(Time Invariant System)
(Time Invariant System)
 
(One intermediate revision by the same user not shown)
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<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.
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<math>x(t-T) = 2e^{t-5} </math> -> [SYSTEM] -> <math>2*e^{35t-5}</math>
 
<math>x(t-T) = 2e^{t-5} </math> -> [SYSTEM] -> <math>2*e^{35t-5}</math>
  
<math>x(t) = 2e^{t} -> [SYSTEM] -> <math>y(t-T)=2*e^{35(t-5)} = 2*e^{35t - 175}</math>
+
<math>x(t) = 2e^{t}</math> -> [SYSTEM] -> <math>y(t-T)=2*e^{35(t-5)} = 2*e^{35t - 175}</math>
  
 
Clearly these two results are not the same, so the system is time variant.
 
Clearly these two results are not the same, so the system is time variant.

Latest revision as of 16:32, 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.

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EISL lab graduate

Mu Qiao