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    <p>These are the steps I would take to solve your problem in a vectorized way, starting with a given vector <code>sig</code>:</p> <ul> <li><p>First, threshold the vector to get a vector <code>tsig</code> of zeros and ones (zeroes where the absolute value of the signal drops close enough to zero, ones elsewhere):</p> <pre><code>tsig = (abs(sig) &gt;= eps); %# Using eps as the threshold </code></pre></li> <li><p>Next, find the starting indices, ending indices, and duration of each string of zeroes using the functions <a href="http://www.mathworks.com/access/helpdesk/help/techdoc/ref/diff.html" rel="nofollow noreferrer">DIFF</a> and <a href="http://www.mathworks.com/access/helpdesk/help/techdoc/ref/find.html" rel="nofollow noreferrer">FIND</a>:</p> <pre><code>dsig = diff([1 tsig 1]); startIndex = find(dsig &lt; 0); endIndex = find(dsig &gt; 0)-1; duration = endIndex-startIndex+1; </code></pre></li> <li><p>Then, find the strings of zeroes with a duration greater than or equal to some value (such as 3, from your example):</p> <pre><code>stringIndex = (duration &gt;= 3); startIndex = startIndex(stringIndex); endIndex = endIndex(stringIndex); </code></pre></li> <li><p>Finally, use <a href="https://stackoverflow.com/questions/2807270/matlab-vectorized-array-creation-from-a-list-of-start-end-indices/2807994#2807994">the method from my answer to the linked question</a> to generate your final set of indices:</p> <pre><code>indices = zeros(1,max(endIndex)+1); indices(startIndex) = 1; indices(endIndex+1) = indices(endIndex+1)-1; indices = find(cumsum(indices)); </code></pre></li> </ul>
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    1. COWas going to suggest this, more or less exactly.
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    2. COhow come I didnt think of using DIFF myself?? thanks
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