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| − | <td>Xk[n]=δ[n-k]</td><td> -> </td><td> Yk[n]=(k+1)2 δ[n-(k+1)] | + | <td>Xk[n]=δ[n-k]</td><td> -> </td><td> Yk[n]=(k+1)2 δ[n-(k+1)] </td> |
</tr> | </tr> | ||
</table> | </table> | ||
For any non-negative integer k | For any non-negative integer k | ||
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==Test for Time Invariance== | ==Test for Time Invariance== | ||
Latest revision as of 14:22, 11 September 2008
Homework 2 Ben Horst: A :: B :: C :: D :: E
Given the system
| Input | Output | |
| X0[n]=δ[n] | -> | Y0[n]=δ[n-1] |
| X1[n]=δ[n-1] | -> | Y1[n]=4δ[n-2] |
| X2[n]=δ[n-2] | -> | Y2[n]=9 δ[n-3] |
| X3[n]=δ[n-3] | -> | Y3[n]=16 δ[n-4] |
| ... | ... | |
| Xk[n]=δ[n-k] | -> | Yk[n]=(k+1)2 δ[n-(k+1)] |
For any non-negative integer k
Test for Time Invariance
Start with X2[n]=δ[n-2]
Delay the signal by 2 => X2[n] = δ[n-4]
Then run the signal through the system:
X2[n] = δ[n-4] -> system -> Y2 = 25δ[n-5]
Repeat in reverse...
Run the signal through the system:
X2[n]=δ[n-2] -> system -> Y2 = 9δ[n-3]
Then delay the signal by 2 => Y2 = 9δ[n-5]
Since 9δ[n-5] ≠ 25δ[n-5], the system is not Time Invariant.
Second Part
Question: Assuming that this system is linear, what input X[n] would yield the output Y[n]=u[n-1]?
X[n] = u[n] would yield Y[n]=u[n-1] since u[n] is simply the sum of shifted delta functions, and linearity dictates that they could be sent through the system (producing δ[n-1]'s) and then summed after the fact to give Y[n]=u[n-1].
