# Resolution and Sampling frequency

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Cecilie Tefre on 28 Apr 2022
Answered: Chandra on 2 May 2022
%BUTTERFILTFILT Design and apply Butterworth filter.
%
% INPUTS
% ------
% x - Data to be filtered.
% Filter operates on first non-singleton dimension.
%
% fc - Cutoffs frequencies (one frequency value for types 'lowpass'
% and 'highpass' and two frequency values for types 'bandpass'
% and 'bandstop').
%
% fs - Data sampling frequency.
%
% order - Order of the Butterworth filter (default: order = 6)
%
% type - Filtering type option. Can be either 'lowpass', 'highpass',
% 'bandpass' or 'bandstop'.
%
% direction - Direction to filter in
% 'forward' | 'backward' | {'both'}
% NB: if this is 'both' the output has zero phase filtering,
% and order must be an even number.
%
% NB: If a bandpass is requested with filter bounds [0 F], a lowpass
% filter is performed instead. If a bandpass is requested with filter
% upper bound Inf or the Nyquist freq, a highpass filter is performed.
%
%
% OUTPUTS
% -------
% x - Filtered data
%
% A - Filter coefficient matrix (denominator)
%
% B - Filter coefficient matrix (numerator)
function [x, A, B] = butterfiltfilt(x, fc, fs, order, type, direction)
% DEFAULT INPUTS ----------------------------------------------------------
if nargin<6 || isempty(direction)
direction = 'both';
end
if nargin<5 || isempty(type)
type = 'bandpass';
end
if nargin<4 || isempty(order)
order = 6;
end
if nargin<3 || isempty(fc)
if all(fc>=0) && all(fc<=1)
fs = 1;
else
error('Sampling frequency not given');
end
end
% INPUT HANDLING ----------------------------------------------------------
if nargin<2; error('Insufficient number of inputs'); end
if ~isnumeric(x); error('Data must be numeric'); end;
if ~isnumeric(fc); error('Filter bounds must be numeric'); end
if ~isnumeric(fs) || ~isscalar(fs); error('Sampling frequency must be a scalar'); end
if ~isnumeric(order) || ~isscalar(order); error('Order must be a scalar'); end
if ~ischar(type); error('Type must be a string'); end
if ~ischar(direction); error('Direction must be a string'); end
% MAIN --------------------------------------------------------------------
% If bidirectional filter, halve the order as filter is done twice
if strcmp(direction,'both')
if mod(order,2)~=0;
error('Filter order must be even if filter is bidirectional');
end
order = order/2;
end
% Change type to lowercase
type = lower(type);
% Move to band notation
switch type
% Lowpass
case {'low','lowpass'}
if numel(fc)~=1; error('Too many bounds given'); end;
fc = [0 fc];
% Highpass
case {'high','highpass'}
if numel(fc)~=1; error('Too many bounds given'); end;
fc = [fc Inf];
% Bandpass
case {'band','bandpass'}
% Do nothing
% Bandstop
case {'bandstop','stop'}
type = 'stop'; % Cannonical
otherwise
end
% Make bounds be relative to Nyquist frequency
fc = fc*2/fs;
% Design the filter -------------------------------------------------------
if numel(fc)~=2
error('Filter requires two cutoff frequencies');
elseif strcmp(type,'stop')
% Do bandstop filter
[B,A] = butter(order, fc, 'stop');
% If not bandstop, must be bandpass.
elseif (fc(1)==0) && (isinf(fc(2)))
% No filter required
B = NaN;
A = NaN;
return;
elseif fc(1)==0
% Lower bound is 0: design lowpass filter
[B,A] = butter(order, fc(2), 'low');
elseif isinf(fc(2)) || fc(2)==1
% Upper bound is Inf or Nyquist freq: design highpass filter
[B,A] = butter(order, fc(1), 'high');
else
[B,A] = butter(order, fc);
end
% Use the filter ----------------------------------------------------------
switch direction
case 'forward'
x = FilterM(B, A, x);
case 'backward'
x = FilterM(B, A, x, [], 'reverse');
case 'both'
if ispc
x = filter(B, A, x);
x(end:-1:1,:) = filter(B, A, x(end:-1:1,:));
else
x = FiltFiltM(B, A, x);
end
otherwise
error('Unrecognised filter direction: %s',direction);
end
end
I want to test what resolution and sampling frequency do to the audio signal by reducing sampling rate and resolution of audio files.
How do I do this?
The information I have gotten is that I should start by reducing sampling rate, since it requires the least programming.
matlab example of downsampling:
x_downsamling = x (1: 3: end)% takes out every third sample of x vector.
Remember to low-pass filters before downsampling. Why?
Also remember to change the playback speed when playing the sound.
Bit resolution reduction:
Let's say you have a signal that goes between -1 and 1 y.The floor (y + 1) function will take out an integer that is either 0 or 1 so that it is now reduced to 1 bit resolution.Then remember to scale so that the signal goes between -1 and 1.
For other bite resolutions:
1./N*floor(N.*y+a)-bThen select a, b, N to fit the desired bit resolution ..Feel free to test the function you create on a synthetic signal that goes between -1 and 1. you want a curve that looks like a step.

Chandra on 2 May 2022
Hi,
To obtain the desired results use sound function with different parameters
>>sound(y,Fs,nBits) % y – input signal, Fs – sampling rate,
% nBits is number of bits here 8,16 and 24 bit is supported
If needed, use the function that mentioned above, to use for obtain coefficients for filter and then use filter function to filter the signal, before using sound function

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