| Line 1: | Line 1: | ||
| + | |||
| + | == A) == | ||
| + | |||
| + | |||
Here we see that when the system input is <math> X_k[n]=\delta[n-k]\,</math> | Here we see that when the system input is <math> X_k[n]=\delta[n-k]\,</math> | ||
we get the following system output | we get the following system output | ||
| Line 15: | Line 19: | ||
'''From (1) and (2) it is clear that it is time invariant.''' | '''From (1) and (2) it is clear that it is time invariant.''' | ||
| + | |||
| + | |||
| + | == B) == | ||
| + | |||
| + | The input <math>u[n]</math> would yield <math>u[n-1]</math> | ||
Revision as of 12:11, 12 September 2008
A)
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.
B)
The input $ u[n] $ would yield $ u[n-1] $
