how should I stem convolution?

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Armin Javidfar
Armin Javidfar 2015 年 12 月 25 日
コメント済み: Image Analyst 2015 年 12 月 26 日
can anyone help me draw this convolution?
y[n] = dirac[n-1] * (U[n] - U[n-6])
I'm very new to matlab and I need some help to figure out my problem. tnx in advance...
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Walter Roberson
Walter Roberson 2015 年 12 月 25 日
What is dirac[n-1] intended to mean?
Is U a continuous function that is being sampled at integer locations? Or is it an array that is being indexed? If it is an array then y[1] through y[6] are not defined.
Armin Javidfar
Armin Javidfar 2015 年 12 月 25 日
this is the question

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回答 (2 件)

Walter Roberson
Walter Roberson 2015 年 12 月 25 日
Under a quite reasonable interpretation of dirac[n-1], the function y[n] would always be 0 . Unless U can assume infinite values, in which case y would be undefined.
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Image Analyst
Image Analyst 2015 年 12 月 25 日
You can convolve with delta functions. It's done all the time. Basically it makes replicates of things. Like the Fourier Transform of a rectangular aperture is a 2D sinc function and so the Fourier transform of 2D array of square holes (one hole convolved with a 2D comb function) is an array of 2D sinc functions (a signle sinc convolved with a 2D comb).

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Image Analyst
Image Analyst 2015 年 12 月 25 日
Hint to get the difference of the U's
kernel = ones(1, 13);
kernel(7) = 1;
kernel(end) = -1;
du = conv(U, kernel, 'full');
  4 件のコメント
Walter Roberson
Walter Roberson 2015 年 12 月 25 日
It has to be time in this context. If you were sampling in space you would need to refer to the content of the location not just the index. For example, delta(X[n-1])
Image Analyst
Image Analyst 2015 年 12 月 26 日
I don't follow. n could be an index representing a location in time with units of seconds, or an index representing a location in space with units of mm or whatever. I believe convolution by the dirac delta simply shifts the difference of U's signal one index to the left. I wonder where this equation came from. Maybe it's a 1-D spatial masking situation, or if it's a time signal maybe it's some kind of feedback reduction/echo cancellation filter.

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