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The system is defiantly not time invariant. When the signal is shifted the output is shifted also but a different magnitude is applied to the output, showing that time affects the output. | The system is defiantly not time invariant. When the signal is shifted the output is shifted also but a different magnitude is applied to the output, showing that time affects the output. | ||
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Sine we are talking about DT a setup of | Sine we are talking about DT a setup of | ||
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| + | <math>X[n]=\sum_{k=-\infty}^{\infty}</math> δ[n-k]. | ||
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| + | However for <math>u[n]</math> we will need to cap off the <math>-\infty</math> to 0, since the for all values less then 0 the unit step would produce 0. | ||
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| + | However <math>u[n-1]</math> is being evaluted so the delta function will need to be offset producing | ||
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| + | X[n]=<math>\sum_{k=0}^{\infty}</math> δ[n-(k+1)]. | ||
Revision as of 14:31, 12 September 2008
The system is defiantly not time invariant. When the signal is shifted the output is shifted also but a different magnitude is applied to the output, showing that time affects the output.
Sine we are talking about DT a setup of
$ X[n]=\sum_{k=-\infty}^{\infty} $ δ[n-k].
However for $ u[n] $ we will need to cap off the $ -\infty $ to 0, since the for all values less then 0 the unit step would produce 0.
However $ u[n-1] $ is being evaluted so the delta function will need to be offset producing
X[n]=$ \sum_{k=0}^{\infty} $ δ[n-(k+1)].
