Source video: Continuous time convolution with impulses - ProfKathleen
Convolution
For example, there is a man keep eating all the time. $f(t)$ is the food eaten at time t, and $h(t)$ is the food mass changing function for a period of digestion time t. For example, after the duration $(t-τ)$ of digestion, the amount of food eaten at time $τ$ becomes $h(t-τ)$.
Therefore, the total food reamined in stomach at time t is f convolving with h:
$$ y(t) = f(t) * h(t) = ∫_{-∞}^{+∞}f(τ)⋅h(t-τ) dτ $$
Intuitively, the value $x(τ)$ is scaled by a factor from τ time ago. Thus, the function $h(t)$ looks like in reverse.
Given $f = x(t)$ and $h = δ(t)$, the existing amount at time t is:
$$ y(t) = f * h = ∫_{-∞}^{+∞} x(τ) h(t-τ) dτ $$
Delta function
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$δ(τ)$:
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$δ(τ-3)$: Right shift $δ(τ)$ by 3.
Where the 0 was becomes -3. Therefore, all the coordinates minus 3, i.e., τ becomes τ-3.
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$δ(-τ-3)$: Rverse $δ(τ-3)$:
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$δ(t-τ-3)$: Shift $δ(τ-3)$ to t, which is a constant.
Where the τ-3 = 0 becomes t, so all coordinate plus t:
Convolve with Impulse
Given $x(t)$
The existing amount of a system containing two functions x(t) and δ(t-3) at time t is:
$$ y(t) = x(t) * δ(t-3) = ∫_{-∞}^{+∞} x(τ) δ(t-τ-3) dτ $$
Multiply x(τ) by the reversed impluse:
Delta function is 0 except for τ = t-3, therefore, only x(t-3) will be computed:
$$ y(t) = x(t) * δ(t-3) = ∫_{-∞}^{+∞} x(t-3) δ(t-τ-3) dτ $$
And in that integral, x(t-3) has nothing to do with τ, so it can be pulled outside the integral:
$$ y(t) = x(t) * δ(t-3) = x(t-3) ∫_{-∞}^{+∞} δ(t-τ-3) dτ \\ = x(t-3)*1 \\ = x(t-3) $$
By convolving with an impulse function, x(t) is shifted (based on the origin) to where the impulse is.
This conclusion can be generalized to any f.
Box function
Two impulses
Convolution is linear. Compute separately and sum together.
Impulse train
Convolving with a infinite sum of delta function: $Σ_{n=-∞}^{+∞} δ(t-n)$
The replicas of the signal overlaps: