# More Signals Help

Discussion in 'Math' started by crazyengineer, Jul 4, 2011.

1. ### crazyengineer Thread Starter Member

Dec 29, 2010
156
2
Hi! I need help understanding this signal

$image=http://latex.codecogs.com/gif.latex?x[n]u[1-n]&hash=7fbcc355c75d186eaf63ab76170cc213$

But it's actually this

Can anyone help me understand how they got this graph? I thought since x[n] is one when n>=0, I pay attention to u[1-t] from 0 to 1.

2. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
This is completely arbitrary.

You need to tell us what x[n] is for us to explain the graph. We assume u[n] is the step function and x[n] is some other arbitrary signal. Without its definition, these plots are completely meaningless

I think you may be misunderstanding the concept of the unit step function

3. ### kevinarms New Member

Jun 29, 2011
6
2
this is just the original signal - x[n] - multiplied by a modified step function. u[1-n] is a step function that is time reversed around n=1. So u[1-n] = 1 for n<=1 and u[1-n] = 0 for n > 1 (instead of one for n>=0). So for g[n] = x[n]u[1-n], g[n] = x[n] for n <= 1 (where u[1-n] = 1), and g[n] = 0 for n>1 (where u[1-n] = 0).

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4. ### crazyengineer Thread Starter Member

Dec 29, 2010
156
2
Opps! I looked in my book and found x[n] *slaps his head* Sorry for the mistake!

This is X[n]

5. ### kevinarms New Member

Jun 29, 2011
6
2
So x[n]u[1-n] is simply a pointwise multiplication of this given x[n] with u[1-n]. u[1-n] has a value of 1 for all n<=1, and a value of 0 for n >1. So with this given x[n], the result will be almost the same as x[n] (since you are multiplying most sample values by 1), except for the resulting function changes to 0 at n = 2,3, since u[1-n] = 0 at these two points. This gives you the result that you originally posted.

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6. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
I think the gentleman is talking about sidebands, as in RF and AM theory. Hard to be sure though.