# Is there a simple way to express this DTFT in polar form?

Consider the discrete-time system $$H(z) = a_0 + a_1 z^{-1} + a_2 z^{-2}$$ To compute the DTFT, let $$z = e^{j\omega}$$ such that $$H(e^{j\omega}) = e^{-j\omega} \left(a_0 e^{j\omega} + a_1 + a_2e^{-j\omega}\right) \label{eq:H_w} \tag{1}$$ If $$a_0 = a_2$$, then $$H(e^{j\omega})$$ can be put into polar form as $$H(e^{j\omega}) = \left\lvert a_1 + 2 \cos \omega\right\rvert e^{-j\omega}$$ However, if $$a_0 \neq a_2$$, then it is not clear how $$H(e^{j\omega})$$ can be put into polar form. More precisely, substituting $$e^{j\omega} = \cos \omega + j \sin \omega$$ into \eqref{eq:H_w} yields \begin{align} H(e^{j\omega}) &= e^{-j\omega} \big(a_0 (\cos \omega + j \sin \omega) + a_1 + a_2(\cos \omega - j \sin \omega)\big) \\ &= e^{-j\omega} \big( (a_0 + a_2)\cos \omega + j(a_0 - a_2) \sin \omega + a_1\big) \end{align} Since $$(a_0 + a_2)\cos \omega + j(a_0 - a_2) \sin \omega$$ is a complex number, then it can be expressed in polar form as $$C(\omega)e^{j\theta(\omega)}$$, where \begin{align} C(\omega) &= \sqrt{((a_0 + a_2)\cos \omega)^2 + ((a_0 - a_2) \sin \omega)^2} \\ \theta(\omega) &= \operatorname{atan2}\big( (a_0 - a_2) \sin \omega,\ (a_0 + a_2)\cos \omega \big) \end{align} Therefore, \begin{align} H(e^{j\omega}) &= e^{-j\omega} \left( C(\omega)e^{j\theta(\omega)} + a_1\right) \end{align} At this point, I can factorize $$e^{j\frac{\theta(\omega)}{2}}$$ out of the parentheses and then simplify further, but I feel that this method is cumbersome. Is there an easier approach?

• There are different ways to slice this but none of them is "easy". Dec 10, 2021 at 15:12
• @Hilmar what are some other ways? If you expand on these other ways in an answer, I'm happy to accept it. Dec 10, 2021 at 17:30
• I have a general biquad frequency response thingie here if that does any good. Dec 10, 2021 at 21:56

Here is one way to do it. Let's call the coefficients "$$b$$" instead of "$$a$$" since "$$a$$" is usually used for the denominator and "$$b$$" for the numerator.

$$H(e^{j\omega}) = b_0 + b_1 e^{-j\omega} + b_2 e^{-j2\omega}$$

The phase is the inverse tangent of the the imaginary part divided by the real part so we get:

$$\phi(\omega) = \operatorname{tan}^{-1} \left( \frac{- b_1 \sin(\omega) - b_2 \sin(2\omega)}{1 + b_1 \cos(\omega) + b_2 \cos(2\omega)} \right)$$

provided the quadrant is chosen properly. I'm not sure if there is a way to simplify this further. You could try to pull out $$e^{-j\omega}$$ to make it more symmetrical, but I don't think this will make it any easier.

For the magnitude squared we just multiply with complex conjugate.

$$\big|H(e^{j\omega})\big|^2 = H(e^{j\omega}) \cdot H^*(e^{j\omega}) = (b_0 + b_1 e^{-j\omega} + b_2 e^{-j2\omega})(b_0 + b_1 e^{j\omega} + b_2 e^{j2\omega})$$

We just have to multiply this out and sort the terms. We get something like this.

$$\big|H(e^{j\omega})\big|^2 = b_0^2 + b_1^2 + b_2^2 + 2(b_0b_1 + b_1b_2)\cos(\omega) + 2b_0b_2\cos(2\omega)$$

None of these expressions is particularly pretty or intuitive, but that's just the way it is.

CAVEAT: I haven't double checked the math, so it's possible that there is an arithmetic error or typo in there.