(New page: == Question 6a == == Question 6b ==) |
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| + | == System == | ||
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| + | Input: <font size = 4><math>X_k[n]=\delta [n-k]</math></font> | ||
| + | |||
| + | Output: <font size = 4><math>Y_k[n]=(k+1)^2 \delta [n-(k+1)]</math></font> | ||
| + | |||
| + | For any non-negative integer k | ||
== Question 6a == | == Question 6a == | ||
| + | <math>x[n] \rightarrow \mbox{Time Delay} \rightarrow y[n]=x[n-n_0] \rightarrow System \rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)]</math> | ||
| + | <math>x[n] \rightarrow System \rightarrow Y_k[n]=(k+1)^2 \delta [n-(k+1)] \rightarrow \mbox{Time Delay}\rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)]</math> | ||
| + | |||
| + | So the system is time invariant. | ||
== Question 6b == | == Question 6b == | ||
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| + | <font size = 4><math>x_0[n]= u[n]</math></font> will yield the output <font size = 4><math>y[n]=u[n-1]</math></font> | ||
Latest revision as of 14:32, 11 September 2008
System
Input: $ X_k[n]=\delta [n-k] $
Output: $ Y_k[n]=(k+1)^2 \delta [n-(k+1)] $
For any non-negative integer k
Question 6a
$ x[n] \rightarrow \mbox{Time Delay} \rightarrow y[n]=x[n-n_0] \rightarrow System \rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)] $
$ x[n] \rightarrow System \rightarrow Y_k[n]=(k+1)^2 \delta [n-(k+1)] \rightarrow \mbox{Time Delay}\rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)] $
So the system is time invariant.
Question 6b
$ x_0[n]= u[n] $ will yield the output $ y[n]=u[n-1] $
