Difference between revisions of "HW2-E Seung Seob Lee ECE301Fall2008mboutin" - Rhea

Revision as of 17:27, 12 September 2008

$\,x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1)\,$

Proof:

$x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1) \to Time Shift(t0) \to z(t)=y(t-t0)$

$\, =(k+1)^{2}x(t-t0-k-1)\,$

$x(t-k) \to Time Shift(t0) \to y(t)=x(t-t0-k) \to System \to z(t)=(k1+1)^{2}y(t-k1-1)$

$\, =(t0+k+1)^{2}x(t-t0-k-1)\,$

It is not time invariant.

@@ Line 5: / Line 5: @@
 '''Proof:'''
-<math>x(t) \to System \to y(t)=e^{x(t)} \to Time Shift(t0) \to z(t)=y(t-t0)</math>
+<math>x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1) \to Time Shift(t0) \to z(t)=y(t-t0)</math>
-<math>\,                                                            =e^{x(t-t0)}\,</math>
+<math>\,                                                            =(k+1)^{2}x(t-t0-k-1)\,</math>
-<math>x(t) \to Time Shift(t0) \to y(t)=x(t-t0) \to System \to z(t)=e^{y(t)}</math>
+<math>x(t-k) \to Time Shift(t0) \to y(t)=x(t-t0-k) \to System \to z(t)=(k1+1)^{2}y(t-k1-1)</math>
-<math>\,                                                            =e^{x(t-t0)}\,</math>
+<math>\,                                                            =(t0+k+1)^{2}x(t-t0-k-1)\,</math>
+It is not time invariant.
-Both cascades yielded the same outputs, thus <math>\,y(t)=e^{x(t)}\,</math> is time invariant.