| (2 intermediate revisions by the same user not shown) | |||
| Line 5: | Line 5: | ||
b.) Assuming the input is linear, an input X[n] that would yield an output of Y[n]=u[n-1] would be: | b.) Assuming the input is linear, an input X[n] that would yield an output of Y[n]=u[n-1] would be: | ||
| − | to get the output that is listed, you need to show how the unit step function is present. In the notes it was mentioned that the unit step function can be shown by a summation of shifted delta functions over a series of - <math> \ | + | to get the output that is listed, you need to show how the unit step function is present. In the notes it was mentioned that the unit step function can be shown by a summation of shifted delta functions over a series of - <math>\infty </math> to +<math>\infty</math>. |
| − | + | next the time shift can be proved by stating that the function is linear and that if you add a deltaT time shift to the input, that the output will see the same shift in its deltaT. which can be shown by (t-to) but in this case it would be [n] -> [n-1]. | |
| − | + | ||
Latest revision as of 12:37, 12 September 2008
Linearity and Time Invariance
a.) The system can not be Time Invariant, because the output is scaled by a factor of the delta T that occurs throughout the function.
b.) Assuming the input is linear, an input X[n] that would yield an output of Y[n]=u[n-1] would be:
to get the output that is listed, you need to show how the unit step function is present. In the notes it was mentioned that the unit step function can be shown by a summation of shifted delta functions over a series of - $ \infty $ to +$ \infty $.
next the time shift can be proved by stating that the function is linear and that if you add a deltaT time shift to the input, that the output will see the same shift in its deltaT. which can be shown by (t-to) but in this case it would be [n] -> [n-1].
