Revision as of 06:27, 12 September 2008 by Dhopper (Talk)

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.

6B - What input X[n] would yield the output Y[n]=u[n-1]?

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010