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<tr> | <tr> | ||
| − | <td>X0[n]= | + | <td>X0[n]=δ[n]</td><td> -> </td><td> Y0[n]=δ[n-1]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>X1[n]= | + | <td>X1[n]=δ[n-1]</td><td> -> </td><td> Y1[n]=4δ[n-2]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>X2[n]= | + | <td>X2[n]=δ[n-2]</td><td> -> </td><td> Y2[n]=9 δ[n-3]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>X3[n]= | + | <td>X3[n]=δ[n-3]</td><td> -> </td><td> Y3[n]=16 δ[n-4]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
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</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>Xk[n]= | + | <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 | ||
| + | |||
| + | ==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]. | ||
Latest revision as of 13:22, 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].
