[EE - Signals] Question about Discrete-Time Impulse/Discrete-Time Delta

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The discussion centers on the representation of a discrete-time impulse function, specifically delta[n+1]. It clarifies that for n = -1, the value of delta[-1+1] equals delta[0], which is 1, confirming that the impulse is indeed shifted one unit to the left. Participants emphasize that discrete-time signals are typically represented as stem plots, where vertical lines indicate non-zero values. The correct plot would feature a vertical line at n = -1 reaching up to y = 1. This understanding reinforces the concept of shifting in discrete-time signals.
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Quick question about a discrete unit impulse.

If I were to draw delta[n+1], at

... etc
n = -2 ... the point would be at location: 0
n = -1 ... the point would be at location: 1
n = 0 ... the point would be at location: 0
n = 1 ... the point would be at location: 0
n = 2 ... the point would be at location: 0
... etc

What I'm asking is that the line that I have in red... is that correct, with the shift of +1 to the left (an advance signal)?

The reason I'm saying it's -1 is because when n is -1, delta[-1+1] is delta[0] = 1.
 
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Yes that's right... just like ordinary continuous function... f(x+a) is f(x) shifted a units to the left.
 
learningphysics said:
Yes that's right... just like ordinary continuous function... f(x+a) is f(x) shifted a units to the left.

Thanks for the reply. Just to make sure, I would have a "1" dot for y where x = -1 then, correct?
 
user101 said:
Thanks for the reply. Just to make sure, I would have a "1" dot for y where x = -1 then, correct?

Yes, that would be correct but usually discrete time signals are as stem plots... ie a vertical line is drawn from the x-axis up to the dot (nothing is drawn when y=0)... eg:

http://en.wikibooks.org/wiki/Digital_Signal_Processing/Discrete_Data#Stem_Plots

So your plot would have a vertical line segment at n=-1 from the x-axis up to the dot at y=1.
 
Ok, great! Thanks
 
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