# Synthesize FIR filters using high-school algebra (Part 2)

**The story so far**

In __ Part 1__ we took a regular FIR filter design and wrote down the filter coefficients in polynomial form to get equation [1]:

Then we found all the roots of this polynomial, and used them to write down the

factorized form of the polynomial:

As a parting shot, I pointed out that there are three quadratic terms there with unity coefficients of z^0 – and there are three deep nulls in the gain response of the filter, as was shown in the figures from Part 1. Let's take a deep breath and examine the responses of all these individual linear and quadratic factors, to see if there are some clues there.

Figure 1 shows the individual responses, treated as two- or three-tap FIR filters, of each of the factors in parentheses in equation [2]; the five quadratic factors marked as q1 to q5 and the four linear factors and L1 to L4. It's quite a jumble of a graph, but you don't have to be very awake to see the major salient detail: three of those quadratic factors have deep notches in the frequency response. These are indeed the three factors whose constant (z^0) coefficient is unity!

So, here's the first takeaway. In an FIR filter whose stopband contains a number of sharp nulls, each one of them comes into being because the response of one of the polynomial's quadratic factors falls to zero at one frequency. Just so that we don't jump to conclusions about the particular form the factor needs to have, let's do some more algebra to make sure. Are we having fun yet?

**Figure 1: The frequency response of all the quadratic and linear factors of equation [2]**