Well hello everyone and welcome to our second class on Laplace transforms. Today we're going to
introduce the inverse Laplace transform and we're going to see how we can use the first
shifting theorem in reverse, in other words use the first shifting theorem to calculate
inverse Laplace transforms and we'll also look at inverse Laplace transforms requiring algebraic
techniques such as partial fractions and complete in the square. Now if you remember in our last
class we saw this diagram here and we said that Laplace transforms are often used to solve problems
like differential equations where we're unable to go directly from the calculus problem as a
differential equation to its solution in the time domain, either because the problem is very difficult
or even impossible using the methods that we know to solve differential equations. So what we do is
we take our time domain problem typically at differential equation as I've said take its Laplace
transform so move from the time domain the t domain over to the s domain and in so doing we convert
the calculus problem to an algebra problem in the s domain also called the Laplace domain.
The algebraic problem can be solved without too much bother to get our solution in the
s domain and then we apply the inverse Laplace transform to go back to the time domain where
the problem was originally set. So as I said in a previous lecture we looked at this step here
going from the time domain to the s domain by calculating the Laplace transform. Today we're
going to go the other way we're going to take our s domain function which could be the solution to
our algebra problem take our s domain function and take its inverse Laplace transform to go back
to the time domain. So that will then have completed the time domain to the s domain calculation
and when we finish today we'll be able to do the s domain back to the t domain and in the next
lecture we'll look at this bit here solving the algebra problem we'll take a differential equation
takes Laplace transforms solve the algebra problem and then invert it so but for now we're going to
focus on calculating inverse Laplace transforms. So the inverse Laplace transform well for a function
f of s in the s domain its inverse Laplace transform is a function f of t in the t domain
such that the Laplace transform of f of t is equal to f of s. So in other words last time we saw how
to calculate this we saw how to calculate the Laplace transform of a time domain function to
obtain f of s today we are going to take f of s and retrieve the f of t for that f of s. So that's
what I say here given a Laplace transform of f of s to determine the corresponding f of t we could
use the inverse transformation and the inverse transformation is given by this integral here
now in the last lecture we saw that when we're going the other way from the time
domain to the s domain there was an integral to calculate the Laplace transform it was an
improper integral with infinity as one of the limits and we saw that it was quite an involved
calculation that would just become harder and harder as our time domain functions got more
complicated and that's why we use tables this is even worse because to evaluate this integral we would
no have to know a topic called complex analysis and it's beyond the scope of this course so what
we're going to do is we are actually going to take the tables that we're using the last time
and we're going to use them in reverse so in other words or in lecture one we took a time
domain function as in this column here and we calculated the Laplace transform using the tables
today we're going to take our estimate function and we're going to go back the other way so we'll
have a function in the s domain which is in this column here on the table and we want to retrieve
the time domain function in the column on the left here now as I've mentioned before this will
undoubtedly require us to do some manipulation on the time that we are given we need to manipulate it
to obtain something that's in the form given in the tables so we'll need to know algebraic
techniques such as completing the square and partial fractions and I'll certainly go into some
examples of these today so first of all before we launch into some examples let's just have a look at
the linearity property of the inverse Laplace transform very similar to the linearity property
of the Laplace transform we saw the other day basically what it's saying is if you've got let's say
two three it doesn't matter how many functions I've just taken two here you've got two functions in
the s domain f of s and g of s and they're both multiplied by some constants and you want the Laplace
of that whole thing well it's perfectly okay to take the constants outside the Laplace the inverse
operators take the Laplace of inverse Laplace of each individual function and add them together
so this is very similar to what we did last time and very similar to what we did when we're doing
differentiation we were differentiating let's say t squared plus sine t we differentiate t squared
we differentiate sine t and would add the derivatives well the same thing applies here
we just take in inverse Laplace of this function the inverse Laplace of that function and add them
together that's all I'm saying here here I've taken the functions the s domain functions
without being multiplied by a constant so I've got the inverse Laplace of the sum of f of s plus g
of s that's equal to the inverse Laplace of f of s plus the inverse Laplace of g of s and I've
got a constant multiplying one of them or both of them we can take the constant outside the operator
just take the inverse Laplace of f of s and multiply the count by the constant c and that's all
summarized in one sentence here by this term up at the top of the slide okay so let's have a look
at calculating some inverse Laplace transforms so the first one is we're given an s domain function
f of s is equal to six over s to the power four and we want to find what was the time domain function
whose Laplace transform is equal to six over s to the four okay so like I said we use the tables for
this so what we do is we look down in this right hand column I focus on the denominator to start
with so I'm going to focus on the denominator here s to the power four is there anything that looks
like that in the table well we've got s to the power one here s to the power two s to the power three
but we've run out there we don't have s to the power four but what we do have is we've got a general
term here with s to the power n plus one and in our example the n plus one must equal four because
it's got s to the four the denominator here with s to the n plus one here so n plus one must equal
four in other words n must equal three so I can now check also what the top line says the top line
must have n factorial or three factorial three factorial is six and that's exactly what we've
got in the top line so I've been able to go to the second column or the third column in the table
the far right column in the table I've compared the denominator and from that I've found that n is equal
to three I've then had to compare the numerator to see if n factorial matches the value in my question
and n factorial with n equal to three n factorial is six and that exactly matches what I've got here
so I can directly use the table here so that tells me that if n is equal to three the inverse
Laplace of six over s to the power four is equal to the inverse Laplace of three factorial over s
to the power four and directly with n equal to three I find that that's equal to t cubed and if you
wanted to check this you could just go and calculate the Laplace transform of t cubed and you should
get six over s to the power four then you can see that t cubed would be three factorial which is six
over s to the three plus one which is four you'd have six or s to the four and that's exactly
what you start with so we're happy that your answer is correct so moving on let's have a look at number two
in this question we're asked to find the inverse Laplace transform over 15 where s to the power six
so we want to find which walk was the function f of t when we took its Laplace transform we obtained
this quantity here this function here so again I would look at my tables focusing on the denominator
to start with so I've got s to the power six well we know we don't have that in our tables we've
run out at s cubed but we know that we can use number four so number four is s to the n plus one
so we match that with the s to the power six so n plus one must equal six so in other words
n must equal five because that would give you s to five plus one which says to the six so we've got
s to the power six and we know it's going to be number four with n equal to five now this table
tells me that if I'm going to use number four that my top line must be five factorial it's
got to be because I've said from the denominator I've said that to give me s to the power six I
must have n equal to five and if I am using number four then I must have five factorial in the top
line so five factorial is 120 and that's not what I've got I've got 15 so I need to do a little bit
of algebraic manipulation and the easiest way to do this is to write down in the top line exactly
what the table would have you use in the top line the table wants a five factorial in the top line
so we write that in and we've got to multiply this by some fractional expression that will
make sure things aren't changed so that we so we've got the function that we started out with
and the easiest way to do that is for a fraction to have in the top line we write the thing that we
want which is 15 and on the bottom line we write what the table has forced us to introduce which is
five factorial and you can see the five factorial there cancels with a five factorial there leave me
15 over s to the six and that's exactly what I started out with so I can now go ahead I can do the
inrational plus of this term here so five factorial or s the power six will just be number four with
n equal to five so that will be t to the power five 15 over five factorial well five factorial is 120
so I get this expression here I can easily tidy this up because 15 goes out into 128 times so I've
got one over eight t to the five and again I could just take the Laplace transform off this
using number four with n equal to five and I should obtain 15 over s to the six if I do that
I could check my answer for me so I'm a little bit of manipulation needed there because the
function we were given isn't exactly as what's given in the table we looked at the denominator
to try and figure out which one we should use and we we reached number four and we said well
number four with n equal to five will work but that means I must have five factorial in the top line
not a problem we wrote that down in the top line and we introduced this factor this important factor
to make sure that we aren't changing any and in this factor on the top line is the thing that we want
which is 15 and down below is what the table forces to introduce that's five factorial as you can see
they cancel to leave me 15 over s to the six which is what I started out with okay moving on let's
have a look at another example here so we've got f of s is equal to eight over s to the minus three
and we want to find the inverse Laplace transform of this function and that will give us the function
f of t which Laplace transform is eight over s to the minus three so f of t is equal to
inverse Laplace of eight over s minus three so I need to look at my table focus on the denominator
to start with s minus three well none of these will work but when I hit this line here looking at
the denominator I've got s plus some number well it doesn't matter that that's a minus because
you know I could write this as s plus minus three so we can certainly handle that so
what do we do here so I would say that this one comparing this with my tabulated value I would
say that this here we would have to have alpha equal to minus three and if alpha is equal to minus
three then I would get a I would get I've got an eight here eight e to the minus minus three t
which is eight e to the positive three t and again I could calculate the Laplace transform of this
to obtain the function f of s that I was given so once again we focus on the denominator
obtain the candidate from the tables which was number five then we compare the denominator
we've got s minus three and s plus alpha and that tells me that alpha must be equal to minus three
so if I put it in here I've got e to the minus minus three t or e to the three t and don't forget
though there's an eight on the outside here and that gives me my answer so I'll just say one other
thing here um our table gives us e to the minus alpha t if you had another set of tables they may
very well have given you e to the plus alpha t and in that case the Laplace transform would be one over
s minus alpha and we could have just used this one directly like so but of course the tables are
limited in how much they can contain just as in this year they stopped at s cubed the denominator
because the tables have to be a sensible size of course so okay so that's another one done so
let's move on and look at this one here now so we've got three over s times s plus six so once
again we focus on the denominator we go to the tables well it's none of these s into s plus some
number well eventually we hit this one here number seven that looks that looks like a possible candidate
there are others of course after this but you know we rule them out this is the first one
that we've reached that we think might work so let's have a look at it so what is this one
required so I want my function f of t will be equal to the inverse Laplace of three over s to the s
plus six now compare the expression we're given with the expression in the final column with
f number seven there's a slight difference s into s plus six that's fine the denominator is fine s
into s plus some number that's exactly what we've got but if you look closely at this
whatever the number of the top line must match the number here so the alphas must match in our case
they don't we've got a six down below and a three up top so we've got to do a little bit of algebraic
manipulation that will allow us to use number seven and we can't really change anything down below
here we can't change anything down below here because this shift value here is fixed in the
