Hello everyone and welcome to our first class on Laplace Transforms.
So first of all, a quick look at what we're going to cover today.
We'll start with a brief introduction on Laplace Transforms, then motivation as to why we are
studying Laplace Transforms and follow that up with some definitions that we'll need throughout
the classes on Laplace. Then we'll move on and look at calculating Laplace Transforms from
the definition. Then we look at calculating Laplace Transforms from the tables and that's
the method we'll use predominantly in this course. We'll do very little as regards calculating Laplace
Transforms from the definition. We'll use the tables for nearly all our work. And after we've
looked at how to calculate Laplace Transforms from basic functions, we'll move on and look at
the first shifting theorem, which is a method for calculating Transforms of special types of
functions. Okay, so Laplace Transforms, well they date back to the time of the French mathematician
Pierre Simon Laplace, who's around from the mid 18th to the early 19th century. He also worked in the
areas of engineering, physics and astronomy. And Laplace Transforms is what's known as an
integral transform and is useful in solving physical problems found in engineering, maths and physics.
The integral transform or integral transforms in general enable us to convert what are complicated
problems into ones that are easier to solve. And other examples of integral transforms that are
widely used in, for example, signal processing and the Fourier transform, Hilbert transform and
the wavelet transform. However, we don't look at any of these in this module. We're only going to focus
on Laplace Transforms. So as you can see, this slide is titled Laplace Transforms and ODEs. And
ultimately that is what we are going to be using Laplace Transforms for. That's for solving ordinary
differential equations. And as we know, ODEs can be used to describe the dynamical behaviour
of systems, mechanical, electrical systems and so on. And traditional calculus methods, as we've seen
previously, are used to solve ODEs for a range of different inputs. We saw when we're looking at
second order differential equations, the inputs were the functions on the right hand side of the
equations. We saw situations where we had constant terms, linear terms, quadratics,
exponentials, sines and cosines. And we used the method of undetermined coefficients to solve these
equations to obtain the output that we would get from the differential equation. However,
when signals become more complicated, the calculus approach can become very difficult indeed. And
the methods that we've seen up to now really don't work very well. In fact, sometimes they can't be
used to solve certain types of differential equations. So what we're going to do is we're going to use
Laplace transforms and that will enable us to convert these time domain calculus problems,
such as differential equations, to problems that only need algebra to solve them. And that's what
the Laplace transform does for us. So let's just summarise this in this diagram, this diagram here.
So we're over here in the time domain, we've got a calculus problem, which could be a differential
equation. We take the Laplace transform of the time domain problem of the differential equation
and we convert in so do it by taking Laplace transforms of the ODE, we convert it from a
calculus problem to an algebra problem. We then solve the algebra problem, which tends to be quite
straightforward. And we have a transform solution in the Laplace domain, but we need to get back to
the time domain because our original problem was a time domain problem and we do that by taking
inverse Laplace transforms. And this is the place or the step that people find the most difficult.
And it's not because of this new mass involved. It tends to be because of the different
types of algebra that we need to know. There's nothing new in this, but it can get quite tricky
at times. So that's really schematically what's going on. Calculus problem in the time domain
transformed to an algebra problem in the Laplace domain. We solve the algebra problem,
get our transform solution, and then we invert it by taking inverse Laplace transforms to obtain
solution. As I said earlier, we would ideally like to be able to go straight from the calculus
problem to the solution, but it can be either very, very difficult to do or even impossible
using the methods that we know for solving time domain problems, differential equations.
Okay, so let's have a definition of the Laplace transform. Well, so we're only going to consider
positive time t greater than zero. So an independent variable from now on will be t time.
So there are Laplace transforms of a function f of t. It's defined by the improper integral.
Now an improper integral, one situation when an integral is improper is when you get infinity
plus infinity or minus infinity or even both as the limits. And you can see that this integral,
we have positive infinity as one of the limits. So how do we get the Laplace transform from the
definition of a function f of t? Well, we take our function f of t, we multiply it by e to the minus
s t. Now in here, s is actually a complex variable, but we're not really going to do anything with
that. We just bear in mind that it is a complex variable, but we're able just to
work this as we would do most other integrals. So we multiply our function f of t by the
exponential e to the minus s t, and then we integrate this product from zero to infinity.
And that will take the notation here, the L means Laplace and curly brackets are function f of t.
This is how we calculate it, and the output will be a function of s. And it's standard practice to
denote that with a capital F. If f of t, little lowercase f of t is a function, the output of Laplace
transform will be uppercase f of s. So we feed in a function of time of t time, the time domain,
and out of that comes a function of s in the Laplace domain or the s domain, as it's also called.
So that's what we're saying down here. So the function of t, it's quite often some form of
time signal, and we talk about moving from the time domain or t domain to the Laplace
domain, that's the s domain, when we're performing a Laplace transformation.
So let's have a look at an example of using that definition to calculate the Laplace transform
of a function f of t. And I'm going to take a very simple function, f of t is equal to one,
just the constant function that's equal to one everywhere from t greater than zero onwards.
