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This is the linear phase condition for FIR filters as expressed by my prof:

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I don't understand why $G(f)$ can be negative. Isn't the Fourier transform expressed in polar form ? So the magnitude is always positive.

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  • $\begingroup$ I edited your question changing $\beta(f)$ to $G(f)$, because I think you're wondering why $G(f)$ (and not $\beta(f)$) can be negative. The phase can be negative anyway. $\endgroup$ – Matt L. Jan 13 at 15:32
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You can describe a frequency response in terms of its real-valued amplitude and its phase:

$$H(f)=A(f)e^{j\phi(f)}\tag{1}$$

Note that $A(f)$ is not the magnitude, but a bipolar amplitude function. Equivalently, you can express $H(f)$ in terms of its magnitude and its phase:

$$H(f)=M(f)e^{j\tilde{\phi}(f)}\tag{2}$$

Now we have $M(f)=|A(f)|\ge 0$, and, consequently, the phase $\tilde{\phi}(f)$ in $(2)$ is generally different from the phase $\phi(f)$ in $(1)$. For frequencies for which $A(f)<0$ is satisfied, we have to add $\pi$ (or $-\pi$) to the phase to compensate for the sign change:

$$\tilde{\phi}(f)=\begin{cases}\phi(f),& A(f)\ge 0\\\phi(f)\pm\pi,&A(f)<0\end{cases}\tag{3}$$

For a linear phase filter of type I or type II (even symmetry), the phase $\phi(f)$ is a linear function of $f$, where $\phi(f)$ is defined as in $(1)$, i.e., with a bipolar amplitude function. The phase $\tilde{\phi}(f)$ as defined in $(2)$ has jumps wherever $A(f)$ changes its sign.

You can extend the definition to type III and type IV linear phase filters (odd symmetry) by expressing $H(f)$ as

$$H(f)=jA(f)e^{j\phi(f)}\tag{4}$$

where $A(f)$ is again a real-valued bipolar amplitude function, and $\phi(f)$ is a linear function of $f$.

Note that by linear function I mean a strictly linear function without an additive term, i.e., $\phi(f)=a \cdot f$, with $a=-2\pi\tau_g$, where $\tau_g$ is the filter's group delay.

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  • $\begingroup$ Okay, let's suppose i calculate $H(f)=A(f)e^{j\phi(f)}$ . Why $H(f)=A(f)e^{j\phi(f)}$ is equal to $H(f)=M(f)e^{j\tilde{\phi}(f)}$ ? $\endgroup$ – themagiciant95 Jan 13 at 20:03
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    $\begingroup$ @themagiciant95: Because I define it to be equal by choosing $M(f)=|A(f)|$ and choosing the phase according to Eq. (3). $\endgroup$ – Matt L. Jan 13 at 20:38

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