Lesson 1 - Laplace Transform Definition (Engineering Math)

Summary

In this comprehensive tutorial series, Jason introduces students to the LL (Laplace) transform, a powerful mathematical tool used in science and engineering to solve differential equations more efficiently. Starting with the basics, the course builds up to using a table of LL transforms and properties for both direct and inverse transformations. The script emphasizes the practicality of the LL transform in various fields and guides learners through the process of transforming functions of time into the S domain, showcasing the transform's utility in simplifying complex problems.

Takeaways

• 📚 The course is designed to teach students how to master the use of the Laplace Transform, starting from scratch and building up to practical applications.
• 🔧 The Laplace Transform is a powerful tool used in science and engineering to solve complex problems, particularly differential equations, more efficiently.
• 📈 The course covers the Laplace Transform properties and how to apply both forward and inverse transforms to functions of time.
• 🎓 Students are expected to have basic knowledge of calculus and algebra, as these are fundamental to understanding and performing the Laplace Transform calculations.
• ⏳ The Laplace Transform takes a function of time and transforms it into a function in the 'S-domain', simplifying the process of solving differential equations.
• 📉 The integral definition of the Laplace Transform involves integrating a function of time multiplied by an exponential function over time from 0 to infinity.
• 📑 A table of common Laplace Transforms is an essential resource that students will learn to apply and derive, starting with simple functions and building up to more complex ones.
• 🌟 The Laplace Transform of an exponential function e^(λt) is particularly important and results in a simple function of 's', which is 1 / (s - λ), provided s > λ.
• 🔄 The Laplace Transform can be applied in a wide range of scientific and engineering disciplines, including electrical circuits, mechanical systems, and aerospace engineering.
• 📝 The process of solving problems with the Laplace Transform involves transforming the problem into the S-domain, solving it there with simpler algebraic methods, and then transforming the solution back to the time domain.
• 📚 The special case of the Laplace Transform when λ = 0, which is the transform of the constant function 1, results in 1 / s, valid for s > 0.

Q & A

What is the main objective of the LLas Transform Tutor course?

The main objective of the LLas Transform Tutor course is to build students' skills so that they can master the use of the LLas transform, including how to use a table of LL transforms and the properties of the LLas transform to perform both direct and inverse transformations.

Why are LLas transforms important in science and engineering?

LLas transforms are important in science and engineering because they provide a practical skill used to solve more complicated problems in various classes, including electrical engineering, mechanics, pure mathematics, and differential equations. They simplify the process of solving differential equations, which are ubiquitous in these fields.

What mathematical background is expected from students before starting the LLas Transform course?

Students are expected to have a basic understanding of calculus, ideally at the level of 'calculus one', and proficiency in algebra. This background will enable them to perform the mathematical operations required for the LLas transform.

What is the basic concept behind the LLas transform?

The basic concept behind the LLas transform is to take a function of time (or any function) and transform it into a new function in a different domain (the S domain), simplifying the process of solving differential equations and other complex problems.

How does the LLas transform help in solving differential equations?

The LLas transform helps in solving differential equations by providing a unified method to convert a complex differential equation into a new form in the S domain, which is often easier to solve using algebraic methods. Once solved, the solution can be transformed back into the time domain.

What is the mathematical expression for the LLas transform of a function F(t)?

The mathematical expression for the LLas transform of a function F(t) is given by the integral from 0 to infinity of e to the minus s t * F of t with respect to t, denoted as F(S) = L{F(t)} = ∫[0,∞] e^(-st) * F(t) dt.

What is an improper integral, and how is it related to the LLas transform?

An improper integral is an integral where at least one of the limits is infinity, or the integrand has an unbounded point in the interval of integration. In the context of the LLas transform, the improper integral is used to transform a time-domain function into the S domain.

What is the significance of the LLas transform of an exponential function e^(Λt)?

The LLas transform of an exponential function e^(Λt) is significant because it results in a simple function of S, which is 1 / (s - Λ). This result is a fundamental part of the LLas transform table and is used as a basis for transforming more complex functions.

What is the LLas transform of the constant function '1'?

