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?

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


Considering an LTI continuous-time system with impulse response $h(t)$, its response to any valid input $x(t)$ can be computed with a convolution integral as:

$$y(t) = x(t) \star h(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau$$ and equivalently $$y(t) = h(t) \star x(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d\tau$$
due to the commutativity of the convolution operator.

Hence, shifting the output $y(t)$ by $k$ could be shown as $$y(t-k) = \int_{-\infty}^{\infty} x(\tau) h(t-k-\tau) d\tau$$ or equivalently as

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

Where, the first one equals $y(t-k) = x(t) \star h(t-k)$ and the second one equals $y(t-k) = h(t) \star x(t-k)$.

  • $\begingroup$ The problem is that one is easily falling into the expression $x(t) \star h(t)$ with $t-k$ replacement. $\endgroup$
    – MathArt
    Apr 27 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 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 at 18:45

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