# Notch filters: why is $H(e^{j0}) = H(e^{j\pi})$ required?

I'm studying for the final exams and many example exercises require to design fir notch filters of length 5, given some passband and some stopband frequencies. You've got to find 3 coefficients α0, α1, α2. You need 3 equations.

• R(passband ω) = 1
• R(stopband ω) = 0

I completely understand why these are here. The final degree of freedom is determined by the equation $$H(e^{j0}) = H(e^{j\pi})$$, because such thing is true for notch filters. I struggle to understand why this is true.

• Could you post the complete exercise? I'm a bit confused about the use of the term "passband frequencies", because for a notch filter any frequency is a passband frequency, except for the notch frequency. Also, if you have a length $5$ filter and just $3$ coefficients, I guess the FIR filter is required to have linear phase (but that's not explicitly stated in your question). – Matt L. Jun 23 '19 at 20:37
• Yes @MattL. that's correct, it's required to have linear phase (they don't tell us that which is a bit weird). Your answer is sufficient. – ntakouris Jun 23 '19 at 21:00

In general, there is no such requirement for notch filters that $$H(e^{j0})=H(e^{j\pi})$$ must be satisfied. You could definitely have a notch filter with $$H(e^{j0})\neq H(e^{j\pi})$$. Having the same gain at DC and at Nyquist is just a practical definition, and if you have a sufficient number of degrees of freedom (i.e., filter coefficients) you might as well make it a requirement. Ideally you want $$|H(e^{j\omega})|=1$$ for all frequencies except for the notch frequency $$\omega_0$$, for which you require $$H(e^{j\omega_0})=0$$.
As a simple example, take the most primitive notch filter, namely a FIR filter of length $$3$$. Its frequency response must be
$$H(e^{j\omega})=c\cdot \left(1-2z^{-1}\cos(\omega_0)+z^{-2}\right)\tag{1}$$
where $$\omega_0$$ is the notch frequency, and $$c$$ is a scaling constant. All degrees of freedom are taken by the requirement of the notch at $$\omega_0$$, so you don't get equal gains at DC and at Nyquist. Yet, it is a notch filter.