Are the eigenvalues of the FIR convolution matrix the zeros of the corresponding FIR filter?

Suppose I have an FIR filter $H(z) = h_{0} + h_{1}z^{-1} + h_{2}z^{-2}$. I want to implement it using a matrix, so I have:

$$ H = \begin{bmatrix} h_{0} & 0 & 0 \\ h_{1} & h_{0} & 0 \\ h_{2} & h_{1} & h_{0} \end{bmatrix} $$

Can I find the zeros of $H(z)$ by finding the eigenvalues of $H$?

I know that in general, the roots of a polynomial are given by the eigenvalues of its companion matrix. However, can I just find the eigenvalues of $H$ for this example instead?


For any convolution matrix (even truncated), your diagonal entries would be $h_0$; hence, the characteristic polynomial, setting $b_0 = h_0-\lambda$,

\begin{align}|H-\lambda I| = \begin{vmatrix} b_0 & 0 & 0\\h_1 & b_0 & 0\\h_2 & h_1 & b_0 \end{vmatrix} &= b_0\,\begin{vmatrix} b_0 & 0\\h_1 & b_0 \end{vmatrix} - 0\,\begin{vmatrix} h_1&0\\h_2&b_0\end{vmatrix} + 0\,\begin{vmatrix} h_1 & b_0\\h_2&h_1 \end{vmatrix} \\ \\ &=b_0\,\begin{vmatrix} b_0 & 0\\h_1 & b_0 \end{vmatrix}\\ &=b_0\left(b_0\,b_0 - 0\,h_1 \right)\\ &=b_0^3 \end{align}

Little induction is necessary to see that for any square matrix of dimension $N\times N$ with an lower left structure and a constant $d$ in each element of the diagonal, the characteristic polynomial is $(d-\lambda)^N$; the Eigenvalues are roots of that polynomial, and hence there's but one Eigenvalue (with $N$ multiplicity): $d$, which in your case is $h_0$.

So, since not all zeros of $H(z)$ are defined by your first filter tap, no, you can't find all zeros of $H(z)$ through Eigenvalues.

| improve this answer | |

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.