# Time shift — two interpretations

In their textbook(Example 1.1), Oppenheim and Willsky provide the following method for obtaining $g(t)=x(-t+1)$ from $x(t)$ : Replace $t$ with $-t$ in $x(t+1)$. In other words, advance the signal by 1 unit and flip the result.

But there could be another interpretation as well. Based on the fact that $-t+1 = -(t-1)$, we can have the desired signal as $h(t)=x(-t+1) = x(-(t-1))$,i.e. Delay the signal by 1 unit and flip the result.

But the two signals $g(t)$ and $h(t)$ are not identical. Why is the latter interpretation incorrect ?

• On your interpretation: When you flip any function $h(t)=x(t-1)$, you flip the abscissa variable, that is $t$ here and the flipped function will then be $x(-t-1)$. So, the interpretation of $x(-(t-1))$ as delay the signal $x(t)$ by 1 unit and flip is not correct. I hope this makes it clear. – Neeks Aug 22 '14 at 4:11

$$h_1(t) = x(t-1) ~~ \text{ for all }~ t, \tag{1}$$ that is, $h_1(t)$ is just $x(t)$ delayed by one unit. Let $h(t)$ be the result of "flipping" $h_1(t)$, that is, $$h(t) = h_1(-t) ~~ \text{ for all }~ t, \tag{2}.$$ But, $(1)$ says that $h_1(-t)$ is the same as $x(-t-1)$ for all $t$, and so we conclude that $h(t)$, the result of delaying $x(t)$ by one time unit and then flipping the result gives us $$h(t) = x(-t-1)~~ \text{ for all }~ t, \tag{3}$$ which of course is not the desired $x(-t+1)$.
So let's try O&W's prescription. Let $g_1(t) = x(t+1)$ for all $t$ be the result of advancing $x(t)$ by one time unit. Now flip $g_1(t)$ to get $$g(t) = g_1(-t) = x(-t+1)$$ which is what is desired, no?