I'm programming a sort of audio plugin, and I get an array of values that represent a step-based signal such as this:
that have these values:
[ 0.27, 0.43, 0.48, 0.51, 0.85, 0.15, 0.48, 0.01, 0.28, 0.84, 0.15, 0.22, 0.11, 0.86, 0.66, 0.92, 0.40, 0.71 ]
I'm looking for to transform those values into a bigger array of interpolated values that represent a smooth signal, such as a sine wave. Somethings like this (sorry for my Paint art):
What kind of math should I use here? Inside my development environment (Ruby based), I have a common number of math functions. But I don't know where to start.
What you want here is a digital filter - specifically a lowpass filter.
There are two types of simple digital filter, Finite Impulse Response and Infinite Impulse Response.
A FIR filter works by summing, with some weighting, the previous n samples of the audio, and using that to generate the output sample. It's called "Finite Impulse Response", because a single impulse in the input can only affect a finite number of output samples.
An IIR filter, in contrast, uses its own previous output in addition to the current sample. It's called "Infinite Impulse Response" because of this feedback property; a single impulse can affect all future samples.
Of the two, the IIR filter is the simplest to implement, and in its most basic form looks like this:
state(N) = state(N - 1) * weighting + sample(N)
output(N) = state(N)
That is, for each input sample, reduce the previous state value by some amount and add the input, then use that as the output. As such, it's basically a moving average filter.
For example, if you set 'weighting' to 0.95, then each output sample is influenced 95% by previous samples and 5% by the current sample, and the output value will shift slowly in response to changing inputs. It will also be scaled up by 20X (1/(1-weighting)), so you should renormalize it accordingly.
Here's how the first few steps would work with your input data:
Start by setting state = 20 * 0.27.
Output state / 20 = 0.27
Update state = state * 0.95 + 0.43 = 26.08
Output state / 20 = 0.278.
Update state = state * 0.95 + 0.48 = 5.76
Output state / 20 = 0.288
And so forth. If you need more output data points than input data points, repeat your input samples n times before feeding into the filter, or interleave input samples with n zero samples. Both are valid, though they have different impacts on the filtered output.
There is a lot of theory behind digital filter design; in practice for a simple implementation you can probably use this first-order filter, and adjust the weighting value to suit.
Related
What does the d3.scale.sqrt() scale do? As per the documentation it is similar to d3.scale.pow().exponent(.5), so the returned scale is equivalent to the sqrt function for numbers; for example:
sqrt(0.25) returns 0.5.
so when we apply a domain similar to this:
d3.scale.sqrt().domain([1, 100]).range([10, 39])
does it signify it takes the value between 1-100 and return the sqrt function which ranges between 10-39? Could anybody clarify and provide more details on how this scale works?
The way scales work in D3 is that they map input values (defined by .domain()) to output values (defined by .range()). So
d3.scale.sqrt().domain([1, 100]).range([10, 39])
maps values from 1 to 100 to the 10 to 39 range. That is, 1 corresponds to 10 and 100 to 39. This has nothing to do with the transformation the scale applies, which only affects the distribution of values within the range. For the sqrt function, the growth is sub-linear, which means that more of the input values will fall into the latter part of the output range.
This is actually more of a theoretical question, but here it goes:
I'm developing an effect audio unit and it needs an equal power crossfade between dry and wet signals.
But I'm confused about the right way to do the mapping function from the linear fader to the scaling factor (gain) for the signal amplitudes of dry and wet streams.
Basically, I'ev seen it done with cos / sin functions or square roots... essentially approximating logarithmic curves. But if our perception of amplitude is logarithmic to start with, shouldn't these curves mapping the fader position to an amplitude actually be exponential?
This is what I mean:
Assumptions:
signal[i] means the ith sample in a signal.
each sample is a float ranging [-1, 1] for amplitudes between [0,1].
our GUI control is an NSSlider ranging from [0,1], so it is in
principle linear.
fader is a variable with the value of the NSSlider.
