(New page: == Part E. Linearity and Time Invariance == A discrete-time system is such that when the input is one of the signals in the left column, then the output is the corresponding signal in the ...) |
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| − | <td>X0[n]=δ[n]</td><td> | + | <td>X0[n]=δ[n]</td><td> </td><td> Y0[n]=δ[n-1]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>X1[n]=δ[n-1]</td><td> | + | <td>X1[n]=δ[n-1]</td><td> </td><td> Y1[n]=4δ[n-2]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>X2[n]=δ[n-2]</td><td> | + | <td>X2[n]=δ[n-2]</td><td> </td><td> Y2[n]=9 δ[n-3]</td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td>X3[n]=δ[n-3]</td><td> | + | <td>X3[n]=δ[n-3]</td><td> </td><td> Y3[n]=16 δ[n-4]</td> |
</tr> | </tr> | ||
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| − | <td>Xk[n]=δ[n-k]</td><td> | + | <td>Xk[n]=δ[n-k]</td><td> </td><td> Yk[n]=(k+1)<math>^{2}</math> δ[n-(k+1)] For any non-negative integer k</td> |
</tr> | </tr> | ||
</table> | </table> | ||
| − | + | ||
| + | == Can This System Be Time Invariant? == | ||
| + | Let the system be defined according to the first line, input: X0[n]=δ[n] and output: Y0[n]=δ[n-1] and time delay of 3. Using the same method as in Part D, we can determine whether this system is time invariant or not. | ||
| + | |||
| + | δ[n] -> time delay -> δ[n-3] -> system -> 16δ[n-4] | ||
| + | |||
| + | δ[n] -> system -> δ[n-1] -> time delay -> δ[n-4] | ||
| + | |||
| + | |||
| + | Since both cascades produce different outputs, this system is NON-time invariant. | ||
| + | |||
| + | |||
| + | == What Input X[n] Would Yield the Output Y[n]=u[n-1]? == | ||
| + | According to the definition of the DT unit impulse that was given in class, u[n]=<math>\sum_{k=0}^{\infty}</math> δ[n-k]. | ||
| + | This means that the unit impulse is nothing more than the sum of all shifted deltas from 0 to infinity. Now, we know that the input must be of this form. So, in order to get the correct time shift, we simply need to subtract k+1 instead of k. So, the input that produces the output Y[n]=u[n-1], is | ||
| + | |||
| + | X[n]=<math>\sum_{k=0}^{\infty}</math> δ[n-(k+1)]. | ||
Latest revision as of 10:26, 12 September 2008
Part E. Linearity and Time Invariance
A discrete-time system is such that when the input is one of the signals in the left column, then the output is the corresponding signal in the right column:
| 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 |
Can This System Be Time Invariant?
Let the system be defined according to the first line, input: X0[n]=δ[n] and output: Y0[n]=δ[n-1] and time delay of 3. Using the same method as in Part D, we can determine whether this system is time invariant or not.
δ[n] -> time delay -> δ[n-3] -> system -> 16δ[n-4]
δ[n] -> system -> δ[n-1] -> time delay -> δ[n-4]
Since both cascades produce different outputs, this system is NON-time invariant.
What Input X[n] Would Yield the Output Y[n]=u[n-1]?
According to the definition of the DT unit impulse that was given in class, u[n]=$ \sum_{k=0}^{\infty} $ δ[n-k]. This means that the unit impulse is nothing more than the sum of all shifted deltas from 0 to infinity. Now, we know that the input must be of this form. So, in order to get the correct time shift, we simply need to subtract k+1 instead of k. So, the input that produces the output Y[n]=u[n-1], is
X[n]=$ \sum_{k=0}^{\infty} $ δ[n-(k+1)].
