Digital signal processing, linear time invariant system,

  • Thread starter Asma
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  • #1
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I really confused, I found in a book that the following system,
y[n]= x[n+1]-x[n], is not causal!
But from the definition of causality that the output y[n0] depends only on the input samples x[n] for n<=n0,,,
So I think that this system is causal...

If you agree with me please tell me that it's correct...

Thanks in advance...
 

Answers and Replies

  • #2
FOIWATER
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x[n+1] is not a past or present sample of the input it is a future sample so that is why it is not causal.

It's actually like this, x[n-k] where k>=0
 
  • #3
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But x[n+1] means that the sample is shifted to the left, (less than 0 or means it is positioned at -1) so it is a past value then the system is causal.!!!!
 
  • #4
FOIWATER
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It's true that x[n+1] would be x[n] shifted to the left by one, but x[n+1] is a future value. Think about the graphs - if we have x[n] = {1,2,3,4..} where 1 has the zero place, x[n+1] would be, as you say, {2,3,4,5..} with 2 having the zero place. Do you see how now the first sample in x[n+1] (which is 2) is the FUTURE value of x[n] at the ones place?

You're right that the graph shifts, this is visual meaning of future values. The graph does shift left.
 
  • #5
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It's true that x[n+1] would be x[n] shifted to the left by one, but x[n+1] is a future value. Think about the graphs - if we have x[n] = {1,2,3,4..} where 1 has the zero place, x[n+1] would be, as you say, {2,3,4,5..} with 2 having the zero place. Do you see how now the first sample in x[n+1] (which is 2) is the FUTURE value of x[n] at the ones place?

You're right that the graph shifts, this is visual meaning of future values. The graph does shift left.

Thanks for your answer... It is clear now.. Thanks
 

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