A small program that I wrote in octave does not yield desired phase spectrum. The magnitude plot is perfect though.
f = 200;
fs = 1000;
phase = pi/3;
t = 0: 1/fs: 1;
sig = sin((2*pi*f*t) + phase);
sig_fft = fft(sig);
sig_fft_phase = angle(sig_fft) * 180/pi;
sig_fft_phase(201)
sig_fft_phase(201) returns 5.998 (6 degrees) rather than 60 degrees. What am I doing wrong? Is my expectation incorrect?
In your example, if you generate the frequency axis: (sorry, I don’t have Octave here, so Python will have to do—I’m sure it’s the same in Octave):
faxis = (np.arange(0, t.size) / t.size) * fs
you’ll see that faxis[200] (Python is 0-indexed, equivalent to Octave’s 201 index) is 199.80019980019981. You think you’re asking for the phase at 200 Hz but you’re not, you’re asking for the phase of 199.8 Hz.
(This happens because your t vector includes 1.0—that one extra sample slightly decreases the spectral spacing! I don’t think the link #Sardar_Usama posted in their comment is correct—it has nothing to do with the fact that the sinusoid doesn’t end on a complete cycle, since this approach should work with incomplete cycles.)
The solution: zero-pad the 1001-long sig vector to 2000 samples. Then, with a new faxis frequency vector, faxis[400] (Octave’s 401st index) corresponds to exactly 200 Hz:
In [54]: sig_fft = fft.fft(sig, 2000);
In [55]: faxis = np.arange(0, sig_fft.size) / sig_fft.size * fs
In [56]: faxis[400]
Out[56]: 200.0
In [57]: np.angle(sig_fft[400]) * 180 / np.pi
Out[57]: -29.950454729683386
But oh no, what happened? This says the angle is -30°?
Well, recall that Euler’s formula says that sin(x) = (exp(i * x) - exp(-i * x)) / 2i. That i in the denominator means that the phase recovered by the FFT won’t be 60°, even though the input sine wave has phase of 60°. Instead, the FFT bin’s phase will be 60 - 90 degrees, since -90° = angle(1/i) = angle(-i). So this is actually the right answer! To recover the sine wave’s phase, you’d need to add 90° to the phase of the FFT bin.
So to summarize, you need to fix two things:
make sure you’re looking at the right frequency bin. For an N-point FFT (and no fftshift), the bins are [0 : N - 1] / N * fs. Above, we just used a N=2000 point FFT to ensure that 200 Hz be represented.
Understand that, although you have a sine wave, as far as the FFT is concerned, it gets two complex exponentials, at +200 and -200 Hz, and with amplitudes 1/(2i) and -1/(2i). That imaginary value in the denominators shifts the phase you expect by -90° and +90° respectively.
If you happened to have used cos, a cosine wave, for sig, you wouldn’t have run into this mathematical obstacle 😆, so pay attention to the difference between sin and cos in the future 💪!
change to t=0:1/fs:1-1/fs; then
sig_fft_phase(201)
ans = -30.000
Related
I am a student working with time-series data which we feed into a neural network for classification (my task is to build and train this NN).
We're told to use a band-pass filter of 10 Hz to 150 Hz since anything outside that is not interesting.
After applying the band-pass, I've also down-sampled the data to 300 samples per second (originally it was 768 Hz). My understanding of the Shannon Nyquist sampling theorem is that, after applying the band-pass, any information in the data will be perfectly preserved at this sample-rate.
However, I got into a discussion with my supervisor who claimed that 300 Hz might not be sufficient even if the signal was band-limited. She says that it is only the minimum sample rate, not necessarily the best sample rate.
My understanding of the sampling theorem makes me think the supervisor is obviously wrong, but I don't want to argue with my supervisor, especially in case I'm actually the one who has misunderstood.
Can anyone help to confirm my understanding or provide some clarification? And how should I take this up with my supervisor (if at all).
The Nyquist-Shannon theorem states that the sampling frequency should at-least be twice of bandwidth, i.e.,
fs > 2B
So, this is the minimal criteria. If the sampling frequency is less than 2B then there will be aliasing. There is no upper limit on sampling frequency, but more the sampling frequency, the better will be the reconstruction.
So, I think your supervisor is right in saying that it is the minimal condition and not the best one.
