Laplace transform 2
Laplace transform of e^at. Created by Sal Khan.
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- How do you compare complex and real number? I mean he writes s>a...(8 votes)
- For those who are wondering this happens at4:37.
Well firstly, 'a' can also be complex, and secondly, in maths you cannot compare complex numbers, either with real numbers or with other complex numbers.
So what he did is a little more in depth that what he wrote down, and I'm sure that most people don't realise this, as he hasn't defined 's' as a complex number yet.
Let's say that 's = a + bi', and 'a = c + di', then that part on the inside of the integral becomes 'exp((a-s)t) = exp((c + di - a - bi)t)'. Grouping the real and imaginary terms we get 'exp((c-a)t+i(d-b)t)'. We can then split the exponent as such: 'exp((c-a)t)exp(i(d-b)t)'.
Looking at this, as t tends to infinity, we know that the complex exponential will just go round in circles (from Euler's identity, 'exp(i*theta)'). But if we look at the real exponential, we can see that it will either go to infinity or to zero, depending on the value of '(c-a)'. So something that goes around in circles times zero will be zero, and something that goes around in circles times something that tends to infinity will also tend to infinity.
So to make a slight correction (which is only necessary if you want to be strictly mathematical), by writing 's>a', he actually means 'Re(s)>Re(a)', which is entirely possible to compare.
I hope this explanation is clear.(46 votes)
- At2:10, why is e^(-st) * e^(at) combine to e^(a-s)t instead of e^(a-s) ?? Wouldn't the -t and t combine to cancel??(4 votes)
- For multiplying terms which have the same base, (x^a) * (x^b) = x^(a + b). So,
e^(-st) * e^(at) = e^(-st + at). Take out the common factor, "t" from (-st + at) =>
t(a-s), which gives us e^(a-s)t(11 votes)
- What about the case when a = s? We get 0*t, when t tends to infinity, do we consider it an indeterminate form in this case?(8 votes)
- in doing these transformations it is assumed s>a by definition otherwise the function diverges.(1 vote)
- What is the difference between ordinary differential equations and partial differential equations?(2 votes)
- Ordinary differential equations involve quantities that vary with respect to a single variable.
Partial differential equations involve quantities that vary with respect to more than one variable. For example the quantity of interest could be thermal energy across a metal surface and the variables are time and space.
Note: When I say quantity I am reffering to the output of a function
Thumbs Up!(5 votes)
- How would you go about the Laplace transform of e^(t+2) ?(1 vote)
- Notice that e^(t+2) could be written as (e^2)*e^t, since e^2 is a constant, you can use the linearity of the Laplace transform, and find (e^2)*L(e^t).
This would equal
- Is 's' here something like the 'plus C' in indefinite integrals? I mean it can be nearly any constant, isn't is?(2 votes)
- Not exactly, it was my understanding that the 's' that appears in laplace transforms is actually another variable in a different dimension or with different units of measure (i.e. t -> time domain, while s -> frequency domain).(3 votes)
- What if s=a? would this mean the limit/Laplace Transform DNE because you would have -1/0?(2 votes)
- yes, correct. The Transform still does not work.(1 vote)
- why does he change the expression to be a limit of a as before?(2 votes)
- It is often a formality to explicitly show why the next steps happen. It is like a justification. And sometimes, with complex systems, it is necessary to show a limit, else the function may behave unexpectedly! :O However, often times we can see what the limit does without putting it down and we don't bother writing it at all! :)(1 vote)
- At4:30I understand what he is doing, but why does he evaluate to infinity? In the first video, he solves for the limit at a constant. Why is this no longer required?(2 votes)
- He directly evaluates to infinity because the limit isn't absolutely necessary to preform the substitution of infinity in the Laplace transform equation.
