(New page: A time invariant system is a system that for any x(t) that goes into the system and has an output y(t) has the same response as a shifted input x(t-T) which has an output of the system of ...) |
|||
| Line 4: | Line 4: | ||
Let the system be: | Let the system be: | ||
y(t) = e^t * x(t) | y(t) = e^t * x(t) | ||
| + | |||
Justify: | Justify: | ||
| − | + | Let T = 3 | |
| − | + | ||
| + | And x(t) = 2t+1 | ||
| + | |||
| + | |||
Then the graph of e^t * x(t) = e^3 * (2t+1) is the same as | Then the graph of e^t * x(t) = e^3 * (2t+1) is the same as | ||
| + | |||
e^(t-3)*x(t-3) = e^(t-3) * (2(t-3)+1) only the second graph is shifted by 3 units. | e^(t-3)*x(t-3) = e^(t-3) * (2(t-3)+1) only the second graph is shifted by 3 units. | ||
== Time Variant System == | == Time Variant System == | ||
Let the system be: | Let the system be: | ||
| + | |||
y(t) = x(2t) | y(t) = x(2t) | ||
| + | |||
Justify: | Justify: | ||
| + | |||
Let T=3 again | Let T=3 again | ||
| + | |||
and x(t) = 2t+1 again | and x(t) = 2t+1 again | ||
| + | |||
Then the graph of x(2t) = 4t+1 | Then the graph of x(2t) = 4t+1 | ||
| + | |||
And the graph of x(2t-3) = 2(2t-3)+1 = 4t-5 | And the graph of x(2t-3) = 2(2t-3)+1 = 4t-5 | ||
| + | |||
The second graph is a completely different graph, not the original shifted along the horizontal axis. Therefore this system is not time invariant. | The second graph is a completely different graph, not the original shifted along the horizontal axis. Therefore this system is not time invariant. | ||
go back to : [[Homework 2_ECE301Fall2008mboutin]] | go back to : [[Homework 2_ECE301Fall2008mboutin]] | ||
Latest revision as of 11:03, 11 September 2008
A time invariant system is a system that for any x(t) that goes into the system and has an output y(t) has the same response as a shifted input x(t-T) which has an output of the system of y(t-T).
Time Invariant System
Let the system be: y(t) = e^t * x(t)
Justify:
Let T = 3
And x(t) = 2t+1
Then the graph of e^t * x(t) = e^3 * (2t+1) is the same as
e^(t-3)*x(t-3) = e^(t-3) * (2(t-3)+1) only the second graph is shifted by 3 units.
Time Variant System
Let the system be:
y(t) = x(2t)
Justify:
Let T=3 again
and x(t) = 2t+1 again
Then the graph of x(2t) = 4t+1
And the graph of x(2t-3) = 2(2t-3)+1 = 4t-5
The second graph is a completely different graph, not the original shifted along the horizontal axis. Therefore this system is not time invariant.
go back to : Homework 2_ECE301Fall2008mboutin
