# FIR filter (from zeros to filter coefficients)

Assume $F=1$ $(T=1)$. A 2nd order FIR filter with real coefficients has a zero in $H(e^{j\omega T})$ at $z_0=e^{j\omega /6}$. In addition, $|H(e^{j\pi T})| = 1$.

What are the coefficients of the FIR filter?

current progress

I'm having trouble understanding the $H(e^{j\omega T})$ function. Tried to start from

$$y[n] = \sum_{i=0}^2 b_i z[n-2].$$

Does that fact that it has a zero in a point means I must apply zero-pole rules here?

• Hi. Is that a homework question? – jojek Apr 3 '16 at 8:45
• If you show us what you did and where you're stuck, people here will help you. It's not enough to just post the exercise. – Matt L. Apr 3 '16 at 9:27
• $H(z)$ is a polynomial. You have the zeros and a scaling factor. What you need to do is convert the representation of a polynomial by its zeros to the representation by polynomial coefficients. – Matt L. Apr 3 '16 at 10:11
• If your last equation describes the input-output relationship of the FIR filter, you should replace the argument $n-2$ by $n-i$, and it would be good to use $x[n]$ instead of $z[n]$ for the input, because $z$ is the independent variable of the Z-transform, e.g. in $H(z)$. – Matt L. Apr 3 '16 at 11:16

If you remember, there is a theorem in calculus saying: for a polynomial to have real coefficients any complex roots which are not on the real line must come in complex conjugate pairs. So if $z_0 = e^{j\omega /6}$ then $z_1 = e^{-j\omega / 6}$ and that gives : $$(z - z_0)(z - z_1)C = (z - e^{j\omega /6})(z - e^{-j\omega /6})C$$
Then to find the constant $C$ solve the equation with the absolute value of the polynomial equals 1 for $z = e^{j\pi}$.