There are not many easy-to-follow Perlin Noise tutorials out there and certainly not in Java or 2D. I followed this tutorial to a point but it doesn't explain 2D noise very much at all. I know you have to generate an array of numbers then interpolate them and everything. My problem is that I do not know how to implement frequency, persistence, or amplitude to help affect the outcome of the numbers. Can anyone give me some basic Perlin Noise functions or a link to a 2D Perlin Noise tutorial in Java or similar languages? Thanks!
EDIT: Can someone just briefly explain the process at least or how one implements the frequency, amplitude, and persistence to influence the generation? Please :)
Amplitude and frequency are no free variables in the Perlin Noise generation. Instead they are parametrized by something called persistence.
The noise function is then the sum over several basic functions.
n(x) = sum( n_i(x*f_i) * a_i, i=0..N-1)
Each function is called octave and therefore numbered by the index i. The values f_i denote the frequencies and a_i the amplitudes. As mentioned before they are completely determined by the index i itself, parametrized by the persistence p:
f_i = 2^i
a_i = p^i
While each noise function n_i(x) is normalized for frequency 1 and amplitude 1, the overall term n_i(x*f_i) * a_i now has frequency and amplitude given by the expressions above.
In other words the noise function n(x) is the sum of octaves where the first one has frequency 1 and amplitude 1, the second one has frequency 2 and amplitude p, the third has frequency 4 and amplitude p^2, and so on.
Related
While trying to understand Fast Fourier Transform I encountered a problem with the phase. I have broken it down to the simple code below. Calculating one period of a 50Hz sinewave, and applying an fft algorithm:
fs = 1600;
dt = 1/fs;
L = 32;
t=(0:L-1)*dt;
signal = sin(t/0.02*2*pi);
Y = fft(signal);
myAmplitude = abs(Y)/L *2 ;
myAngle = angle(Y);
Amplitude_at_50Hz = myAmplitude(2);
Phase_at_50Hz = myAngle(2);
While the amplitude is ok, I don't understand the phase result. Why do I get -pi/2 ? As there is only one pure sinewave, I expected the phase to be 0. Either my math is wrong, or my use of Matlab, or both of them... (A homemade fft gives me the same result. So I guess I am stumbling over my math.)
There is a similar post here: MATLAB FFT Phase plot. However, the suggested 'unwrap' command doesn't solve my problem.
Thanks and best regards,
DanK
The default waveform for an FFT phase angle of zero is a cosine wave which starts and ends in the FFT window at 1.0 (not a sinewave which starts and ends in the FFT window at 0.0, or at its zero crossings.) This is because the common nomenclature is to call the cosine function components of the FFT basis vectors (the complex exponentials) the "real" components. The sine function basis components are called "imaginary", and thus infer a non-zero complex phase.
That is what it should be. If you used cosine, you would have found a phase of zero.
Ignoring numerical Fourier transforms for a moment and taking a good old Fourier transform of sin(x), which I am too lazy to walk through, we get a pair of purely imaginary deltas.
As for an intuitive reason, recall that a discrete Fourier transform is averaging a bunch of points along a curve in the complex plane while turning at the angular frequency of the bin you're computing and using the amplitude corresponding to the sample. If you sample a sine curve while turning at its own frequency, the shape you get is a circle centered on the imaginary axis (see below). The average of that is of course going to be right on the imaginary axis.
Plot made with wolfram alpha.
Fourier transform of a sine function such as A*sin((2*pi*f)*t) where f is the frequency will yield 2 impulses of magnitude A/2 in the frequency domain at +f and -f where the associated phases are -pi/2 and pi/2 respectively.
You can take a look at its proof here:
http://mathworld.wolfram.com/FourierTransformSine.html
So the code is working fine.
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.
let us assume,
I have a vector t with the times in seconds of my samples. (These samples are not equally distributed on the time domain.
Also I have a vector data containing the samplevalues at the time t.
t and data have the same length.
If I plot the graph some sort of periodical signal is obtained.
now I could perform: abs(fft(data)) to get my spectrum, which is then plotted over the amount of data points on the x-axis.
How can I obtain my spectrum regarding the times in vector t and plot it?
I want to see which frequencies in 1/s or which period in s my signal contains.
Thanks for your help.
[Not the OP's intention]: FFT will give you the spectrum (global) for any number of input data points. You cannot have a specific data point (in time) associated with parts (or the full) spectrum.
What you can do instead is use spectrogram and obtain the Short-Time Fourier Transform (STFT). This will give you a NxM discrete grid of time-frequency FT values (N: FT frequency bins, M: signal time-windows).
