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?
First of all a discrete filter having a zero somewhere means it's Z-transform (which is a polynomial) acquires the value 0 there.
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 $z = e^{j\pi}$.