denominator we can't change that but we can manipulate the top line a little so we use
exactly the same trick as we did for the previous example we write down here we write down exactly
what the table would have if i'm going to use number seven so on the denominator i've got s
into s plus alpha which in our case is s into s plus six now the top line the alpha and the
top must match the alpha on the bottom so in our case we must have a six in the top line
and then the two values the alphas in our case which is equal to six will match but in doing that
we have actually changed what we started out with we started out with a three so we do the same
trick as in the previous example we multiply by a fraction whose top line is the thing that we want
which is a three and whose bottom line includes what the table forced us to introduce and that was
the six so if i do that i can now directly apply the table to this expression here so that is just
number seven with alpha equal to seven so i get one minus e to the minus seven t for this part here
this is the inverse part here one minus e to the minus six t sorry one minus e to the minus six t
because alpha is equal to six and on the outside don't forget we multiplied by this fraction which
simplifies to a half so that is the answer there that is the function f of t which is a plus transform
is given by big f of s up here and once again if you wanted to you could calculate the la plus
transform of this one here from these tables and you would find that you get exactly what's given
in the question i mean that is fairly straightforward here because if you compare let's just ignore the
half for now but you look at this part here that's exactly the same as number seven here with alpha
equal to six so you would get six over s into s plus six multiplied by a half and the six
the half multiplied again to give you three and that's exactly what you've started out with
okay so moving on let's have a look at yet another example so what do we have here we've got nine
over s into s squared plus nine so that's our f of s we want the inverse Laplace of that to see
which function f of t has this as its Laplace transform so again we focus on the denominator
and look down the column in our table and see if there's any that looks like this well you know you
can think of this as s squared plus some number squared you know s squared plus omega squared
if you like and you know that would give you that omega squared is nine so omega is three
so let's have a look at the first one is there any of this form in the table well you could work
all your way all the way down to here that doesn't work it's just s squared plus a number
that doesn't work it's the same thing but we reached this one here that one could work that
one could work it's s into s squared plus some number squared called the number omega and that's
good because we can rewrite this as s into s squared plus three squared meaning that omega is equal to
three so that's good so we're down let's say we're down about here now we're s into s squared plus
three squared what is the top line need the top line says that we're going to use number 10 then we
must have the omega squared we must have this quantity matching the quantity in the denominator
here and if we look we actually do it's nine up top and it's nine down here they do match they're
both three squared so omega is equal to three in this case and the inverse Laplace of this function
the function we started out with is just one minus cos three t that's using omega equal to three and
again you can check this and I did that here just so that you can see it's the Laplace transform
of one minus cos three t this is what we're doing in our last lecture you can take because of the
linearity property you can take the Laplace transform of each individual term and then
perform the subtraction so if I do that that's one over s that's number one in the table this
is number nine in the table cos three t gives me that it's s over s squared plus three squared s
over s squared plus nine I then write this over a common denominator so my common denominator would
be s times s squared plus nine so if I multiply that by s squared plus nine I must multiply the top
by s squared plus nine that's where that comes from and here I'm multiplying the denominator by s
so I must multiply the top by s so I get s squared I simplify all of that and I get back exactly what I started out with
so one minus cos three t is the function f of t which Laplace transform is nine over s into s squared plus nine
okay um
moving on let's have a look what we've got here
okay 13 over s squared plus three right so again we'll focus on the denominator and see is there
anything in the tables that looks like that well I would reach number eight I've got s squared plus
omega squared so it's basically s squared plus a number and that's exactly what I've got so it could
be number eight I move on ah it could also be number nine s squared plus a number so
it's going to be one of those two and how do I know which one it is well I now look at the top
line I look at the numerator and first of all when I look at number nine I see number nine
includes the Laplace variable s whereas number eight in the top line just includes a constant a number
the Laplace variable s doesn't appear and I look at my question and I see there's no s up top so that
tells me with just a number in the top line that tells me that I should use number eight so it's
going to be number eight but I need to do a little bit of work first of all it needs a little bit of
manipulation so my function f of t is the inverse Laplace of 13 over s squared plus three that's what
I've been given and I've identified that it's going to be number eight that I'm going to use here
and that should actually read that omega squared is equal to three I'll just make that correction
right now I think I maybe missed out the square on this term here so that should read
omega squared is equal to three so let's do a little bit of manipulation to get
this correct so I've actually got it written correctly in green here omega squared is equal to
so that means that omega will be root three so I can rewrite this term here as s squared plus
omega squared or in other words s squared plus root three squared and that'll give me my s squared
plus three because when you square the square root it just gives you three now what about the top
line I've got to be careful here my top line at 13 that's not equal remember to omega omega we said
was root three so what I'm what I need to have is I must have a root three in the top line here so
that's what I'm going to do I'm going to write down exactly what the table would force me to have
in the top line and now I've got the form omega over s squared plus omega squared exactly this form
but I've changed things so I must have my correction factor in here again it's this fraction
whose top line is the thing that I want and that's a 13 and down below here is what the table
forced me to introduce and that was root three so as you can see if I cancel out the root threes
I'll get 13 on the top line it's exactly what I started out with and I can now use the tables
