you should try increasing the window length - the intuition is that it gives the algorithm a better chance at finding slowly changing periodic features in the data.
I am working in research where we are using smart-phone camera to monitor users heart-rate using color variation as signal.
What I did is getting the red color channel every 0.1 second (10Hz).
The problem is that I am trying to use an FFT to get different frequencies that exist in the extracted signal and I used this Java code where the FFT function get as input two arrays (one for real part and one for img part of complex numbers).
I saw also from this post that I can compute frequencies from the FFT function's results by using the formula:
freq = i * Fs / N
where Fs is the sampling rate and N is the number of points(input).
The problem is that my sampling rate Fs, is too low (10Hz) and if I use above formula I am getting very low frquencies. Is there any other way to get frequencies?
I am using fft function from std.numeric
Complex!double[] resultfft = fft(timeDomainAmplitudeVal);
The parameter timeDomainAmplitudeVal is audio amplitude data. Sample rate 44100 hz and there is 131072(2^16) samples
I am seeing that resultfft has the same size as timeDomainAmplitudeVal(131072) which does not fits my project(also makes no sense) . I need to be able to divide FFT to N equally spaced frequencies. And I need this N to be defined by me .
Is there anyway to implement this with std.numeric.fft or can you have any advices for fft library?
Ps: I will be glad to hear if some DSP libraries exist also
That's just how Fourier transforms work in the practical number-crunching world. Give S samples of signal, get S amplitudes. (Ignoring issues with complex numbers and symmetries.)
If you want N amplitudes, you'll have to interpolate the S-points amplitudes you get from FFT. Your biggest decision is to choose between linear, cubic, truncated sinc, etc.
Altnernative: resample the original audio signal to have your desired N samples in the same overall time interval. Then FFT it.
take a look at pfft, a fast FFT written in D.
http://jerro.github.io/pfft/doc/pfft.pfft.html
or numpy & Pyd
http://docs.scipy.org/doc/numpy/reference/routines.fft.html
http://pyd.dsource.org/
HTH
This is absolutely normal that the FFT gives the same data length.
Here some C++ code to perform windows FFT analysis with overlap and optional "zero-phase" ordering. http://pastebin.com/4YKgbed1
What do FFT coefficients mean?
Question: "OK so I've done the FFT and I'm said I can recover the original signal. Now, what are these coefficients."
Answer: "You can think of coefficient i as representing the phase and amplitude of frequencies from SR*i/(2*N) to SR*(i+1)/(2*N). This is a helpful metaphor. But a more accurate view is that coefficient i is the contribution of a sine of frequency SR*i/(2*N) in a reconstruction of the original input chunk."