So that's what it looks like. The red line is just the function f of t is equal to one.
So we want to calculate the Laplace transform of this using our definition. So the definition
is given here in the green box, the function f of t is equal to one. So in here, I've just
replaced a one. So the notation here is that I want the Laplace transform of the function one. So
it's L curly brackets for the one inside. And that's equal to the function f of t is equal to one
times e to the minus st integrated between zero and infinity. Now to perform this integration,
there's no such number as infinity. And it's really not correct to substitute that in as one
of the limits directly. What we do is we introduce a w variable u and we let u tend to infinity.
That's how we handle these improper integrals. So the function is just one. So that just means
we've just got to integrate this exponential e to the minus st with respect to t. And we know that
when we do that, when we integrate an exponential like over here in the red type, what happens is
it'll just be one over whatever's in front of the t. So one over k e to the minus kt. And in
our case, what's in front of the t is the minus s. So we're going to have minus one over s e to the
minus st. I've taken the minus one over s outside because it doesn't depend on t. So we can move
it outside of this bracket just to make things a little easier for her. I'm going to take the limit
as u tends to infinity between t equals zero and t equals u. Low limit is zero. As you can see,
the upper limit is u because we've introduced this w variable. So we just substitute in these
values. u goes in at the upper limit and zero goes in at the lower limit. So we get this expression
here. Now e to the zero is one. So that's where that comes from. e to the minus s u. So what's
happening here as u goes off to infinity? Well, I can write e to the minus s u as one over e to the
s u. And I'm going to let u tend to infinity. I'm going to let u become massive, become huge.
And as u becomes bigger and bigger and bigger, this quantity here will get closer and closer
and closer to zero because you're dividing one by something that's getting increasingly larger. So
this term as u goes to infinity, this term here, which is equivalent to this term, will go to
zero. And all that's left is minus one over s times minus one, which is one over s. So the Laplace
transform of our constant function f of t is equal to one is one over s. So that's our time
domain function f of t equals one gets fed into the Laplace integral. And the output is a function
of s, big f of s is equal to one over s. Now that was relatively straightforward,
but it doesn't take very much of big change in the function that we're calculating in a Laplace
transform of for things to get quite tricky. So just imagine, you can think of one as t to the zero,
just ended the power zero, this one. What happens if we rather than having t to zero, we've got t
to the power one, we've got t there. So we want the Laplace transform of t, but we would need to
use integration by parts. Things start to get a bit messy. If we want the Laplace of t squared,
we would need integration by parts twice. So calculating from the definition can lead to
an awful lot of work. And thankfully, all that has been done for us. And we've got a set of a table
of Laplace transforms for a range of functions. It just depends on which course you're doing,
which book you're looking at, how many functions are given. We've got maybe about two or three
pages of functions that we supply the Laplace transforms for. But as I said, some books will
give you more, some books will give you less. So we start off with one which is t to zero. We've
just seen that the Laplace of that is one over s. So I should say that in this table here, I number
all of these down the left hand side in the first column. In the next column, I've got my function
at time. That's the function that I want to take the Laplace transform of. And in the third column,
I've got the actual Laplace transform, big F of s, which is just the Laplace transform of ft. So
we've already seen that one over s is the Laplace transform of one. For the Laplace transform of t,
it's one over s squared. For t squared, it's two over s cubed. And then we could carry on,
we could have t cubed, t to four, but you've got to stop somewhere. And what they do is they just
give a general expression for t to the power n. The Laplace of that is n factorial over s to the
n plus one. And I will see an example of that shortly. Then we've got some exponentials that
we've got the Laplace transform of. And we'll be seeing them a lot because exponentials, as you know,
feature a lot in the solutions of differential equations. So these ones will be quite common.
We've got the trig functions then, sine and cosine and variations of that. And the table goes on,
we've got the hyperbolic functions here, the hyperbolic sine, the hyperbolic cosine. We don't
use them so much in our course. And then there's a range of other functions here. Interestingly,
here, 2425 are the Laplace transforms of derivatives. And that's what's going to help us when we're
going to be solving differential equations using Laplace transforms. And there's some other
functions here, which will come across in due course, such as the unit step function
and the delta impulse function. Okay, so before we move on to look at some examples,
we should just get familiar with one of the properties of the Laplace transform.