Actually, you and your supervisor are both wrong. The minimum sampling rate required to faithfully represent a real-valued time series whose spectrum lies between 10 Hz and 150 Hz is 140 Hz, not 300 Hz. I'll explain this, and then I'll explain some of the context that shows why you might want to "oversample", as it is referred to (spoiler alert: Bailian-Low Theorem). The supervisor is mixing folklore into the discussion, and when folklore is not properly-contexted, it tends to telephone tag into fakelore. (That's a common failing even in the peer-reviewed literature, by the way). And there's a lot of fakelore, here, that needs to be defogged.
For the following, I will use the following conventions.
There's no math layout on Stack Overflow (except what we already have with UTF-8), so ...
a^b denotes a raised to the power b.
∫_I (⋯x⋯) dx denotes an integral of (⋯x⋯) taken over all x ∈ I, with the default I = ℝ.
The support supp φ (or supp_x φ(x) to make the "x" explicit) of a function φ(x) is the smallest closed set containing all the x-es for which φ(x) ≠ 0. For regularly-behaving (e.g. continuously differentiable) functions that means a union of closed intervals and/or half-rays or the whole real line, itself. This figures centrally in the Shannon-Nyquist sampling theorem, as its main condition is that a spectrum have bounded support; i.e. a "finite bandwidth".
For the Fourier transform I will use the version that has the 2π up in the exponent, and for added convenience, I will use the convention 1^x = e^{2πix} = cos(2πx) + i sin(2πx) (which I refer to as the Ramanujan Convention, as it is the convention I frequently used in my previous life oops I mean which Ramanujan secretly used in his life to make the math a whole lot simpler).
The set ℤ = {⋯, -2, -1, 0, +1, +2, ⋯ } is the integers, and 1^{x+z} = 1^x for all z∈ℤ - making 1^x the archetype of a periodic function whose period is 1.
Thus, the Fourier transform f̂(ν) of a function f(t) and its inverse are given by:
f̂(ν) = ∫ f(t) 1^{-νt} dt, f(t) = ∫ f̂(ν) 1^{+νt} dν.
The spectrum of the time series given by the function f(t) is the function f̂(ν) of the cyclic frequency ν, which is what is measured in Hertz (Hz.); t, itself, being measured in seconds. A common convention is to use the angular frequency ω = 2πν, instead, but that muddies the picture.
The most important example, with respect to the issue at hand, is the Fourier transform χ̂_Ω of the interval function given by χ_Ω(t) = 1 if t ∈ [-½Ω,+½Ω] and χ_Ω(t) = 0 else:
χ̂_Ω(t) = ∫_[-½Ω,+½Ω] 1^ν dν
= {1^{+½Ω} - 1^{-½Ω}}/{2πi}
= {2i sin πΩ}/{2πi}
= Ω sinc πΩ
which is where the function sinc x = (sin πx)/(πx) comes into play.
The cardinal form of the sampling theorem is that a function f(t) can be sampled over an equally-spaced sampled domain T ≡ { kΔt: k ∈ ℤ }, if its spectrum is bounded by supp f̂ ⊆ [-½Ω,+½Ω] ⊆ [-1/(2Δt),+1/(2Δt)], with the sampling given as
f(t) = ∑_{t'∈T} f(t') Ω sinc(Ω(t - t')) Δt.
So, this generally applies to [over-]sampling with redundancy factors 1/(ΩΔt) ≥ 1. In the special case where the sampling is tight with ΩΔt = 1, then it reduces to the form
f(t) = ∑_{t'∈T} f(t') sinc({t - t'}/Δt).
In our case, supp f̂ = [10 Hz., 150 Hz.] so the tightest fits are with 1/Δt = Ω = 300 Hz.
This generalizes to equally-spaced sampled domains of the form T ≡ { t₀ + kΔt: k ∈ ℤ } without any modification.
But it also generalizes to frequency intervals supp f̂ = [ν₋,ν₊] of width Ω = ν₊ - ν₋ and center ν₀ = ½ (ν₋ + ν₊) to the following form:
f(t) = ∑_{t'∈T} f(t') 1^{ν₀(t - t')} Ω sinc(Ω(t - t')) Δt.
In your case, you have ν₋ = 10 Hz., ν₊ = 150 Hz., Ω = 140 Hz., ν₀ = 80 Hz. with the condition Δt ≤ 1/140 second, a sampling rate of at least 140 Hz. with
f(t) = (140 Δt) ∑_{t'∈T} f(t') 1^{80(t - t')} sinc(140(t - t')).
where t and Δt are in seconds.
There is a larger context to all of this. One of the main places where this can be used is for transforms devised from an overlapping set of windowed filters in the frequency domain - a typical case in point being transforms for the time-scale plane, like the S-transform or the continuous wavelet transform.
