# Experimental Physics - Defining convolution in terms of equipment resolution

So I think this video:

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)?