And that's the linearity property. Now, this is something you've actually seen before in both
integration and differentiation. So what's it saying? It says that, for example, the Laplace
transform, C1 and C2 here are constants. The Laplace transform of a sum of functions is equal
to the sum of the individual Laplace transforms. So that's what I'm saying here. The Laplace transform
of the sum or difference or two or more time domain functions is equal to the sum or difference of
the Laplace transform of the individual functions. So for example, if you remember when you were
doing differentiation, if we had a function t squared plus sine t, well, if you wanted to
differentiate that d by dt of that, you would say that that's just d by dt of t squared plus d by dt
of sine t. So this is all I'm saying here with Laplace. It's exactly the same as this. So the
derivative of a sum is the sum of the individual derivatives. So you'd get 2t plus cos t. So in
other words, you just go along, do differentiating term by term. We did the same thing with integration
and we'll do the same thing with Laplace. When we've got terms added or subtracted, we just go
and do them one at a time and add or subtract the result. So we'll see some examples shortly
of that. So here's a range of functions. We've got eight different examples here to highlight
different properties. So let's just get started. So our first function that we want to take the Laplace
transform is 7t to the power 6. So this is the general notation. Quite often they write the L
as a curly L like that. And then curly bracket is 7t to the 6. Now we know from the linearity
property we just saw that the constant can be taken outside. You don't have to, but it just
makes things a little easier. So out comes the constant and then we've got to take the Laplace
transform of 6. Now if we look at our tables, look at this column in our tables, here's our function of
time f of t is t to the 6. Well we run out after t squared, but we've got a general expression
for t to any power. t to the power n, the Laplace transform of that is n factorial over s to the
n plus 1. So all I have to do here is identify the value of n, which is pretty obvious. N is equal
to 6. So I'll use number 4 with n equal to 6 to give me 6 factorial over s to the 6 plus 1 or 6
factorial over s to the power 7. So that's what I want. And don't forget of course we took the 7
outside so that must feature there. So 7 times 6 factorial over s to the 6 plus 1, s to the 7 in
other words. Now 6 factorial, the factorial, what does function, what does that do? Well basically
all that saying is whatever number you take the factorial of you just multiply it by every number
smaller than it down to 1. So for example 6 factorial, so just get the pen, 6 factorial will be 6 times
5 times 4 times 3 times 2 times 1 and if you work that out you find that it's 720. And then of course
we've got a 7 on the outside so we must multiply that and 5 0 4 0 over s to the power 7 is the Laplace
transform of 7t to the power 6. Okay move on now. 6 e to the minus 8t. Well we could go, we would
go through our tables and we saw that all the way 1 2 and 3, 1 2 3 and 4, none of these involve the
exponential function. None of these involve the exponential function but then we hit number 5.
Number 5 is e to the minus alpha t. Well we've got e to the minus 8t so let's just start. So I want to
take the Laplace transform of this function. I can take the constant outside, the 6 can get taken
outside. Again you don't have to but it's a little easier to work with, a little neater. Out comes the 6
and I want the Laplace transform of e to the minus 8t. So then I just need to identify which one we've
already said. It's going to be number 5 in the tables and I've got to compare what I've got. I've
got e to the minus 8t and in the table, in the table I've got e to the minus alpha t. Okay so that
would mean that minus 8t must correspond to minus alpha t. The t's can cancel so minus 8 must correspond
to minus alpha so in other words alpha must equal 8. Okay so that's my value of alpha identified.
So what do I want to do now? Okay so I've identified that alpha is equal to 8 so the Laplace transform
of e to the minus 8t alpha is equal to 8 is 1 over s plus 8 and that's what I've got here and don't
forget the 6 on the outside. We must multiply by that to give me 6 over s plus 8 as the Laplace
transform of f of t equal to 6e to the minus 8t. So there's our time domain function. We applied the
Laplace transform I want to do s domain or Laplace domain function here. Okay moving on let's have
another well here's another exponential and again we just compare it with number 5. I can take
4 outside of course. 4 can get taken outside and I've got to calculate the Laplace transform
e to the 9t. Just got to be a little bit careful here because I've got e to the 9t but in the table
number 5 is e to the minus alpha t. So in other words 9 must correspond to minus alpha
or in other words alpha is equal to minus 9. Okay so alpha is equal to minus 9 so I can use
number 5 and that would give me
that the Laplace transform of e to the minus e to the 9t is 1 over s minus 9.
Don't forget I've got that 4 on the outside that I can multiply by so I get 4 over s minus 9.
And that's the Laplace transform of 4 e to the 9t. Okay so that's fine we can move on to
look at another example though. So this again is a different type of function f of t equals 9 cos 4 t
minus 4 sin 9t. So this involves trig functions cos and sin. Again now we could go back
numbers 1 2 3 and 4 are only involved powers of t so they won't work. Number 5 6 7 well they
involve the exponential functions they're not going to work either but when I reach number 8 and 9
they involve sin and cos so that would tell me that these are the ones to use. So here I go I'm
setting up my problem here I want to take the Laplace transform of this the difference of these
two functions and if you remember the linearity property said that I can go in and just take the
Laplace transform of the individual functions and then do the subtraction at the end. So I'm going to
do that I'm going to use the linearity property so I will take the Laplace transform of this first
term minus the Laplace transform of the second term and in both cases I've got a constant which I
can take outside. Again you don't have to if you don't want to but I find that it just it's a little
bit a little less messy to work with than you take the constant outside. So that just leads to define
the Laplace of cos 4t okay so cos 4t now it's this one with omega equal to 4 so the Laplace
transform of cos 4t that's this one here will be s over s squared plus 4 squared there we are
down here and of course don't forget you got a 9 multiplying that term. Then we've got minus
4 times the Laplace transform of sin 90 so omega here is equal to 9 so I'm going to have the Laplace
transform of sin 90 would be 9 over s squared plus 9 squared so there you are it's that term there
and if I simplify this I get the solution at the bottom 9s over s squared plus 16 minus 36
over s squared plus 81 so that is the Laplace transform of this function f of t here.
