# Is the following convolution property true?

Is the following convolution property true?

$$\text{If} \ y(t)=x(t)*h(t) , \text{then} \ y(t)=\int_{-\infty}^{t} [x'(\tau)*h(\tau)] \,d\tau$$

My proof :

Let's denote by $$g$$ the integrand $$g(\tau)=x'(\tau)*h(\tau)$$, then $$y(t)=\int_{-\infty}^{t} g(\tau) \,d\tau=g(t)*u(t)=x'(t)*h(t)*u(t)$$

(where $$u$$ denotes the heaviside step function).

We know that $$x'(t)*u(t)=x(t)$$, then we would have $$y(t)=x'(t)*u(t)*h(t)=x(t)*h(t)$$, so the property is true.

Is this correct? I'm not sure if I did the integral right.

• * means convolution, also, u(t) is the heaviside step function
– user1115701
Commented Nov 4, 2022 at 23:40
• I have taken the liberty to re-write some parts of your text in order for it to be more readable. I wish you agree with it. Besides, your property isn't exact. I am going to explain why by a direct differentiation + integration proof. Commented Nov 5, 2022 at 7:26

In fact, your first line should be:

$$\text{If} \ y(t)=x(t)*h(t) , \text{then} \ y(t)=\int_{-\infty}^{t} [x'(\tau)*h(\tau)] \,d\tau \color{red}{+ constant}$$

Let us first recall 2 facts (1) and (2):

$$\text{differentiation of a conv. product:} \ \ \ (\varphi \star \psi)' = \varphi'\star \psi = \varphi \star \psi' \tag{1}$$

$$\frac{d}{dt}\int_{-\infty}^{t} \varphi(\tau)d\tau = \varphi(t)\tag{2}$$

Therefore, starting from the hypothesis $$y=x \star h$$, let us first differentiate (using (1)), then integrate (using (2)):

$$y'=x' \star h \ \ \text{implying} \ \ y(t)=\int_{-\infty}^t [x' \star h] d\tau + \text{constant}$$

• Any comment ?... Commented Nov 5, 2022 at 17:30