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Old 2nd September 2002, 09:06 AM   #1
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Question How does digital attenuation work?

Browsing through the data sheets for some of Burr Brown's digital filters (specifically the DF1706 digital filter and the DSD1608 8 channel 24bit DAC with 8x digital filter, I have samples of both), I see both implement a digital volume control, ranging from 0dB loss to -127dB in 0.5dB steps.

Is this implemented by reducing the no. of bits used for the output? If so, is this equilivant to data loss and reduced dynamic range, with the signal not being able to use the full 24bits output, or is it implemented by a digital potentiometer with its associated distortion.

Any thoughts anyone?

Thanks, Adrian
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Old 7th September 2002, 05:40 AM   #2
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If the digital volume is before the DAC, then you are losing bits. If the effective gain is 0.5, then you lose one bit, G = 0.25, 2 bits, etc. You can lose "fractional bits" with smaller gain increments.

If you do it in the DAC or after it is possible to do it in the analog domain with out losing bits. I haven't seen such a setup in a DAC, though. Even if you do do the volume reduction in analog, the S/N ratio reduces with reducing gain.

I have been using the attenuator in my PCM1716s. I know I am losing bits, though. Can't complain about the sound.
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Old 7th September 2002, 11:44 PM   #3
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Yep, digital attenuation is just a multiplication of all the data samples by a fixed coefficient. Thus, the lower the volume setting, the more effective resolution you will lose. There are methods of doing "digital" attenuation at the DACs, which is usually more an analog technique really, but depending on the implementation, can preserve the full resolution of the DAC and incoming data.
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