By localizing the (overlapping) STFT windows on your data samples of interest you will get N frequency magnitude values, thus the distribution of short-term spectrum estimates as the signal changes in time.
See also the possibly relevant answer here: https://stackoverflow.com/a/12085728/651951
EDIT/UPDATE:
For unevenly spaced data you need to consider the Non-Uniform DFT (and Non-uniform FFT implementations). See the relevant question/answer here https://scicomp.stackexchange.com/q/593
The primary approaches for NFFT or NUFFT, are based on creating a uniform grid through local convolutions/interpolation, running FFT on this and undoing the convolutional effect of the interpolation filter.
You can read more:
A. Dutt and V. Rokhlin, Fast Fourier transforms for nonequispaced data, SIAM J. Sci. Comput., 14, 1993.
L. Greengard and J.-Y. Lee, Accelerating the Nonuniform Fast Fourier Transform, SIAM Review, 46 (3), 2004.
Pippig, M. und Potts, D., Particle Simulation Based on Nonequispaced Fast Fourier Transforms, in: Fast Methods for Long-Range Interactions in Complex Systems, 2011.
For an implementation (with an interface to MATLAB) try NFFT and possibly its parallelized version PNFFT. You may find a nice walk-through on how to set-up and use here.
You can resample or interpolate your sample points to get another set of sample points that are equally spaced in t. The chosen spacing or sample rate of the second set of equally spaced sample points will allow you to infer frequencies to the result of an FFT of that second set.
The results may be noisy or include aliasing unless the initial data set is bandlimited to a sufficiently low frequency to allow interpolation. If bandlimited, then you might try something like cubic splines as an interpolation method.
Although it may look like one can get a high FFT bin frequency resolution by resampling to a larger number of data points, the actual useful resolution accuracy will be more related to the original number of samples.
I am wondering if I am using Fourier Transformation in MATLAB the right way. I want to have all the average amplitudes for frequencies in a song. For testing purposes I am using a free mp3 download of Beethovens "For Elise" which I converted to a 8 kHz mono wave file using Audacity.
My MATLAB code is as follows:
clear all % be careful
% load file
% Für Elise Recording by Valentina Lisitsa
% from http://www.forelise.com/recordings/valentina_lisitsa
% Converted to 8 kHz mono using Audacity
allSamples = wavread('fur_elise_valentina_lisitsa_8khz_mono.wav');
% apply windowing function
w = hanning(length(allSamples));
allSamples = allSamples.*w;
% FFT needs input of length 2^x
NFFT = 2^nextpow2(length(allSamples))
% Apply FFT
fftBuckets=fft(allSamples, NFFT);
fftBuckets=fftBuckets(1:(NFFT/2+1)); % because of symetric/mirrored values
% calculate single side amplitude spectrum,
% normalize by dividing by NFFT to get the
% popular way of displaying amplitudes
% in a range of 0 to 1
fftBuckets = (2*abs(fftBuckets))/NFFT;
% plot it: max possible frequency is 4000, because sampling rate of input
% is 8000 Hz
x = linspace(1,4000,length(fftBuckets));
bar(x,fftBuckets);
The output then looks like this:
Can somebody please tell me if my code is correct? I am especially wondering about the peaks around 0.
For normalizing, do I have to divide by NFFT or length(allSamples)?
For me this doesn't really look like a bar chart, but I guess this is due to the many values I am plotting?
Thanks for any hints!
Depends on your definition of "correct". This is doing what you intended, I think, but it's probably not very useful. I would suggest using a 2D spectrogram instead, as you'll get time-localized information on frequency content.
There is no one correct way of normalising FFT output; there are various different conventions (see e.g. the discussion here). The comment in your code says that you want a range of 0 to 1; if your input values are in the range -1 to 1, then dividing by number of bins will achieve that.
Well, exactly!
I would also recommend plotting the y-axis on a logarithmic scale (in decibels), as that's roughly how the human ear interprets loudness.
Two things that jump out at me:
I'm not sure why you are including the DC (index = 1) component in your plot. Not a big deal, but of course that bin contains no frequency data
I think that dividing by length(allSamples) is more likely to be correct than dividing by NFFT. The reason is that if you want the DC component to be equal to the mean of the input data, dividing by length(allSamples) is the right thing to do.
However, like Oli said, you can't really say what the "correct" normalization is until you know exactly what you are trying to calculate. I tend to use FFTs to estimate power spectra, so I want units like "DAC / rt-Hz", which would lead to a different normalization than if you wanted something like "DAC / Hz".
