1
$\begingroup$

$u[n]$ is the unit step function in discrete form. I want to know which one of the following two is the correct waveshape for $u[2-n]$, where $k$ is a constant.

What's the answer? The top one is $u[n]$, the unit step function. So, what's the graph for $u[2-n]$, middle or bottom?

enter image description here

$\endgroup$
  • $\begingroup$ You know that $u[1]=1$ and $u[-1]=0$. Plug values of $n$ from your second and third axis so that the function argument is 1 and -1, and you'll see which one is right. $\endgroup$ – MBaz Jan 25 '16 at 3:08
  • $\begingroup$ The second one is the right one -(n-2) = 2-n $\endgroup$ – Moti Jan 25 '16 at 6:12
  • $\begingroup$ For me, I will do the your first suggested operation: delay 2 samples then reversal, but the graph should be the bottom one. $\endgroup$ – Tony Tan Aug 25 at 12:05
6
$\begingroup$

HINT: For which value of $n$ does the argument of $u[2-n]$ become zero? That's where the step occurs. For which values of $n$ is the argument non-negative? That's where your unit step equals $1$. If you think about it for a minute, it should become really easy.

$\endgroup$
  • $\begingroup$ Really?????????????? Need 15 characters... $\endgroup$ – Moti Jan 26 '16 at 1:26
  • $\begingroup$ @Moti: My comment was a reaction to your statement that the "second one" is the correct solution. I thought you meant the second of the two "solutions" (i.e. the last figure), so I thought you might want to reconsider your comment. Now I realize that you meant the first solution, i.e. the second figure from top, which is of course correct. $\endgroup$ – Matt L. Jan 26 '16 at 7:46
3
$\begingroup$

Basically, discrete unit step signal may be defined as: $$u[n] = \begin{cases} \begin{align} &1 &\scriptstyle{\text{for n > 0}}\\ &0 &\scriptstyle{\text{for n < 0}}\\ \end{align} \end{cases}$$ Doing time reversing (inverse) $$u[-n] = \begin{cases} \begin{align} &1 &\scriptstyle{\text{for -n > 0}}\\ &0 &\scriptstyle{\text{for -n < 0}}\\ \end{align} \end{cases}$$ $$u[-n] = \begin{cases} \begin{align} &1 &\scriptstyle{\text{for n < 0}}\\ &0 &\scriptstyle{\text{for n > 0}}\\ \end{align} \end{cases}$$ Now doing shifting by (+2) $$u[-n+2]=u[2-n] = \begin{cases} \begin{align} &1 &\scriptstyle{\text{for n < 2}}\\ &0 &\scriptstyle{\text{for n > 2}}\\ \end{align} \end{cases}$$ $Which\ gives\ you\ the\ second\ signal\ you\ stated\ in\ your\ question$

$\endgroup$

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.