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ADS1282: ADS1282 HPF Transfer function

Vitaliy Shevkunov
Prodigy 30 points
Part Number: ADS1282

Hello all,

I am trying to evaluate HPF response of ADS1282. The equation 14 in datasheet shows HPF(Z) = (2 - a) / 2 * (1 - Z^-1) / (1 - bZ^-1).

What is the meaning of the 'a' in this equation.

Thank you

Vitaly

  • 0
    TI__Mastermind 44680 points

    Hi Vitaliy,

    Welcome to the TI E2E forums!

    The 'a' term applies a gain error to the transfer function. I've spent a few hours trying to solve this equation for the values of 'a' and 'b' in the past and eventually ended up just adding the gain error term (from Equation 13) to a typical single-pole HPF response. You can find my results in the attached Excel file:

    ADS1282 HPF Response.xlsx

    Best regards,
    Chris

  • 0 in reply to Christopher Hall
    Prodigy 30 points

    Hello Chris,

    Thank you for the spreadsheet. It is very helpful.

    One more question. How would you compute the zeroes and poles of that filter? Say, I need them for 0.1Hz and 0.01Hz corner frequency at 250 SPS.

    Thank you

    Vitaly

  • 0 in reply to Vitaliy Shevkunov
    TI__Mastermind 44680 points
    Hi Vitaly,

    I'm not sure I fully understand your question... The HPF is a single pole filter with a pole at the filter's cutoff frequency. To program the pole/cutoff frequency you set the HPF[1:0] register bytes to corresponded to a cutoff frequency between 0 - 10.24% of the data rate.

    For the 250 SPS data rate and a HPF cutoff frequency of 0.1 Hz, set HPF[1:0] = 0x00A5;
    For the 250 SPS data rate and a HPF cutoff frequency of 0.01 Hz, set HPF[1:0] = 0x0010;

    Does that answer your question?

    Note that the HPF's settling time will be inversely proportional to the cutoff frequency. Refer to e2e.ti.com/.../979018

    Best regards,
    Chris
  • 0 in reply to Christopher Hall
    Prodigy 30 points

    Hello Chris,

    Sorry for the confusion. Here is what I meant.

    The Z Transfer function of the first order filter is H(Z) = (Z - z0) / (Z - p0), where z0 and p0 are zero and the pole.

    From the equation 14 I can see that z0 = 1 and p0 = b, where b = (1 + (1 - a)^2) / 2. 

    How do I compute 'a' for different cut off frequencies and sample rates?

    Thanks,

    Vitaly

  • 0 in reply to Vitaliy Shevkunov
    TI__Mastermind 44680 points

    Hi Vitaly,

    Sorry for the delay. I believe you would just calculate the coefficients like so:

    Best regards,
    Chris