The LLas transform of the constant function '1' is 1 / s, which is derived from the special case of the exponential function e^(Λt) when Λ equals zero.

Why is it necessary to consider the condition 'S greater than Λ' when discussing the LLas transform of e^(Λt)?

The condition 'S greater than Λ' ensures that the denominator in the resulting function of S from the LLas transform does not become zero, thus avoiding undefined or infinite values in the transform.

Can you provide an example of how the LLas transform simplifies the solution of a real-world problem?

While the script does not provide a specific real-world example, the LLas transform simplifies real-world problems by allowing engineers and scientists to convert complex time-domain differential equations into simpler algebraic equations in the S domain. Solving these equations is typically more straightforward, and the solutions can then be transformed back to the time domain to provide practical solutions to real-world problems.

Outlines

00:00

📚 Introduction to LLas Transforms

Jason introduces the LLas Transform Tutor series, aiming to build students' skills in mastering the LLas transform, a tool used in science and engineering. He emphasizes that no prior knowledge is expected but by the end, students will understand and be able to apply the transform, its table, and its properties to solve complex problems. The LLas transform is a calculus-based technique for converting functions of time into a different domain, simplifying the solution of differential equations, which are ubiquitous in various fields of study.

05:01

🔍 Understanding the LLas Transform Concept

The script delves into the definition of the LLas transform, explaining it as an integral from 0 to infinity of a function of time multiplied by an exponential term. This process transforms a time-domain function into a new function in the S-domain. The explanation includes the mathematical notation and the significance of each component in the transform formula. It also touches on the concept of improper integrals and how limits of integration are applied in the context of the LLas transform.

10:02

📈 Calculating the LLas Transform of Exponentials

The tutorial focuses on calculating the LLas transform of exponential functions, a fundamental skill in mastering the transform. The process involves substituting the exponential function into the LLas transform formula and performing an integral with respect to time. The explanation includes a step-by-step guide on how to simplify the integral, apply a substitution to make it more manageable, and evaluate the limits of integration to arrive at a function of S, which is the result of the transform.

15:04

📘 Special Case: LLas Transform of One

The script highlights a special case of the LLas transform where the exponential function becomes a constant one, leading to the conclusion that the LLas transform of one is 1/s, with the condition that S is greater than zero to avoid undefined expressions. This result is emphasized as an important takeaway from the tutorial, marking the simplicity of the transform for constant functions.

20:04

🔧 LLas Transform as a Problem-Solving Tool

The tutorial concludes by reinforcing the practical applications of the LLas transform, particularly in solving differential equations across various scientific and engineering disciplines. It positions the LLas transform as a valuable skill that simplifies complex problem-solving by transforming time-domain functions into a more algebraically manageable domain. The script encourages students to follow through with the tutorial to gain proficiency in applying the LLas transform to real-world problems.

Keywords

💡LLas Transform

The LLas Transform, also known as the Laplace Transform, is a mathematical technique used to simplify the solving of differential equations by converting a function of time into a function of a complex variable, often denoted as 's'. In the video, it's the central theme, with the instructor guiding viewers through the process of understanding and applying this transform to various functions, illustrating its practical applications in science and engineering.

💡Differential Equations

Differential Equations are equations involving derivatives of a function with respect to an independent variable. They are foundational in understanding rates of change and are omnipresent in fields like physics, engineering, and economics. The video emphasizes the utility of the LLas Transform in simplifying the process of solving these equations, which can otherwise be quite complex.

💡Time Domain

The Time Domain refers to the representation of a function or signal as it varies over time. In the context of the video, functions of time are those that change with time, such as signals in electrical engineering. The instructor explains that the LLas Transform takes a function from the time domain and transforms it into a different domain, simplifying its analysis.

💡S Domain

The S Domain, also known as the Complex Frequency Domain, is the domain into which the LLas Transform converts a time-domain function. It is a complex number domain that simplifies the analysis of dynamic systems. The video script mentions that after transforming a problem into the S domain, it becomes easier to solve, often through algebraic methods rather than the more complex calculus required in the time domain.

