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So i have the following rectangular pulse:

$$h(t)=\begin{cases} 1 & 0<t< T \\ 0 & \text{otherwise} \end{cases}$$

Now, i am supposed to create $h(T-t)$, and the problem is that i do it two different ways, and get two different results, and it is obvious that this is not possible. Anyway, here is what i have.

My first attempt is step by step building of the $h(T-t)$ from the $h(t)$, firstly, i "flipped" variable $t$. so i got $$h(-t)=\begin{cases} 1 & 0<-t< T \\ 0 & \text{otherwise} \end{cases}=\begin{cases} 1 & -T<t< 0 \\ 0 & \text{otherwise} \end{cases}$$

Now, i add T to this, and i have $$h(T-t)=\begin{cases} 1 & -T<T+t< 0 \\ 0 & \text{otherwise } \end{cases}=\begin{cases} 1 & -2T<t< -T \\ 0 & \text{otherwise} \end{cases}$$

However, i've done this directly:

$$h(T-t)=\begin{cases} 1 & 0<T-t< T \\ 0 & \text{otherwise} \end{cases}=\begin{cases} 1 & -T<-t< 0\\ 0 & \text{otherwise } \end{cases}=\begin{cases} 1 & 0<t< T \\ 0 & \text{otherwise} \end{cases}$$

Now, there is something wrong here, but i don't see any mistakes in any of the two cases. Any help appreciated!

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Clearly, in this example you have $h(t)=h(T-t)$. The easiest way to see this is to check for which value of $t$ the argument becomes zero. So the point $t=0$ of the original function is mapped to the point $t=T$, and because you have $-t$ in the argument, the function is inverted on the time axis.

The mistake in your first approach is the way you added the value of $T$. What you should have done is replace $-t$ by $T-t$ (which is what's happening on the left-hand side of the equation), to obtain $h(T-t)=1$ for $0<T-t<T$, which is equivalent to $0<t<T$, corresponding to the correct solution.

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They must yield the same answers. When you are in doubt, it's helpful to define an intermediate signal like $$y(t) = h(-t)$$ especially when manipulating the arguments. Then you have :

$$ y(t) = h(-t)=\begin{cases} 1 & 0<-t< T \\ 0 & \text{ otherwise } \end{cases} = \begin{cases} 1 & -T<t< 0 \\ 0 & \text{ otherwise } \end{cases} $$

then since $h(-t+T) = y(t-T)$, you get : $$ h(T-t) = y(t-T) = \begin{cases} 1 & -T< t-T < 0 \\ 0 & \text{ otherwise } \end{cases} = \begin{cases} 1 & 0< t < T \\ 0 & \text{ otherwise } \end{cases} $$

Which yields the correct result now.

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