First Observation:
We perceive amplitude in a logarithmic way. So if we have a linear fader and merely adjust a signal's amplitude by doing: signal[i] * fader what we are perceiving (hearing, regardless of the math) is something along the lines of:
This is the so-called crappy fader-effect: we go from silence to a drastic volume increase across the leftmost segment in the slider and past the middle the volume doesn't seem to get that louder.
So to do the fader "right", we instead either express it in a dB scale and then, as far as the signal is concerned, do: signal[i] * 10^(fader/20) or, if we were to keep or fader units in [0,1], we can do :signal[i] * (.001*10^(3*fader))
Either way, our new mapping from the NSSlider to the fader variable which we'll use for multiplying in our code, looks like this now:
Which is what we actually want, because since we perceive amplitude logarithmically, we are essentially mapping from linear (NSSLider range 0-1) to exponential and feeding this exponential output to our logarithmic perception. And it turns out that : log(10^x)=x so we end up perceiving the amplitude change in a linear (aka correct) way.
Great.
Now, my thought is that an equal-power crossfade between two signals (in this case a dry / wet horizontal NSSlider to mix together the input to the AU and the processed output from it) is essentially the same only that with one slider acting on both hypothetical signals dry[i] and wet[i].
So If my slider ranges from 0 to 100 and dry is full-left and wet is full-right), I'd end up with code along the lines of:
Float32 outputSample, wetSample, drySample = <assume proper initialization>
Float32 mixLevel = .01 * GetParameter(kParameterTypeMixLevel);
Float32 wetPowerLevel = .001 * pow(10, (mixLevel*3));
Float32 dryPowerLevel = .001 * pow(10, ((-3*mixLevel)+1));
outputSample = (wetSample * wetPowerLevel) + (drySample * dryPowerLevel);
The graph of which would be:
And same as before, because we perceive amplitude logarithmically, this exponential mapping should actually make it where we hear the crossfade as linear.
However, I've seen implementations of the crossfade using approximations to log curves. Meaning, instead:
But wouldn't these curves actually emphasize our logarithmic perception of amplitude?
The "equal power" crossfade you're thinking of has to do with keeping the total output power of your mix constant as you fade from wet to dry. Keeping total power constant serves as a reasonable approximation to keeping total perceived loudness constant (which in reality can be fairly complicated).
If you are crossfading between two uncorrelated signals of equal power, you can maintain a constant output power during the crossfade by using any two functions whose squared values sum to 1. A common example of this is the set of functions
g1(k) = ( 0.5 + 0.5*cos(pi*k) )^.5
g2(k) = ( 0.5 - 0.5*cos(pi*k) )^.5,
where 0 <= k <= 1 (note that g1(k)^2 + g2(k)^2 = 1 is satisfied, as mentioned). Here's a proof that this results in a constant power crossfade for uncorrelated signals:
Say we have two signals x1(t) and x2(t) with equal powers E[ x1(t)^2 ] = E[ x2(t)^2 ] = Px, which are also uncorrelated ( E[ x1(t)*x2(t) ] = 0 ). Note that any set of gain functions satisfying the previous condition will have that g2(k) = (1 - g1(k)^2)^.5. Now, forming the sum y(t) = g1(k)*x1(t) + g2(k)*x2(t), we have that:
E[ y(t)^2 ] = E[ (g1(k) * x1(t))^2 + 2*g1(k)*(1 - g1(k)^2)^.5 * x1(t) * x2(t) + (1 - g1(k)^2) * x2(t)^2 ]
= g1(k)^2 * E[ x1(t)^2 ] + 2*g1(k)*(1 - g1(k)^2)^.5 * E[ x1(t)*x2(t) ] + (1 - g1(k)^2) * E[ x2(t)^2 ]
= g1(k)^2 * Px + 0 + (1 - g1(k)^2) * Px = Px,
where we have used that g1(k) and g2(k) are deterministic and can thus be pulled outside the expectation operator E[ ], and that E[ x1(t)*x2(t) ] = 0 by definition because x1(t) and x2(t) are assumed to be uncorrelated. This means that no matter where we are in the crossfade (whatever k we choose) our output will still have the same power, Px, and thus hopefully equal perceived loudness.
