(What input will yield an output of Y[n]=u[n-1]?)
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==What input will yield an output of Y[n]=u[n-1]?==
 
==What input will yield an output of Y[n]=u[n-1]?==
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The system seems to work specifically on delta functions, so I take the approach of describing u[n] as an infinite sum of shifted deltas:
  
Now since, the output of the system is a step function shifted time, this means that the input must also be a step function.
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<math>\ u[n] = \sum_{k=0} \delta [n-k]</math>
Since we know that <math> \ (n-1) = (n-(k+1)) </math> solving for k we find <math>\ k = 0 </math>
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Therefore the input is <math>\ X[n]= u[n-k] = u[n] </math>
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So we find that the input must be a regular step function in order to produce the an output step function shifted 1 unit in time.
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Revision as of 13:57, 12 September 2008

Is the system time invariant?

No, the system is not time invariant because the output of an input signal shifted some value in time does not equal the output of the original signal, shifted the same value in time. In this system the coefficient or amplitude of the shifted output signal changes with time.

What input will yield an output of Y[n]=u[n-1]?

The system seems to work specifically on delta functions, so I take the approach of describing u[n] as an infinite sum of shifted deltas:

$ \ u[n] = \sum_{k=0} \delta [n-k] $

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