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appropriate cut-off frequency for 2kHz sampling data rate in ADS1298

mohsen hamun
Prodigy 180 points
Other Parts Discussed in Thread: ADS1298

Hi .

I bought the ADS1298 chip and used it with ARM MCU and transferred data to a computer in short input mode. I want to use external electrodes to capture data in normal mode.

Should I place a simple RC anti-aliasing low-pass filter before the input pins of ADS1298 ?
If I place this filter, regard to ADS1298 internal modulator frequency, how can I calculate the filter's which is the appropriate cut-off frequency for this filter? My desired Data rate in ADS1298 is 2kHz..

Thanks.

  • 0
    TI__Guru 57296 points

    Hi Mohsen,

    The modulator frequency on the ADS1298 is either 512kHz for High-Resolution Mode or 256kHz for Low-Power Mode, assuming a master clock of 2.048MHz.

    The digital filter in the ADS1298 is a third-order sinc low-pass filter with a passband of 0.262*Data Rate (in your case, BW = 524Hz). At frequencies that are integer multiples of fMOD, the digital filter does not provide the attenuation necessary to suppress signals in this frequency range:

    Therefore, the cutoff of the Anti-Aliasing filter should be chosen such that it provides enough attenuation at fMOD so that you can achieve the desired performance. An example diagram is shown below:

    A single-pole RC filter at 2.5kHz, for example, would provide about -40dB of attenuation at fMOD = 256kHz without interfering with your signal bandwidth of interest.

    Hope this helps!

    Best Regards,

  • 0 in reply to Ryan Andrews
    Prodigy 180 points

    Thanks for your reply.

    I want use High resolution mode, so fmod is 512kHz.. If I want attenuation approximately -50dB, then -3dB cut-off frequency will be  512*3/50 = 30.72kHz. is this frequency calculation method right ?

    And my second question is about enough attenuation in fmod frequency. In the datasheet I cannot see any number for that. is -40 dB good value for attenuation ? 

  • 0 in reply to mohsen hamun
    TI__Guru 57296 points

    Hi Mohsen, 

    I do not understand your calculation, but the resulting cutoff frequency is not correct.

    The slope of a single-pole RC filter is -20dB/decade. So to hit -40dB at 512kHz, the filter cutoff would have to start two decades earlier = 512k / 102 = 5.12kHz.

    -40dB is a factor of 100x attenuation. If you require more than that, you may consider adding a second filter with the same cutoff frequency. The resulting double-pole filter would provide -40dB/decade, giving you 2x the attenuation at 512kHz.

    Best Regards,

  • 0 in reply to Ryan Andrews
    Prodigy 180 points

    Thanks again Ryan.

    I assumed filter response is linear, so if we have -40dB in 512kHz, for calculation cut-off frequency in -3dB ( for R and C calculation), we can use fitting and consequently cut-off frequency is 512*-3/-40 = 38.4kHz.  is equal my calculation ?

    for my application which DR=2kHz (DR=100), cut-off frequency equal to 5.18kHz is correct ?