# Inverse Fourier Transform and Convolution Theorem

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

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 $$H(\tau-x)H(\tau)=H(\tau)$$

• Thank you, it is indeed very simple when explained...but I could not get it until then
– SAK
Apr 19 '18 at 17:34
• Sure thing, just watch out for the exact definition of your Heaviside function. Apr 19 '18 at 17:52
• @SAK PS if your question has sufficiently well been answered you could mark it as “answered”, this let’s other people see that you don’t need any further assistance (and also gives the answerer some reputation points) Apr 20 '18 at 15:15
• Noted. I was not aware of this..will certainly do it for this and in future.. but I am unable mark it "answered". How do I do that?
– SAK
Apr 28 '18 at 15:36
• On the left side of any answer you can find a checkmark and up/down arrows. You can upvote or downvote with the arrows, if you click on the checkmark it will become green and accepted Apr 28 '18 at 15:39