intensity of sound/speech

"stromhau" <stromhau@stud.ntnu.no> wrote in message 
news:Lq6dneTCCJl2RT_YnZ2dnUVZ_uuqnZ2d@giganews.com...
>
>>
>>Tommy,
>>
>>OK - well I'm not at all sure that you really care about absolute levels
> nor
>>an absolute reference for this application.  dB, being a ratio, can be
>>scaled by adding or subtracting a constant - in which case all measures
>>remain the same distance apart in dB.  So, an arbitrary reference is
> quite
>>likely fine for what you're doing.  As long as the sensor is the same
> then
>>intensities will vary according to their absolute levels and the
> attenuation
>>from the source to the sensor - which it sounds like you don't really
> care
>>about and only care about the levels at the sensor (microphone or ....
> ).
>>
>>The simplest reference is no reference at all - that is use an intensity
> of
>>1.0 on whatever scale you're using.
>>So, 20*log(I1/I2) = 20*log(I1) - 20*log(I2).
>>If I2=1.0 then log(I2)=0 and you're left with 20*log(I1).
>>
>>Part of the problem you're having is that there can be a sound pressure
>>level reference which will translate to volts according to the sensor's
>>sensitivity (gain or transfer function) in volts per micropascal for
>>example.  At the output of a microphone you get volts and the only way to
> go
>>backwards to sound pressure level is to inversely apply the sensitivity
> to
>>the output voltage: volts/(volts/micropascal).  The other way is to
> simply
>>relate the microphone output to 1.0v as the zero dB level and forget
> about
>>absolute sound pressure levels altogether - since it's related to the
> volts
>>by a constant factor.
>>
>>You could calculate the rms level in the frame time - that's a measure of
>
>>energy.
>>Square the values, add them together, divide by the number of them and
> take
>>the square root for relative intensity (as micropascals or volts) or
> don't
>>take the square root for relative power.  In either case, the dB level
> will
>>be the same as 20* log10(ratio) if using intensities or 10*log10(ratio)
> if
>>using power.
>>
>>If you care about comparing speakers (people) in a room then intensity
>>differences will be more difficult because of distance / attenuation for
>
>>each speaker and the need to differentiate from one speaker to another.
>>That would be a much tougher problem.
>>
>>Fred
>>
>>
>>
>>
>
>
> Thank you very much for helping me out here!
>
> However i have another thing on my mind if you or someone else would be so
> nice to help.
>
> I see that when people calculate short time evergy they use a time
> frame(about 10 ms) and the multiply with a window function(often hanning)
> and then they have a overlap between frames. Could anyone explain why ?
> What i dont understand is the windowing function and the overlap of
> frames.
>
> Let me dig this algorithm back :) :
>
> "The algorithm used there says ; The values in the sound are first
> squared,
> then convolved with a Gaussian analysis window (Kaiser-20; sidelobes
> below
> -190 dB)"
>
> Nearly the same thing but here a convolution with the windowing function
> is used. Why ?

It seems this is a quote out of context.
If all you need is an intensity measure then why window at all? And, why 
have a spectral analysis at all?  Windows are more likely used when there is 
spectral analysis.

The window will tend to smooth out sharp intensity changes - which may or 
may not be what you want.

Fred