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I am having trouble understanding the concept of convolution. I get the mathematical idea behind it. I am able to evaluate that the convolution of the following problem enter image description here is $$f*g(t) = {0,\ \ t <= -1,\ \ t >= 1;\\ t+1, -1 <= t <= 0;\\ 1 - t,\ \ 0 <= t <=1}$$ However, what exactly is happening to the signal on more physical level?

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    \$\begingroup\$ Overlay the two graphs, and align the y axes. Keep the f(t) graph stationary and slide the g(t) graph to the right, calculating the area under the product of the two graphs as you go. You produce a triangle which starts at zero at t=0; has an apex height of unity that at t=1; and reduces to zero at t=2. The resultant graph is the convolution of f(t) and g(t). \$\endgroup\$ – Chu Mar 15 '16 at 23:08
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Convolution basically tells you how similar two signals are as one of them is shifted in the x-axis and reflected. So consider taking your signal \$g\$ and shifting it by various amounts. The convolution will have a peak when the two signals are mirror images across the y-axis. In this case, they already are, so the peak convolution occurs with a shift of 0.

The most important thing to understand is that the meaning of the x and y axes have changed from (time, value) to (shift, alignment).

I've ignored the meaning of the y-axis magnitude. When I've used the convolution before it's been with normalized signals so that the values ranged from 0 to 1, but that is certainly not always the case.

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  • \$\begingroup\$ You talk about correlation when you talk about how similar two signal are.... \$\endgroup\$ – MathieuL Mar 17 '16 at 1:09
  • \$\begingroup\$ @MathieuL - Yes, that does make more sense. They are closely related, but I'd like to know more about the difference. Maybe the last sentence of Chu's answer (that the convolution is the response of a system given the input and impulse response) is a more direct use for convolution. \$\endgroup\$ – Justin Mar 17 '16 at 11:32
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Here is one example I use to help me understand some of it. By no means is it complete.

In x-ray imaging, the x-ray source is a distributed source. You can consider it as many point sources next to each other. If there was only one point source, the shadow that is cast from the object (patient) would be very sharp. If you placed your hand on the xray bed, there wouldn't be any fuzzy edges. However, because the x-ray source is comprised of many point sources, you get a little fuzziness. It turns out the detected x-ray image is a convolution between the source and the object (3-dimensional convolution, but still).

Try it with your room lighting. If you hold your hand close to the ceiling lamp, its shadow is very blurry, but if you place your hand closer to the table, the shadow is sharper.

Signal convolution is similar. If you convolve a signal with a delta function, you get the exact same signal as a result. If you convolve it with two delta functions, it would be similar to the shadow of your hand with two ceiling lights (two images of your hand with a darker region in the middle). When you convolve a signal with a million delta functions, all of which have different magnitudes, well it becomes harder to intuitively understand. You should probably then refer to the other responses here.

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Convolution can be interpreted geometrically as: folding, sliding, multiplying, and adding. From the mathematical analysis in the OP, it appears that the g(t) graph has already been folded about its vertical axis, therefore:

  • align the vertical axes of the two graphs
  • keep the f(t) graph stationary and slide the g(t) graph to the right
  • calculate the area under the product of the two graphs as g(t) slides over f(t)

This produces a triangle that starts at zero at t=0; has an apex height of unity at t=1; and reduces to zero at t=2.

The resultant graph is the convolution of f(t) and g(t), in other words it's the response of a system, whose unit impulse response is g(t), to an input signal, f(t).

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