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I have designed an FIR filter to have linear phase response using odd-symmetry design. The coefficients of this filter are {2,1,3,1,0,-1,-3,-1,-2}. I am now being asked to prove it has linear phase response.

I don't think saying odd symmetry design is deemed a sufficient answer.

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  • $\begingroup$ so, why you think that saying the odd symmetry is sufficient? That's a well-known fact among people that deal with FIRs. Anyway, it's easy to solve this homework-style problem: What's the definition of linear phase response? $\endgroup$ – Marcus Müller May 7 at 19:00
  • $\begingroup$ Because the previous task was to design a filter with odd symmetry design. I know linear phase is a filter proerty whereby the phase is a linear function with frequency, im just wondering how i would show this from just the coefficients $\endgroup$ – Rebecca Meagher May 7 at 19:39
  • $\begingroup$ as said, it's easy to show if you just consider what the definition of linear phase response is, and how you can calculate the thing that you read that from based on FIR taps. $\endgroup$ – Marcus Müller May 8 at 5:16
  • $\begingroup$ if its easy to show please do it. If I could do it easily I obviously would $\endgroup$ – Rebecca Meagher May 8 at 8:01
  • $\begingroup$ That's why I'm giving you hints to solve your task! So, what's the definition of linear phase response? From which kind of graph can you see that? How do you generate that graph? $\endgroup$ – Marcus Müller May 8 at 8:52
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Remember that there are four types of linear-phase FIR filters. The filter in your example is a type III filter: odd filter length and odd symmetry. The frequency response of such a system has the form

$$H(e^{j\omega})=A(\omega)je^{-j\omega(N-1)/2}\tag{1}$$

where $N$ is the filter length (number of taps), and $A(\omega)$ is real-valued and odd, i.e., $A(\omega)=-A(-\omega)$. Now it's up to you to show that your filter has the form given by $(1)$. Make use of

$$\sin(x)=\frac{e^{jx}-e^{-jx}}{2j}\tag{2}$$

Note that because of the factor $j$ in $(1)$ and because $A(\omega)$ is odd, there's a phase jump of $\pi$ at $\omega=0$, so the phase is not strictly linear. Such a phase response is called generalized linear phase.

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