# FIR Linear Phase Condition

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

• 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. – Matt L. Jan 13 at 15:32

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.

• 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 $H(f)=M(f)e^{j\tilde{\phi}(f)}$ ? – themagiciant95 Jan 13 at 20:03
• @themagiciant95: Because I define it to be equal by choosing $M(f)=|A(f)|$ and choosing the phase according to Eq. (3). – Matt L. Jan 13 at 20:38