directly on this expression here with number eight and in that case omega is equal to root three so
the inverse Laplace of this thing is just sine root three t and of course don't forget the factor
on the outside 13 over root three so that is the function f of t which Laplace transform is given
by 13 over s squared plus three okay next problem so our next problem will be given our function
f of s is equal to four over s seven s plus three so we're looking for the function f of t whose Laplace
transform is given by four over seven s plus three now again we look through the table focusing on
the denominator and I don't I won't find this I will not find this in the table because as I've said
before the tables of course have to be a manageable size and we can't include every
single type of function in the table so we only include some general terms and we need to do some
manipulation to get this into a tabulated form so searching down the table the only one that looks
like it would be this one but s plus alpha where the coefficient of s is one and if you look in the
table the coefficient of s in virtually all cases is equal to one we've got a seven front
bs but that's not a problem we can manipulate this and get it into the tabulated form so it's
number five we're going to use here and what we're going to have to do is we're going to have to write
as a set s plus something it'll all be multiplied by something but that factor doesn't matter let's
get it written as s plus a number first of all so what I'm going to do here is I'm going to take the
seven out as a common factor so I'll just get the pin I'll take the seven out as a common factor I've
got seven s plus three I'll take the seven out as a common factor now to get seven s I need an s there
plus now plus what well if I wrote three there that wouldn't that wouldn't be right because seven times
three would be 21 and that's not what I've got so what I'm going to do is if I'm going to multiply
the three by seven so that I don't change anything I just divide by seven and you can check that it
can be seven s plus 21 over seven 21 over seven is three and so I haven't changed anything and I've
got it written here as well so that will be the manipulation that I'll do here I'll have
four over seven into s plus three over seven so that will be that will be number five and I've
got the s plus alpha form in the denominator now I can break this up as four over seven times that
and now I can then have four over seven think it's a constant so I can take it outside the
inverse Laplace operator I can pull that outside so I just got to get the inverse Laplace of that
term there that's it I've repeated it down here but that's easy that's easy that's just number five
with alpha equal to three over seven and the inverse Laplace of that is e to the minus three
over seven t and don't forget we've got the four sevenths on the outside multiplying that
so I use number five with alpha equal to three sevenths to give me e to the minus three over seven
t and then I had my multiplying factor on the outside which was four sevenths and again if you
wanted to check that that answer is correct you would take the Laplace transform of this function
here okay so we can move on and look at the next set of calculations of the inverse Laplace transform
now in our last class we introduced what we called the first shifting theorem or the FST for
short and we said that the first shifting theorem was used to calculate the Laplace
transforms of a product involving an exponential function we also said that it was really important
to remember that if you got the Laplace transform if you're asked for the Laplace transforms of two
functions of t f t times g of t the Laplace transform of a product it is not equal to the
Laplace trans the product of the individual Laplace transforms that is incorrect just the same way as
when we're doing differentiation and you had to differentiate something like t cubed cos t you
wouldn't just take the Laplace the derivative of t cubed and multiply it by the derivative of cos t
you had to use the product rule similar things apply for Laplace transforms in that the Laplace
transform of a product as I've just finished saying is not equal to the product of the individual
Laplace transforms that's completely wrong and we saw in the last lecture that for a specific
product where we had an exponential function multiplying some other function at time that we
had to use the first shifting theorem to obtain the Laplace transform of that specific
product type and what it amounts to is when you calculate the Laplace transform of this time
domain function what you get is you get the a shifted function in the estimate basically what I mean
by that is if your raw function f of t has Laplace transform big f of s then if you multiply your
function f of t by an exponential function e to the minus alpha t what you need to do to the Laplace
transform of f of s is shift it by an amount alpha and we saw that this table here allowed us or
helped us to calculate the Laplace transform of a function that was a product of an exponential
and some other time domain function so we start down here with the function we're giving we've got
an e to the minus alpha t multiplying some function f of t it could be t to the power six or it could
be sine 3t or cos 9t or anything like that so if you remember what we did was we ignored the exponential
to give us the function f of t we calculated the Laplace transform of f of t which gave us f of s
so the Laplace transform of f t is f of s and then we had to do the final step wherever we saw an s
in f of s we had to shift it by the amount alpha we replaced the s by all occurrence of s in f of s
we replaced by s plus or minus alpha whatever the value of alpha was and that is summarized in number
22 in our table so I'll do an example this is an example from last time just to remind ourselves
of what was involved here today what we're actually going to do is we're going to use
the first shift in theorem to calculate inverse Laplace transforms and all that is this whole
procedure in reverse but let's recap what we did last time so here's an example in the time domain
we want the Laplace transform of this function we want to go to the s domain so we're going to use
the diagram here we've got our function e to the minus 4t t to the power six and I know that because
I'm multiplying the function t to the power six by an exponential I know that's the first
shift in theorem that I need to use so my diagram is here and I compare my starting function
e to the minus alpha t f of t with e to the minus 4t t to the six and from that I can easily
identify that alpha is equal to four just all it is is pattern matching so alpha must equal four
and f of t well f of t just to grow the exponential f of t is t to the power six
and I then go to the tables that my diagram tells me that I must now calculate the Laplace
transform of f of t to give me big f of s the tables would give me that with a little bit of work
so what have I got I've got t to the power six that's going to be number four with n equal to six
so it'll give me six factorial over s to the six plus one in other words six factorial over s to the