💡Improper Integral

An Improper Integral is an integral that has one or both limits of integration extending to infinity or where the integrand has an undefined point in the interval of integration. In the video, the instructor explains that the LLas Transform involves calculating an improper integral from 0 to infinity, emphasizing the need to take limits as part of the calculation process.

💡Exponential Function

An Exponential Function is a mathematical function of the form e^(λt), where 'e' is the base of the natural logarithm, 'λ' is a constant, and 't' represents time. The video script discusses the LLas Transform of exponential functions, showing how they transform into a simple form in the S domain, which is crucial for solving differential equations involving exponential growth or decay.

💡Algebraic Simplification

Algebraic Simplification refers to the process of making algebraic expressions more compact and easier to understand or solve. The video highlights that after applying the LLas Transform, what might have been a complex differential equation can be simplified to an algebraic form that is easier to solve.

💡Inverse LLas Transform

The Inverse LLas Transform is the process of converting a function in the S domain back into the time domain. It is the reverse of the LLas Transform and is essential for finding the original time-domain function after solving the transformed equation. The video mentions the importance of understanding both the forward and inverse transforms for complete problem-solving.

💡Table of LLas Transforms

A Table of LLas Transforms is a compilation of common functions and their corresponding transforms in the S domain. It serves as a reference for quickly finding the transformed version of standard functions without having to derive them from scratch. The video script mentions building a table of transforms as a part of mastering the LLas Transform technique.

💡Engineering Applications

Engineering Applications refer to the practical uses of mathematical and scientific principles in the design, construction, and maintenance of various structures, machines, systems, etc. The video script emphasizes the wide applicability of the LLas Transform in different engineering disciplines, such as electrical, mechanical, and aerospace engineering, for solving complex problems involving dynamic systems.

Highlights

Introduction to the LLas Transform Tutor series, aimed at building practical skills for mastering the LLas transform.

Expectation of no prior knowledge of the LLas transform, with the course aiming to teach both its concept and application.

The LLas transform is a practical skill used in science and engineering to solve complex problems.

The LLas transform is a unified method for solving differential equations more easily than traditional methods.

The LLas transform involves calculus and algebra, requiring proficiency in these areas.

The LLas transform takes a function of time and transforms it into a new function in the S domain.

The usefulness of the LLas transform is in its ability to simplify the process of solving differential equations.

The LLas transform is integral to various branches of science and engineering, including electrical, mechanical, and aerospace.

The LLas transform formula is introduced, explaining the process of transforming a time function into the S domain.

Explanation of the improper integral involved in the LLas transform and its mathematical rigor.

The process of creating a table of LLas transforms for common functions to simplify problem-solving.

Detailed walkthrough of calculating the LLas transform of an exponential function e^(Λt).

The result of the LLas transform of e^(Λt) is a function of s, specifically 1/(s - Λ), highlighting its significance.

Special case analysis when Lambda equals zero, revealing the LLas transform of one to be 1/s.

Emphasis on the importance of understanding the LLas transform for its wide applicability in real-world problems.

The LLas transform is positioned as a valuable tool for anyone in engineering or science due to its broad utility.

Encouragement to follow the course for mastering the LLas transform and its applications in solving differential equations.

