(New page: == 6A - Is the system time invariant? == System: <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br> Time-delay, then system: <math>T_k[n]=\delta[n-(k+1)]</math><br> <math>Y_k[n]=(k+1)^2\delta...) |
(→6A - Is the system time invariant?) |
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== 6A - Is the system time invariant? == | == 6A - Is the system time invariant? == | ||
System: <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br> | System: <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br> | ||
+ | Time-delay: <math>n-n_0</math><br> | ||
Time-delay, then system: | Time-delay, then system: | ||
− | <math>T_k[n]=\delta[n-(k+1)]</math><br> | + | <math>T_k[n]=\delta[(n-n_0)-(k+1)]</math><br> |
− | <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br> | + | <math>Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)]</math><br> |
System, then time-delay: | System, then time-delay: | ||
<math>T_k[n]=(k+1)^2\delta[n-k]</math><br> | <math>T_k[n]=(k+1)^2\delta[n-k]</math><br> | ||
− | <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br> | + | <math>Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)]</math><br> |
Both <math>Y_k</math> yield the same output; therefore, the system is time invariant. | Both <math>Y_k</math> yield the same output; therefore, the system is time invariant. |
Revision as of 06:26, 12 September 2008
6A - Is the system time invariant?
System: $ Y_k[n]=(k+1)^2\delta[n-(k+1)] $
Time-delay: $ n-n_0 $
Time-delay, then system:
$ T_k[n]=\delta[(n-n_0)-(k+1)] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
System, then time-delay:
$ T_k[n]=(k+1)^2\delta[n-k] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
Both $ Y_k $ yield the same output; therefore, the system is time invariant.