I deal with the problem of the synthesis of the filter with the requirements of both the frequency and time characteristics.

Now there is a filter that meets the requirements. Roughly speaking the requirements are as follows:

Frequency response

FIR Low Pass

  • Fpass ~ 0.0(3) (* pi rad/sample)
  • Fstop ~ 0.1(6) (* pi rad/sample)
  • bandwidth irregularity of ~ 1e-3 (maybe ~ 1e-2) dB
  • ~ -40 dB suppression

Time requirements

  • Overshoot in step response ~ 4% (transition time ~ 1e-2 ms)

I would like to know about the possibilities of such a synthesis (I’m with filterDesigner in the matlab failed, tried free tools - Iowa Hills) or at least about the transformation of the finished filter into a narrower (Fpass = 0.025, Fstop = 0.125) (Matlab can do this, but only IIR is obtained).

  • $\begingroup$ What kinda filter is this? High-pass? I don't grok the specs. Particularly the numbers in (parenths). $\endgroup$ Commented May 14, 2019 at 6:25
  • $\begingroup$ It's Low Pass filter. Numbers in (parenths): k.(m) = k.mmmmmmmmm... is Normalized Frequency. $\endgroup$
    – Alexey K
    Commented May 14, 2019 at 6:28
  • $\begingroup$ This being a LP-Filter, I am confused how Fstop(0.16666) can be smaller than Fpass (0.3333)? Are you missing a zero in the Fpass (0.033333)? $\endgroup$ Commented May 14, 2019 at 7:13
  • $\begingroup$ My fault. Fpass 0.03333. $\endgroup$
    – Alexey K
    Commented May 14, 2019 at 7:15

1 Answer 1


The most straightforward way to impose constraints in the frequency domain as well as in the time domain for the design of (linear phase) FIR filters is to use linear programming.

Constraints on the step response or on the impulse response are naturally linear, and constraints on the frequency response can also be formulated as linear constraints in the case of linear phase FIR filters.

Take a look at this document for some examples. More details on the linear programming formulation of the FIR filter design problem can be found in this paper.

  • $\begingroup$ Thank you. That's what I need. $\endgroup$
    – Alexey K
    Commented May 14, 2019 at 12:32

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