Since you want the filters to be smoothly-windowed functions, without sharp corners, then in order for them to provide a complete set that adds up to a finite non-zero value over all of the frequency spectrum (so that they can all be normalized, in tandem, by dividing out by this sum), then their respective supports have to overlap.
(Edit: Generalized this example to cover both equally-spaced and logarithmic-spaced intervals.)
One example of such a set would be filters that have end-point frequencies taken from the set
Π = { p₀ (α + 1)ⁿ + β {(α + 1)ⁿ - 1} / α: n ∈ {0,1,2,⋯} }
So, for interval n (counting from n = 0), you would have ν₋ = p_n and ν₊ = p_{n+1}, where the members of Π are enumerated
p_n = p₀ (α + 1)ⁿ + β {(α + 1)ⁿ - 1} / α,
Δp_n = p_{n+1} - p_n = α p_n + β = (α p₀ + β)(α + 1)ⁿ,
n ∈ {0,1,2,⋯}
The center frequency of interval n would then be ν₀ = p_n + ½ Δp₀ (α + 1)ⁿ and the width would be Ω = Δp₀ (α + 1)ⁿ, but the actual support for the filter would overlap into a good part of the neighboring intervals, so that when you add up the filters that cover a given frequency ν the sum doesn't drop down to 0 as ν approaches any of the boundary points. (In the limiting case α → 0, this produces an equally-spaced frequency domain, suitable for an equalizer, while in the case β → 0, it produces a logarithmic scale with base α + 1, where octaves are equally-spaced.)
The other main place where you may apply this is to time-frequency analysis and spectrograms. Here, the role of a function f and its Fourier transform f̂ are reversed and the role of the frequency bandwidth Ω is now played by the (reciprocal) time bandwidth 1/Ω. You want to break up a time series, given by a function f(t) into overlapping segments f̃(q,λ) = g(λ)* f(q + λ), with smooth windowing given by the functions g(λ) with bounded support supp g ⊆ [-½ 1/Ω, +½ 1/Ω], and with interval spacing Δq much larger than the time sampling Δt (the ratio Δq/Δt is called the "hop" factor). The analogous role of Δt is played, here, by the frequency interval in the spectrogram Δp = Ω, which is now constant.
Edit: (Fixed the numbers for the Audacity example)
The minimum sampling rate for both supp_λ g and supp_λ f(q,λ) is Δq = 1/Ω = 1/Δp, and the corresponding redundancy factor is 1/(ΔpΔq). Audacity, for instance, uses a redundancy factor of 2 for its spectrograms. A typical value for Δp might be 44100/2048 Hz., while the time-sampling rate is Δt = 1/(2×3×5×7)² second (corresponding to 1/Δt = 44100 Hz.). With a redundancy factor of 2, Δq would be 1024/44100 second and the hop factor would be Δq/Δt = 1024.
If you try to fit the sampling windows, in either case, to the actual support of the band-limited (or time-limited) function, then the windows won't overlap and the only way to keep their sum from dropping to 0 on the boundary points would be for the windowing functions to have sharp corners on the boundaries, which would wreak havoc on their corresponding Fourier transforms.
The Balian-Low Theorem makes the actual statement on the matter.
https://encyclopediaofmath.org/wiki/Balian-Low_theorem
And a shout-out to someone I've been talking with, recently, about DSP-related matters and his monograph, which provides an excellent introductory reference to a lot of the issues discussed here.
A Friendly Guide To Wavelets
Gerald Kaiser
Birkhauser 1994
He said it's part of a trilogy, another installment of which is forthcoming.
I have implemented a very simple linear regression with gradient descent algorithm in JavaScript, but after consulting multiple sources and trying several things, I cannot get it to converge.
The data is absolutely linear, it's just the numbers 0 to 30 as inputs with x*3 as their correct outputs to learn.
This is the logic behind the gradient descent:
train(input, output) {
const predictedOutput = this.predict(input);
const delta = output - predictedOutput;
this.m += this.learningRate * delta * input;
this.b += this.learningRate * delta;
}
predict(x) {
return x * this.m + this.b;
}
I took the formulas from different places, including:
Exercises from Udacity's Deep Learning Foundations Nanodegree
Andrew Ng's course on Gradient Descent for Linear Regression (also here)
Stanford's CS229 Lecture Notes
this other PDF slides I found from Carnegie Mellon
I have already tried:
normalizing input and output values to the [-1, 1] range
normalizing input and output values to the [0, 1] range
normalizing input and output values to have mean = 0 and stddev = 1
reducing the learning rate (1e-7 is as low as I went)
having a linear data set with no bias at all (y = x * 3)
having a linear data set with non-zero bias (y = x * 3 + 2)
initializing the weights with random non-zero values between -1 and 1
Still, the weights (this.b and this.m) do not approach any of the data values, and they diverge into infinity.
I'm obviously doing something wrong, but I cannot figure out what it is.
Update: Here's a little bit more context that may help figure out what my problem is exactly:
I'm trying to model a simple approximation to a linear function, with online learning by a linear regression pseudo-neuron. With that, my parameters are:
weights: [this.m, this.b]
inputs: [x, 1]
activation function: identity function z(x) = x
As such, my net will be expressed by y = this.m * x + this.b * 1, simulating the data-driven function that I want to approximate (y = 3 * x).
What I want is for my network to "learn" the parameters this.m = 3 and this.b = 0, but it seems I get stuck at a local minima.
My error function is the mean-squared error:
error(allInputs, allOutputs) {
let error = 0;
for (let i = 0; i < allInputs.length; i++) {
const x = allInputs[i];
const y = allOutputs[i];
const predictedOutput = this.predict(x);
const delta = y - predictedOutput;
error += delta * delta;
}
return error / allInputs.length;
}
My logic for updating my weights will be (according to the sources I've checked so far) wi -= alpha * dError/dwi
For the sake of simplicity, I'll call my weights this.m and this.b, so we can relate it back to my JavaScript code. I'll also call y^ the predicted value.
From here:
error = y - y^
= y - this.m * x + this.b
dError/dm = -x
dError/db = 1
And so, applying that to the weight correction logic:
this.m += alpha * x
this.b -= alpha * 1
But this doesn't seem correct at all.
I finally found what's wrong, and I'm answering my own question in hopes it will help beginners in this area too.
First, as Sascha said, I had some theoretical misunderstandings. It may be correct that your adjustment includes the input value verbatim, but as he said, it should already be part of the gradient. This all depends on your choice of the error function.
Your error function will be the measure of what you use to measure how off you were from the real value, and that measurement needs to be consistent. I was using mean-squared-error as a measurement tool (as you can see in my error method), but I was using a pure-absolute error (y^ - y) inside of the training method to measure the error. Your gradient will depend on the choice of this error function. So choose only one and stick with it.
Second, simplify your assumptions in order to test what's wrong. In this case, I had a very good idea what the function to approximate was (y = x * 3) so I manually set the weights (this.b and this.m) to the right values and I still saw the error diverge. This means that weight initialization was not the problem in this case.
After searching some more, my error was somewhere else: the function that was feeding data into the network was mistakenly passing a 3 hardcoded value into the predicted output (it was using a wrong index in an array), so the oscillation I saw was because of the network trying to approximate to y = 0 * x + 3 (this.b = 3 and this.m = 0), but because of the small learning rate and the error in the error function derivative, this.b wasn't going to get near to the right value, making this.m making wild jumps to adjust to it.
Finally, keep track of the error measurement as your network trains, so you can have some insight into what's going on. This helps a lot to identify a difference between simple overfitting, big learning rates and plain simple mistakes.
I am using a nice FFT library I found online to see if I can write a pitch-detection program. So far, I have been able to successfully let the library do FFT calculation on a test audio signal containing a few sine waves including one at 440Hz (I'm using 16384 samples as the size and the sample rate at 44100Hz).
The FFT output looks like:
433.356Hz - Real: 590.644 - Imag: -27.9856 - MAG: 16529.5
436.047Hz - Real: 683.921 - Imag: 51.2798 - MAG: 35071.4
438.739Hz - Real: 4615.24 - Imag: 1170.8 - MAG: 5.40352e+006
441.431Hz - Real: -3861.97 - Imag: 2111.13 - MAG: 8.15315e+006
444.122Hz - Real: -653.75 - Imag: 341.107 - MAG: 222999
446.814Hz - Real: -564.629 - Imag: 186.592 - MAG: 105355
As you can see, the 441.431Hz and 438.739Hz bins both show equally high magnitude outputs (the right-most numbers following "MAG:"), so it's obvious that the target frequency 440Hz falls somewhere between. Increasing the resolution might be one way to close in, but that would add to the calculation time.
How do I calculate the exact frequency that falls between two frequency bins?
