2
\$\begingroup\$

I have been experimenting with several digital filters for my thesis recently, and while learning about the FIR and IIR kind, they seem to have similar output expressions.

While using a Java class I found online to design a Butterworth IIR Filter's coefficients, I realized that the 'a0' value turns out to be 1 and, consequentially, that the resulting expression looks like a FIR filter.

Is this the case? Or is the fact that the 'y[n-i]' terms are present on the final implementation enough to warrant it not being FIR and definitely IIR?

\$\endgroup\$
  • 1
    \$\begingroup\$ Ther is no way to answer this without the exact equations for your particular IIR filter. Think about it. How are we supposed to know what "a0" is in your algorithm? \$\endgroup\$ – Olin Lathrop Mar 19 '13 at 13:20
  • \$\begingroup\$ Probably a simple confusion here. If you zero'd the coefficients of all of the y terms rather than the x, there would be a resemblance, though in most cases to an ineffectively short one. \$\endgroup\$ – Chris Stratton Mar 19 '13 at 13:22
  • 1
    \$\begingroup\$ That was what I was hoping to confirm: IIR implies output feedback, no matter what value 'a0' contains. \$\endgroup\$ – ravemir Mar 19 '13 at 13:45
3
\$\begingroup\$

It's unlikely to be true. IIR filters use feed-back paths and recirculate a fraction of the output (ever-diminishing hopefully). Hence they have the name Infinite Impulse Response meaning that an impulse on the input would cause an output that continues to decay to infinite.

FIR filter do not have feedback paths and hence the name Finite Impulse Response because the output following an inputted impulse changes and then restores to "normality" after a finite length of time.

The resulting expression may look like an IIR expression but if it contains y[n-i] then it can't be FIR

\$\endgroup\$
  • \$\begingroup\$ So it is as I was imagining. Unless the rest of the 'ai' coefficients turn out to be 0, and IIR filter is an IIR filter. Just out of curiosity, I was asking this because if it turned out to be so, my IIR filter would have a delay expression equal to that of a FIR filter. \$\endgroup\$ – ravemir Mar 19 '13 at 13:49
  • \$\begingroup\$ The feedback path of an IIR could itself be an FIR, but as it is a feedback rather than feedforward, it makes the overall filter IIR. \$\endgroup\$ – Chris Stratton Mar 19 '13 at 13:54
1
\$\begingroup\$

If one has a "quadratic" equation of the form Ax²+Bx+C, and the A term is zero, the resulting curve will look like a straight line--because it will be a straight line. On the other hand, equations of that form where A is zero are generally termed "linear equations" with the A term omitted.

Likewise, it is possible to express a general form of linear filter which will be an IIR filter if any of the coefficients which describe the exponentially-decaying parts of the response are non-zero, or an FIR filter if all such coefficients are zero.

One added slight wrinkle is that even with an IIR filter it's possible for a precise sequence of impulse stimuli to cause the output to quickly go to and remain at exactly zero even though a single impulse stimulus would cause the output to be forever non-zero.

\$\endgroup\$
0
\$\begingroup\$

If there are y[n-i] terms its an IIR. That's because the output will asymptotically approach zero, hence "Infinite Impulse Response".

\$\endgroup\$

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.