Transcripts

hello I'm Jason welcome to the llas

transform tutor uh and here what we're

going to do in this lesson and in this

whole set of lessons is really build

your skills so that you can Master how

to use the llas transform now I expect

you to know absolutely nothing about the

llas transform going into this course

but at the end of it you'll not only

know what the transform is U but also

how to use it and how to use a table of

LL transforms and also the llas

transform properties to take lla

transforms and also inverse lla

transforms so it's a very practical

skill that we use in science and

engineering classes uh you know in order

to really solve some more complicated

problems and you'll find that lots and

lots of different classes that you'll

take as you go on in your engineering

sequence or in your science sequence

will use the LL transform so if you ever

get to a class in uh electrical

engineering or mechanics or uh uh pure

mathematics or differential equations

where LL transforms are used uh then

what you'll be able to do is come into

here and get that whole sequence of

information on how to use this very

powerful tool along with lots of

problems to practice your skills so uh

I'm going to obviously write some math

on the board in a second but in the uh

General scheme of things what the heck

is this thing called a lla transform uh

well first of all it is uh it is is

calculus so you have to know some

calculus in order to be able to

calculate the transform so I expect you

up until this point to know a little bit

of calculus really not much more than

calculus one to be honest with you will

enable you to be able to do the math

here but you will have to be proficient

with it also obviously your algebra

skills because I know that you all know

how to do basic algebra but when you get

into taking some of these transforms it

can um it can quickly explode into a

bunch of things to simplify none of it's

hard it's just a lot of little terms to

keep track of and so we'll be doing that

here and I'll show you as I go but it's

basically calculus and uh what you're

basically doing is you're using a

technique to take a function so here

we're going to be talking about

functions of time but really it's it's

any function you know F ofx you normally

study an algebra um but you know when

you get into engineering and studying

real systems you're usually talking

about functions of time things that

change with time for instance in

electrical engineering you might have an

input signal you know like a sign you

know wave or some kind of U random voice

uh recording of a voice might give you

an amplitude as a function of time that

would be a function of time right and so

the LL transform takes a function of

time and transforms it to a new function

that's uh in a new uh new way of

representing that original function and

you might say well why would you ever

care about doing that and the the truth

is the reason why it works is beyond the

scope of what I'm talking about right

now but basically llas transforms let

you solve a lot of different types of

differential equations in an easier way

than doing them by hand so in in the

original way so if you think back to my

differential equations tutor or any

class that you may have taken in

differential equations you should agree

with me that it's kind of a nightmare to

solve a lot of differential equations

there's a lot of math involved a lot of

theory and differential equations really

have a lot of different techniques so

you may do method of undetermined

coefficient coefficients in differential

equations you might do integrating

factors in different equations you might

use exact differential equations um

there's lots of different ways to solve

differential equations and a lot of it

boils down to recognizing what the

equation is figuring out the method that

works and then applying it and so it

seems to be disconnected though because

there's lots of different techniques for

different styles of differential

equations LL transform is a unified

method that really allows you to take a

differential equation and apply the

transform to it and change it into a new

form that's easier to solve

and then when you get the answer in the

LL domain or the S domain we'll talk

about in a minute then you can transform

the answer back into the time domain

that's why it's called a transform we

take the problem or the function or

whatever it is we transform it into the

llao domain which we're going to call

the S domain and a lot of times doing

the solution in that domain just becomes

algebra or some other easier Simple

Solution method than differential

equations brute force method so we solve

it over here in the llas domain or the S

Dom domain we get an answer and then we

can take that an inverse transform it

back into a function of time um so

that's kind of without getting into the

details that's basically why it's useful

and as you know differential equations

are used in all branches of science and

engineering we use them in electric

circuits we use them in mechanical

systems use them in control systems

theory use them in chemical engineering

aerospace engineering differential

equations are everywhere and because of

that and because LL transform is kind of

a general method um to to make these

types of problems easier you're going to

find that LL transforms pop up in all

kinds of different situations and so

that's why I want to put this tutorial

in its own little course because you can

use it if you're mechanical engineer

electrical engineer uh physicist you

know anything so let's just dive into it

