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So I think this video:

https://www.youtube.com/watch?v=N-zd-T17uiE&t=67s

By Faculty of Khan does a wonderful job in explaining what convolutions are. We basically consider two pulses $f(\tau)$ and $g(\tau)$ and "sweep" $g(t-\tau)$ from $- \infty$ to $\infty$ (We do this by taking $\int_{-\infty}^{\infty}$). At points $f(\tau)$ and $g(\tau)$ overlaps, the value of $(f*g)(t)$ become non-zero.

However, I would like to quote a paragraph from KF Riley 3rd Edition (page 447). Where the explanation of convolution is in terms of equipment resolution.

The probability that a true reading lying between $x$ and $x+dx$, and so having probability $f(x)dx$ of being selected by the experiment, will be moved by the instrumental resolution by an amount $z-x$ into a small interval of width $dz$ is $g(z-x)dz$. Hence the combined probability that the interval dx will give rise to an observation appearing in the interval $dz$ is $f(x)dxg(z-x)dz$. Adding together the contributions from all values of $x$ that can lead to an observation in the range $z$ to $z+dz$, we find that the observed distribution is given by: $$h(z) = \int_{-\infty}^{\infty}f(x)g(z-x)dx$$

I think this is confusing. How exactly are the bold sentences linked to the idea of two signals affecting one another (as described by the faculty of khan)?

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So I manage to work it out and I am going to answer my own question just in case someone in the future needs it.

My answer

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  • $\begingroup$ Sir I'm exactly stuck at this point i hope you can help me. what exactly is resolution function? is it noise? why the probability that an output y=0 will be recorded instead as between y and y+dy is given by g(y)dy. what does it mean to say the output y=0. and what does it mean to say the probability the true reading lying between x and x+dx and having so, the probability of f(x) being selected (selected why?) by experiment will be moved by instrumental resolution by an amount z-x? $\endgroup$
    – Hello
    Commented May 24 at 11:10

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