Zero Crossing Counter

In some DSP applications, it can be very helpful to know how many times your signal has crossed the zero-line (amplitude origin). 

How is this helpful?  Well, zero-crossings can tell you very quickly if your signal is composed of high frequency content or not.  Let's say your sample rate is 50kHz and over a small window of 1,000 samples there are 500 zero-crossings.  That would mean that every two samples crosses the zero-line (i.e. 12.5kHz)

In speech processing, the zero-crossing counts can help distinguish between voiced and un-voiced speech.  Un-voiced sounds are very noise-like ('Shh' and 'Sss' for example).  In addition, zero-crossings could also be used to determine if your signal has a DC offset.  If you signal is 'muted' and you are not seeing alot of zero-crossings might mean that your signal is offset from the zero-line

One nice thing about the matlab code below is that it is implemented in a very DSP-friendly way.  It ports very easily into C-Code and does minimizes the amount of conditional statements for faster processing time.

function count = zero_crossings(x)
% Count the number of zero-crossings from the supplied time-domain
% input vector.  A simple method is applied here that can be easily
% ported to a real-time system that would minimize the number of
% if-else conditionals.
%
% Usage:     COUNT = zero_crossings(X);
%
%            X     is the input time-domain signal (one dimensional)
%            COUNT is the amount of zero-crossings in the input signal
%
% Author:    sparafucile17 06/27/04
%

% initial value
count = 0;

% error checks
if(length(x) == 1)
    error('ERROR: input signal must have more than one element');
end

if((size(x, 2) ~= 1) && (size(x, 1) ~= 1))
    error('ERROR: Input must be one-dimensional');
end
    
% force signal to be a vector oriented in the same direction
x = x(:);

num_samples = length(x);
for i=2:num_samples

    % Any time you multiply to adjacent values that have a sign difference
    % the result will always be negative.  When the signs are identical,
    % the product will always be positive.
    if((x(i) * x(i-1)) < 0)
        count = count + 1;
    end
    
end