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6 a) The system cannot be time-invariant. | 6 a) The system cannot be time-invariant. | ||
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<math>X_k[n] = \delta[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 \delta[n - (k + 1)]</math> | <math>X_k[n] = \delta[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 \delta[n - (k + 1)]</math> | ||
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| − | <math>\delta[n - k] \rightarrow system \rightarrow (k + 1)^2 \delta[n - (k + 1)] \rightarrow time-delay \rightarrow (k + 1)^2 \delta[n - n_0 -(k + 1)] </math> | + | Let us apply a time-delay of <math>n_0</math> to the system. |
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| + | System followed by time-delay: | ||
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| + | <math>\delta[n - k] \rightarrow system \rightarrow (k + 1)^2 \delta[n - (k + 1)] \rightarrow time-delay \rightarrow (k + 1)^2 \delta[n - n_0 -(k + 1)] = (k + 1)^2 \delta[n -(k + 1 +n_0)] </math> | ||
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| + | Time-delay followed by system: | ||
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| + | <math>\delta[n - k] \rightarrow time-delay \rightarrow \delta[n-(k + n_0)] \rightarrow system \rightarrow (k + n_0 + 1)^2 \delta[n - (k + n_0 + 1)]</math> | ||
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| + | Since the outputs don't match, the system is not time-invariant. | ||
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| + | 6 b) Since the system is linear, the input needed to yield Y[n] = u[n - 1] is X[n] = u[n]. | ||
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| + | <math>X_k[n] = u[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 u[n - (k + 1)]</math> | ||
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| + | In this case k = 0. | ||
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| + | Therefore, | ||
| + | <math>X_0[n] = u[n] \rightarrow system \rightarrow Y_0[n] = u[n - 1)]</math> | ||
Latest revision as of 16:52, 11 September 2008
6 a) The system cannot be time-invariant.
$ X_k[n] = \delta[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 \delta[n - (k + 1)] $
Let us apply a time-delay of $ n_0 $ to the system.
System followed by time-delay:
$ \delta[n - k] \rightarrow system \rightarrow (k + 1)^2 \delta[n - (k + 1)] \rightarrow time-delay \rightarrow (k + 1)^2 \delta[n - n_0 -(k + 1)] = (k + 1)^2 \delta[n -(k + 1 +n_0)] $
Time-delay followed by system:
$ \delta[n - k] \rightarrow time-delay \rightarrow \delta[n-(k + n_0)] \rightarrow system \rightarrow (k + n_0 + 1)^2 \delta[n - (k + n_0 + 1)] $
Since the outputs don't match, the system is not time-invariant.
6 b) Since the system is linear, the input needed to yield Y[n] = u[n - 1] is X[n] = u[n].
$ X_k[n] = u[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 u[n - (k + 1)] $
In this case k = 0.
Therefore, $ X_0[n] = u[n] \rightarrow system \rightarrow Y_0[n] = u[n - 1)] $
