making pink noise (1/f) using list of frequencies
I would like to see what type of noise I would get if I used just the frequency in my voice. I created a matlab/octave array using fft to get the [frequency,amplitude,phase] to reproduce my vocal signal.
I would like to take this file/data and use it to create pink noise (1/f). Of course when I use 1/f for the frequency the numbers become very small does anyone have any ideas how to use my own vocal frequencies that I get from doing a fft in matlab to create pink noise (1/f).
Thanks
If I am correct, what you are doing is generating noise based on a 1/f frequency. However if you read the following article: http://en.wikipedia.org/wiki/Pink_noise the frequencies are the same except that the power spectral density is S is proportional to 1/f. Hence you should not be generating noise of frequency 1/f.
I would suggest reading this page for the necessary algorithms.
However if the problem you are facing is that the volume is too low, try amplifying the synthesized noise by multiplying the result by a factor ex.: pinkNoise = pinkNoise * 100
This might do the trick: compute the mean power in your spectrum from the amplitude A = A(f), where f is the frequency.
P = mean(A.^2);
Spread that over your frequency range:
N = length(f);
invfnorm = 1./[1:N];
Anew = sqrt(P*invfnorm/sum(invfnorm));
Anew has the property of having the same total power density as the original spectrum, and power decaying as 1/f.
Substitute Anew for A and inverse FFT your new spectrum to generate the new waveform.
Related
I am currently working on a project for my Speech Processing course and have just finished making a time waveform plot as well as both wide/narrow band spectrograms for a spoken word in Spanish (aire).
The next part of the project is as follows:
Make a 3-D plot of each word signal, as a function of time, frequency and power spectral density. The analysis time step should be 20ms, and power density should be computed using a 75%-overlapped Hamming window and the FFT. Choose a viewing angle that best highlights the signal features as they change in time and frequency.
I was hoping that someone can offer me some guidance as to how to begin doing this part. I have started by looking here under the Spectrogram and Instantaneous Frequency heading but was unsure of how to add PSD to the script.
Thanks
I am going to give you an example.
I am going to generate a linear chirp signal.
Fs = 1000;
t = 0:1/Fs:2;
y = chirp(t,100,2,300,'linear');
And then, I am going to define number of fft and hamming window.
nfft=128;
win=hamming(nfft);
And then I am going to define length of overlap, 75% of nfft.
nOvl=nfft*0.75;
And then, I am performing STFT by using spectrogram function.
[s,f,t,pxx] = spectrogram(y,win,nOvl,nfft,Fs,'psd');
'y' is time signal, 'win' is defined hamming window, 'nOvl' is number of overlap, 'nfft' is number of fft, 'Fs' is sampling frequency, and 'psd' makes the result,pxx, as power spectral density.
Finally, I am going to plot the 'pxx' by using waterfall graph.
waterfall(f,t,pxx')
xlabel('frequency(Hz)')
ylabel('time(sec)')
zlabel('PSD')
The length of FFT, corresponding to 20ms, depends on sampling frequency of your signal.
EDIT : In plotting waterfall graph, I transposed pxx to change t and f axis.
I was studying for a signals & systems project and I have come across this code on high and low pass filters for an audio signal on the internet. Now I have tested this code and it works but I really don't understand how it is doing the low/high pass action.
The logic is that a sound is read into MATLAB by using the audioread or wavread function and the audio is stored as an nx2 matrix. The n depends on the sampling rate and the 2 columns are due to the 2 sterio channels.
Now here is the code for the low pass;
[hootie,fs]=wavread('hootie.wav'); % loads Hootie
out=hootie;
for n=2:length(hootie)
out(n,1)=.9*out(n-1,1)+hootie(n,1); % left
out(n,2)=.9*out(n-1,2)+hootie(n,2); % right
end
And this is for the high pass;
out=hootie;
for n=2:length(hootie)
out(n,1)=hootie(n,1)-hootie(n-1,1); % left
out(n,2)=hootie(n,2)-hootie(n-1,2); % right
end
I would really like to know how this produces the filtering effect since this is making no sense to me yet it works. Also shouldn't there be any cutoff points in these filters ?
