Can the magnitude of a discrete-time Fourier transform be negative?

Question

Consider the discrete-time system $$ H(z) = 1 + z^{-1} + z^{-2} + z^{-3} $$ To obtain the magnitude of the discrete-time Fourier transform, I substitute $z = e^{j\omega}$ to get \begin{align} H(\omega) &= 1 + e^{-j\omega} + e^{-2j\omega} + e^{-3j\omega} \\ &= e^{-\frac{3}{2}j\omega} \cdot \left(e^{\frac{3}{2}j\omega} + e^{\frac{1}{2}j\omega} + e^{-\frac{1}{2}j\omega} + e^{-\frac{3}{2}j\omega}\right) \\ &= e^{-\frac{3}{2}j\omega} \cdot \left(2 \cos\left(\frac{3}{2}\omega\right) + 2 \cos\left(\frac{1}{2}\omega\right)\right) \end{align} As $H(\omega)$ is now in polar form, such that \begin{align} r(\omega) &= 2 \cos\left(\frac{3}{2}\omega\right) + 2 \cos\left(\frac{1}{2}\omega\right) \\ \theta(\omega) &= -\frac{3}{2}\omega \end{align} Then the magnitude of $H(\omega)$ is just $r(\omega)$. However, $r(\omega)$ can be negative for some $\omega$, which does not make sense to me, as the magnitude of a complex number can't be negative. What am I missing?

Answer 1

You can't conclude that the magnitude response is $r(\omega)$ and the phase response is $\theta(\omega)$.

Note that $e^{j\pi}=-1$, at the frequencies that $r(\omega) < 0$, let $\phi(\omega) = \theta(\omega) + \pi$ and you get the non-negative magnitude response.

The real magnitude response is $|r(\omega)|$ and the phase response is

$$ \phi(\omega)=\left\{ \begin{aligned} &\theta(\omega) , & r(\omega)\geq 0 \\ &\theta(\omega) + \pi , & r(\omega)<0 \end{aligned} \right. $$

Answer 2

Then the magnitude of H(ω) is just r(ω).

No.

By definition, the magnitude of a complex number is always non-negative.

Let's look at something simple: $z = -1$. The real part is -1, the imaginary part is 0, the magnitude is 1 and the phase is $\pi$ (or $\pi + k \cdot 2\pi , k \in\mathbb{Z} $ to be precise), i.e

$$z = -1 + j \cdot 0 = 1 \cdot e^{j \pi} $$