(New page: 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.) |
|||
| (3 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
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. | ||
| + | |||
| + | |||
| + | |||
| + | Sine we are talking about DT a setup of | ||
| + | |||
| + | <math>X[n]=\sum_{k=-\infty}^{\infty}</math> δ[n-k]. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | X[n]=<math>\sum_{k=0}^{\infty}</math> δ[n-k]. | ||
Latest revision as of 14:34, 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.
X[n]=$ \sum_{k=0}^{\infty} $ δ[n-k].
