Revision as of 12:11, 12 September 2008 by Mgoklani (Talk)

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 $ <math>u[n-1] $</math>

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett