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== Part b == | == Part b == | ||
| + | |||
| + | <font size="3">In order for <math>Y[n]=u[n-1]</math> to be true, <math>X[n]=u[n]</math> must also be true. | ||
| + | |||
| + | Proof: | ||
| + | |||
| + | <math>u[n]=\delta[n]-\delta[n-1]</math> | ||
| + | |||
| + | <math>\delta[n] \to sys \to \delta[n-1] \to</math> | ||
| + | <math>- \to \delta[n-1]-4\delta[n-2]</math> | ||
| + | <math>\delta[n-1] \to sys \to 4\delta[n-2] \to</math> | ||
Revision as of 10:42, 11 September 2008
Part a
System: $ X_{k}[n]=\delta[n-k] \to Y_{k}[n] = (k+1)^2 \delta [n-(k+1)] $
Time-delay: $ X_{k}[n]=\delta[n-k] \to X_{k}[n-N]=\delta[n-N-k] $
$ X_{k}[n] \to timedelay \to sys \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $
$ X_{k}[n] \to sys \to timedelay \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $
Since $ (k+1)^2 \delta [n-N-(k+1)] $ is equal to $ (k+1)^2 \delta [n-N-(k+1)] $, the system is time-invariant.
Part b
In order for $ Y[n]=u[n-1] $ to be true, $ X[n]=u[n] $ must also be true.
Proof:
$ u[n]=\delta[n]-\delta[n-1] $
$ \delta[n] \to sys \to \delta[n-1] \to $ $ - \to \delta[n-1]-4\delta[n-2] $ $ \delta[n-1] \to sys \to 4\delta[n-2] \to $
