# Which one is the correct graph for $u[2-n]$?

$u[n]$ 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?

• 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. – MBaz Jan 25 '16 at 3:08
• The second one is the right one -(n-2) = 2-n – Moti Jan 25 '16 at 6:12
• For me, I will do the your first suggested operation: delay 2 samples then reversal, but the graph should be the bottom one. – Tony Tan Aug 25 at 12:05

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.
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} $Which\ gives\ you\ the\ second\ signal\ you\ stated\ in\ your\ question$