power seven six factorial 720 so I've got 720 over s to the power seven that's my f of s the final
step in my diagram tells me that I must replace every occurrence of s obtained at step two this
expression here I must replace every occurrence of s by s plus alpha alpha was equal to four so I've
got to replace s by s plus four and that gives me my Laplace transform of this function here e to the
minus four t t to the power six of this product involving an exponential function using the first
shift of theorem I obtain its Laplace transform so one thing I think I said last time that you've
got to be very careful about a common mistake when people get to the last step here they write down
that the Laplace transform was 720 over and they look at step two the big f of s and they write s to
the seven plus four that's not that's incorrect because what you're actually supposed to do is
you if s is raised to the power seven and you're replacing s by s to the power four then you must
raise the s to the power four to the power seven so make sure that you know the difference between
this and this because that's a fairly common mistake to make okay so that was what we did last week
and today we're going to apply this process in reverse so that when we're given a function
that's shifted in the s domain and you know it's shifted if you see that you've got an s
plus or minus some number in your big f of s when you get a function like that with a shift in the s
then you use the first shifted theorem in reverse and all it is is you do the steps backwards you go
around the diagram in the anticlockwise direction so first of all it tells you ignore the alpha
so I would do that so ignore the four out of 720 over s to the power seven and then I would go to get
by time domain function I would do the inverse Laplace of this one here and that would give me
t to the six and then the final step would tell me multiply t to the six by e to the minus alpha t
alpha we know is four so that would retrieve my time domain function so let's just have a look at
some examples of that before we do just we'll have a couple of words about it again
so this really just covers what I've been saying so calculate the inverse Laplace transform
of a functional form f of s plus alpha in other words f of s as in this case in step two that's
been shifted in the s domain because you've got s plus some number so this is it really this is what
we want to get we've got Laplace transform e to the minus alpha t f of t gives us a shifted function
in the s domain so we want to retrieve this function so we want the inverse Laplace of f of s plus alpha
to give us our time domain function where we've got some function of time multiplied by an
exponential so if I see a function like this with a shift in it I know it's going to be this type of
function some exponential multiplying some other time domain function and there's a video you can
watch here talking about the first shift in theorem being used in reverse so the procedure
as I've already mentioned is just what we did last week and in the example I just covered
but in the opposite direction going in reverse three steps we go around the diagram anti-clockwise
today rather than clockwise when we're going from the time domain to the s domain last time
so let's just go straight into an example and have a look at what's involved here to determine the
inverse Laplace transform of it would be f of s is equal to seven over s plus two to the four
so that would be sorry that would be my s f of s plus alpha so it's f of s plus two so I've seen the
shift so I'm pretty sure it's going to be the first shift in theorem in reverse so I go to my
diagram and what does that tell me to do well that tells me to ignore the alpha that will leave me f of s
so if I ignore the alpha what's the value of alpha here well it's whatever is added or subtracted
from the s in this case we've added a two so basically I ignore the two and that leaves me
seven over s to the power four so that's my f of s so up here I want to invert this one and to do that
I go to my table I look down this far right column here looking focusing on the denominator well I
know actually I don't have I stop at s cubed so s to the four I need to use this one with n equal to
three okay so I need n equal to three so I would have three factorial over s to the four if I was
going to use this one here that's what I've got to have if I've got s to the four in the denominator
that means n is equal to three so to give me three factorial over s to the four that's not exactly
what I've got three factorial is six but I've got a seven in the top line so I do the same as I did
with the earlier examples I write down in the top line exactly what the table would have me use and that
would be three factorial so on the outside I'm multiplied by this fraction whose top line is
the thing that I want which is a seven and on the bottom I've got what the table forced me to
use and that was three factorial and you can see the three factorial is cancelled to give me seven
over s to the four which is exactly where I've got so I can use this directly now the tables that's
just number four in the tables with n equal to three so that'll be t cubed and here three factorial is
seven over six times t cubed so that's it there so that's my f of t the last step says multiply
that f of t by e to the minus alpha t alpha we said was equal to two so I get seven over six
e to the minus two t times t cubed or if you prefer seven over six t cubed times e to the minus
the order there doesn't matter you know that's that's absolutely fine that's absolutely fine
okay so let's have a look at another one now we could use number 14 in the tables for this one
you'll find this or you know something similar to this in the tables but as I've said before
not every set of tables will give you that so really you need to know how to use the first
shifted theorem for this type of function so first of all how do I know it's the first
shift in theorem well I've got s plus or minus some number so that's a pretty good indication that
at least we should be trying the first shift in theorem anyway so what have I got so this is going
to be f of s plus alpha what's the alpha well it's minus two so it's f of s minus two so I want to
find f of s well that's easy I just ignore the alpha so it'll give me three over s squared plus nine
that's if I ignore the minus two I get three over s squared plus nine what's the value alpha well
it's minus two okay for now it's not two because in this setup it's s plus alpha we're using here
and we've got s minus two here so alpha must be minus two so we've got alpha we've got f of s
now there's f of s there so I want to um get the inverse Laplace of that term that's f of s to give
me f of t so the inverse Laplace of this one here if we looked in our tables we would find that this
will be number eight if I could go back what am I going I've got three over s squared plus nine
so if I've got a number eight anywhere here there should be one somewhere yeah I've I've got three
over s squared plus nine nine is three squared so what I've got is three over s squared plus three