One thing before we move on I'll say here these two have exactly the same denominator it's s squared
plus omega squared whatever your omega value was and this one it's 4 and this one it's 9. The way
the Laplace of the sin and cosine differ is that for the sin in the numerator we've just got the
constant value whatever omega is that's what goes in the numerator for the cosine however the numerator
includes the Laplace variable s that's the difference between the Laplace of sin and cos
that will be useful when we come to calculate inverse Laplace transforms later on.
Okay moving on to another problem we've now got one of the hyperbolic functions
and we don't treat this any differently we just basically look in our table in this column of
our table to see if we can find a cos of 6t or cos of some number times t we won't find this
exactly with a 6 but we'll find some general expression in our table so if we do that we'll
reach down number 18 which is the hyperbolic sign so that's close but not not the one we want this
one isn't the one we want either but then when we reach number 20 we find that we've got the
hyperbolic cosine or cos of beta t so beta can match up with our 6 so we've just got this one here
with beta equal to 6 and that'll be fairly easy because we just replace beta with 6 so the Laplace
transform of cos 6t will just be s over s squared minus 6 squared which gives me in the blue box
here s over s squared minus 36 so that's all that's to it really now another point to note here is if
we look at our regular trig functions sin and cos as we saw on the previous slide both functions
have a Laplace transform with s squared plus omega squared in the denominator that is the Laplace
transform of both sin and cos the denominators are the same with s squared plus omega squared the
tops differ with sin having just the omega the constant value but cos having the Laplace variable
s in the numerator how do these compare with the hyperbolic functions well if you look at number 18
and number 20 you can see that something very similar happens here the two of these 18 and 20
have the same denominator and once again the sine version or the hyperbolic sine has a constant
just as the regular sine did while the hyperbolic cosine like the regular cos had the variable
the Laplace variable s in the numerator now where the regular trig functions sine and cos
differ from the hyperbolic since if you look at the denominator the regular trig functions
have s squared plus omega squared s squared plus some number whereas the hyperbolic functions have
s squared minus some number so the only real difference between them is the hyperbolic have
a minus in the denominator whereas the regular trig functions have a plus and again that will come
in useful when we come to calculating inversions later on okay so moving on we've got a different
thing here now we've got a product of two functions we've got an exponential multiplying a trig
f of t is equal to e to the 8t sine 2t now we'll see later on in this class that when we've got this
type of product involving an exponential function and some other function that we use the first
shift in theorem but this particular case is actually given in our tables this is actually
given in our tables so we don't need to use the first shift in theorem for this one but
depending on what tables you're issued with you may very well have to your table tables from
l squared might not involve might not include this type of function in the tables so let's
have a quick look at how we do this so can we find n that looks like it we would look down
this column in our table and eventually we would hit number 14 e to the minus alpha t
sin omega t number 15 is the same or similar except and it's got the cosine instead of the
sign so we've decided it's number 14 it doesn't matter that that's a positive and that's a negative
we can handle that we saw that already when we're using number five in probably questions two and
part b and c i think possibly we can handle that that's a positive eight and the table has a minus
alpha there that's that's not a problem for us and we know from comparison that omega will be equal
to two so i'm going to use number 14 with we've got to figure out what alpha is and we know that
omega is equal to two so let's just compare this we know from our tables we've got e to the minus
alpha t and here we've got e to the eight so these two things must be equivalent e to the minus alpha
t must equal e to the eight t right so in other words minus alpha t is equal to eight t and then
if we go up to the the green type here that must mean that minus alpha is equal to eight
or alpha is equal to minus eight you might be able to spot that whether you're having to work all
this out but there's no harm in doing this so alpha is equal to minus eight so i'm going to use
number 14 with alpha equal to minus eight and omega omega is equal to two because it's e to the minus
alpha t sine omega t i've got e to the at alpha minus eight so that works it's going to be e to the
minus minus eight or e to the at just as i've got omega equal to two will give me sine two t and that's
exactly what my question has so if that if we're going to use this one then we have omega so i'm
going to have two over s minus eight all squared because alpha is equal to eight minus eight plus
two squared because omega is equal to two multiple or yeah that just two squared is four
and then i can take another step and open out these brackets to obtain my solution at the foot of
the slide here so this is actually given in the tables but we're shortly going to see how we handle
this sort of product of two functions where one is an exponential we use the first shift theorem
we'll see that shortly okay let's have a look at another example of calculating the past transforms
from the tables now this one f of t is equals cos squared t minus sine squared t well we could look
in our tables and we wouldn't find cos squared or sine squared so we've got the problem of how are
we going to manipulate this into a form that's given in the tables now you may know this relationship
this is a trig identity cos squared t minus sine squared t you could actually write this if i just
get my tablet if i write this out in full you know this is equal to cos t cos t minus sine t
