Why Is n = 3 Not Considered in Convolution?

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Discussion Overview

The discussion revolves around the convolution of two discrete signals, specifically addressing why the value at n = 3 for the input signal x[n] is not included in the resulting convolution output y[n]. The context is primarily homework-related, focusing on the mathematical reasoning behind convolution operations involving delta functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about the convolution result, questioning why the value at n = 3 for x[n] appears to be ignored in the solution.
  • Another participant suggests that the output y[n] is longer than shown in the answer, but most values are zero, implying that the product at n = 3 does not contribute to the result.
  • A further reply encourages visualizing the convolution by graphing the signals, indicating that the overlap at n = 3 results in a product of zero, which explains its absence in the final output.

Areas of Agreement / Disagreement

Participants generally agree that the value at n = 3 does not contribute to the convolution result due to it being zero, but the initial confusion about its omission indicates some uncertainty in understanding the convolution process.

Contextual Notes

The discussion highlights the importance of visualizing convolution and understanding how the properties of delta functions affect the output, particularly in terms of which values contribute to the final result.

Who May Find This Useful

This discussion may be useful for students learning about convolution in signal processing, particularly those working with discrete signals and delta functions.

Larrytsai
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Homework Statement


δ = dirac delta
x[n] = δ[n] + 2δ[n-1] - δ[n-3]
h[n] = 2δ[n+1] + 2δ[n-1]

y[n] = x[n]*h[n]


Homework Equations



y[n] = x[n]h[n] = [tex]\sum[/tex]h[k]x[n-k]

The Attempt at a Solution



I have graphed the x[-k] and h[n], the solution saids

y[n] = h[-1]*x[n+1] + h[1]*x[n-1]
= 2x[n+1] + 2x[n-1]

I do not understand why, it seems like the solution forgot about the value at n = 3 for x[n].

can anyone help explain to me what happened in the solution and why n = 3 is not considered in the convolution.
 
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It is hard to visualize convolution problems. Did you solve it by hand?
 
y[n] is longer than what's shown in the answer but most of them are zero. that's why n=3 is not in the answer because that prduct is zero. you can solve the product of sum with [itex]-1 \leq n \leq 4[/itex] and see what you get
 
Would you mind posting your solution? Graphing convolution problems - particular with a system consisting of delays - generally shows the solution relatively simply. Try showing the x, h, and then as they overlap. You'll see, as Jaynte pointed out, that the product at n=3 is zero, therefore negating the need to write it out.
 

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