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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.

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  • $\begingroup$ * means convolution, also, u(t) is the heaviside step function $\endgroup$
    – user1115701
    Commented Nov 4, 2022 at 23:40
  • $\begingroup$ 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. $\endgroup$
    – Jean Marie
    Commented Nov 5, 2022 at 7:26

1 Answer 1

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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)):

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

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  • $\begingroup$ Any comment ?... $\endgroup$
    – Jean Marie
    Commented Nov 5, 2022 at 17:30

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