squared so it's just sine three t so let's go back to my example there we are it's just sine
three t because I've got omega over s squared plus omega squared where omega is equal to three
so there we are so that's fine sine three t so that's f of t my last step says multiply the f of t
I got step two by e to the minus alpha t I've already identified that alpha was minus two so just be
careful with it it's going to be e to the minus minus two t in other words e to the two t so my answer
is e to the two t sine three t so that is the function of time that's the function f of t
use Laplace transform is given by this expression up here okay so that was the first shift in
theorem being used to calculate inverse Laplace transforms let's now move on and look at
some other functions that require algebraic manipulation before we can use the tables and
this will require us first of all the types of functions we'll look at will involve using partial
fractions so we'll have three different types of these and then to finish off with we look at
functions that involve us completing the square in the denominator mainly because the denominator
doesn't factorize easily to use partial fractions but I'll discuss that a little bit more shortly
so first of all we've got um partial fractions and as I said we're going to look at three
important different types of partial fractions all based on the type of function that's appearing in
the denominator of the function we want to find the inverse Laplace transform of so um in this case
here we are here's my function here now one way and the best way to go about this is first of all
let's see if that denominator is a quadratic function let's see if we can make things a
little easier for ourselves by factorizing this denominator well if you you know if you either
use trial and error or whatever you could find that this term here factorizes into two linear
terms a product of two linear terms s minus two s plus one and just checking it s times s is s squared
s times one is s minus two s gives me minus s minus two times one is minus two so that's fine
that's the correct factorization for that expression in the denominator now most of you
will have seen partial fractions before but I'll go into a little bit of detail as we're working
through them so in this case in the denominator we've been able to factorize the quadratic into two
distinct linear terms distinct just mean that they're different s minus two times s plus one so
they're two different two distinct linear terms so what I'm going to need is I'm going to need a
fraction for each term and with partial fractions I'm going to break this into two separate fractions
I'm going to have a over s minus two plus b over s plus one so a fraction for each term now with
partial fractions the numerator always has degree one less than the denominator now I said these are
two distinct linear terms linear meaning that the highest power of s is one so they're both
s to the power one so the numerator will have degree one less so it'll be s to the zero in other
words s to zero is one so that's why you have a constant in the top line because you get a linear
factor in the denominator so that means a constant goes in the top line so that's our fractions and
we need to find a and b and then once we've broke once we've identified a and b these will become
quite easy because if you remember there'd be just be a number over s plus a number and that's
very like you know tables one over s plus alpha number five in our tables but let's look at a
little bit of work to do before we finally get to a and b so first of all the first thing to do here
is I want to get an equation where I can find a and b so I'm going to multiply both sides of this
equation here by the denominator of the left hand side so I multiply this by its own denominator I
get five s minus one move it over just if I get that multiply the first fraction here the s minus
two is cancelled to give me a into s plus one and then move it over to here and you can see the s
plus one's cancelled to give me b into s minus two okay so once we get to this stage we've got one
equation with two unknowns which might seem a bit of a problem but we can set values of s such
that we end up with one equation with one unknown and basically you just look at what will make
either bracket equal to zero so first of all I'm going to take s equal to minus one if I set s equal
to minus one I'll have a times minus one plus one or a times zero so that'll vanish and s minus one
I'll have b times minus one minus two minus three so I minus three b there s minus one here I'll have
minus five minus one which is minus six so I'll have minus six is equal to well that's zero
minus three b in other words b is equal to two okay so I'm done with that one um
any other values that I can set s equal to two to make something vanish well yes I can set s equal
to two and I'll set s equal to two this then vanishes and just leave so in equation five times two is
ten minus one is nine there's the nine there s is equal to two i'm three a nine is equal to three a
that part is zero so I just got three a is equal to nine or a is equal to three so the partial
fractional representation of this term here the thing that I started out with is given at the
foot of the slide here okay so that now just leaves me to find the inverse Laplace transform
of these two individual fractional terms I don't have anything that looks like that in my tables
don't have anything that looks remotely like it but partial fractions have allowed me to obtain
two terms that I can easily find the inverse Laplace transform of so here I am so there's my
original expression that I want the inverse Laplace transform of I broke it up into its partial
representation by calculating these two fractions the linearity property says that I can just take
the inverse Laplace transform of each one individually and add them together any constants I can take
outside that's all down to the linearity property I've got one over s minus two look at the bottom
line and I hit this one s plus a constant well you can take it with a constant b minus two it's
s plus minus two so that's just going to give me that alpha is minus two so this becomes e to the
minus minus two t for e to the two t for this part here don't forget though it's multiplied by three
and for this one I've got one over s plus alpha one over s plus one that again it's number five
but this time with alpha equal to one and i've e to the minus one t don't forget there's a two
outside there so I get two e to the minus t so that's um that's the answer to that one so that
was the case when we were given a function whose denominator was a quadratic who were able that
were able to factorize quite easily into the product of two linear terms and so we then
got two fractions one for each term in the denominator and with a little bit of algebra
we were able to solve for the unknown constants a and b and that left us a fairly straightforward
task using number five in the tables to find the inverse Laplace transform of our original function
as given here and I think I've mentioned before that number five will be seeing quite a lot of this
this one because the solutions to differential equations tend to involve exponential functions
