Hello,
I am working on the implementation of a digitally controlled dual loop AGC for a
speech processing system.
My question regards the dynamic range. Out of the documentation on the web I
have thet:
DR lg(Vmax/Vmin).
The microphone I am using has an output DC voltage between 0.2 - 0.7 V.
Typically 0.5 V.
How would an input 80 dB Dynamic range translate into output Voltage levels?
Thax for your help
Paul
Speech Dynamic Range
Started by ●March 24, 2009
Reply by ●March 25, 20092009-03-25
Ok... my problem regards the ac levels.
Mybe I should describe first my agc:
I have a variable gain amplifier made up from 2 OTA. The first OTA is the actual amplifier and the second one implements an active resistance. So by modification of the resistance and gain I should have the AGC. Rezistance and amplification will be controlled in a digital loop.
Now for the tranzistor dimensioning within the OTA I would be intersted in the minimum and maximum input , respectively output.
Out of the microphone datasheet I know thet the DC voltage is 0.5 V, that will be my input for the OTA. I also know that the dynamic range is 80 dB. What I don't know is how to translkate this into ac voltage levels. 10000 ration is a little too much :P
I have done some rough calculation, for orientation, but for my transistor dimensions, the vga only performs well as a follower, and only a very very slight amplificatin/attenuation. Problem: the tranzistors saturate very fast because of the poor dimensioning.
And again: what does 60 db output instantaneous dynamic range translate into voltage leveles. What should the amplification/attenuation of my sistem be?
Thanks for your help,
Paul
Mybe I should describe first my agc:
I have a variable gain amplifier made up from 2 OTA. The first OTA is the actual amplifier and the second one implements an active resistance. So by modification of the resistance and gain I should have the AGC. Rezistance and amplification will be controlled in a digital loop.
Now for the tranzistor dimensioning within the OTA I would be intersted in the minimum and maximum input , respectively output.
Out of the microphone datasheet I know thet the DC voltage is 0.5 V, that will be my input for the OTA. I also know that the dynamic range is 80 dB. What I don't know is how to translkate this into ac voltage levels. 10000 ration is a little too much :P
I have done some rough calculation, for orientation, but for my transistor dimensions, the vga only performs well as a follower, and only a very very slight amplificatin/attenuation. Problem: the tranzistors saturate very fast because of the poor dimensioning.
And again: what does 60 db output instantaneous dynamic range translate into voltage leveles. What should the amplification/attenuation of my sistem be?
Thanks for your help,
Paul
Reply by ●March 26, 20092009-03-26
Paul:
Decibels are relative measurements. See: >
20 decibels is a voltage or current relationship of multiplying or
dividing by ten.
A voltage increase of 20 decibels means multiply the original voltage
by ten; an increase of 40 decibels means multiply the voltage by 100.
60 dB increase is 1000, and 80 dB is 10,000.
Your specification for the microphone is probably its maximum output,
so 80 decibels BELOW this would be 1/10000 of 0.5V, or 0.05
millivolts, a VERY low level signal.
One must distinguish between the dynamic range of the equipment versus
the dynamic range of the subject matter. Practically, you do not need
80 dB of compression to analyze human speech, as the dynamic range of
speech is much smaller.
Good luck with your project!
David Reaves
On Tue Mar 24, 2009 8:14 am ((PDT)) "f...@yahoo.co.uk" wrote:
>
> Hello,
>
> I am working on the implementation of a digitally controlled dual
> loop AGC for a speech processing system.
> My question regards the dynamic range. Out of the documentation on
> the web I have thet:
>
> DR lg(Vmax/Vmin).
>
> The microphone I am using has an output DC voltage between 0.2 - 0.7
> V. Typically 0.5 V.
> How would an input 80 dB Dynamic range translate into output Voltage
> levels?
>
> Thax for your help
> Paul
Decibels are relative measurements. See: >
20 decibels is a voltage or current relationship of multiplying or
dividing by ten.
A voltage increase of 20 decibels means multiply the original voltage
by ten; an increase of 40 decibels means multiply the voltage by 100.
60 dB increase is 1000, and 80 dB is 10,000.
Your specification for the microphone is probably its maximum output,
so 80 decibels BELOW this would be 1/10000 of 0.5V, or 0.05
millivolts, a VERY low level signal.
One must distinguish between the dynamic range of the equipment versus
the dynamic range of the subject matter. Practically, you do not need
80 dB of compression to analyze human speech, as the dynamic range of
speech is much smaller.
Good luck with your project!
David Reaves
On Tue Mar 24, 2009 8:14 am ((PDT)) "f...@yahoo.co.uk" wrote:
>
> Hello,
>
> I am working on the implementation of a digitally controlled dual
> loop AGC for a speech processing system.
> My question regards the dynamic range. Out of the documentation on
> the web I have thet:
>
> DR lg(Vmax/Vmin).
>
> The microphone I am using has an output DC voltage between 0.2 - 0.7
> V. Typically 0.5 V.
> How would an input 80 dB Dynamic range translate into output Voltage
> levels?
>
> Thax for your help
> Paul