okay and that is actually equal to using the trig identity it's cos of t plus t because remember
when you've got cos of a plus b or t plus t it's cos cos minus sine sine the sine changes when we're
using this identity and that's just cos of cos of 2t you know that's a fairly well known trig
identity when we've got cos squared t minus sine squared t it's equivalent to cos of 2t so whereas
we were not able to use our tables directly for this function here we were able to apply
a well known trig identity to rewrite cos squared t minus sine squared t as cos 2t and then if i look
at my table when i reach number nine i've got i've got cos of omega t and this is just cos of omega t
with omega equal to two so it's a straightforward matter of using number nine with omega equal to two
so this is what i'm saying here that the Laplace of the function as it's given cos
squared t minus sine squared t is the same as the Laplace of cos 2t i use number nine with omega equal
to two i get s or s squared plus two squared or s over s squared plus four so that's an example of
using a trig identity to manipulate the expression we were given and that doesn't appear in our tables
to a form that actually does appear in our tables and we are able then quite easily to obtain
its Laplace transform okay so here's another one again you know we could look in our tables we need
something that's got the product sine t cos t but we're not going to find this this doesn't appear
in our tables we may then remember the trig identity that says two sine t cos t is equal to
sin 2t and that's if i can just write here that's because if i write sine 2t
if i write sine 2t i'll just get the pen going sine 2t is equal to sine t plus t
and if you remember this trig identity that's sine t cos t plus cos t sine t and all that is
it's the same thing sine t cos t plus cos t sine t is just two sine t cos t and our trig identity
as we know it says that that's equal to sine 2t so i can replace the sin t cos t by sine 2t but
remember i need a two here so my sine t cos t would just be a half sine t you know there's my
trig identity so i've got to divide both sides by two to get an expression for sine t cos t because
that's what we're given in the question okay so i've been able to identify a trig identity that
i can use here so let's have a look at it so i have the Laplace transform of sine t cos t
using my trig identity i know that that's the same as the Laplace of a half sine 2t
i can take the half outside if i want to i don't have to but again it's just things are a little
less cluttered so it's just a direct Laplace transform of sine 2t it's number eight with omega
equal to two so that becomes a half don't forget the half on the outside times omega over s squared
plus omega squared it's uh a half times two over s squared plus two squared that two cancels with
that one and the answer would be one over s squared plus four so that's the Laplace transform of our
function here so one important point that i that i should make here is that in this example
our function was a product of two time domain functions of sine t and cos t
this did not appear in our tables so we had to find a method of doing some sort of manipulation
of this term into a form that's given in the tables and we were able to do that using the
trig identity that sine 2t is equal to two sine t cos t and a little bit of rearrangement gave us
this expression so we could directly calculate the Laplace of this term now one thing here that you
do not do this is a product and just the same way as when we did differentiation when we're
differentiating a product you do not do the product of the first thing the derivative of the first
thing times the derivative of the second so you know if i just go to the next slide that's what i'm
saying here an important point to notice while the function f of t is the product of two functions
sin t and cos t it's not true that the Laplace transform of our function f of t
is the product of sine it's the product of the Laplace of sine t and the Laplace of cos t so
in other words this here does not hold that Laplace it's not the Laplace of sine t times the Laplace
of cos t so in our case we used the trig identity we'll go to shortly see some functions that involve
products we've already seen one of them i was seeing already seen an example but we're able to use
the tables on it because it was given in the tables that was e to the at i can't remember what it was
was cos t or something like that but we'll soon see the first shifting theorem now i mentioned
uh about differentiation if you remember when we had a function f of t was equal to t squared
sin t when we went to calculate the derivative of that f dashed of t we did not just differentiate
the first term and multiply it by the second term that's completely wrong that's completely wrong
the derivative of a product is not the product of the individual derivatives if you remember
we had to use the product rule we had to use the product rule where you know depending you could
use maybe u and v for example that f dashed was equal to u dashed v plus u v dashed that was the
product rule and similar things happen with integration as well although it's a bit more
complicated there's no one rule that works there sometimes integration by part sometimes
integration by substitution and there are other other methods which you know we wouldn't have seen but
you know the whole purpose of me saying this is that really in general the Laplace transform of a
product is not equal to the product of the individual Laplace transforms and that's what this
is saying here if you take Laplace of f of t times g of t it's not the Laplace of f of t times the
Laplace of g of t so let's have a look at one special case and what we do when we come across it
so suppose we've got some function f of t and we're multiplying it by an exponential function
then as I've said earlier we use the first shift in theorem so here's an example suppose or here's
the definition sorry of the first shift in theorem which is also called the frequency shift theorem
fst for sure suppose the function f of t has Laplace transform f of s now we know all this you
know we've seen this f of t was the second column you know in our tables we've numbered the Laplace
