Difference between revisions of "HW2-E Nicholas Browdues ECE301Fall2008mboutin" - Rhea

Latest revision as of 15:07, 12 September 2008

The system is time variant because of the following example:

Dealing with the following system

$X_k[n]=Y_k[n] \,$

where

$X_k[n]=\delta[n-k]\,$

and

$Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,$

Consider the input and output of the system when k = 0

$X_0[n]=\delta[n]\,$

and

$Y_0[n]=\delta[n-1] \,$

If I time shift the input by $n_0\,$ , then run it through the system I obtain:

$X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1],$

but this isn't equal to running the input through the system, then time shifting the output by $n_0\,$

$X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\,$

Assuming this system is linear, an input $X_0[n]=u[n]\,$ would result in an output $Y_0[n]=u[n-1]\,$ .

@@ Line 1: / Line 1: @@
 == Question 6a ==
+The system is time variant because of the following example:
-I'm assuming k is the variable representing any fo.
+Dealing with the following system
-<math> X_k[n]=X_k[n] \,</math>
+<math> X_k[n]=Y_k[n] \,</math>
-where <math> X_k[n]=\delta[n-k]\,</math>     and <math> Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math>
-Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math>
+where
-Under this assumption the following system cannot possibly be time invariant because of the <math>(k+1)^2</math> term.
+<math> X_k[n]=\delta[n-k]\,</math>
+and
+<math> Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math>
+Consider the input and output of the system when k = 0
+<math> X_0[n]=\delta[n]\,</math>
+and
+<math> Y_0[n]=\delta[n-1] \,</math>
+If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain:
+<math>X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1],</math>
+but this isn't equal to running the input through the system, then time shifting the output by <math> n_0\,</math>
+<math>X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\,</math>
+== Question 6b ==
+Assuming this system is linear, an input
+<math> X_0[n]=u[n]\,</math>
+would result in an output
+<math> Y_0[n]=u[n-1]\,</math>.