Hi Folks, I read that the inverse of an FIR filter is always an IIR filter. I have no idea how to prove this mathematically. If anybody could tell me where to start, I'd be more than happy. I'm a DSP newbie, please do excuse me if the question is a little stupid. Thanks, Karthik.
inverse of an FIR filter
Started by ●January 18, 2005
Reply by ●January 19, 20052005-01-19
easy. let U(z),Y(Z) and H(z) denote the input, output Z-transforms, and the transfer function of the FIR filter, respectively. Then Y(z)=H(z)U(z) and equivalently U(z)=1/H(z) Y(z). hence the inverse of the FIR filter is IIR.
Reply by ●January 19, 20052005-01-19
Karthik wrote:> Hi Folks, > > I read that the inverse of an FIR filter is always an IIR filter. I > have no idea how to prove this mathematically. If anybody could tellme> where to start, I'd be more than happy.Step 1: Derive the frequency-domain expression for the transfer function of a FIR filter, given the filter coefficients. In other words, use the coefficients of the time-domain difference equation to find the frequency domain function. Step 2: Find the filter inverse to the FIR in frequency domain Step 3: Find a general expression for the digital filter in frequency domain, given both feed-forward coefficients and feedback coefficients (both the a's and b's in the difference equation) Step 4: Compare the filter you found in step 2 with the general expression you found in step 3. How do the non-zero coefficients from step 2 fit into the expression of step 3? Rune
Reply by ●January 19, 20052005-01-19
"Karthik" <karthik301176@gmail.com> writes:> Hi Folks, > > I read that the inverse of an FIR filter is always an IIR filter. I > have no idea how to prove this mathematically.That's good, because it's false. Here's a counterexample: Consider the FIR H(z) = z^{-1}. The inverse filter is G(z) = z. The impulse response of G(z) is g[n] = d[n+1], where d[] is the unit impulse function. This is a finite impulse response, hence the claim is disproven. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Reply by ●January 19, 20052005-01-19
bryant_j_j@yahoo.com wrote:> easy. let U(z),Y(Z) and H(z) denote the input, output Z-transforms, and > the transfer function of the FIR filter, respectively. Then > Y(z)=H(z)U(z) and equivalently U(z)=1/H(z) Y(z). hence the inverse of > the FIR filter is IIR.Continuing the discussion, there isn't a requirement that an IIR filter actually have an infinite impulse response, only that it may have one. That is, FIR are a subset of IIR. For an example, consider H(z)=1. -- glen
Reply by ●January 19, 20052005-01-19
Randy Yates wrote:> "Karthik" <karthik301176@gmail.com> writes: > > >>Hi Folks, >> >>I read that the inverse of an FIR filter is always an IIR filter. I >>have no idea how to prove this mathematically. > > > That's good, because it's false. Here's a counterexample: Consider > the FIR H(z) = z^{-1}. The inverse filter is G(z) = z. The impulse > response of G(z) is g[n] = d[n+1], where d[] is the unit impulse > function. This is a finite impulse response, hence the claim is > disproven.That's neat! But I think it's fair to say that the original assertion is true PROVIDED THAT both impulse responses are causal. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 19, 20052005-01-19
Jerry Avins wrote:> Randy Yates wrote: > > >>"Karthik" <karthik301176@gmail.com> writes: >> >> >> >>>Hi Folks, >>> >>>I read that the inverse of an FIR filter is always an IIR filter. I >>>have no idea how to prove this mathematically. >> >> >>That's good, because it's false. Here's a counterexample: Consider >>the FIR H(z) = z^{-1}. The inverse filter is G(z) = z. The impulse >>response of G(z) is g[n] = d[n+1], where d[] is the unit impulse >>function. This is a finite impulse response, hence the claim is >>disproven. > > > That's neat! But I think it's fair to say that the original assertion is > true PROVIDED THAT both impulse responses are causal. > > JerryGlen's H(z) = 1 is a counterexample. Are there others? jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 19, 20052005-01-19
Jerry Avins wrote: (someone wrote)>>>>I read that the inverse of an FIR filter is always an IIR filter. I >>>>have no idea how to prove this mathematically.(snip)> Glen's H(z) = 1 is a counterexample. Are there others?Yes, but I also claimed that FIRs are a subset of IIRs. Consider the problem of the reciprocals of integers. 1/3 is a repeating decimal, 1/5 is not. I used to know some math people who claimed that you could write 1/5 instead of 0.2, as 0.199999999999999999999...., that is, as a repeating decimal. Of all possible FIRs only a very small fraction have an FIR as their inverse, but you might get lucky. -- glen
Reply by ●January 19, 20052005-01-19
glen herrmannsfeldt wrote:> Jerry Avins wrote: > (someone wrote) > >>>>> I read that the inverse of an FIR filter is always an IIR filter. I >>>>> have no idea how to prove this mathematically. > > > (snip) > >> Glen's H(z) = 1 is a counterexample. Are there others? > > > Yes, but I also claimed that FIRs are a subset of IIRs. > > Consider the problem of the reciprocals of integers. > 1/3 is a repeating decimal, 1/5 is not. I used to knowThat seems artificial to me, as it depends on the representation base. A number is divisible by 7 if the sum of its digits are, provided you work in octal.> some math people who claimed that you could write > 1/5 instead of 0.2, as 0.199999999999999999999...., that is, > as a repeating decimal. > > Of all possible FIRs only a very small fraction have an > FIR as their inverse, but you might get lucky. > > -- glen >Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●January 19, 20052005-01-19
