00:01
Given the transfer function below, what is y as t approaches infinity for the subsequent inputs? 2 .1, which is the step input of height m.
00:08
2 .2, which is the unit impulse input delta t.
00:11
2 .3 is sine t.
00:13
And 2 .4 is the unit rectangle pulse where h is equal to 1.
00:19
Okay, for 2 .1, where the step input is of height m, we can do this we can mathematically define this the step input as u of t is equal to m and 0 and this is m for t greater than or equal to 0 and then it's 0 for t less than 0 what we then do is take the laplace transform of this input.
00:55
So the laplace transform of u of t, u of t, this would then equal to being m over s.
01:16
Then what we can do is use the final value theorem and plug this in.
01:21
The final value theorem states that y of infinity is equal to the limit as s approaches 0 as this is in the laplace domain or the s domain as s times y of s this is equal to the limit as s approaches 0 of s times u of s times g of s g of s being the the transfer function and u of s being the input in the lab estimate we just found earlier.
02:00
We can plug those in.
02:02
So this becomes the limit as s approaches 0 of s times m over s times the input, which is 10 over 5s plus 1 times 3s plus 1.
02:23
And so what we can do is just solve this.
02:28
And so as it approaches zero, this then becomes, this would become 10m.
02:49
So that's the solution for 2 .1.
02:51
Now for 2 .2, 2 .2, which is the unit impulse input for delta t.
02:59
Again, we can define what delta t is as an input.
03:04
That would be 1 and 0, and it's 1 when t is equal to 0, and it's 0 for when t is not equal to 0.
03:17
Again, we find the laplace transform of delta t, and that is equal to just 1.
03:30
We can then plug this into the final value theorem again.
03:34
Again, it's the same theorem as we used above, so i'm just going to go skip into this final section down here...