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So I have this equation to demonstrate: $$ x(t)*u(t)= \int_{-\infty}^t x(\tau)d\tau $$ , where $u(t)=\int_{-\infty}^t \delta(\tau)d\tau$

I opened the convolution as $ \int_{-\infty}^\infty x(\tau)u(t-\tau)d\tau $ but when i try to use the $u(t)$ equation i can't figure out if the $u(t-\tau)$ argument is going to turn the delta into $\delta(t-\tau)$ or if it is only going to change the upper integration limit to $t-\tau$. In reality, i suspect i have to use fourier transforms property of convolution, but i'm not sure. Thanks.

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This really is a problem to do with confusing notation. Even though it may not seem so, the $\tau$ in the definition of $u(t)$ is not the same as the one in the convolution definition. If $t \mapsto t-\tau$, then the upper integration limit on your definition of $u(t)$ is changed to $t-\tau$, not anything else. Try writing $u(t)=\int_{-\infty}^t \delta (s) ds $. Knowing that this is the Heaviside step function, you should have no problem figuring the rest.

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  • $\begingroup$ Yes! In the end it had to do with the dirac filtering property, thanks! $\endgroup$ Commented Nov 28, 2014 at 23:12

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