so number five maybe number six seven and so on they tend to feature quite a lot that's the ones
that involve exponential functions so it's good to be familiar exactly with with these ones with
the exponential involving the exponentials okay moving on now here's another one and this is a
different type of partial fraction again you know you would look at this and you might be tempted to
say well I could write that that fraction there that denominator I could write it as the product
of three linear terms because that's s to the one s to the one s minus six and I could just
get my partial fraction a over s plus b over s plus c over s minus six that's one fraction
for each term but be very careful that will not work this is what we call this is a repeated
linear factor we've got s times s it's a repeated linear factor and using this formulation will not
work we'll end up with total nonsense when we try and get to this stage here when we're trying to
simplify it so how do we get around that well the way we get around that is we set up a partial
fraction square one fraction for the s minus six term that's fine there it is we've got one fraction
for s b over s and rather than writing another fraction like a c over s we write it as c over s
squared okay so that's what we do we write it like that okay so uh let's try and get an equation
that we can work with out of that well again we do the same as we did in the last example we multiply
both sides by the denominator of the left hand side and as you can see here multiplying this first
term by this denominator which I just copied over here in red the s minus six cancel to give me a
times s squared moving this on to the next term on the right hand side I multiply this this term by
the denominator s squared into s minus six that s will cancel one of them giving me b times s into
s to the minus six that second term and finally I multiply my third fraction the right hand side
by the denominator of the left hand side the s squared this time cancel to give me c into s minus
six that's it there okay so we've now got one equation with three unknowns which on the face of
it would almost seem impossible to solve but again we can set values of s so that some of these
unknowns disappear and leave us simple equations to solve so for example let's choose a value of s
that will eliminate some of these terms so clearly if I said s equal to zero that'll vanish with s
equal to zero that will also vanish and this will just be minus six c there's no s there so that just
stays as 18 so I'll get 18 is equal to minus six c that's with s equal to zero in this equation here
that vanishes that vanishes and I've just got c times zero minus six or minus six c and that's
18 on the right hand so c is equal to minus three okay that's fine can anything else be
eliminated well yes with setting s equal to six then that term will completely go because this bracket
will be six minus six which is zero so that'll all go so to you will this because you get six
minus six times c so it's c times zero so that'll completely vanish so if I just get my tablet uh
what am I got I've got that this term has gone and I've got that this term has gone
s was equal to six so I've got 18 is equal to 36 a so in other words a must be equal to 18 over 36
or a is equal to a half so that's okay that's that one solved now can I eliminate anything else
any other value of s that would help me well not really because I've said s equal to zero
and I've said s equal to six and that's the only things that could directly eliminate any of these
three expressions and I've still left to find b I found c and I found a I've still got to find b
but the way you do that is you just choose any value for s that you haven't used before
and try and keep it as a simple value we've used zero so we can't use that we use six we can't use
that either so let's just choose the most sensible value and that's s equal to one so let's say s
equal to one here well that still gives me 18 on the left hand side so that'll be 18 there we are
there it is there and with s equal to one that'll just give me a s equal to one just be careful b
times one times one minus five so that'll give me minus five b there it is there and again I'll have
one minus six remember s is equal to one so I'll have minus five c now I know what a is a half
c is minus three so I can work all of this out I can 18 is equal to a half minus five b plus 15
if I simplify this a little and solve it for b I find that b is equal to minus a half so I've
now been able to find a b and c I've been able to find a b and c and I can substitute them back in here
so that's what I'm going to do on the next slide so if I just copy my expression over
so let's see exactly what we've done so just down here I've got my expression there's the left hand
side that's it up there I had that a was equal to a half so I get one over two into s minus six there
it is I had that b was equal to minus a half so what do I have b is minus a half so I get
minus one over two s there it is there and I had that c was equal to minus three so I get minus three
over s squared there it is there so that's just substituting the a b and c values I found in the
previous slide into my general partial fraction expression here and I've obtained this expression
up here so now I can try and use my tables to find the inverse Laplace transform so the
inverse Laplace transform of my original function 18 over s squared into s minus six is equal to
the inverse Laplace transform of all of that but by the linearity property I can just take
inverse Laplace of each individual term and then do my subtractions right so this one first of all
well I think we know this one by now it's just the two we can ignore that for now to look after
itself it's just one over s plus alpha that's number five so that's all I've done here is I've
taken the one over two outside the operator outside the inverse Laplace operator to give me
this and I've just said it's number five with alpha equal to minus six so it'll be e to the
60 don't forget multiplied by a half here I've thought of this as minus one over minus a half
times one over s so I've taken the minus a half outside one over s the inverse Laplace of that's
just one so I get minus a half this one I've just taken the three outside and I've got minus three
inverse Laplace of one over s squared which is just number two which is t so I get minus three t
put it all together and that is the inverse Laplace transform of my original function there it is
given here okay so that's the second type of partial fraction we're looking at this was different
from the first one the first one we had distinct linear terms in the denominator in this case
we've got a repeated linear term in the denominator so we had to treat it slightly different when we
set up our partial fraction representation we had to take account of the repeated linear factor
and we did that by including the c over s squared term rather than having
plus b over s plus c over s I had to have an s squared here right okay so the third
the third case that we're going to look at is when we've got a quadratic factor in the denominator
