# Why a pair of complex-conjugate zeros provide a nulling filter? (FIR filter case)

FIR filter having a pair of complex-conjugate zeros that lie on the unit circle, with zeros of the form: $$z_1 = e^{j\omega_i}\qquad\text{and}\qquad z_2 = e^{-j\omega_i}$$ And transfer function: $$H(z) = \left(1 - z_1z^{-1}\right)\left(1 - z_2z^{-1}\right) = 1 - 2\cos\left(\hat{\omega}_i\right)z^{-1} + z^{-2}$$

Has the following pole-zero map, and magnitude & frequency response graph:

But what is the intuitive idea behind that? And how the locations of zeros will affect the result? Thank you!

The unit circle on the z-plane represents the frequency axis, similar to the imaginary axis $$j\Omega$$ on the s-plane for the Laplace Transform in the continuous time case. So the frequency response of the system is given by $$H(z)$$ when $$z= e^{j\omega}$$ with $$\omega$$ going from $$0$$ to $$2\pi$$ representing the normalized fractional radian frequency (which is the continuous time radian frequency $$2\pi f$$ divided by the sampling rate $$f_s$$.

That said, any zero on the unit circle will create a null in the frequency response. With the OP's case of complex conjugate zeros (resulting in a real response), two nulls would result as shown. The location of the zero if on the unit circle, is the fractional radian frequency where $$H(z) = 0$$, thus called a "zero".

If the zero is not on the unit circle, the null will not be zero but will be lower the closer the zero is to the unit circle for that frequency.

This may be clearer from the plot below showing the frequency response for a 2 point moving average filter, which has a zero at $$z= -1$$. The frequency response is $$H(z)$$ as z sweeps over the unit circle, thus giving a numerator magnitude as the difference between z at any point on the unit circle and the zero location: $$z-q_z$$ (or the multiplication of multiple such magnitudes if there is more than one zero), and a denominator magnitude given by the same for the pole locations: $$z-q_p$$. In this case the pole is at the origin, so $$z-q_p=1$$ for all $$z=e^{j\omega}$$. What also should be now clear is how the resulting phase response is formed since the net phase will be the difference between the phase of the numerator and the phase of the denominator (phases subtract in the division of complex numbers).

This type of nulling filter (zero-only) is not very effective given the gradual roll-off in frequency. To achieve very sharp nulls, place a pole very close to the zero; the closer the pole, the sharper the response! Given all poles must be inside the unit circle for a stable causal linear time invariant system, the magnitude of the pole would therefore be less than but close to 1.

This IIR approach is further detailed here: Transfer function of second order notch filter

Also this is an excellent write-up by Richard Lyons on linear phase nulling (or notch) filters that do provide a sharp notch with an FIR appraoch. This could similarly be translated to provide a notch at any frequency: https://www.dsprelated.com/showarticle/58.php

• Thank you very much for the answer. – Ece Su Ildiz Jun 7 '20 at 13:52