Find the Laplace Transformation of $H(\pi-t)$.

Question

I know how to find the Laplace Transformation of $H(t-\pi)$, but what about if the $t$ is negative. Any help is much appreciated.

Answer 1

The Heaviside function $H$ is defined as $1$ when its argument is positive and zero otherwise. Thus, for $H(\pi-t)$, the function is $1$ when $t<\pi$ and $0$ otherwise. Thus,

$\int_0^{\infty}H(\pi-t)e^{-st}dt=\int_0^{\pi}e^{-st}dt=\frac1 s (1-e^{-\pi t})$

Thus, the result is identical to that from the sum $H(t)-H(t-\pi)$.

Answer 2

Assuming the unilateral Laplace transform, you can see that ${\cal L} (t \mapsto H(\pi-t)) = {\cal L} (1_{[0,\pi]})$. So you need to compute the Laplace transform of a pulse of length $\pi$.

Similarly, you could note that ${\cal L} (t \mapsto H(\pi-t)) = {\cal L} (t \mapsto H(t)-H(t-\pi))$.