first of all uh and talk about it a

little bit so we have what we call the

llas

transform okay and I'm not going to beat

around the bush too much we're just

going to write it on the board and we'll

talk about it so the llas transform

gives you a function in the lass domain

which is the S domain that's why you're

getting a function capital f of x of s

that's what you're arriving at when you

perform this thing called the LL

transform and it's the integral from 0o

up to Infinity of e to the minus s t * F

of t d t so this ladies and gentlemen is

the famous L PLO transform all right so

what you're doing all right is if I give

you a function of time think about any

function of time could be a square wave

could be a s of T could be a saw too

could be any function of time you want

could be X squ or some function of time

squared you stick that function of time

in F of T then you multiply by this

exponential e to the minus St right and

so you're getting something inside of

the integral sign you integrate it over

time right and then you apply the limits

of integration to the answer that you

get which are obviously going into your

uh into your time spots because you're

integrating over time so the limits of

integration is from zero to Infinity in

terms of time right then when you do all

of that you're going to get just a

function of s back because if you

integrate over time and then you plug in

limits of integration over time there's

no T anymore left in what you've done

there's just the other uh uh aspects of

it here so there's no T anymore here

there's no T anymore here because you've

plugged in limits of integration only s

remains so when you do a ll transform

properly you get purely a function of s

and so that's the notation here you put

F of T little F of T is your input

function and out of this computation

comes another function we call it

capital F we use the same letter capital

F and little F to imply that little F

and capital F are related to one another

we we put a little F into the LL

transform and we get a capital f out

which is a function of s all right so we

can also

write we can also

write

as the following F capital F of s which

is the the pl transform is Curly L I

don't know how else you would say that

right F of T so so this is the shorthand

way of saying hey I'm going to take the

LL transform of some function of time or

some function and when I do that I'm

just going to get a pure function of s

so these two things are exactly the same

thing this is the shorthand way of

writing what you're doing you're taking

the the pl transform of a function of

time this is one of the details this is

what you actually do you have to

multiply this you have to integrate it

you have to plug in the limits of

integration so what we're doing is we're

transforming a function from the time

domain to the S domain

that's why I'm telling you that because

the answers that you get are just

functions of what we call S um and so we

get a new function as a result of that

and so I know that you've seen this this

kind of action before where you have

zero and infinity up on an integral you

probably saw that in calculus but just

to refresh your memory uh that's called

an improper integral right so when you

have an improper

integral and you've studied this once

probably a long time ago but basically

when you have any kind of uh 0 to

Infinity e to Theus St F of T DT what

you really are doing is you do the

integration but what you do is you take

the limit so what I'm going to do is

I'll say let me write it in a different

color what you're really doing here is

you're taking the

limit and you'll see what I mean in a

minute as H goes to

Infinity of the integral from 0 to h eus

s t uh F of T

DT I'm just mostly explaining what

you're doing with the calculus when you

see an integral like this with one of

the limits of integration is an Infinity

then what you're really doing is you're

taking the integral and you pretend that

this top number is is just some variable

you calculate the integral and then you

take the limit as H goes to Infinity

that's how you would write it

mathematically but in practice what you

really do is you enter you evaluate the

interal you plug Infinity in one limit

and zero in the other limit and then you

know how to evaluate Infinities you know

if they're on the denominator if they're

on the numerator they might drive your

answer a different way uh and so that's

basically what you're doing but

mathematically rigorously what you're

doing is you're integrating the guy and

then you're taking the limit as this uh

limit goes to

Infinity all right so what I would like

to do at this

point is um calculate a llas transform

basically what you have is we have the

definition that was on the board a

minute ago that's um really the the

bread and butter of it if you understood

how to how to apply that to all

situations then you wouldn't need me you

would just do the Theo transform all the

time but the truth is that as you do

some of these things um there's a little

bit of Tricks along the way to help make

it comprehensible to you and I'm going

to show you those here and what you're

going to find pretty quickly in any book

that deals with the llas transform is

they're going to give you a table of

transforms in other words there are some

pretty basic functions that are pretty

easy to apply this definition to that

you get the answer and that answer is

very useful going forward um because

some functions pop up in nature all the

time you know exponentials pop up all

the time um for instance so we want to

learn how to take the LL transform of