The frequency response for a filter can be roughly estimated using a pole-zero plot. How this works can be found on the internet, for example in this link. The filter can be for example be a so called Finite Impulse Response (FIR) filter, or an Infinite Impulse Response (IIR) filter. The FIR-filters properties is determined only from the input signal (no feedback, open loop), while the IIR-filter uses the previous signal output to control the current signal output (feedback loop or closed loop). The general equation can be written like,
a_0*y(n)+a_1*y(n-1)+... = b_0*x(n)+ b_1*x(n-1)+...
Applying the discrete fourier transform you may define a filter H(z) = X(z)/Y(Z) using the fact that it is possible to define a filter H(z) so that Y(Z)=H(Z)*X(Z). Note that I skip a lot of steps here to cut down this text to proper length.
The point of the discussion is that these discrete poles can be mapped in a pole-zero plot. The pole-zero plot for digital filters plots the poles and zeros in a diagram where the normalized frequencies, relative to the sampling frequencies are illustrated by the unit circle, where fs/2 is located at 180 degrees( eg. a frequency fs/8 will be defined as the polar coordinate (r, phi)=(1,pi/4) ). The "zeros" are then the nominator polynom A(z) and the poles are defined by the denominator polynom B(z). A frequency close to a zero will have an attenuation at that frequency. A frequency close to a pole will instead have a high amplifictation at that frequency instead. Further, frequencies far from a pole is attenuated and frequencies far from a zero is amplified.
For your highpass filter you have a polynom,
y(n)=x(n)-x(n-1),
for each channel. This is transformed and it is possble to create a filter,
H(z) = 1 - z^(-1)
For your lowpass filter the equation instead looks like this,
y(n) - y(n-1) = x(n),
which becomes the filter
H(z) = 1/( 1-0.9*z^(-1) ).
Placing these filters in the pole-zero plot you will have the zero in the highpass filter on the positive x-axis. This means that you will have high attenuation for low frequencies and high amplification for high frequencies. The pole in the lowpass filter will also be loccated on the positive x-axis and will thus amplify low frequencies and attenuate high frequencies.
This description is best illustrated with images, which is why I recommend you to follow my links. Good luck and please comment ask if anything is unclear.
I am working with biological signal data, and am trying to count the number of regions with a high density of high amplitude peaks. As seen in the figure below, the regions of interest (as observed qualitatively) are contained in red boxes and 8 such regions were observed for this particular trial. The goal is to mathematically achieve this same result in near real time without the intervention or observation of the researcher.
The data seen plotted below is the result of raw data from a 24-bit ADC being processed by an FIR filter, with no other processing yet being done.
What I am looking for is a method, or ideally code, to help me detect such regions as identified while subsequently ignoring some of the higher amplitude peaks in between the regions of interest (i.e. between regions 3 and 4, 5 and 6, or 7 and 8 there is a narrow region of high amplitude which is not of concern). It is worth noting that the maximum is not known prior to computation.
Thanks for your help.
Data
https://www.dropbox.com/s/oejyy6tpf5iti3j/FIRData.mat
can you work with thresholds?
define:
(1) "amplitude threshold": if the signal is greater than the threshold it is considered a peak
(2) "window size" : of a fixed time duration
algorithm:
if n number of peaks was detected in a duration defined in "window size" than consider the signal within "window size" as cluster of peaks.(I worked with eye blink eeg data this way before, not sure if it is suitable for your application)
P.S. if you have data that are already labelled by human, you can train a classifier to find out your thresholds and window size.
Does it make sense in your problem to have some sort of "window size"? In other words, given a region of "high" amplitude, if you shrink the duration of the region, at what point will it become meaningless to your analysis?
