# Convolution of some function with Heaviside

Suppose there is a function, $$f(t)$$, not explicitly defined. Is the convolution of $$f(t)$$ with Heaviside Eq.1 or Eq.2? $$\int_{0}^{t}f(\tau)H(\tau-t)d\tau = -\int_{0}^{t}f(t)dt \ \Leftarrow Eq.1$$

$$\int_{0}^{t}f(\tau)H(\tau-t)d\tau= \int_{0}^{t}f(t)dt \ \Leftarrow Eq.2$$

I have doubts because doesn't it depend on whether $$f(t)$$ is odd or even? and My knowledge of convolution is sadly very limited. My knowledge of convolution mainly comes from this document.

If the answer is neither, then I wish someone can explain to me what convolution with Heaviside can be simplified into, especially if the integral's limit is from 0 to some finite t.

Your problem is that you don't write convolution in the convenient way ; you should write:

$$\int_{0}^{t}f(\boxed{\tau})H(\boxed{t-\tau})d\tau = \int_{0}^{t}f(\tau)d\tau \tag{1}$$

Mnemonic : the sum of the two boxed expressions must simplify into the value of the "final" variable, here $$t$$:

$$\boxed{\tau}+\boxed{t-\tau}=\boxed{t}$$

Remark 1: please note that, on the RHS, I have modified the "dumb" variable into $$\tau$$ in order it is not the same as the "final" variable $$t$$.

Remark 2: if we write (1) under the form

$$f \star H = F$$ where $$F$$ is the primitive function that takes value $$0$$ in $$0$$, we can apply to it the formula for the differentiation of a convolution product:

$$(f \star g)'= f \star (g')$$

giving:

$$(f \star H)' = f \star \delta = f$$

which is a proof of (1).

• Any comment ?... Commented Jun 29, 2022 at 17:03
• Yes, I have mixed up the dummy variable with t. Just a follow up question though, if i was to implement in Simulink, i think it will have to have the minus sign at the front of integral because the integrator cant do backward integration and the step function is defined to be positive 1 at time > 0. Is this correct? Commented Jun 29, 2022 at 23:28
• i reliaze my follow up question is more to Simulink implementation. But at the end of the day i have to implement this convolution in Simulink/MATLAB, and it doesnt support continous online convolution Commented Jun 29, 2022 at 23:30
• I realize that the operation with $\tau$ and $\tau-t$ as you have written it, is not a convolution but nevertheless has a name: it is the correlation operation. Commented Jul 1, 2022 at 21:03