things like exponentials and things like

cosiness and signs because those things

pop up all the time so what we're going

to do in this lesson in the next few

lessons is we're going to apply that

definition of the laas transform this

this full-blown definition of the LL

transform to a few core um functions and

then we're going to assemble our own

table ofl transforms which in many cases

you'll just find them listed in a book

but I'm going to derive how they get how

we get there so that you'll understand

how we're applying this once you have a

basic table of La transforms of common

functions then what you typically do is

you use that table to explore more

complicated functions so we're kind of

getting our footing we first learn what

the real definition of laass transform

is then we're going to apply it to some

simple functions we're going to assemble

our table of common functions that we

are useful to know the lla transform and

then we'll apply that to many problems

going forward so the most important uh

function that you could probably know

how to or or want to know how to take

the LL transform of would be the LL

transform of the function e to the

Lambda t e to the Lambda T now this is

just an exponential it's e to the T it's

a function of time right but there's a

constant in front we're calling it

Lambda in your book it might say e to

the a t or it might say e to the BT or I

might say e to the alpha t e to the beta

T it doesn't matter but there is some

number in front of t t it could be one

it could be two it's left open-ended

because a lot of times exponentials pop

up in solutions to differential

equations they pop up in lots of

physical systems right so it's an

exponential if it's e to the T the ual

value of Lambda just changes the slight

shape of what it looks like so we're

leaving that as a constant left open in

there but we want to find out in general

what would this LL transform be so the

way you do it is you just apply it

directly 0 to infinity and you apply

that LL transform which if you remember

was e to Theus s t f of T DT this was

the General equation uh for the LL

transform so then what we do is we say 0

to Infinity eus S T then we put our

function of time in here which is e to

the Lambda t d t and this is what we

want to integrate and we suspect and we

claim and I'm telling you that once you

do this properly all you get is a

function of s and we say that that LL

transform function of s is inexorably

tied to the function of time through

this thing called the transform and you

have to trust me on faith that once you

know how to do these transforms and once

you get proficient at them that they

help you solve real problems you're

going to have to take that part on faith

we're going to get to that part a little

bit later all right so what we're going

to do is begin to evaluate this interval

I could just give you the answer of

course I could do that but I want to

walk you through it so that you can

really feel like you understand what's

really going on so notice this is uh two

exponentials so we can simply combine

the exponents at the top that's

something we can do because these

exponentials are multiplied together

right

so we can say that we'll

have s- Lambda DT make sure you agree

that there's nothing different here if

you distribute the negative n you get

Negative s which is what we have here uh

we have a t here as well right neg s t

and then here this will be positive

Lambda T positive Lambda T so basically

uh we are adding these guys together in

the exponent we're basically adding

negative St plus Lambda T we're adding

that together but I pull the negative

sign out so but it still becomes St plus

Lambda T in the exponential uh there all

right so what you need to do is is

integrate this right integrate this this

is an exponential integrals of

exponentials are relatively simple the

only thing is um notice we're

integrating over time here is your time

variable Lambda is just a constant s is

going to end up being a variable that we

are going to have in our transform in

the end but since we're not integrating

over s we're integrating over T you

basically treat S as a constant and

that's really important for you to

understand you know anytime you take

derivatives or integrals you have to

look at what variable you're taking it

with respect to if I give you an

expression I say take derivative with

respect to X if you're talking about

derivatives then you pretend everything

else besides X is a constant in that uh

in that function that's how you take

partial derivatives right well in

integrals it's the same thing if we're

integrating over time then this is the

variable we care about every other

letter or symbol in there you pretend as

a constant so for the purposes of this

integral you pretend that s is a

constant uh and Lambda is also a

constant which means that everything in

front of the T is really just a constant

so it's like a giant number here so this

integral looks intimidating but really

since it's all just an constant it's not

that intimidating so many of you can

look at this and write the answer to the

integral down but a lot of times we have

to make a substitution to make it

absolutely clear and so what you would

do is you would say U is equal to minus

um s minus

Lambda right T so we want to make a

substitution because we want to make it

an easy integral to solve so when I do

this off to the side I'm doing a

substitution I say du

DT is equal to now notice if I'm taking

a derivative with respect to time this

whole thing is a constant so the

derivative is just going to be - s minus