UPDATE:
I tried out Barry Quinn's "Second Estimator" discussed on the DSPGuru website and got excellent results. The following shows the result for 440Hz square wave - now I'm only off by 0.003Hz!
Here is the code I used. I simply adapted this example I found, which was for Swift. Thank you everyone for your very valuable input, this has been a great learning journey :)
To calculate the "true" frequency, once I used parabola fit algorithm. It worked very well for my use case.
This is the way I followed in order to find the fundamental frequency:
Calculate DFT (WOLA).
Find peaks in your DFT bins.
Find Harmonic Product Spectrum. Not the most reliable nor precise, but this is a very easy way of finding your fundamental frequency candidates.
Based on peaks and HPS, use parabola fit algorithm to find fundamental pitch frequency (and amplitude if needed).
For example, HPS says the fundamental (strongest) pitch is concentrated in bin x of your DFT; if bin x belongs to the peak y, then parabola fit frequency is taken from the peak y and that is the pitch you were looking for.
If you are not looking for fundamental pitch, but exact frequency in any bin, just apply parabola fit for that bin.
Some code to get you started:
struct Peak
{
float freq ; // Peak frequency calculated by parabola fit algorithm.
float amplitude; // True amplitude.
float strength ; // Peak strength when compared to neighbouring bins.
uint16_t startPos ; // Peak starting position (DFT bin).
uint16_t maxPos ; // Peak location (DFT bin).
uint16_t stopPos ; // Peak stop position (DFT bin).
};
void calculateTrueFrequency( Peak & peak, float const bins, uint32_t const fs, DFT_Magnitudes mags )
{
// Parabola fit:
float a = mags[ peak.maxPos - 1 ];
float b = mags[ peak.maxPos ];
float c = mags[ peak.maxPos + 1 ];
float p = 0.5f * ( a - c ) / ( a - 2.0f * b + c );
float bin = convert<float>( peak.maxPos ) + p;
peak.freq = convert<float>( fs ) * bin / bins / 2;
peak.amplitude = b - 0.25f + ( a - c ) * p;
}
Sinc interpolation can be used to accurately interpolate (or reconstruct) the spectrum between FFT result bins. A zero-padded FFT will produce a similar interpolated spectrum. You can use a high quality interpolator (such as a windowed Sinc kernel) with successive approximation to estimate the actual spectral peak to whatever resolution the S/N allows. This reconstruction might not work near the DC or Fs/2 FFT result bins unless you include the effects of the the spectrum's conjugate image in the interpolation kernel.
See: https://ccrma.stanford.edu/~jos/Interpolation/Ideal_Bandlimited_Sinc_Interpolation.html and https://en.wikipedia.org/wiki/Whittaker%E2%80%93Shannon_interpolation_formula for details about time domain reconstruction, but the same interpolation method works in either domain, frequency or time, for bandlimited or time limited signals respectively.
If you require a less accurate estimator with far less computational overhead, parabolic interpolation (and other similar curve fitting estimators) might work. See: https://www.dsprelated.com/freebooks/sasp/Quadratic_Interpolation_Spectral_Peaks.html and https://mgasior.web.cern.ch/mgasior/pap/FFT_resol_note.pdf for details for parabolic, and http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.555.2873&rep=rep1&type=pdf for other curve fitting peak estimators.
I need to discretise a 3rd order Bezier curve with points equally distributed along the curve. The curve is defined by four points p0, p1, p2, p3 and a generic point p(t) with 0 < t < 1 is given by:
point_t = (1 - t) * (1 - t) * (1 - t) * p0 + 3 * (1 - t) * (1 - t) * t * p1 + 3 * (1 - t) * t * t * p2 + t * t * t * p3;
My first idea was to discretise t = 0, t_1, ... t_n, ..., 1
This doesn't work as, in general, we don't end up with a uniform distance between the discretised points.
To sum up, what I need is an algorithm to discretise the parametric curve so that:
|| p(t_n) - p(t_n_+_1) || = d
I thought about recursively halving the Bezier curve with the Casteljau algorithm up to required resolution, but this would require a lot of distance calculations.
Any idea on how to solve this problem analytically?
What you are looking for is also called "arc-length parametrisation".
In general, if you subdivide a bezier curve at fixed interval of the default parametrisation, the resulting curve segments will not have the same arc-length. Here is one way to do it http://pomax.github.io/bezierinfo/#tracing.