Note that for completely correlated signals, you can achieve constant output power by doing a "linear" fade - using and two functions that sum to one ( g1(k) + g2(k) = 1 ). When mixing signals that are somewhat correlated, gain functions between those two would theoretically be appropriate.
What you're thinking of when you say
And same as before, because we perceive amplitude logarithmically,
this exponential mapping should actually make it where we hear the
crossfade as linear.
is that one signal should perceptually decrease in loudness as a linear function of slider position (k), while the other signal should perceptually increase in loudness as a linear function of slider position, when applying your derived crossfade. While your derivation of that seems pretty spot on, unfortunately that may not the best way to blend your dry and wet signals in terms of consistency - often, maintaining equal output loudness, regardless of slider position, is the better thing to shoot for. In any case, it might be worth trying a couple different functions to see what is most usable and consistent.
I am trying to understand the FFT algorithm and so far I think that I understand the main concept behind it. However I am confused as to the difference between 'framesize' and 'window'.
Based on my understanding, it seems that they are redundant with each other? For example, I present as input a block of samples with a framesize of 1024. So I have byte[1024] presented as input.
What then is the purpose of the windowing function? Since initially, I thought the purpose of the windowing function is to select the block of samples from the original data.
Thanks!
What then is the purpose of the windowing function?
It's to deal with so-called "spectral leakage": the FFT assumes an infinite series that repeats the given sample frame over and over again. If you have a sine wave that is an integral number of cycles within the sample frame, then all is good, and the FFT gives you a nice narrow peak at the proper frequency. But if you have a sine wave that is not an integral number of cycles, there's a discontinuity between the last and first sample, and the FFT gives you false harmonics.
Windowing functions lower the amplitudes at the beginning and the end of the sample frame, to reduce the harmonics caused by this discontinuity.
some diagrams from a National Instruments webpage on windowing:
integral # of cycles:
non-integer # of cycles:
for additional information:
http://www.tmworld.com/article/322450-Windowing_Functions_Improve_FFT_Results_Part_I.php
http://zone.ni.com/reference/en-XX/help/371361B-01/lvanlsconcepts/char_smoothing_windows/
http://www.physik.uni-wuerzburg.de/~praktiku/Anleitung/Fremde/ANO14.pdf
A rectangular window of length M has frequency response of sin(ω*M/2)/sin(ω/2), which is zero when ω = 2*π*k/M, for k ≠ 0. For a DFT of length N, where ω = 2*π*n/N, there are nulls at n = k * N/M. The ratio N/M isn't necessarily an integer. For example, if N = 40, and M = 32, then there are nulls at multiples of 1.25, but only the integer multiples will appear in the DFT, which is bins 5, 10, 15, and 20 in this case.
Here's a plot of the 1024-point DFT of a 32-point rectangular window:
M = 32
N = 1024
w = ones(M)
W = rfft(w, N)
K = N/M
nulls = abs(W[K::K])
plot(abs(W))
plot(r_[K:N/2+1:K], nulls, 'ro')
xticks(r_[:512:64])
grid(); axis('tight')
Note the nulls at every N/M = 32 bins. If N=M (i.e. the window length equals the DFT length), then there are nulls at all bins except at n = 0.
When you multiply a window by a signal, the corresponding operation in the frequency domain is the circular convolution of the window's spectrum with the signal's spectrum. For example, the DTFT of a sinusoid is a weighted delta function (i.e. an impulse with infinite height, infinitesimal extension, and finite area) located at the positive and negative frequency of the sinusoid. Convolving a spectrum with a delta function just shifts it to the location of the delta and scales it by the delta's weight. Therefore when you multiply a window by a sinusoid in the sample domain, the window's frequency response is scaled and shifted to the frequency of the sinusoid.