transforms in the first column the second column gives the time domain functions we've been
using that up to now and in our third column we had f of s which was just the Laplace transform
of f of t so if a function f of t is Laplace transform f of s then if we multiply this function f
of t by an exponential then what this corresponds to is it corresponds to what's called a shift in
the s domain will either be adding alpha or subtracting alpha from the s in here and the alpha
is this value in the original function we'll crack in their Laplace transform of so basically what we
do is we we we ignore the exponential to start with we take f of t and we find its Laplace transform
big f of s and then wherever we see an s in the Laplace in big f of s we replace it by s plus or
minus alpha so that's what the first shift in theorem does and I've summarized that here in three
steps so that's what's involved in it but easier to remember I've got this diagram here that will
hopefully help us so we've been given a function the bottom corner here f of t multiplied by some
exponential so the first thing we do is we ignore the exponential to leave us f of t
we can calculate the Laplace transform of f of t using the tables might need to do a little
manipulation beforehand but we can certainly calculate the Laplace of f of t from the tables
and then as I said earlier we take a big f of s that we've got from the tables and wherever we see an
s we replace it by s plus alpha where alpha is the value that appears here in our original function
so that's what we're going to have a look at now okay so and again it's summarized here I think I'll
say one one more time and this diagram here gives you something hopefully to remember what this
appear is saying so this says that if the Laplace transform of a function f of t is f of s then if
you multiply that function f of t by an exponential then the Laplace transform of this new function
will be the Laplace transform of the raw function f of t that's the Laplace of that is big f of s
where s is replaced by s plus alpha okay so let's have a look at that in operation then
so here's the question I've been given determine the Laplace transform of e to the minus 2 t
times t cubed okay so um the first thing we have to do is we have to decide how
we're going to go about doing this problem um suppose we didn't know that we had to use the
first shifting theorem um well if I look at this function I can see there's two separate functions
once an exponential and the other's a power of t so I've got t cubed multiplied by an exponential
function now that tells me that I need the first shifting theorem I've got some function in this
case t cubed and we've got it's been multiplied by an exponential function and that's the key for
identifying when to use the first shifting theorem if you've got some time domain function like a power
of t here it could be sin t it could be cos 4t something like that but if it's multiplied by an
exponential we use the first shifting theorem so um to to use the first shifting theorem we need to
check of course that it's in the form e to the minus alpha t f of t well yes it is there's the
exponential and f of t is equal to t cubed so I've been able to identify that f of t is equal to t
cubed now what about alpha what's the value of alpha well in this box here if I look here's the
general expression from the definition of the first shifting theorem e to the minus alpha t f of t
and here's what I've got e to the minus 2t t cubed well as we've said we can easily identify that f
of t is t cubed alpha I think it's pretty obvious must be equal to 2 it's not minus 2 if alpha was
minus 2 you would have e to the minus minus 2 and you would have e to the 2t if you just compare
this term directly with this term you can see that alpha must equal 2 and that's what I say here at
step one so in my diagram what I've done is um I've identified I've ignored the exponential first of
all and I identified what f of t is it's t cubed I then went in and I identified what alpha was alpha
is equal to 2 the next step going around this diagram in a clockwise direction the next step is to find
the Laplace transform of f of t in other words find the Laplace transform of t cubed so is that
given my tables well no it isn't I've run out after t squared but I've got a general expression t to
the power n whether Laplace is n factorial over s to the n plus one so that'll be with n equal to 3
because t cubed it'll be 3 factorial over s to the 3 plus one that's s to the fourth so it'll become
6 over s to the power 4 so that's big f of s there we are we're up in the top right hand corner
one more step to do and that says that I must replace every occurrence of s in f of s that are
found in step two by s plus alpha where alpha's the value I found at step one in this case alpha is
equal to 2 so what this is telling me to do is replace every occurrence of s whether it's only
appears once by s plus 2 because alpha was equal to 2 so s plus alpha becomes s plus 2
so if I do that I obtain 6 over s plus 2 all to the power 4 that's my f of s plus alpha so that
there is my solution now one thing to be very careful not to do what people sometimes do here
is they get to they get step two correctly and then they say well I've got to replace
it's s plus alpha so they've got to have a plus 2 in it so what they write is 6 over s to the 4
plus 2 that's wrong what's what's been done here is that s is getting replaced by s plus 2 so if s
was raised to the power 4 then what you're replacing it by must get raised to the power 4 and that's why
it's s plus 2 all to the power 4 so that one is correct this one is definitely wrong so bear that
in mind when when we're replacing s by s plus alpha that if it's raised to a power then the s
plus alpha in this case s plus 2 must also be raised to the power and that was power 4 in this case
okay so let's just do another example now so what do we got here well here's my little diagram to
remind me of what the steps that are involved the three steps we go around the diagram in a clockwise
direction here's the expression from the first shifted theorem that I need to compare the
expression I've been given so my first step here says that I've got to identify f of t well I can do
