Kindly explain me abt this command

Can anybody explain me this command

ss = (mean(conv(tones(:,1),hh).^2) > mean(tones(:,1).^2)/5);

This command the book is using for DTMF decoding Plz explain me this

Answer 1

[moonman]

Yes you are exactly right. This portion is related to detection for presence of specific tone. Thanks for u reply I just want to ask now why book has scaled it by factor of 1/5

If i make it 1/20 or by 1/2 what will be the impact.

I have one more query. In equation

h[n] = (2/L) cos ( 2 π fb n / fs ), 0≤ n < L

where L is filter length, fb center frequency of the filter (frequency to be filtered), n is number of filter points, and fs is sampling frequency on which filter is implemented.

What we mean by Filter Points. Can u kindly explain

[moonman]

What is the difference between L and n

[moonman]

Is the length of Filter and Number of Tabs in same thing

Answer 2

[Wayne King]

Hi, This is comparing the mean value of the squared values of a convolution with the mean of the squared values of the waveform scaled by 1/5.

I'm assuming this is part of the detection algorithm to decide which tone is present.

Filtering the waveform (convolving) with a frequency-selective filter should remove a lot of the frequency content. If most of the signal's energy is in the passband of the filter, you will still be left with a sizable proportion of the total energy, if not, then you will not.

ss is a 1 or 0 which tests that hypothesis. It's 1 if most of the energy is in the passband of the filter, and 0 if it is not.

Answer 3

[Wayne King]

I'm not sure exactly why they picked 5 here. Perhaps it's an empirical finding?

If you increased it to say 20, then that will reduce the value of

mean(tones(:,1).^2)/N);

where N is the specific value, that would mean that ss would be 1 in more cases. In other words, that would result in many more decisions that the tone was present and perhaps too many false positives.

On the other hand, making it 1/2 would have the opposite effect.