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!
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.
I am using fmcw radar to find out distance and speed of moving objects using stm32l476 microcontroller. I transmit the modulation signal as sawtooth waveform and I read the recieved signal in the digital form using ADC function available. Then, I copy this recieved ADC data into fft_in array(converting it into float32_t)(fft_in array size = 512). After copying this fft_in array, I apply fft on this array and process it for finding out range of the object. Until here everything works fine.
Now, in order to find velocity of the object, first, I copy this arrays(fft_in) as rows of the matrix for 64 chirps(Matrix size[64][512]). Then, I take Peak range bin column and apply fft for this column array. So while processing this column array by applying fft, its length reduce to half[32 elements]. Then finding out peak value bin multiplied by frequnecy resolution gives the phase differnce 'w' from which velocity can be calculated as "𝐯=𝛌𝛚/𝟒𝛑𝐓 𝐜".
while running this algorithm, I find that when object is stationery, I get peak value at 22th element(out of 32 elements). what does this imply?
I have sampling frequency for ADC as 24502hz. So per bin value for range estimation is 47.8566hz (24502/512).
I have 64 chirps and Tc is 0.006325s. So 1/0.006325 gives 158.10Hz.What would be per velocity bin resolution, Is it 2.47Hz(158.10/64)? I have bit confusion in this concept.How does 2nd fft works for finding out velocity in fmcw radar?
Infineon has excellent resources on this topic, see this FAQ for the basics: https://www.infineon.com/dgdl/Infineon-Radar%20FAQ-PI-v02_00-EN.pdf?fileId=5546d46266f85d6301671c76d2a00614
If you want to know more details, check out the P2G Software User Manual:
https://www.infineon.com/dgdl/Infineon-P2G_Software_User_Manual-UserManual-v01_01-EN.pdf?fileId=5546d4627762291e017769040a233324 (Chapter 4)
There is even the software available with all the algorithms (including FMCW). How to get the software with the "Infineon Toolbox" is described here: https://www.mouser.com/pdfdocs/Infineon_Position2Go_QS.pdf
Some hints from me:
I suggest applying a window function before the fft https://en.wikipedia.org/wiki/Window_function and remove the mean.
Read about frequency mixers https://en.wikipedia.org/wiki/Frequency_mixer
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 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.
I'm able to read a wav files and its values. I need to find peaks and pits positions and their values. First time, i tried to smooth it by (i-1 + i + i +1) / 3 formula then searching on array as array[i-1] > array[i] & direction == 'up' --> pits style solution but because of noise and other reasons of future calculations of project, I'm tring to find better working area. Since couple days, I'm researching FFT. As my understanding, fft translates the audio files to series of sines and cosines. After fft operation the given values is a0's and a1's for a0 + ak * cos(k*x) + bk * sin(k*x) which k++ and x++ as this picture
http://zone.ni.com/images/reference/en-XX/help/371361E-01/loc_eps_sigadd3freqcomp.gif
My question is, does fft helps to me find peaks and pits on audio? Does anybody has a experience for this kind of problems?
It depends on exactly what you are trying to do, which you haven't really made clear. "finding the peaks and pits" is one thing, but since there might be various reasons for doing this there might be various methods. You already tried the straightforward thing of actually looking for the local maximum and minima, it sounds like. Here are some tips:
you do not need the FFT.
audio data usually swings above and below zero (there are exceptions, including 8-bit wavs, which are unsigned, but these are exceptions), so you must be aware of positive and negative values. Generally, large positive and large negative values carry large amounts of energy, though, so you want to count those as the same.
due to #2, if you want to average, you might want to take the average of the absolute value, or more commonly, the average of the square. Once you find the average of the squares, take the square root of that value and this gives the RMS, which is related to the power of the signal, so you might do something like this is you are trying to indicate signal loudness, intensity or approximate an analog meter. The average of absolutes may be more robust against extreme values, but is less commonly used.
another approach is to simply look for the peak of the absolute value over some number of samples, this is commonly done when drawing waveforms, and for digital "peak" meters. It makes less sense to look at the minimum absolute.
Once you've done something like the above, yes you may want to compute the log of the value you've found in order to display the signal in dB, but make sure you use the right formula. 10 * log_10( amplitude ) is not it. Rule of thumb: usually when computing logs from amplitude you will see a 20, not a 10. If you want to compute dBFS (the amount of "headroom" before clipping, which is the standard measurement for digital meters), the formula is -20 * log_10( |amplitude| ), where amplitude is normalize to +/- 1. Watch out for amplitude = 0, which gives an infinite headroom in dB.
If I understand you correctly, you just want to estimate the relative loudness/quietness of an audio digital sample at a given point.
For this estimation, you don't need to use FFT. However your method of averaging the signal does not produce the appropiate picture neither.
The digital signal is the value of the audio wave at a given moment. You need to find the overall amplitude of the signal at that given moment. You can somewhat see it as the local maximum value for a given interval around the moment you want to calculate. You may have a moving max for the signal and get your amplitude estimation.
At a 16 bit sound sample, the sound signal value can go from 0 up to 32767. At a 44.1 kHz sample rate, you can find peaks and pits of around 0.01 secs by finding the max value of 441 samples around a given t moment.
max=1;
for (i=0; i<441; i++) if (array[t*44100+i]>max) max=array[t*44100+i];
then for representing it on a 0 to 1 scale you (not really 0, because we used a minimum of 1)
amplitude = max / 32767;
or you might represent it in relative dB logarithmic scale (here you see why we used 1 for the minimum value)
dB = 20 * log10(amplitude);
all you need to do is take dy/dx, which can getapproximately by just scanning through the wave and and subtracting the previous value from the current one and look at where it goes to zero or changes from positive to negative
in this code I made it really brief and unintelligent for sake of brevity, of course you could handle cases of dy being zero better, find the 'centre' of a long section of a flat peak, that kind of thing. But if all you need is basic peaks and troughs, this will find them.
lastY=0;
bool goingup=true;
for( i=0; i < wave.length; i++ ) {
y = wave[i];
dy = y - lastY;
bool stillgoingup = (dy>0);
if( goingup != direction ) {
// changed direction - note value of i(place) and 'y'(height)
stillgoingup = goingup;
}
}