If you can come up with a window, just apply this window to your data as it comes in and compute the energy within the window. Then, you can define some energy threshold and perform simple peak detection on the energy signal.
By inspection of your data, the regions with high amplitude peaks are repeated at what appears to be fairly uniform intervals. This suggests that you might fit a sine or cosine wave (or a combination of the two) to your data.
Excuse my crude sketch but what I mean is something like this:
Once you make this identification, you can use the FFT to get the dominant spatial frequencies. Keep in mind that the spatial frequency spectrum of your signal may be fairly complex, due to spurious data, but what you are after is one or two dominant frequencies of your data.
For example, I made up a sinusoid and you can do the calculation like this:
N = 255; % # of samples
x = linspace(-1/2, 1/2, N);
dx = x(2)-x(1);
nu = 8; % frequency in cycles/interval
vx = (1/(dx))*[-(N-1)/2:(N-1)/2]/N; % spatial frequency
y = sin(2*pi*nu*x); % this would be your data
F = fftshift(abs(fft(y))/N);
figure; set(gcf,'Color',[1 1 1]);
subplot(2,1,1);plot(x,y,'-b.'); grid on; xlabel('x'); grid on;
subplot(2,1,2);plot(vx,F,'-k.'); axis([-1.3*nu 1.3*nu 0 0.6]); xlabel('frequency'); grid on;
Which gives:
Note the peaks at ± nu, the dominant spatial frequency. Now once you have the dominant spatial frequencies you can reconstruct the sine wave using the frequencies that you have obtained from the FFT.
Finally, once you have your sine wave you can identify the boxes with centers at the peaks of the sine waves.
This is also a nice approach because it effectively filters out the spurious or less relevant spikes, helping you to properly place the boxes at your intended locations.
Since I don't have your data, I wasn't able to complete all of the code for you, but the idea is sound and you should be able to proceed from this point.
I have 2 arrays of 800000 input and output data samples of a system. The system in a kind of oven that works among 0 and 10 volts. The sample time is 0.001s.
I have to identify the model of this system, but first of all, given that the data are clearly dirty, I would like to filter the noise.
How can I do it with the System Identification Toolbox of Matlab?
Moreover, how can I estimate the cutoff frequency to remove the noise?
Thank you in advance.
PS: given that this is a bit out of topic, please, post your answer here thank you.
The cutoff frequency is directly given by you sampling time or sampling frequency.
you sampling frequency is 1/(sampling time) and must at least 2 the factor of the highest frequency of interest:
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
f_s = 1/T_s >= 2*f_cutOff
You can then simply to same frequency domain processing in the case you sampling frequency is realy high enough. The easiest way would to have a look at the frequency domain (with function fft() ). And check first where you have high noise components. Then filter out these components (zeroing) and then transform it back into time domain ( with function ifft() ).
Noise is modeled as a white Gaussian distribution in the simplest case. If you estimate the noise energy, you can make a dummy noise by calling
noise = A*randn(1,N);
Here, A is the amplitude and N is the sample count. then just take the fft of this signal and subtract it from the fft of input signal and take the inverse fft (ifft)
I am trying to prove that the white noise has constant power spectral density using matlab
but the amplitude of the spectrum looks like random amplitude.
can anyone tell me why?
here is my code.
noise = randn(1,10000);
fft_noise=fft(noise);
plot(abs(fft_noise(1:5000)))
thanks.
You need to average a bunch (law of large numbers) of FFTs of white noise to approach the average power spectral density.
If you take the FFT of an independent set of random variables from the same distribution, then you'll get an independent set of random variables from the same distribution since the inverse Fourier transform is (more or less) the same as the Fourier transform. The point is that the expected value for each frequency is the same.
you need to multiple the fft by the complex conjugate of the fft to show a flat PSD. i.e. change
fft_noise=fft(noise);
to
fft_noise=fft(noise).*conj(fft(noise));