[SOLVED] Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it right?)

Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it right?)

Hello All--

I'm learning DSP as I go while trying to implement a Lock-In Amplifier on an FPGA. Given the situation illustrated in the attached diagram, is it reasonable to say that the filter stop-band attenuation should be sufficient to reduce a full amplitude signal at the 16-bit ADC to one half the step size of the 16-bit DAC? In other words:

Minimum Atten. = 10*log(0.5*DAC-step / ADC-Range) = 10*log(0.5 / 65536) = -51.2dB

I know there are rounding and fixed-point noise considerations within the system, but I've not yet learned how I should or should not factor those into my high-level filter specs. In short, I could use some advice. Thanks in advance for the help.

--Brad

Re: Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it righ

Since you are dealing with input and output voltages, you must multiply the log by 20 to get dB. The correct answer will be twice your calculated value. Otherwise, your math looks OK, assuming that your DAC and ADC steps are equal (represent. If the steps are not equal for equal changes in voltage, then the DAC-step and ADC-range must be given in terms of analog voltages.

Re: Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it righ

Kral--

Thanks a bunch for your reply. The steps/ranges on the DAC and ADC are indeed equal, so it's good to hear I've at least got the overall logic correct.

If I might trouble you for one more push in the right direction, I think you may have pointed up a flaw in my understanding of decibels in this context. I had it in my head that the 20*log definition was for use with comparison of quantities squared, since:

10*log(a/b) = 20*log(a^2/b^2)

I assumed that since I was considering amplitudes (full range at ADC, half-step at DAC), I should use 10*log. So I guess my general question is, what rule of thumb should I be applying to select this as a case where I'm concerned with powers (or squared quantities) as opposed to amplitudes. I know that's 101 level, but I've obviously not learned it correctly.

In any case, I do appreciate your taking time for a new guy. Be well, and have a great day.

--Brad

Re: Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it righ

You're correct that 20*log refers to squared quantities. In your case the ADC range is in terms of voltage steps. So to compare that to something that is power (the required filter attenuation) the voltage quantity must be squared.

For instance, to attenuate the voltage by a factor of 2, you need to attenuate the power by a factor of 4 (6 dB).

Re: Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it righ

bcro,
pstuckey is correct. The decibel is a measure of a power ratio. Since power is proportional to voltage squared you must take the log of the (voltage squared) ratio.
Regards,
Kral

Re: Estimating Attenuation for Low Pass Filter in ADC->DAC System (am i doing it righ

Aha...the old 40W bulb just came on. I was focusing on decibels in relation to the I/O ranges, not to the filter itself. It makes sense now.

Thanks a bunch guys for sparing your time. Perhaps later I'll have some 201-level questions to post on the forum.

Be well.