There are a couple of scenarios to examine regarding the length of a rectangular window. First let's look at the case where the window length is an integer multiple of the sinusoid's period, e.g. a 32-sample rectangular window of a cosine with a period of 32/8 = 4 samples:
x1 = cos(2*pi*8*r_[:32]/32) # ω0 = 8π/16, bin 8/32 * 1024 = 256
X1 = rfft(x1 * w, 1024)
plot(abs(X1))
xticks(r_[:513:64])
grid(); axis('tight')
As before, there are nulls at multiples of N/M = 32. But the window's spectrum has been shifted to bin 256 of the sinusoid and scaled by its magnitude, which is 0.5 split between the positive frequency and the negative frequency (I'm only plotting positive frequencies). If the DFT length had been 32, the nulls would line up at every bin, prompting the appearance that there's no leakage. But that misleading appearance is only a function of the DFT length. If you pad the windowed signal with zeros (as above), you'll get to see the sinc-like response at frequencies between the nulls.
Now let's look at a case where the window length is not an integer multiple of the sinusoid's period, e.g. a cosine with an angular frequency of 7.5π/16 (the period is 64 samples):
x2 = cos(2*pi*15*r_[:32]/64) # ω0 = 7.5π/16, bin 15/64 * 1024 = 240
X2 = rfft(x2 * w, 1024)
plot(abs(X2))
xticks(r_[-16:513:64])
grid(); axis('tight')
The center bin location is no longer at an integer multiple of 32, but shifted by a half down to bin 240. So let's see what the corresponding 32-point DFT would look like (inferring a 32-point rectangular window). I'll compute and plot the 32-point DFT of x2[n] and also superimpose a 32x decimated copy of the 1024-point DFT:
X2_32 = rfft(x2, 32)
X2_sample = X2[::32]
stem(r_[:17],abs(X2_32))
plot(abs(X2_sample), 'rs') # red squares
grid(); axis([0,16,0,11])
As you can see in the previous plot, the nulls are no longer aligned at multiples of 32, so the magnitude of the 32-point DFT is non-zero at each bin. In the 32 point DFT, the window's nulls are still spaced every N/M = 32/32 = 1 bin, but since ω0 = 7.5π/16, the center is at 'bin' 7.5, which puts the nulls at 0.5, 1.5, etc, so they're not present in the 32-point DFT.
The general message is that spectral leakage of a windowed signal is always present but can be masked in the DFT if the signal specrtum, window length, and DFT length come together in just the right way to line up the nulls. Beyond that you should just ignore these DFT artifacts and concentrate on the DTFT of your signal (i.e. pad with zeros to sample the DTFT at higher resolution so you can clearly examine the leakage).
Spectral leakage caused by convolving with a window's spectrum will always be there, which is why the art of crafting particularly shaped windows is so important. The spectrum of each window type has been tailored for a specific task, such as dynamic range or sensitivity.
Here's an example comparing the output of a rectangular window vs a Hamming window:
from pylab import *
import wave
fs = 44100
M = 4096
N = 16384
# load a sample of guitar playing an open string 6
# with a fundamental frequency of 82.4 Hz
g = fromstring(wave.open('dist_gtr_6.wav').readframes(-1),
dtype='int16')
L = len(g)/4
g_t = g[L:L+M]
g_t = g_t / float64(max(abs(g_t)))
# compute the response with rectangular vs Hamming window
g_rect = rfft(g_t, N)
g_hamm = rfft(g_t * hamming(M), N)
def make_plot():
fmax = int(82.4 * 4.5 / fs * N) # 4 harmonics
subplot(211); title('Rectangular Window')
plot(abs(g_rect[:fmax])); grid(); axis('tight')
subplot(212); title('Hamming Window')
plot(abs(g_hamm[:fmax])); grid(); axis('tight')
if __name__ == "__main__":
make_plot()
If you don't modify the sample values, and select the same length of data as the FFT length, this is equivalent to using a rectangular window, in which case the frame and the window are identical. However multiplying your input data by a rectangular window in the time domain is the same as convolving the input signal's spectrum with a Sinc function in the frequency domain, which will spread any spectral peaks for frequencies which are not exactly periodic in the FFT aperture across the entire spectrum.
