0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Hi. Is that a homework question? $\endgroup$ – jojek Apr 3 '16 at 8:45
  • 1
    $\begingroup$ 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. $\endgroup$ – Matt L. Apr 3 '16 at 9:27
  • 2
    $\begingroup$ $H(z)$ 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. $\endgroup$ – Matt L. Apr 3 '16 at 10:11
  • $\begingroup$ 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 $H(z)$. $\endgroup$ – Matt L. Apr 3 '16 at 11:16
1
$\begingroup$

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 $z = e^{j\pi}$.

$\endgroup$

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.