I'll be using a NXP DSC with 56800Ex core and my aim is to implement a Elliptic IIR Filter in it.

The User Reference Manual of the 'General Digital Filters Library' (which is available here) has functions to implement IIR Filters.

Page 3-19

But on page 3-19 shown above, it says

The filter coefficients are calculated using Butterworth approximation.

  • Does this mean that we can only use Butterworth Filter? Also, as per my knowledge, we have to give the coefficients and therefore other filters can also be implemented.

  • So what exactly does the function do?

  • Can I implement a Elliptic filter using this function or not? If not, then what should I do use Elliptic filter?

1 Answer 1


The function you are refering to only implements an order-1 filter. There is another function for order-2 filter.

If you want to implement an order-3 filter, you will have to split it in an order-1 filter and an order-2 filter. There are many ways to do it, but cascading the filters, as we would do in the analog world, is a good solution, albeit not optimal IIRC.

As to your original question, since the function refers to an order-1 IIR filter, an order-1 Elliptic or Butterworth, will have the same coefficients if you design them with the same cut-off frequency.

However, starting with order 2 or more, the coefficients will not be the same.

  • $\begingroup$ So is the IIR4 function capable of implementing the elliptic filter or not? $\endgroup$ Jun 15, 2017 at 5:32
  • $\begingroup$ Yes, but I'm not sure it is such a good idea to use a function that implements an order-4 IIR filter directly. Especially since you use integers, quantization effects could come back to haunt you. You can implement an elliptic filter using a combination of order-2 IIR and order-1 IIR. Say your filter is an order-3 elliptic IIR filter. Split in 2 cascaded-filter. You will have an order-2 filter and an order-1 filter. Use the appropriate function for each filter. I would have written the same thing had your filter been a Butterworth, or Chebyshev. The IIR type doesn't matter. $\endgroup$
    – Ben
    Jun 15, 2017 at 13:17

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