Non-rectangular windows are often used so the the resulting FFT spectrum is convolved with something a bit more "focused" than a Sinc function.
You can also use a rectangular window that is a different size than the FFT length or aperture. In the case of a shorter data window, the FFT frame can be zero padded, which can result in an smoother looking interpolated FFT result spectrum. You can even use a rectangular window that is longer that the length of the FFT by wrapping data around the FFT aperture in a summed circular manner for some interesting effects with the frequency resolution.
ADDED due to a request:
Multiplying by a window in the time domain produces the same result as convolving with the transform of that window in the frequency domain.
In general, a narrower time domain window with produce a wider looking frequency domain transform. This is the reason that zero-padding produces a smoother frequency plot. The narrower time domain window produces a wider Sinc with fatter and smoother curves in relation to the frame width than would a window the full width of the FFT frame, thus making the interpolated frequency results look smoother than an non-zero padded FFT of the same frame length.
The converse is also true to some extent. A wider rectangular window will produce a narrower Sinc, with the nulls closer to the peak. Thus you might be able to use a carefully chosen wider window to produce a narrower looking Sinc to null a frequency closer to a bin of interest than 1 frequency bin away. How do you use a wider window? Wrap the data around and sum, which is identical to using FT basis vectors that are not truncated to 1 FFT frame in length. However, since when doing this the FFT result vector is shorter than the data, this is a lossy process which will introduce artifacts, and introduce some new novel aliasing. But it will give you a sharper frequency selection peak at each bin, and notch filters that can be placed less than 1 bin away, say halfway between bins, etc.
I want to know the frequency of data. I had a little bit idea that it can be done using FFT, but I am not sure how to do it. Once I passed the entire data to FFT, then it is giving me 2 peaks, but how can I get the frequency?
Thanks a lot in advance.
Here's what you're probably looking for:
When you talk about computing the frequency of a signal, you probably aren't so interested in the component sine waves. This is what the FFT gives you. For example, if you sum sin(2*pi*10x)+sin(2*pi*15x)+sin(2*pi*20x)+sin(2*pi*25x), you probably want to detect the "frequency" as 5 (take a look at the graph of this function). However, the FFT of this signal will detect the magnitude of 0 for the frequency 5.
What you are probably more interested in is the periodicity of the signal. That is, the interval at which the signal becomes most like itself. So most likely what you want is the autocorrelation. Look it up. This will essentially give you a measure of how self-similar the signal is to itself after being shifted over by a certain amount. So if you find a peak in the autocorrelation, that would indicate that the signal matches up well with itself when shifted over that amount. There's a lot of cool math behind it, look it up if you are interested, but if you just want it to work, just do this:
Window the signal, using a smooth window (a cosine will do. The window should be at least twice as large as the largest period you want to detect. 3 times as large will give better results). (see http://zone.ni.com/devzone/cda/tut/p/id/4844 if you are confused).
Take the FFT (however, make sure the FFT size is twice as big as the window, with the second half being padded with zeroes. If the FFT size is only the size of the window, you will effectively be taking the circular autocorrelation, which is not what you want. see https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Circular_convolution_theorem_and_cross-correlation_theorem )
Replace all coefficients of the FFT with their square value (real^2+imag^2). This is effectively taking the autocorrelation.
Take the iFFT
Find the largest peak in the iFFT. This is the strongest periodicity of the waveform. You can actually be a little more clever in which peak you pick, but for most purposes this should be enough. To find the frequency, you just take f=1/T.
Suppose x[n] = cos(2*pi*f0*n/fs) where f0 is the frequency of your sinusoid in Hertz, n=0:N-1, and fs is the sampling rate of x in samples per second.
Let X = fft(x). Both x and X have length N. Suppose X has two peaks at n0 and N-n0.
Then the sinusoid frequency is f0 = fs*n0/N Hertz.
