0
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Neeks Aug 22 '14 at 4:11
1
$\begingroup$

Let

$$h_1(t) = x(t-1) ~~ \text{ for all }~ t, \tag{1}$$ 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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.