Lambda because it's almost like this is

just the number three or the number five

or the number seven out in front of the

time the T just disappears for our first

derivative now since we're going to end

up substituting it back here we want us

we want to solve for DT and so what

we're going to get is -1 over s minus

Lambda duu all we've done here is move

the DT over here and then we've taken

all of this stuff and moved it over here

so we could solve for DT getting 1 / s-

Lambda

du all right so now that we have that we

want to take this and stick it back in

and substitute into our integral so

we'll change colors again and for now

we're going to leave the limits of

integration 0 to Infinity we're not

going to change them right now because

we're going to you know we'll see how to

handle that in a minute it becomes e to

the power of U because this is exactly

coming from that and DT just becomes

what we have found here so it's going to

be NE 1/ s- Lambda du U right so that's

sub substitution just goes in now

remember this is an integral over du now

right so everything in here is again

just a constant it's just a constant so

what we're going to have here we pull

the whole thing out- 1/ s minus Lambda

comes out we integrate from 0 to

Infinity e to the U du now we did this

whole thing so that we could get an

integral into a form that we know how to

solve very

easily right and

so uh what we're going to do then is say

we have -1/ s- Lambda the integral of e

to the U Still Remains e to the U and

again I'm leaving my limits of

integration 0 to Infinity for now

because what I'm going to do before I

apply the limits of integration is I'm

going to substitute back in for U so

what I'm going to have is -1/ s minus

Lambda and then U is going to come right

back substituted in as it was before E

minus s minus Lambda

T now goes from zero to Infinity so you

see I I realize um that as you have the

limits of integration here this really

If This Were a true statement you would

have to transform the limits of

integration into limits of U and this

would have to be limits of U but I'm not

going to substitute the limits in until

after I've plugged in for U so I end up

having a function of time here's a

limits in terms of T and so then I can

substitute everything in as exactly as I

want to and then what I'm going to get

is -1/ s- Lambda right and then we have

to apply this in so what this basically

becomes if you think about it this is s

minus Lambda time T and I'm putting

Infinity in for T so really it becomes e

to the neg negative Infinity it doesn't

matter what s and t are if I put

Infinity here it's going to be negative

infinity and then you do a subtraction

because you're evaluating the limits e

to the 0 because 0 goes in there now

this becomes very tractable because we

know that e to the infinity it's like

taking that

limit what you're going to have e to the

negative Infinity is like 1 over e to

the infinity so what you're going to

have this guy is going to be zero and

this guy anything to the power of zero

is just a one so you have a negative one

here and this whole thing evaluates to

negative 1 which makes a positive one

when you multiply this whole thing here

so you get 1 / s - Lambda 1 / s-

Lambda all right so what I'm going to do

then is show you that through this whole

situation all we did was we found the

laage transform of this guy and said

that it was equal to that so let me go

to the next board and summarize that

because it's a very important result so

what we found that is that the LL

transform of e to the lamb * T which is

just e uh to the power of some number

time T is just equal to 1 s minus Lambda

so for instance if it was e to 2T then

it would be 1 / s - 2 if it were e to

the 7t it would be 1 / s - 7 if it were

e to the 4T it would be 1/ S - 4 so you

see what we get as a result is just a

function of s Lambda is just a constant

it's going to be locked down by whatever

we start with right so the answer that

we get F of S capital F of s remember we

said let me go back we said that when

you do this LL transform you should get

just a function of s and that's why we

call it capital F of s and that's

exactly what we got we got a single

function pretty simple looking function

just a function of s this guy is just a

constant uh there now since it's in the

denominator and we don't want any zeros

to be in our denominator because then

you get undefined you know Port parts of

the transform or Infinity parts then

what you can also say is that this is

valid for S greater than Lambda that

just ensures that the denominator is not

going to be zero it ensures the

denominator is going to be positive this

is something that mathematically you

write down just to lock it down you

don't have any zeros in the in the

bottom but realistically you don't

really use this fact all that terribly

much when you're solving essential basic

problems with a ll transform now this is

an important result and it's an

important enough that I'm going to

circle it for you right and circle the

whole thing in fact and it's also

important enough that we want to draw

your attention to something else let me

change to Blue here in the special

case let's say Lambda is equal to zero

so let's

say

that Lambda is just equal to zero

then you would have e to the 0 * T right

which would be e to the 0 that would be

my function and of uh time if Lambda

were 0 be e 0 t and e to the 0 is just

one because anything to the power of

zero is just one so because of this we

can draw kind of another conclusion

that's already here but we can kind of

write it ourself we can say that the LL