He uses a constant in the first video to make it easier to understand.(1 vote)
- can any one tell me how -1/s-a = 1/s+a what is the math behind this ?(2 votes)
- Multiply -1 on the numerator and denominator(1 vote)
Let's keep doing some Laplace transforms. For one, it's good to see where a lot of those Laplace transform tables you'll see later on actually come from, and it just makes you comfortable with the mathematics. Which is really just kind of your second semester calculus mathematics, but it makes you comfortable with the whole notion of what we're doing. So first of all, let me just rewrite the definition of the Laplace transform. So it's the L from Laverne & Shirley. So the Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function. Times our function of t, and that's with respect to dt. So let's do another Laplace transform. Let's say that we want to take the Laplace transform-- and now our function f of t, let's say it is e to the at. Laplace transform of e to the at. Well we just substituted it into this definition of the Laplace transform. And this is all going to be really good integration practice for us. Especially integration by parts. Almost every Laplace transform problem turns into an integration by parts problem. Which, as we learned long ago, integration by parts is just the reverse product rule. So anyway. This is equal to the integral from 0 to infinity. e to the minus st times e to the at, right? That's our f of t. dt. Well this is equal to just adding the exponents because we have the same base. The integral from 0 to infinity of e to the a minus stdt. And what's the antiderivative of this? Well that's equal to what? With respect to C. So it's equal to-- a minus s, that's just going to be a constant, right? So we can just leave it out on the outside. 1/a minus s times e to the a minus st. And we're going to evaluate that from t is equal to infinity or the limit as t approaches infinity to t is equal to 0. And I could have put this inside the brackets, but it's just a constant term, right? None of them have t's in them, so I can just pull them out. And so this is equal to 1/a minus s times-- now we essentially have to evaluate t at infinity. So what is the limit at infinity? Well we have two cases here, right? If this exponent-- if this a minus s is a positive number, if a minus s is greater than 0, what's going to happen? Well as we approach infinity, e to the infinity just gets bigger and bigger and bigger, right? Because it's e to an infinitely positive exponent. So we don't get an answer. And when you do improper integrals, when you take the limit to infinity and it doesn't come to a finite number, the limit doesn't approach anything, that means that k the improper integral diverges. And so there is no limit. And to some degree, we can say that the Laplace transform is not defined with a minus s is greater than 0 or when a is greater than s. Now what happens if a minus s is less than 0? Well then this is going to be some negative number here, right? And then if we take e to an infinitely negative number, well then that does approach something. That approaches 0. And we saw that in the previous video. And I hope you understand what I'm saying, right? e to an infinity negative number approaches 0, while e to an infinitely positive number is just infinity. So that doesn't really converge on anything. So anyway. If I assumed that a minus s is less than 0, or a is less than s, and this is the assumption I will make, just so that this improper integral actually converges to something. So if a minus s is less than 0, and this is a negative number, e to the a minus s times-- well t, where t approaches infinity will be 0. Minus this integral evaluated at 0. So when you value this at 0, what happens? T equals 0. This whole thing becomes e to the 0 is 1. And we are left with what? Minus 1/a minus s. And that's just the same thing as 1/(s-a). So we have our next entry in our Laplace transform table. And that is the Laplace transform. The Laplace transform of e to the at is equal to 1/(s-a) as long as we make the assumption that s is greater than a. This is true when s is greater than a, or a is less than s. You could view it either way. So that's our second entry in our Laplace transform table. Fascinating. And actually, let's relate this to our previous entry in our Laplace transform table, right? What was our first entry in our Laplace transform table? It was Laplace transform of 1 is equal to 1/s, right? Well isn't 1 just the same thing as e to the 0? So we could have said that this is the Laplace-- I know I'm running out of space, but I'll do it here in purple. We could have said Laplace transform of 1 is the same thing as e to the 0 times t, right? And that equals 1/s. And luckily it's good to see that that is consistent. And actually, remember, we even made the condition when s is greater than 0, right? We assumed that s is greater than 0 this example. Here again, you say s is greater than 0. This is completely consistent with this one, right? Because if a is equal to 0, then the Laplace transform of e to the 0 is just 1/s minus 0. That's just 1/s. And we have to assume that s is greater than zero. So really these are kind of the same entry in our Laplace transform table. But it's always nice in mathematics when we see that two results we got in trying to do slightly different problems actually are, in some ways, connected or the same result. Anyway I'll see you in the next video and we'll keep trying to build our table of Laplace transforms. And maybe three or four videos from now I'll actually show you how these transforms are extremely useful in solving all sorts of differential equations. See you soon.