that just by ignoring the exponential so f of t is t to the power 6 or I can just match it up like
this I can see f of t is t to the 6 I then must identify the value of alpha and I've got e to the
minus alpha t must be equivalent to e to the minus 4t so clearly alpha must equal 4 okay so it's not
minus 4 just as in the last example because it would be e to the minus minus 4t which is e to the
4t and that's not what I've got I've got e to the minus 4t so alpha in this case just by simple
comparison alpha is equal to 4 okay so I'm ignoring the exponential to give me f of t I know what f of
t is it's t to the power 6 so I'm up here I now want to calculate the Laplace transform of f of t
t to some power well we know it certainly would be one of the early ones in the table we don't have
t to the power 6 in the table but we've got the general expression t to the end and we can take
n is equal to 6 and the Laplace transform will be 6 factorial over s to the 6 plus 1 or 6 factorial
over s to the power 7 so that's what I've got 6 factorial 720 because it's 6 times 5 times 4 times
3 times 2 times 1 that's 720 so that's big f of s so there we are we've got we're up in this corner
and I just these one last thing to do is replace every occurrence of s in f of s by s plus alpha
alpha is equal to 4 so I must replace s by s plus 4 that's just that's this final step here
s gets replaced by s plus alpha alpha in this case is 4 so what I'm going to have in the denominator
is s plus 4 getting slotted in here and so if s was raised to the power 7 it's replacement
s plus 4 must get raised to the power 7 and I end up with my solution at the foot of the slide here
so that's the answer to that question so let's move on and have a look at another example here now
so we've got a function here again I would look at it and I would say well okay I've got a function
t to the power 8 I'm multiplying that by an exponential so that tells me that I need the
first shift in theorem because I'm multiplying this power of t t to the power 8 by an exponential
that's the first shift in theorem that's needed to obtain the Laplace transform of the whole function
okay so let's do a usual let's get the general expression for the first shift in theorem is
given here let's have a look we've been given this function here well we can find f of t either by
direct comparison or simply by just ignoring the exponential that tells me that f of t is equal to
8 now I've got to be a little careful here I've got to identify alpha and again I compare this I've
got e to the minus alpha t is equal to e to the 70 that corresponds to e to the 70 so in other words
minus alpha t must correspond to e correspond to 70 so that's what I've got in here okay so for this
to hold then minus alpha must equal 7 you know think of it if you're afraid think of it just cancel
the t's out minus alpha is equal to 7 or in other words alpha is equal to minus 7 so that's what I've
got here alpha is equal to minus 7 so that's step one so the next step well step one I've
ignored the exponential I identified f of t at step two I've got to get the Laplace transform
of f of t f of t was t to the power 8 well I can do that from the tables and I once again it's number
four this time with n equal to 8 so the Laplace transform of t to the 8 is 8 factorial over
s to the 8 plus 1 so it's 8 factorial over s to the 9 so 8 factorial is quite a large number
and quite often it's okay just to leave it in this form or you could work it out if you prefer
it's the factorial button is on any on most calculators anyway so big f of s is 8 factorial
over s to the power 9 that's using number four on the table with n equal to 8 and I'm using n equal
to 8 because we've got t to the power 8 that's what we've got here okay so I'm up here now in my table
my diagram I've got f of s and my final step says replace s in f of s by s plus alpha alpha is minus
seven so I've got to replace s in f of s by s minus seven there's f of s so s minus seven
gets replaced replaces s so if s is raised to the power 9 then s minus 7 must be raised to the power
9 so that is the final answer so just be careful it's not s to the power 9 minus 7 as I spoke about
a couple of examples ago it's s minus 7 all to the power 9 so that's yet another example of the first
shifted theorem now this is perhaps the example we saw earlier on this is given in the tables but in
many many tables this will not be given and we would have to figure out how to calculate the Laplace
Transword and again if I look at this example I would see a function a trig function sin 2t
been multiplied by an exponential so that tells me I've got to use the first shift in theorem
here's my diagram that I use I set up my comparison here the general expression from the first
shift in theorem is here and the expression I've got is here f of t is easy to get either by just
direct comparison or just by ignoring exponential that tells me f of t is equal to sin 2t what about
alpha well alpha just like in this example we can compare and we can find that alpha is minus eight
you know if you if I just get the tablet if you compare this you've got e to the minus alpha t e to
the minus alpha t must match e to the 8t so in other words minus alpha t must be equivalent to
8t I could cancel the t so if I like minus alpha is equal to 8 or alpha is minus 8 so
that's me identified f of t is sin 2t and I've also identified that alpha is equal to minus eight
so I can go ahead now and use the first shift in theorem following my diagram here starting in the
bottom corner with a function I was given I first of all ignore the exponential to give me f of t
I've done that f of t is sin 2t I get the Laplace transform of it well I've just got sin 2t so I
would look down my second column my table till I hit sin omega t and clearly just by comparison
omega is equal to 2 so the Laplace of sin 2t is 2 over s squared plus 2 squared or in other words
2 over s squared plus 4 so that's what that gives me here so I'm now up in the top right of my table
I've found f of s the final step the final step just involves me replacing any occurrence of s in f of s
by s plus alpha so let's have a look so f of s here it is there's an s here so I've got to replace