Example: fs = 8000 samples per second, N = 16000 samples. Therefore, x lasts two seconds long.
Suppose X = fft(x) has peaks at 2000 and 14000 (=16000-2000). Therefore, f0 = 8000*2000/16000 = 1000 Hz.
If you have a signal with one frequency (for instance:
y = sin(2 pi f t)
With:
y time signal
f the central frequency
t time
Then you'll get two peaks, one at a frequency corresponding to f, and one at a frequency corresponding to -f.
So, to get to a frequency, can discard the negative frequency part. It is located after the positive frequency part. Furthermore, the first element in the array is a dc-offset, so the frequency is 0. (Beware that this offset is usually much more than 0, so the other frequency components might get dwarved by it.)
In code: (I've written it in python, but it should be equally simple in c#):
import numpy as np
from pylab import *
x = np.random.rand(100) # create 100 random numbers of which we want the fourier transform
x = x - mean(x) # make sure the average is zero, so we don't get a huge DC offset.
dt = 0.1 #[s] 1/the sampling rate
fftx = np.fft.fft(x) # the frequency transformed part
# now discard anything that we do not need..
fftx = fftx[range(int(len(fftx)/2))]
# now create the frequency axis: it runs from 0 to the sampling rate /2
freq_fftx = np.linspace(0,2/dt,len(fftx))
# and plot a power spectrum
plot(freq_fftx,abs(fftx)**2)
show()
Now the frequency is located at the largest peak.
If you are looking at the magnitude results from an FFT of the type most common used, then a strong sinusoidal frequency component of real data will show up in two places, once in the bottom half, plus its complex conjugate mirror image in the top half. Those two peaks both represent the same spectral peak and same frequency (for strictly real data). If the FFT result bin numbers start at 0 (zero), then the frequency of the sinusoidal component represented by the bin in the bottom half of the FFT result is most likely.
Frequency_of_Peak = Data_Sample_Rate * Bin_number_of_Peak / Length_of_FFT ;
Make sure to work out your proper units within the above equation (to get units of cycles per second, per fortnight, per kiloparsec, etc.)
Note that unless the wavelength of the data is an exact integer submultiple of the FFT length, the actual peak will be between bins, thus distributing energy among multiple nearby FFT result bins. So you may have to interpolate to better estimate the frequency peak. Common interpolation methods to find a more precise frequency estimate are 3-point parabolic and Sinc convolution (which is nearly the same as using a zero-padded longer FFT).
Assuming you use a discrete Fourier transform to look at frequencies, then you have to be careful about how to interpret the normalized frequencies back into physical ones (i.e. Hz).
According to the FFTW tutorial on how to calculate the power spectrum of a signal:
#include <rfftw.h>
...
{
fftw_real in[N], out[N], power_spectrum[N/2+1];
rfftw_plan p;
int k;
...
p = rfftw_create_plan(N, FFTW_REAL_TO_COMPLEX, FFTW_ESTIMATE);
...
rfftw_one(p, in, out);
power_spectrum[0] = out[0]*out[0]; /* DC component */
for (k = 1; k < (N+1)/2; ++k) /* (k < N/2 rounded up) */
power_spectrum[k] = out[k]*out[k] + out[N-k]*out[N-k];
if (N % 2 == 0) /* N is even */
power_spectrum[N/2] = out[N/2]*out[N/2]; /* Nyquist freq. */
...
rfftw_destroy_plan(p);
}
Note it handles data lengths that are not even. Note particularly if the data length is given, FFTW will give you a "bin" corresponding to the Nyquist frequency (sample rate divided by 2). Otherwise, you don't get it (i.e. the last bin is just below Nyquist).
A MATLAB example is similar, but they are choosing the length of 1000 (an even number) for the example:
N = length(x);
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(Fs*N)).*abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(x):Fs/2;
In general, it can be implementation (of the DFT) dependent. You should create a test pure sine wave at a known frequency and then make sure the calculation gives the same number.
