I'm learning about filters and obviously the first kind of filters I came across were low pass and high pass filters:

$$\frac{1}{2}(1+z^{-1})$$ is a low pass filter and $$\frac{1}{2}(1-z^{-1})$$ is a high pass filter

which got me thinking are high order low pass and high pass fir filters possible?

  • 2
    $\begingroup$ yes of course. That's what filter design is all about. $\endgroup$
    – Hilmar
    Aug 27 at 15:09
  • $\begingroup$ You may even find orders in the thousands for FIRs. $\endgroup$ Aug 27 at 16:25

Yes, you can design FIR filter of any finite order.

You can re-write the examples by multiplying Z in the numerator and denominator to get:


Where it is clearer that you have only one zero and one pole. Therefore, this filter is of first order. In stable and causal FIR filters the order is given by the number of poles, and these poles are always located at the origin. You may add as many poles as you wish and get any order. And you may also locate as many zeros as you can (for causal filter the number of zeros should be less-equal to the number of poles) in any arbitrary position.

The FIR filter design theory explains where exactly these zeroes should be placed to obtain certain desirable features such as lineal phase, real impulse response, or to actually have a low, high, band-pass or band-reject filter.

  • $\begingroup$ So designing fir filter is bout knowing where you want to put the zeros so that those frequencies can be rejected? $\endgroup$
    – Orpheus
    Aug 28 at 3:54
  • $\begingroup$ Yes. In fact, the closer to the unit circle you place the zeros, the more you attenuate that particular frequency. If you come to place one zero exactly over the unit circle (and lets say that you do the same for the negative frequencies) then you will completely remove that frequency from your original signal. However, notice that even these zeroes are close to their neighbor frequencies, so they will affect them as well. $\endgroup$ Aug 30 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.