I am trying to understand the following question on Inverse Fourier Transform and use of convolution theorem. I am unable to understand the step boxed in red. Will be grateful if some guidance is provided to understand it
[Math] Inverse Fourier Transform and Convolution Theorem
fourier transform
Best Answer
It is actually straight forward (if this is your only problem):
$H(x)$ is the Heaviside step function, which is
$$ H:x\mapsto \begin{cases}0:&x<0\\1:&x\geq 0\end{cases} $$
and sometimes different values for $H(0)$, but this is surely defined in your book. Anyhow, now just look at the first case where $x>0$ and hence $H(\tau-x)=0$ until $\tau\geq x$, then we have $\tau>\tau-x\geq 0$ (if $\tau-x>0\implies \tau>0$) and therefore
$$ H(\tau-x)H(\tau)=H(\tau-x) $$
In the second case where $x<0$ you the exact same reasoning, but this time $\tau<\tau-x$ and hence the argument switches (if $\tau>0\implies \tau-x>0$) and you get $$ H(\tau-x)H(\tau)=H(\tau) $$