I'm going over an exercise of designing an simple FIR filter, I have the solution in front of me, but I struggle to understand some parts of it.

I'm asked to have an FIR filter with only 3 coefficients.

I have reached this form: $$ H(\theta ) = (2\alpha_0 \cos(\theta)+\alpha_1)e^{-j\theta} $$ where $\theta$ is the frequency variable (DTFT).

The magnitude is easy: $$ \left | H(\theta ) \right |=\left | 2\alpha_0 \cos(\theta)+\alpha_1 \right | $$

But I struggle with the phase, it seems easy: $\phi(H(\theta))=-\theta$ , but I guess this is wrong ? why ?

The solution that is given is:

$$\phi(H(\theta)) = -\theta +\beta$$ $$\beta = 0 \text{ for } 2\alpha_0 \cos(\theta) + \alpha_1 > 0$$

$$\beta = \pi \text{ for } 2\alpha_0 \cos(\theta) + \alpha_1 < 0$$


I think I just answers figured it out, if that expression is negative we can write it as positive with a phase of π. What happens if $\alpha_1$ is larger than $4\alpha_0$ thus the expression is always positive ? does $\beta =0$ ?

  • $\begingroup$ The answer to your last question is yes. It is sufficient if $\alpha_1$ satisfies $\alpha_1>2|\alpha_0|$. $\endgroup$
    – Matt L.
    Feb 13, 2015 at 13:31

2 Answers 2


Your frequency response is of the form


where $A$ is a real-valued number that can take either a positive or a negative sign. $M\ge 0$ is the magnitude, and $\phi(\theta)$ is the phase. For positive $A$ you have $A=M$, and consequently $\phi(\theta)=-\theta$. However, when $A$ is negative you have $A=-M$ and


because $e^{\pm j\pi}=-1$. So the additional value of $\pi$ in the phase occurs because of the sign changes in $A$. If $A$ satisfied $A\ge 0$, your solution would be correct. The latter is the case if $\alpha_1>2|\alpha_0|$.


Your answer does not take into account the possible sign of the magnitude term in the polar description of the frequency response $H(\theta) = A(\theta)e^{j\phi(\theta)}$. I guess you mix this definition with the more usual definiton of $H(\theta) = |A(\theta)|e^{j\phi(\theta)}$ where the magnitude will always be positive and therefore phase will be what you predicted, however for your example $A(\theta)$ may assume a negative value and you shall reflect this as an addition of $\pi$ into the phase where $e^{j(\theta + \pi)} = -e^{j\theta} $


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