Factorization of transfer function using its roots

Question

I'm missing a step to understand the factorization of the FIR filter transfer function: $$H(z)=\sum\limits _{k=0}^{M}b_{k}z^{-k} \tag{1}$$

From DSP First:

The $z$-transform of a finite-length signal, such as $H(z)$ for an FIR filter, is a function of the complex variable $z$, and it is also an $M^{th}$-degree polynomial in the variable $z^{−1}$. Therefore, $H(z)$ has exactly $M$ roots according to the fundamental theorem of algebra. When $H(z)$ is expressed in terms of its roots, it can be written as the product of first-order factors that completely define the polynomial to within a multiplicative constant, that is, $$H(z)=G\prod_{k=1}^{M}(1-z_{k}z^{-1})=G\prod_{k=1}^{M}\frac {(z-z_{k})}{z}\tag{2}$$

The point I don't understand is the $z^{-1}$ factor. I expected to see only factors corresponding to roots:

$$H(z)=G\prod_{k=1}^{M}(z-z_{k}) \tag{3}$$

I know the book is correct (perhaps a function with negative exponents is not exactly a regular polynomial though), but I need a little help to make sense of how the sentence "When $H(z)$ is expressed in terms of its roots, it can be written..." can lead to equation (2) instead of (3).

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Update with a partial understanding:

1/ I found intriguing the author used the word polynomial to describe the sum, and mentioned the use of the roots to factorize it. I think I got a clue... The author actually transforms the Laurent's series (negative exponents) into a polynomial (positive exponents) without explicitly saying it. My guess:

$$H(z)=\sum\limits _{k=0}^{M}b_{k}z^{-k}=\sum\limits _{k=0}^{M}b_{k}z^{M-k}z^{-M} \tag{4}$$

2/ This polynomial has roots, and they can be used for factorization:

$$H(z)=\sum\limits _{k=0}^{M}b_{k}z^{M-k}z^{-M}=G\prod_{k=1}^{M}(z-z_{k})z^{-M} \tag{5}$$

However at this stage I'm lost since the additional factor is not the one in the book: $z^{-M} \ne z^{-1}$

Answer 1

The z transform of a unit sample delay is $z^{-1}$. The z transform of a delay one sample in the future is $z^1=z$ which is non-causal. For example, a simple two element FIR filter with unity gain coefficients, which is the sum of the current sample with the previous sample is given as $H(z) = 1+z^{-1}$.

If we divide the numerator and denominator both by $z$, we get:

$$H(z) = 1+z^{-1} = \frac{z+1}{z}$$

We see here how all causal Nth order FIR filters must have $N$ poles (the roots of the denominator) at the origin (where $z=0$), since a causal filter can only be constructed with delays ($z^{-1}$), not "aheads" ($z^1$).

We see how, to the OP’s question, the solution cannot be written as the product of first order factors in the form of $(z-z_k)$ with positive exponents of $z$ alone, as that would be non-causal as described above and must instead include an additional $z$ in the denominator for each factor to be causal (which means multiply by $z^{-1}$). Thus, the solution can be written as the product of first order factors in the form of $1-z_k z^{-1}$.

Regarding the OP’s update, equation 5 is incorrect; this would have a $z^{-1}$ term and not $z^{-M}$ as shown in the final equation given; $z^{-1}$ is multiplied $M$ times which will provide the $z^{-M}$ term given in the OP’s summation form:

$$H(z)=\sum\limits _{k=0}^{M}b_{k}z^{M-k}z^{-M}=G\prod_{k=1}^{M}(z-z_{k})z^{-1}=Gz^{-M}\prod_{k=1}^{M}(z-z_{k})$$

Answer 2

The most basic description of an FIR filter is it's difference equation:

$$y[n] = \sum_{k=0}^{N}h[k]\cdot x[n-k] \tag{10}$$

Taking the z-transform gives you directly your equation (1) since the z-transform of time domain delay of m samples is simply $z^{-m}$.

That's why you end up with a polynomial in $z^{-1}$. The rest is just algebra. In order to turn this into a polynomial in z you need to multiply both numerator and denominator with $z^{N}$.

What that all means at the end is that an FIR filter does not only have N zeros, it has also N poles at $z=0$.