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* Send through system first then time shift
 
* Send through system first then time shift
x(t) <math>\to</math> (system) y(t) = 3+2*x(t) <math>\to</math> (time shift by T) z(t) = y(t-T) = 3+2*x(t-T)
+
** x(t) <math>\to</math> (system) y(t) = 3+2*x(t) <math>\to</math> (time shift by T) z(t) = y(t-T) = 3+2*x(t-T)
  
 
* Time shift first, then send through system
 
* Time shift first, then send through system
x(t) <math>\to</math> (time shift by T) y(t) = x(t-T) <math>\to</math> (system) w(t) = 3+2*y(t) = 3+2*x(t-T)
+
** x(t) <math>\to</math> (time shift by T) y(t) = x(t-T) <math>\to</math> (system) w(t) = 3+2*y(t) = 3+2*x(t-T)
  
Since the two outputs are equal in this case, then it is safe to say that the system is time invariant.
+
*Since the two outputs are equal in this case, then it is safe to say that the system is time invariant.
  
 
== Example of Time Variant System ==
 
== Example of Time Variant System ==
  
input signal x(t) with a system of y(t) = x(2t) and time shift of T
+
Input signal x(t) with a system of y(t) = x(2t) and time shift of T
 +
 
 +
*Send through system first and then time shift
 +
** x(t) <math>\to</math> (system) y(t) = x(2t) <math>\to</math> (Time shift by T) z(t) = y(t-T) = x(2(t-T)) = x(2t-2T)
 +
 
 +
*Time shift first, then send through system
 +
** x(t) <math>\to</math> (Time shift by T) y(t) = x(t-T) <math>\to</math> (system) w(t) = y(2t) = x(2t-T)
 +
 
 +
*Since the two outputs of z(t) and w(t) are NOT EQUAL, then the system is considered to be time variant.

Latest revision as of 18:42, 10 September 2008

Definition

A system is called 'time invariant' if for any input signal x(t) and for any time to that is a real number, the response to the shifted input x(t-T) is the shifted output y(t-T).

This is saying that for order for a signal to be considered 'time invariant' i must be able to put any signal through the system that has gone through a time shift, and i should get out another signal with the same time shift.

Another way to look at time invariance is that if I had a signal x(t) and i put i through a time delay of T, then through the system, I should get the same output if i put x(t) through the system first, and then shifted the output function of the system by T.

Example of Time Invariant System

Input signal x(t) and output which equals 3+2*x(t-T)

  • Send through system first then time shift
    • x(t) $ \to $ (system) y(t) = 3+2*x(t) $ \to $ (time shift by T) z(t) = y(t-T) = 3+2*x(t-T)
  • Time shift first, then send through system
    • x(t) $ \to $ (time shift by T) y(t) = x(t-T) $ \to $ (system) w(t) = 3+2*y(t) = 3+2*x(t-T)
  • Since the two outputs are equal in this case, then it is safe to say that the system is time invariant.

Example of Time Variant System

Input signal x(t) with a system of y(t) = x(2t) and time shift of T

  • Send through system first and then time shift
    • x(t) $ \to $ (system) y(t) = x(2t) $ \to $ (Time shift by T) z(t) = y(t-T) = x(2(t-T)) = x(2t-2T)
  • Time shift first, then send through system
    • x(t) $ \to $ (Time shift by T) y(t) = x(t-T) $ \to $ (system) w(t) = y(2t) = x(2t-T)
  • Since the two outputs of z(t) and w(t) are NOT EQUAL, then the system is considered to be time variant.

Alumni Liaison

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Mu Qiao