A while ago, I was playing around with a bit of code (curvature flow) that needed the points to be as uniformly separated as possible. Here is a comparison (without proper labeling on axes! ;)) using linear interpolation and monotone cubic interpolation from the same set of quadrature samples (I used 20 samples per curve, each evaluated using a 24 point gauss-legendre Quadrature) to reparametrise a cubic curve.
[Please note that, this is compared with another run of the algorithm using a lot more nodes and samples taken as ground truth.]
Here is a demo using monotone cubic interpolation to reparametrise a curve. The function Curve.getLength is the quadrature function.
I'm looking for a way to get the treble and bass data from a song for some incrementation of time (say 0.1 seconds) and in the range of 0.0 to 1.0. I've googled around but haven't been able to find anything remotely close to what I'm looking for. Ultimately I want to be able to represent the treble and bass level while the song is playing.
Thanks!
Its reasonably easy. You need to perform an FFT and then sum up the bins that interest you. A lot of how you select will depend on the sampling rate of your audio.
You then need to choose an appropriate FFT order to get good information in the frequency bins returned.
So if you do an order 8 FFT you will need 256 samples. This will return you 128 complex pairs.
Next you need to convert these to magnitude. This is actually quite simple. if you are using std::complex you can simply perform a std::abs on the complex number and you will have its magnitude (sqrt( r^2 + i^2 )).
Interestingly at this point there is something called Parseval's theorem. This theorem states that after performinng a fourier transform the sum of the bins returned is equal to the sum of mean squares of the input signal.
This means that to get the amplitude of a specific set of bins you can simply add them together divide by the number of them and then sqrt to get the RMS amplitude value of those bins.
So where does this leave you?
Well from here you need to figure out which bins you are adding together.
A treble tone is defined as above 2000Hz.
A bass tone is below 300Hz (if my memory serves me correctly).
Mids are between 300Hz and 2kHz.
Now suppose your sample rate is 8kHz. The Nyquist rate says that the highest frequency you can represent in 8kHz sampling is 4kHz. Each bin thus represents 4000/128 or 31.25Hz.
So if the first 10 bins (Up to 312.5Hz) are used for Bass frequencies. Bin 10 to Bin 63 represent the mids. Finally bin 64 to 127 is the trebles.
You can then calculate the RMS value as described above and you have the RMS values.
RMS values can be converted to dBFS values by performing 20.0f * log10f( rmsVal );. This will return you a value from 0dB (max amplitude) down to -infinity dB (min amplitude). Be aware amplitudes do not range from -1 to 1.
To help you along, here is a bit of my C++ based FFT class for iPhone (which uses vDSP under the hood):
MacOSFFT::MacOSFFT( unsigned int fftOrder ) :
BaseFFT( fftOrder )
{
mFFTSetup = (void*)vDSP_create_fftsetup( mFFTOrder, 0 );
mImagBuffer.resize( 1 << mFFTOrder );
mRealBufferOut.resize( 1 << mFFTOrder );
mImagBufferOut.resize( 1 << mFFTOrder );
}
MacOSFFT::~MacOSFFT()
{
vDSP_destroy_fftsetup( (FFTSetup)mFFTSetup );
}
bool MacOSFFT::ForwardFFT( std::vector< std::complex< float > >& outVec, const std::vector< float >& inVec )
{
return ForwardFFT( &outVec.front(), &inVec.front(), inVec.size() );
}
bool MacOSFFT::ForwardFFT( std::complex< float >* pOut, const float* pIn, unsigned int num )
{
// Bring in a pre-allocated imaginary buffer that is initialised to 0.
DSPSplitComplex dspscIn;
dspscIn.realp = (float*)pIn;
dspscIn.imagp = &mImagBuffer.front();
DSPSplitComplex dspscOut;
dspscOut.realp = &mRealBufferOut.front();
dspscOut.imagp = &mImagBufferOut.front();
vDSP_fft_zop( (FFTSetup)mFFTSetup, &dspscIn, 1, &dspscOut, 1, mFFTOrder, kFFTDirection_Forward );
vDSP_ztoc( &dspscOut, 1, (DSPComplex*)pOut, 1, num );
return true;
}
It seems that you're looking for Fast Fourier Transform sample code.
It is quite a large topic to cover in an answer.
The tools you will need are already build in iOS: vDSP API
This should help you: vDSP Programming Guide
And there is also a FFT Sample Code available
You might also want to check out iPhoneFFT. Though that code is slighlty
outdated it can help you understand processes "under-the-hood".
Refer to auriotouch2 example from Apple - it has everything from frequency analysis to UI representation of what you want.