(New page: ==Time Invariance== A system is <b>time invariant</b> if a the time shifted input signal <math>x(t-T)</math> implies an output with equal time shift, meaning <math>x(t-T)\rightarrow y(t-T)...) |
|||
| Line 10: | Line 10: | ||
==A Time Variant System== | ==A Time Variant System== | ||
| − | What if we multiply the | + | What if we multiply the system <math>y(t)</math> by <math>t</math>? |
| − | This gives us the new system <math>y(t)=sin[ | + | This gives us the new system <math>y(t)=t*sin[x(t)]</math>. |
| − | Let's use the same inputs <math>x(t)</math> as we did in the first example. When we apply an input <math>x(t)=t</math>, the output is the expected | + | Let's use the same inputs <math>x(t)</math> as we did in the first example. When we apply an input <math>x(t)=t</math>, the output is the expected <math>y(t)=sin(t)</math>. |
| + | |||
| + | The problem arises when the input is time-shifted. When we apply an input <math>x(t)=t-1</math> to our new system, the output becomes <math>y(t)=(t-1)sin(t-1)</math>. This means that the gain, or multiplier, of the sine function depends on time. Therefore, the system is not time invariant. | ||
Latest revision as of 08:38, 9 September 2008
Time Invariance
A system is time invariant if a the time shifted input signal $ x(t-T) $ implies an output with equal time shift, meaning $ x(t-T)\rightarrow y(t-T) $.
A Time Invariant System
Consider the system $ y(t)=sin[x(t)] $. Suppose we apply an input $ x(t)=t $; we get an output $ y(t)=sin(t) $.
Now suppose we apply an input $ x(t)=t-1 $. If this system is time invariant, we would expect an output time-shifted from the original by an amount equal to the input; therefore, we expect an output of $ sin(t-1) $. When we apply the shifted input signal, this is exactly the output of the system.
Therefore, we can conclude that the system $ y(t)=sin[x(t)] $ is time invariant.
A Time Variant System
What if we multiply the system $ y(t) $ by $ t $?
This gives us the new system $ y(t)=t*sin[x(t)] $.
Let's use the same inputs $ x(t) $ as we did in the first example. When we apply an input $ x(t)=t $, the output is the expected $ y(t)=sin(t) $.
The problem arises when the input is time-shifted. When we apply an input $ x(t)=t-1 $ to our new system, the output becomes $ y(t)=(t-1)sin(t-1) $. This means that the gain, or multiplier, of the sine function depends on time. Therefore, the system is not time invariant.