transform of the number one

right would be 1 / s - 0 right because

the way you would come up with that is

you would

say uh which would be 1 / s right the

way you would come up with that is you

would say all right I can take the

special case when Lambda is zero in that

case this exponential just becomes a one

so it's the same as taking the lla

transform of one and then we' be putting

Lambda equals 0 in here getting an

answer of 1 / s so the laage transform

form of the number one is just 1/ s and

that's important enough while I will

also Circle it and of course it's for S

greater than zero because you don't want

any denominator uh driving the whole

thing to Infinity so this is the first

important conclusion of what we have

have done here I'll go quickly through a

a

um a brief history of what we've done we

basically said there's this thing called

A Plus transform it's an integral notice

it's not a double integral or a triple

integral or spherical coordinates or

anything crazy like you get into

calculus 3 territory it's really a

calculus one maybe a calculus 2 type of

thing but the implications of how you

use it is really why it's interesting

and we're going to get to that later but

basically you just stick a function of

time in here you evaluate the integral

evaluate the limits of integration and

if you do it correctly you should get

just a a function of s and we label it

capital F of s because capital F and

little f are related to one another by

this thing we call a transform we say

that cap that little F yields capital F

of s and later on we'll find out that

you can go in the reverse Direction and

start with a ll transform function of s

and get the corresponding function of

time so these things are kind of linked

by an invisible chain and that's the

chain which being is is the transform

that's why we can take a problem which

has which has functions of time

transform it into the LL domain which is

function of s solve it usually

algebraically so you don't you don't

have to deal with differential equations

you deal with algebra and then you get a

function of s and then we're going to

learn how to transform that back to the

time domain and you get your answer in

terms of time and if you do it right it

should be simpler we also say that the

llas transform F of s is squiggly L uh

operating on the function of time and we

just talked about improper integral

saying that it's basically like taking

the limit but really what you're doing

is plugging in the limits of integration

uh there and then we do real problem so

this is how a lot of these problems are

going to go you take your function of

time you put it in and you simplify and

then you realize the critical step here

is that since you're integrating over T

everything here is a constant right so

I've done the details here we did the

substitution of U putting it all in

there and pulling this junk out because

it's a constant giving us an integral

that everyone watching this should know

how to solve right so we do that we plug

the limits of integration in the limits

of ation greatly simplify what's

happening because this becomes a zero

this becomes a one uh because it's

negative Infinity remember e to the

negative Infinity is like 1 over e to

the positive Infinity which means 1 over

infinity giving you zero so all this

stuff drops away giving you negative 1

giving you positive 1 / s minus Lambda

so the conclusion block here is the

appli transform of e to the Lambda T is

just a function of s this Lambda is lock

down with whatever your specific

function is all right and then as a

special case of that we say hey what

happens if Lambda is zero then what you

should get if Lambda is zero is 1 / s

right and if Lambda is zero then it's

the whole thing just goes to one so what

we're saying is the L transform of the

number one is 1 / s if you remember back

to calculus one however many years ago

that was for you first thing we talked

about besides limits but for as far as

derivatives the first thing we talked

about was how to take simple derivatives

how to take the derivative of a constant

remember you had to learn that at one

point in the past how to take the

derivative of x how to take the

derivative of x s then you learn how to

take derivatives of pols then you learn

how to do it when there's a giant

fraction then you learn about signs and

cosines you build those skills learning

how to take those derivatives we're

doing the same thing with lass

transforms we're taking very simple LL

transforms first and we're recording

these answers um which are useful then

as we go on we're going to take more and

more complicated LL transforms to the

point where you can actually uh get

pretty proficient then I'm going to show

you how to use it to solve real problem

so right now you look at this and you're

like what's it for you know well I can't

get into that until you know a little

bit so just trust me follow me on to the

next lesson we'll build our skills

deriving these essential transforms uh

and then we'll we'll get a a repertoire

going so that you have some skills that

we can apply to real problems and

whenever you see how how um much simpler

it makes solving certain kinds of

differential equations you'll understand

that it's worth its weight in gold just

for that application but also since

differential equations are used in all

branches of science and engineering um

they really lend the llas transform to

lots and lots of different uh uh uh

situations in real Math Science and

Engineering so follow me on to the next

lesson in mastering the llas transform

here in the llas transform

tutor