(Time Variant System)
(Time Invariant System)
 
Line 13: Line 13:
 
<math>x(t-T) = 2e^{t-5} </math> -> [SYSTEM] -> <math>35*2*e^{t-5}</math>
 
<math>x(t-T) = 2e^{t-5} </math> -> [SYSTEM] -> <math>35*2*e^{t-5}</math>
  
<math>x(t) = 2e^{t} -> [SYSTEM] -> <math>y(t-T)=35*2*e^{t-5}</math>
+
<math>x(t) = 2e^{t}</math> -> [SYSTEM] -> <math>y(t-T)=35*2*e^{t-5}</math>
  
 
As you can see these two outputs are the same, so the system is time invariant.
 
As you can see these two outputs are the same, so the system is time invariant.

Latest revision as of 15:32, 11 September 2008

Time Invariance

A system is time invariant if for a certain x(t) that produces an output y(t) if you shift the input to x(t-T) it just yields the same output shifted by the same T. y(t-T).

Time Invariant System

I propose that a system where

$ x(t) $ -> [SYSTEM] -> $ y(t) = 35x(t) $ is time invariant. Let's check.

Let $ x(t)=2e^t $ and $ T=5 $

$ x(t-T) = 2e^{t-5} $ -> [SYSTEM] -> $ 35*2*e^{t-5} $

$ x(t) = 2e^{t} $ -> [SYSTEM] -> $ y(t-T)=35*2*e^{t-5} $

As you can see these two outputs are the same, so the system is time invariant.

Time Variant System

I propose that a system where

$ x(t) $ -> [SYSTEM] -> $ x(35t) $

Is time variant. Let's use the same T and x(t) for this example.

$ x(t-T) = 2e^{t-5} $ -> [SYSTEM] -> $ 2*e^{35t-5} $

$ x(t) = 2e^{t} $ -> [SYSTEM] -> $ y(t-T)=2*e^{35(t-5)} = 2*e^{35t - 175} $

Clearly these two results are not the same, so the system is time variant.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood