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# Aliasing and Sampling

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#### Syukri

##### Full Member level 5
Can I please know how do B gets 5kHZ out of the question below?

if we have a 2ksps ADC (we use the Nyquist limit of 1kHz in the following formula) and a signal that is 1kHz, with a 1kHz digital filter following the ADC, the processing is
given by: -10 log (1kHz/1kHz) = 0dB. If we increase the sample rate to 10ksps, the processing gain is now -10 log (1kHz/5kHz) = 7dB

Hi

What do you mean by 1kHz signal?
Is it BW or center frequency?

it's the center freq.

and what i mean by B is log A/B. the frequency on denomerator.

Thanks

Hi

Process gain is defined as improvement in signal to quantization ratio when using oversampling.
PG = 10 * log10(fs / (2 * BW))
BW = output BW after decimation.

To calculate PG you need to know BW, so specify the output BW or decimation ratio then i can help you.

Regards

### Syukri

Points: 2
I think your elaboration is enough

Many thanks

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