What is the result of convolving a triangle signal with a delta function signal?

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SUMMARY

The convolution of a triangle signal with a delta function results in a constant value of 1. This occurs because convolution in one domain translates to multiplication in the frequency domain. The triangle signal can be represented as the integral of two box functions, which transform to sinc functions. When convolving with a delta train, the zeros of the sinc function align with the delta functions, resulting in a single impulse at the origin, which transforms back to a constant value.

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stanigator
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For the question stated in the pictures attached to this message, for part d of the question, I just can't picture why the convolution of the two signals is 1. Can someone please explain how this would be the case as the triangular signal is sweeping through all the delta functions? Thanks.
 

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stanigator said:
For the question stated in the pictures attached to this message, for part d of the question, I just can't picture why the convolution of the two signals is 1. Can someone please explain how this would be the case as the triangular signal is sweeping through all the delta functions? Thanks.

Without looking at the problem, I will take a wild guess. Convolution is one domain is multiplication in the other. The transform of a delta-train is another delta-train with different spacing. I don't know what the convolution of a triangle is, but I'm guessing it's something close to a sinc function, because a triangle can be thought of as the integral of two box functions which each transform to sinc functions. So now you multiply an impulse-train with something sinc-esque. I'm guessing the zeros of the sinc-esque function will hit the impulse-train everywhere except at the origin, so the result will be a single impulse at 0. Then transform that back and you get a constant.
 

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