Ultimately there's no substitute for thinking about exacty what you want to get out of the FFT (including units), and working out for yourself what the correct normalization should be (starting from the definition of the FFT if necessary).
You should also be aware that MATLAB's fft has no requirement to use an array length that is a power of 2 (though doing so will presumably lead to the FFT running faster). Because zero-padding will introduce some ringing, you need to think about whether it is the right thing to do for your application.
Finally, if a periodogram / power spectrum is really what you want, MATLAB provides functions like periodogram, pwelch and others that may be helpful.
1) Besides the negative frequencies, which is the minimum frequency provided by the FFT function? Is it zero?
2) If it is zero how do we plot zero on a logarithmic scale?
3) The result is always symmetrical? Or it just appears to be symmetrical?
4) If I use abs(fft(y)) to compare 2 signals, may I lose some accuracy?
1) Besides the negative frequencies, which is the minimum frequency provided by the FFT function? Is it zero?
fft(y) returns a vector with the 0-th to (N-1)-th samples of the DFT of y, where y(t) should be thought of as defined on 0 ... N-1 (hence, the 'periodic repetition' of y(t) can be thought of as a periodic signal defined over Z).
The first sample of fft(y) corresponds to the frequency 0.
The Fourier transform of real, discrete-time, periodic signals has also discrete domain, and it is periodic and Hermitian (see below). Hence, the transform for negative frequencies is the conjugate of the corresponding samples for positive frequencies.
For example, if you interpret (the periodic repetition of) y as a periodic real signal defined over Z (sampling period == 1), then the domain of fft(y) should be interpreted as N equispaced points 0, 2π/N ... 2π(N-1)/N. The samples of the transform at the negative frequencies -π ... -π/N are the conjugates of the samples at frequencies π ... π/N, and are equal to the samples at frequencies
π ... 2π(N-1)/N.
2) If it is zero how do we plot zero on a logarithmic scale?
If you want to draw some sort of Bode plot you may plot the transform only for positive frequencies, ignoring the samples corresponding to the lowest frequencies (in particular 0).
3) The result is always symmetrical? Or it just appears to be symmetrical?
It has Hermitian symmetry if y is real: Its real part is symmetric, its imaginary part is anti-symmetric. Stated another way, its amplitude is symmetric and its phase anti-symmetric.
4) If I use abs(fft(y)) to compare 2 signals, may I lose some accuracy?
If you mean abs(fft(x - y)), this is OK and you can use it to get an idea of the frequency distribution of the difference (or error, if x is an estimate of y). If you mean abs(fft(x)) - abs(fft(y)) (???) you lose at least phase information.
Well, if you want to understand the Fast Fourier Transform, you want to go back to the basics and understand the DFT itself. But, that's not what you asked, so I'll just suggest you do that in your own time :)
But, in answer to your questions:
Yes, (excepting negatives, as you said) it is zero. The range is 0 to (N-1) for an N-point input.
In MATLAB? I'm not sure I understand your question - plot zero values as you would any other value... Though, as rightly pointed out by duffymo, there is no natural log of zero.
It's essentially similar to a sinc (sine cardinal) function. It won't necessarily be symmetrical, though.
You won't lose any accuracy, you'll just have the magnitude response (but I guess you knew that already).
Consulting "Numerical Recipes in C", Chapter 12 on "Fast Fourier Transform" says:
The frequency ranges from negative fc to positive fc, where fc is the Nyquist critical frequency, which is equal to 1/(2*delta), where delta is the sampling interval. So frequencies can certainly be negative.
You can't plot something that doesn't exist. There is no natural log of zero. You'll either plot frequency as the x-axis or choose a range that doesn't include zero for your semi-log axis.
The presence or lack of symmetry in the frequency range depends on the nature of the function in the time domain. You can have a plot in the frequency domain that is not symmetric about the y-axis.
I don't think that taking the absolute value like that is a good idea. You'll want to read a great deal more about convolution, correction, and signal processing to compare two signals.
result of fft can be 0. already answered by other people.
to plot 0 frequency, the trick is to set it to a very small positive number (I use exp(-15) for that purpose).
already answered by other people.
if you are only interested in the magnitude, yes you can do that. this is applicable, say, in many image processing problems.
Half your question:
3) The results of the FFT operation depend on the nature of the signal; hence there's nothing requiring that it be symmetrical, although if it is you may get some more information about the properties of the signal
4) That will compare the magnitudes of a pair of signals, but those being equal do no guarantee that the FFTs are identical (don't forget about phase). It may, however, be enough for your purposes, but you should be sure of that.