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*Example of a '''time invariant''' system:
 
*Example of a '''time invariant''' system:
<math>y(t) = x(t) \mapsto y(t - t_0) = x(t - t_0)</math>
+
<math>y_1(t) = x_1(t) \mapsto y_2(t - t_0) = x_2(t - t_0)</math>
  
 
*Example of a '''time variant''' system:
 
*Example of a '''time variant''' system:
<math>y(t) = \sin(t) x(t) \mapsto y(t - t_0) \neq \sin(t) x(t - t_0)</math>
+
<math>y_1(t) = \sin(t) x_1(t) \mapsto y_2(t - t_0) = \sin(t) x_2(t - t_0)</math>
 +
 
 +
In the first example, <math>y_2</math> is the shifted version of <math>y_1</math>. This is not true of the second example.
  
 
==[[Linearity_OldKiwi]]==
 
==[[Linearity_OldKiwi]]==

Revision as of 22:11, 17 June 2008

The six basic properties of Systems_OldKiwi

Memory_OldKiwi

A system with memory has outputs that depend on previous (or future) inputs.

  • Example of a system with memory:

$ y(t) = x(t - \pi) $

  • Example of a system without memory:

$ y(t) = x(t) $

Invertibility_OldKiwi

An invertible system is one in which there is a one-to-one correlation between inputs and outputs.

  • Example of an invertible system:

$ y(t) = x(t) $

  • Example of a non-invertible system:

$ y(t) = |x(t)| $

In the second example, both x(t) = -3 and x(t) = 3 yield the same result.

Causality_OldKiwi

A causal system has outputs that only depend on current and/or previous inputs.

  • Example of a causal system:

$ y(t) = x(t) + x(t - 1) $

  • Example of a non-causal system:

$ y(t) = x(t) + x(t + 1) $

Stability_OldKiwi

There are many types of stability, for this course, we first consider BIBO_OldKiwi (Bounded Input Bounded Output) stability.

A system is BIBO stable if, for all bounded inputs ($ \exist B \epsilon \Re, x(t) < B $), the output is also bounded ($ y(t) < \infty $)

Time Invariance_OldKiwi

A system is time invariant if a shift in the time domain corresponds to the same shift in the output.

  • Example of a time invariant system:

$ y_1(t) = x_1(t) \mapsto y_2(t - t_0) = x_2(t - t_0) $

  • Example of a time variant system:

$ y_1(t) = \sin(t) x_1(t) \mapsto y_2(t - t_0) = \sin(t) x_2(t - t_0) $

In the first example, $ y_2 $ is the shifted version of $ y_1 $. This is not true of the second example.

Linearity_OldKiwi

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