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I'm currently learning about FIR filtering from one of my modules and went through topics such as transverse implementation(or structures since I could not find transverse implementation on the net) and was confused at a lab question that had asked for the transverse implementation structure for a filter defined by the following:

enter image description here In which I got this as the difference equation: y(n) = 0.010x + 0.220x(n-1) + 0.538x(n-2) + 0.220x(n-3) + 0.010x(n-4)

All that was shown was the transverse diagram, and that was it. Now I did try searching Google for answers, but I could not find any "useful" ones as some would be videos on how to draw the diagram and that would be it, nothing about implementing any FIR filters on it.

Therefore I'm genuinely unsure as to what I'm actually supposed to do in order to answer such a question, am I just supposed to attach the coefficients to where the b's are? I honestly do not know and any pointers/explanations regarding this would be greatly appreciated.

enter image description here Example Diagram of Transverse Structure

Apart from this was another topic that had me a little unsure of, which is the calculation of output samples(specifically five) with a provided FIR filter. From this filter:

enter image description here I obtained this: 𝑦(π‘š)=0.0265(π‘₯)+0.4735π‘₯(π‘šβˆ’1)+0.4735π‘₯(π‘šβˆ’2)+0.0265π‘₯(π‘šβˆ’3)

When plugging this on Excel, I ended up getting the below: enter image description here

Is this correct? Or have I missed something? If I have left out details about this then please do let me know.

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1 Answer 1

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It seems like you are doing everything correct. Let me point out that your first difference equation should be the following in case the missing first $x(n)$ was throwing you off:

$$y(n) = 0.010x(n) + 0.220x(n-1) + 0.538x(n-2) + 0.220x(n-3) + 0.010x(n-4)$$

From that you should readily see that your diagram describes this difference equation directly; it is simply a sum of products. The output $y(n)$ is the sum of $x(n)$ times $0.020$ plus the rest of the terms. Observe in the block diagram that your output $y(n)$ is the sum of $x(n)$ times $b(0)$ plus the rest of the terms.

As far as solving for the filter results using Excel (which wouldn't be the typical analysis tool but may help provide some immediate insight), at time zero each of the filter terms can be assumed to have zero values (unless there was some other initial condition). So your approach would be correct if you include -1, -2 and -3 for n with values 0 for $x(n)$.

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  • $\begingroup$ Alright got it, thank you so much for your response! $\endgroup$
    – Thanos
    May 3, 2022 at 8:11

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