that by s plus alpha alpha is minus 8 so this s here must get replaced by s minus 8 and once again
the s squared so the thing that's replacing it the s minus 8 must be squared so that's where
that comes from so it's not s squared minus 8 it's s minus 8 all squared and I get this answer down
here which if I wanted to I could expand the denominator the Braggs and the denominator obtain
this solution both are perfectly correct okay so that's an example another example of the first
shifted theorem let's do another one now slightly different or very slightly different here's my
function e to the 60 times t minus 3 all squared again I compared with a general expression for
the first shifted theorem e to the minus alpha t f of t some function of time f of t multiplied by
an exponential function is that what I've got well yes it is because I've got my exponential
function e to the 6t and this part here would be f of t okay I could easily identify f of t again
either by direct comparison or just by ignorantly exponential in the given function t minus 3
all squared that's where f of t is so there's f of t we've identified it now when we come at step 2
to calculate the Laplace transform of this I know this I will not have this in my tables but what I'm
going to do is I'm going to expand it because I know I've got t squared I've got I've got t in the
tables and I've got a constant in the table so if I expand this bracket I can certainly get the Laplace
of these three terms from the table so it'll just be number three number two number one in the tables
so that's fine I could certainly do that what's the value of alpha well again you know if we
compare what have we got we've got in our question we've got e to the 6t let me just write that down
and in the general expression for the first shifted theorem is e to the minus alpha t
so in other words 6t must be equivalent to minus alpha t the t's can cancel if you like
6 is equal to minus alpha or alpha is equal to minus 6 so that's where that comes from
okay so we're just about ready to to start using the diagram to go around the diagram in the clockwise
direction so that's the function we're given down in the bottom left hand corner we ignore the
exponential to leave us f of t f of t we're already identified as this function here so that's f of t
so I now want to take the Laplace transform of this function which involves three separate terms
so but we know because the linearity property that the Laplace transform of this entire function
is equivalent to just taking the Laplace transform of each individual term and adding or subtracting
them together depending on what's required so let's do that so I've got the Laplace transform of t
squared well that's going to be number three in the table minus well I can take the constant out
if I like minus six times the Laplace transform of t that's just number two plus nine times the
Laplace transform of one and that's just number one in our table so I can get these quite easy
from the table t squared it's two over s cubed it's the Laplace transform of t squared minus six
times the Laplace transform of t so it's minus six times one over s squared plus nine times the Laplace
transform of one well it's just nine times one over s we've seen that before anyway and then we
can tidy all of this up and there is the expression for big f of s that's the Laplace transform of f of t
and I'm now up in the top right in my diagram and the last step that I have to do is I've got to replace
every occurrence of s by s plus alpha now up to now I think we've only had one occurrence of s in f of s
but if we look at this expression here we can see that s appears three times it appears in each of the
denominators and that being the case we must replace every occurrence of s by s plus alpha alpha is
minus six so each of these gets replaced by s minus six going all to the next slide that's what I've
got for f of s I've got to replace it by s each s by s minus six so let's just do that so I'll have two
over s minus six all cubed so s minus six is going in here for that s and if the original s was cubed
then the thing that replaces it which was s minus six that too must be cubed similarly here s was
squared so it's going to be s minus six all squared and simply here s gets replaced by s minus six so
that is the Laplace transform of my original function and that's the answer to the question
so just to summarize then today we've introduced Laplace transforms and looked at the motivation
for using them and we've looked at the definition of the Laplace transform that it's actually
an improper integral and it would be a lot of work to calculate if we didn't have the Laplace tables
so we looked at some examples of calculating Laplace transforms of standard functions using the tables
and then we looked at the first shifted theorem which was able to handle a product of two functions
where one of them was an exponential so we saw that in that case we used the first shifted theorem
and very importantly we saw or we know now that the Laplace transform of a function like this is not
the Laplace transform of the first function times the Laplace transform of the second function
that is not how to do that is incorrect so in this case because we multiplied a function by an
exponential we use the first shifted theorem in the next class we're going to look at inverse Laplace
transforms so if today we went from the time domain to the s domain or the Laplace domains it's also
called in our next class we're going to look at going the other way from the s domain from the Laplace
domain back to the time domain and that involves taking inverse Laplace transforms so after this
class then you should be able to attempt question one from the Mobius questions and just remember
that each individual question Mobius question has a large database of questions associated with it so
you can keep on just try you click in the try another button and to get a different question
of the same type so you could practice these once you've seen the Mobius questions and I also
included here from the lecture notes which questions you should also be able to attempt
so we'll leave it at that then for today and as I said the next time we're going to continue Laplace
transforms looking at calculating inverse Laplace transforms