Frequency = speed/wavelength.
Wavelength is the distance between the two peaks.
Graphics and audio editing and processing software often contain functions called "High-Pass Filter" and "Low-Pass Filter". Exactly what do these do, and what are the algorithms for implementing them?
Here is how you implement a low-pass filter using convolution:
double[] signal = (some 1d signal);
double[] filter = [0.25 0.25 0.25 0.25]; // box-car filter
double[] result = new double[signal.Length + filter.Length + 1];
// Set result to zero:
for (int i=0; i < result.Length; i++) result[i] = 0;
// Do convolution:
for (int i=0; i < signal.Length; i++)
for (int j=0; j < filter.Length; j++)
result[i+j] = result[i+j] + signal[i] * filter[j];
Note that the example is extremely simplified. It does not do range checks and does not handle the edges properly. The filter used (box-car) is a particularly bad lowpass filter, because it will cause a lot of artifacts (ringing). Read up on filter design.
You can also implement the filters in the frequency domain. Here is how you implement a high-pass filter using FFT:
double[] signal = (some 1d signal);
// Do FFT:
double[] real;
double[] imag;
[real, imag] = fft(signal)
// Set the first quarter of the real part to zero to attenuate the low frequencies
for (int i=0; i < real.Length / 4; i++)
real[i] = 0;
// Do inverse FFT:
double[] highfrequencysignal = inversefft(real, imag);
Again, this is simplified, but you get the idea. The code does not look as complicated as the math.
Wikipedia:
High-pass filter
Low-pass filter
Band-pass filter
These "high", "low", and "band" terms refer to frequencies. In high-pass, you try to remove low frequencies. In low-pass, you try to remove high. In band pass, you only allow a continuous frequency range to remain.
Choosing the cut-off frequency depends upon your application. Coding these filters can either be done by simulating RC circuits or by playing around with Fourier transforms of your time-based data. See the wikipedia articles for code examples.
Here is a super simple example of a low pass filter in C++ that processes the signal one sample at a time:
float lopass(float input, float cutoff) {
lo_pass_output= outputs[0]+ (cutoff*(input-outputs[0]));
outputs[0]= lo_pass_output;
return(lo_pass_output);
}
Here is pretty much the same thing, except it's high pass:
float hipass(float input, float cutoff) {
hi_pass_output=input-(outputs[0] + cutoff*(input-outputs[0]));
outputs[0]=hi_pass_output;
return(hi_pass_output);
}
They are generally Electrical circuits that tend to pass parts of analog signals. High pass tends to transmit more of the high frequency parts and low pass tends to pass more of the low frequency parts.
They can be simulated in software. A walking average can act as a low pass filter for instance and the difference between a walking average and it's input can work as a high pass filter.
High-pass filter lets high-frequency (detailed/local information) pass.
Low-pass filter lets low-frequency (coarse/rough/global information) pass.
Filtering describes the act of processing data in a way that applies different levels of attenuation to different frequencies within the data.
A high pass filter will apply minimal attentuation (ie. leave levels unchanged) for high frequencies, but applies maximum attenuation to low frequencies.
A low pass filter is the reverse - it will apply no attenuation to low frequencies by applies attenuation to high frequencies.
There are a number of different filtering algorithms that are used. The two simplest are probably the Finite Impulse Response filter (aka. FIR filter) and the Infinite Impulse Response filter (aka. IIR filter).
The FIR filter works by keeping a series of samples and multiplying each of those samples by a fixed coefficient (which is based on the position in the series). The results of each of these multiplications is accumulated and is the output for that sample. This is referred to as a Multiply-Accumulate - and in dedicated DSP hardware there is a specific MAC instruction for doing just this.
When the next sample is taken it's added to the start of the series, and the oldest sample in the series is removed, and the process repeated.
The behavior of the filter is fixed by the selection of the filter coefficients.
One of the simplest filters that is often provided by image processing software is the averaging filter. This can be implemented by an FIR filter by setting all of the filter coefficients to the same value.