# Generalized linear-phase filter

I know I can write the frequency response of a Generalized linear-phase system as: $$H(e^{j\omega}) = A(\omega) e^{-j\left(\alpha \omega - \beta\right)}$$ where $A(\omega)$ is real.

I need to prove that $A(\omega)$ is real and I can't find anywhere the demonstration.

• where does $H(e^{j \omega})$ come from? Commented Jul 6, 2016 at 2:40
• Robert, that's the definition that the book from Oppenheim and Schafer use (Discrete-time Digital Processing Signal) Commented Jul 6, 2016 at 3:06
• So, look at what would happen if $A$ isn't real.
– Peter K.
Commented Jul 6, 2016 at 3:07
• Using Euler's relation between $exp$, $cos$ and $sin$, I have that $H(e^{j\omega})$ is complex, so I don't know why $A(\omega)$ can't be complex Commented Jul 6, 2016 at 3:29
• do you know how to get from $H(e^{j \omega})$ to $$|H(e^{j \omega})| \, e^{j \phi(\omega)}$$ (where $\phi(\omega)=\arg\{ H(e^{j \omega}) \}$)? can you get yourself there? Commented Jul 6, 2016 at 3:48

Let's assume that $A(\omega)$ is complex valued:

$$A(\omega)=B(\omega)e^{j\phi_A(\omega)}\tag{1}$$

where $B(\omega)$ is a (possibly bipolar) real-valued function, and $\phi_A(\omega)$ is a (real-valued) phase function.

The phase of

$$H(e^{j\omega})=A(\omega)e^{-j(\alpha\omega-\beta)}=B(\omega)e^{j\phi_A(\omega)}e^{-j(\alpha\omega-\beta)}=B(\omega)e^{j\phi(\omega)}\tag{2}$$

is then

$$\phi(\omega)=\phi_A(\omega)-(\alpha\omega-\beta)\tag{3}$$

(where I define the phase such that jumps by $\pi$ are taken care of by the sign of $B(\omega)$, not by $\phi(\omega)$). For the phase $\phi(\omega)$ to be linear, $\phi_A(\omega)$ must be constant or linear:

$$\phi_A(\omega)=a\omega+b\tag{4}$$

With ($4$) the total phase becomes

$$\phi(\omega)=(a-\alpha)\omega+(b-\beta)=-\alpha'\omega+\beta'\tag{5}$$

Combining $(2)$ and $(5)$ we get

$$H(e^{j\omega})=B(\omega)e^{j\phi(\omega)}=B(\omega)e^{-j(\alpha'\omega-\beta')}\tag{6}$$

which has the same form as the equation in your question. Note that in $(1)$ we have defined $B(\omega)$ to be real-valued.

This shows that $(6)$ (or, equivalently, the equation in your question) is the most general form of the frequency response of a linear phase filter. If $A(\omega)$ were chosen to be complex-valued, then $A(\omega)$ must have a linear phase (otherwise $H(e^{j\omega})$ wouldn't have a linear phase), and that linear phase can be absorbed by the constants $\alpha$ and $\beta$, such that the expression for the frequency response can always be reduced to the form $(6)$ with a real-valued function $B(\omega)$.