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A system is memoryless if for any <math> t \in \mathbb{R} </math> the output at <math> t_0 \, </math> depends only on the input at <math> t_0 \, </math>(not past or future samples or informations). | A system is memoryless if for any <math> t \in \mathbb{R} </math> the output at <math> t_0 \, </math> depends only on the input at <math> t_0 \, </math>(not past or future samples or informations). | ||
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For example: y(t)=2x(t),y(t)=t-1+x(t) | For example: y(t)=2x(t),y(t)=t-1+x(t) | ||
| − | + | *Question: Is y(t)=1 (or any constant number) memoryless? | |
--Personally I think it is, but it somehow violates the definition given in lecture. Since it doesn't even depend on <math>t_0\,</math>) | --Personally I think it is, but it somehow violates the definition given in lecture. Since it doesn't even depend on <math>t_0\,</math>) | ||
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A system is with memmory if for some <math> t \in \mathbb{R} </math> the output at <math> t_0 \, </math> doesn't only depend on the input at <math> t_0 \, </math>, it also depends on past or future samples or informations. | A system is with memmory if for some <math> t \in \mathbb{R} </math> the output at <math> t_0 \, </math> doesn't only depend on the input at <math> t_0 \, </math>, it also depends on past or future samples or informations. | ||
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| + | For example, | ||
| + | y(t)=x(t)+x(t-1), y(t)=t+1 | ||
Latest revision as of 17:17, 17 September 2008
Memoryless System
A system is memoryless if for any $ t \in \mathbb{R} $ the output at $ t_0 \, $ depends only on the input at $ t_0 \, $(not past or future samples or informations).
For example: y(t)=2x(t),y(t)=t-1+x(t)
- Question: Is y(t)=1 (or any constant number) memoryless?
--Personally I think it is, but it somehow violates the definition given in lecture. Since it doesn't even depend on $ t_0\, $)
System with Memory
A system is with memmory if for some $ t \in \mathbb{R} $ the output at $ t_0 \, $ doesn't only depend on the input at $ t_0 \, $, it also depends on past or future samples or informations.
For example, y(t)=x(t)+x(t-1), y(t)=t+1
