1

I try to calculate the convolution of $\theta(t)\cdot(e^t\theta(1-t))$.

Using the formula

\begin{equation} f*g(t)=\int_{-\infty}^\infty f(t-u)g(u)du \end{equation}

I set $f(t-u)=\theta(t-u)$

and

$g(u)=e^t\theta(1-t)$

So the integral would be:

\begin{equation} \int_{-\infty}^{-1} e^t\theta(1-t)\theta(t-u)du+\int_{-1}^{0}e^t\theta(1-t)\theta(t-u)du+\int_{0}^{\infty}e^t\theta(1-t)\theta(t-u)du \end{equation}

But this gives 1, which is wrong.

Then I tried the formula for casuality:

\begin{equation} f*g(t)=\int_{0}^t f(t-u)g(u)du \end{equation}

and here I got: $-te^t$

which is also wrong.

Where is the error here, in the integral boundaries, or the procedure itself?

Thanks

Luthier415Hz
  • 2,739
  • 6
  • 22
  • What is $\theta(t)$, the Heaviside step function? It also looks like your product is already the integrand of a convolution, so did you mean to compute $\int_{-\infty}^{\infty} e^t \theta(t) \theta(1-t) \ dt?$ – Sean Roberson Mar 04 '22 at 12:22
  • Yes, but the formula has u in it. Shouldn't that be used in replacing $t\rightarrow u$? and doing that gives 0. If I use t as variable as you say, I get 0 there as well. – Luthier415Hz Mar 04 '22 at 12:32

1 Answers1

1

You're mixing up your variables. The convolution integral should be evaluated as follows:$$\left[e^t\,\theta(1-t)\cdot\theta(t)\right](u)=\int\limits_{-\infty}^{\infty} e^t\,\theta (1-t)\,\theta(u-t)\,dt=e^u+\left(e-e^u\right)\,\theta(u-1)$$

Steven Clark
  • 7,363
  • But according to WA this result does not seem correct, and since $\theta(t)$ and $e^t(\theta(1-t)$ are causual, the integral should have boundaries from 0 to t. https://www.wolframalpha.com/input?i=convolution+calculator&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22convolvefunction1%22%7D+-%3E%22theta%28t%29%22&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22variable1%22%7D+-%3E%22t%22&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22convolvefunction2%22%7D+-%3E%22theta%28-t%29e%5Et%22&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22variable2%22%7D+-%3E%22y%22 – Luthier415Hz Mar 09 '22 at 08:48
  • @Luthier You used $\theta(-t)$ instead of $\theta(1-t)$ and reversed the order of the two functions. Accounting for these two differences WoframAlpha gives the same result as in my answer above. – Steven Clark Mar 09 '22 at 14:28
  • @Luthier See https://www.wolframalpha.com/input?i=convolution+calculator&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22convolvefunction1%22%7D+-%3E%22theta%281-t%29+e%5Et%22&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22variable1%22%7D+-%3E%22t%22&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22convolvefunction2%22%7D+-%3E%22theta%28t%29%22&assumption=%7B%22F%22%2C+%22ConvolveCalculator%22%2C+%22variable2%22%7D+-%3E%22u%22. – Steven Clark Mar 09 '22 at 14:29
  • 1
    @Luthier I wrote the integral using the definition of a Fourier convolution, but you can truncate the upper integration limit to $1$ since $\theta(1-t)$ is zero for $t>1$ (see https://www.wolframalpha.com/input?i=Integrate%5B++E%5Et+HeavisideTheta%5B1+-+t%5D+HeavisideTheta%5Bu+-+t%5D%2C+%7Bt%2C+-Infinity%2C+1%7D%5D), but you can't truncate the lower integration limit to $0$ since you're integrating $\theta(u-t)$ instead of $\theta(t)$. – Steven Clark Mar 09 '22 at 14:56