(New page: Here we see that when the system input is <math> X_k[n]=\delta[n-k]\,</math> we get the following system output <math> Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math> Hence if the input is ...) |
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Hence if the input is | Hence if the input is | ||
*<math> X_k[n-n0]=\delta[n-n0-k]\,</math> then the output shall be as follows | *<math> X_k[n-n0]=\delta[n-n0-k]\,</math> then the output shall be as follows | ||
| − | *<math> Y_k[n-n0]=(k+1)^2 \delta[n-n0-(k+1)] \,</math> | + | *<math> Y_k[x[n-n0]]=(k+1)^2 \delta[n-n0-(k+1)] \,</math>......................................(1) |
| + | |||
| + | Now suppose we pass the signal through the system first and then delay it | ||
| + | Therefore for input <math> X_k[n]=\delta[n-k]\,</math> we get | ||
| + | <math> Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math> | ||
| + | |||
| + | Now, we delay Y_k[n] by n0 | ||
| + | *Therefore the output will be <math> Y_k[n-n0]=(k+1)^2 \delta[n-n0-(k+1)] \,</math>..............(2) | ||
| + | |||
| + | '''From (1) and (2) it is clear that it is time invariant.''' | ||
Revision as of 13:07, 12 September 2008
Here we see that when the system input is $ X_k[n]=\delta[n-k]\, $ we get the following system output $ Y_k[n]=(k+1)^2 \delta[n-(k+1)] \, $
Hence if the input is
- $ X_k[n-n0]=\delta[n-n0-k]\, $ then the output shall be as follows
- $ Y_k[x[n-n0]]=(k+1)^2 \delta[n-n0-(k+1)] \, $......................................(1)
Now suppose we pass the signal through the system first and then delay it Therefore for input $ X_k[n]=\delta[n-k]\, $ we get $ Y_k[n]=(k+1)^2 \delta[n-(k+1)] \, $
Now, we delay Y_k[n] by n0
- Therefore the output will be $ Y_k[n-n0]=(k+1)^2 \delta[n-n0-(k+1)] \, $..............(2)
From (1) and (2) it is clear that it is time invariant.
