I memorized the convolution formula, all good. I started doing examples and in one of them I had to calculate the convolution of a shifted function with another one.

I know the property $$H(x)=f(x)\ast g(x)\\H(x-k)=f(x)\ast g(x-k)=f(x-k)*g(x)$$

So I had no trouble finding the result but I had trouble writing the integral. Which of the following is correct?

$$\int_{-\infty}^\infty f(τ)g(t-k-τ)dτ\\\int_{-\infty}^\infty f(τ)g(t+k-τ)dτ$$


2 Answers 2


Considering a continuous-time LTI system with impulse response $h(t)$, its output to any valid input $x(t)$ is described by a convolution integral:

$$y(t) = x(t) \star h(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau \tag{1.a}$$ and equivalently $$y(t) = h(t) \star x(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d\tau \tag{1.b}$$
(convolution operator is commutative .)

Hence, a shifted output $y(t-k)$ is shown as $$y(t-k) = \int_{-\infty}^{\infty} x(\tau) h(t-k-\tau) d\tau \tag{2.a}$$ or equivalently as

$$y(t-k) =\int_{-\infty}^{\infty} h(\tau) x(t-k-\tau) d\tau \tag{2.b}$$

Eqs.2a & 2b are $x(t) \star h(t-k)$ and $ h(t) \star x(t-k)$ respectively.

Note that, Eqs.1 uses the convention that the second operand of convolution is shifted inside the integral, which is also the case in Eqs.2. However, one can also write the shifted output $y(t-k)$ as $x(t-k) \star h(t)$ or $h(t-k) \star x(t)$ which yield the integrals:

$$y(t-k) = \int_{-\infty}^{\infty} x(\tau-k) h(t-\tau) d\tau \tag{3.a}$$ and $$y(t-k) = \int_{-\infty}^{\infty} h(\tau-k) x(t-\tau) d\tau \tag{3.b}$$

  • $\begingroup$ The problem is that one is easily falling into the expression $x(t) \star h(t)$ with $t-k$ replacement. $\endgroup$
    – MathArt
    Apr 27, 2021 at 9:58
  • $\begingroup$ Corollary: if as above $y(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d\tau$, and as it has now been shown that $y(t-k) = \int_{-\infty}^{\infty} x(\tau) h(t-k-\tau) d\tau$, could somebody please reassure me that $y(t-k) = \int_{-\infty}^{\infty} x(\tau-k) h(t-\tau) d\tau$ also holds? $\endgroup$
    – ngiann
    May 26, 2021 at 18:40
  • $\begingroup$ I would also like to point out that the question here is related to the question in math.stackexchange.com/q/2007956/691427 $\endgroup$
    – ngiann
    May 26, 2021 at 18:45

(I'm not sure about this, but that is how i think it works)

$$y(t-k)=\int_{-\infty}^{\infty}h(\tau)h(t-k-\tau)d\tau \\\\\\ \text{Let's call $\ \ \ \ \tau^{'}= \tau + k \ \ \ \ \ \ $then}$$ $$y(t-k)=\int_{-\infty}^{\infty}h(\tau^{'}-k)h(t-\tau^{'}) d\tau^{'}$$


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