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equals to Y(k+t)[n]=<math>(k+t+1)^2</math> &delta;[n-(k+t+1)]
 
equals to Y(k+t)[n]=<math>(k+t+1)^2</math> &delta;[n-(k+t+1)]
 +
 +
while
 +
 +
Xk[n]=&delta;[n-k] -> system -> time delay
 +
 +
equals to Yk[n]=<math>(k+1)^2</math> &delta;[n-(k+1)]->time delay

Revision as of 15:37, 10 September 2008

The system is not time invariant

Let time delay = t

then

Xk[n]=δ[n-k] -> time delay -> system

equals to X(k+t)[n]=δ[n-(k+t)]->system

equals to X(k+t)[n]=δ[n-k-t]->system

equals to Y(k+t)[n]=$ (k+t+1)^2 $ δ[n-(k+t+1)]

while

Xk[n]=δ[n-k] -> system -> time delay

equals to Yk[n]=$ (k+1)^2 $ δ[n-(k+1)]->time delay

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood