0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ where does $H(e^{j \omega})$ come from? $\endgroup$ – robert bristow-johnson Jul 6 '16 at 2:40
  • $\begingroup$ Robert, that's the definition that the book from Oppenheim and Schafer use (Discrete-time Digital Processing Signal) $\endgroup$ – Euler Jul 6 '16 at 3:06
  • $\begingroup$ So, look at what would happen if $A$ isn't real. $\endgroup$ – Peter K. Jul 6 '16 at 3:07
  • $\begingroup$ 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 $\endgroup$ – Euler Jul 6 '16 at 3:29
  • $\begingroup$ 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? $\endgroup$ – robert bristow-johnson Jul 6 '16 at 3:48
1
$\begingroup$

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 $B(\omega)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.