00:02
In this question, we're asked to solve a circuit question using laplace transforms.
00:08
All of the parameters are given below.
00:11
In order to do this, what we're going to do is we first plugged in all of the values into the equation, and then we're going to now apply the laplace transform.
00:24
Notice that because of that piecewise function, i had to express this in terms of unit step functions.
00:30
And because i did that, i'm going to, i wrote this as follows.
00:36
I wrote this turning off at t equals pi, just like that.
00:43
Taking the laplace transform on both sides, what that will do is that will give us s times i of s minus i of zero plus 1 ,000 over s times i of s will be equal to the laplace transform of 100 sign of 10t.
01:10
We know the laplace transform of the sign is just going to be a, or will be omega, over s squared plus omega squared.
01:21
So that would be the first term.
01:24
So the first term, it's going to have something like 100 times s squared plus 10 squared, s squared plus 100 over will be the denominator, and the numerator would just be a 10, minus, and then now we have to use the laplace transform of e of the unit step function times any arbitrary function.
01:50
What we do instead is we shift this function forward by pi, if that makes sense, because we're again going to take the laplace transform of 100 times the sign of 10t, so that we would have here 100 times e to the minus pi s, e to minus pi s times 10 over s plus pi squared plus 100.
02:22
And as we know, i of 0 is just 0, so this term will just vanish.
02:28
Now what we need to do is we need to multiply, is we need to solve for i of s.
02:35
So what we have here is we're going to have that i of s times s plus 1000 over s will be equal to 1 ,000 over s squared plus 100 minus 1000 times e to minus pi s times, and this will all be over s plus pi squared plus 100.
03:12
And finally, we will now, what we're going to do here in this other term is i'm just going to multiply both sides by s, or both the top and the bottom by s, so i can get a common denominator.
03:27
This would result in us having s squared plus 1 ,000 over s, which means that when i solve for i of s, what we will get is we will have that i of s will be equal to s over s squared plus 1 ,000 times 1 ,000 over s squared plus 100 minus 1 ,000 e to the minus pi s over s plus pi squared plus 100.
04:01
Okay, that's what we have here.
04:20
Now what could you do otherwise? is there anything you can cancel? not really.
04:33
We would have to figure out a different method of doing this.
04:40
A good start is to multiply.
04:47
You can begin by multiplying out all of the terms here, and it would look something like that, and you can do some convolution integrals.
05:23
That might actually be a better trick for us to do, if we could just do some convolution integrals here.
05:30
So let's use that fact.
05:33
We'll use a convolution integral.
05:35
So if we have two laplace transforms multiplied together, then the inverse laplace transform of this, is just going to be the following convolution integral.
06:04
The f of t starred with g of t, and it's going to be written as the integral from zero to t, of f of t minus tau times g of tau d tau.
06:23
So that's what we have here.
06:25
And what we would have to do is we would have to figure out the laplace transforms of all three of our current elements here.
06:34
So if we do that now, we have the inverse laplace transform of s over s squared plus 1 ,000 is equal to, let's look this up really quickly.
07:05
That's just the cosine of 10 root 10, the inverse laplace transform of 1 ,000 over s squared plus 100...