that's what's known as irreducible basically you cannot factorize it into two linear terms
without using complex numbers it does not factorize easily you know and you can try and factorize this
but but you won't be able to do that without using complex numbers it does actually factorize to become
s plus 2j times s minus 2j but we don't want to be using that with partial fractions or our Laplace
tables we want this you know expressed in terms that are given in our tables or similar to our
terms in our tables so so we've got to consider this as the product in the denominator of a quadratic
because we've got an s squared and a linear term well we've seen linear terms before and we know it's
just going to be a constant over whatever the linear term is so you know that's where that comes
from c over s minus 4 what about the quadratic term well if you remember earlier on I said that with
partial fractions the order of the numerator is always one less than the denominator so you could
a linear term a power one term in the denominator then you need a power zero term in the top line
the order of the numerator in terms of s is always one less than the denominator so an s to the one
down below means you get an s to the zero on top s to the zero is just one so that's why you get a
constant up top now this other term is not a linear term it's a quadratic term the denominator so
that means if the function that goes up top has degree one less this I've ordered one less than
this then it must be an s to the power one term a linear term and the most general linear function
in s would be as plus b so um that's what we're going to use that's what we're going to use so
there are two terms one this one for the linear term and this one for the quadratic term that appears
in the denominator again we multiply both sides by the denominator left hand side so you multiply
that by its own denominator just gives me three s squared plus 28 move it over now to multiply
the first term on the right hand side the s squared plus four cancels with that s squared plus four
it gives me as plus b times s minus four so that's where that comes from and then finally
I multiply the second term on the right hand side by the denominator of the left hand side
and the s minus four cancels to give me c times s squared plus four so that gives me one equation
this time again with three unknowns a b and c which I've got to find and again I try and choose my s
values wisely so that I can make this equation simpler by eliminating terms is there anything that I
can do here to eliminate terms well I'm looking looking at the right hand side of course yes I can
set s equal to four and that will eliminate the a and the b because I'll have as plus b times four
minus four that's where s equal to four four minus four is zero so that'll all go so if I said s
equal to four on the left hand side I'll have three times 16 is 48 minus 28 that gives me that's where
the 20 there comes from that has we've just said zero four squared 16 plus four is 20 so I'll have
20 is equal to 20 c in other words c is equal to one okay so move on can we choose another value for s
that will make something disappear or vanish well no there isn't actually because you know we've got
nothing we can't set a value here s squared plus four equals zero well you need to have a complex
number it'd be either 2g or minus 2g that that's not going to work for us that's not going to work for us
so the thing to do here is you choose a value for s that you haven't used before
you choose a value for s that we haven't chosen before and make it a sensible value so I'm going
to say that s equal to zero so if I do that my left hand side becomes minus 28 there it is
my right hand side with s equal to zero well the a will actually go here and here i'll have minus four
so I just have minus four b to this part and here i'll have four c so i've got minus 28 is equal to
minus 4b plus 4c there's my equation there it is now I know what c is c is actually equal to one
so that'll be that'll become four moving you over I get minus 32 is equal to minus 4b
or minus 4b is equal to minus 32 or b is equal to eight so that is the solution for b so I've got
c and I've got b now I need to find a so again I choose a sensible value for s that I haven't used
before so I'm just going to set s equal to one so my left hand side becomes minus 25 three times
one minus 28 three minus 28 minus 25 there it is with s equal to one I'll get a plus b times minus
three so I'll have minus 3a minus 3b s equal to one will give me plus 5c so there's my equation down
here I know what c is and I know what b is and that will allow me to find a and when I do that
I'm going to substitute the b and the c values into this equation here fairly easy I find that a is
equal to two so I've got a b and c and I can substitute them back into my partial fracture
representation up here and I obtain the expression at the foot of the slide so this is the thing that
I'm going to now try and find the partial fracture representation of I'm going to try and find the
in-virtual-a-plus transform of this is the partial fracture representation of this original
expression which no matter how much I looked in the tables I would never find an expression that
looks like this but I hope to find some terms that look like the partial fraction representation
that I've got so if I move over to the next slide I've got this thing here now this is an easy one
this again is number five one over s plus alpha that one there in this case alpha minus four so it'd
be e to the four t so that one's easy to deal with this one isn't quite as easy s squared plus four
well I would search my table it's s squared plus four just be careful it's not this one
because this this is s plus a number all squared whereas this one is just s squared plus a number
so sometimes people do actually choose this one but it's different that is different this is
the new denominator everything is squared in it here only the s squared so I would keep looking
keep looking and I would eventually hit number eight and nine s squared plus some number that's
called omega squared s squared plus omega squared now number eight has that the denominator so two
actually does number nine so the decision is which one of these am I going to use well I look at the
numerator and again unfortunately my numerator has more terms in it it doesn't exactly match either
but I can actually be a bit clever with this one and it's perfectly okay to split up the
numerator and have this thing as a sum of two fractions two s over s squared plus four plus
eight over s squared plus four okay so this should help me now I will now go and just check these
with the tables I know let's have a look at this one first of all s squared plus a number so it's
either going to be number eight or nine we've certainly decided on that what's in the top line
well this one the Laplace variable s appears in the top line here so that will tell me that it should
be number nine it's number nine not number eight because number eight only has a number
there is no s in the numerator in number eight so it's definitely number nine and I can take the
two outside if I like so I've just got to get this so that's pretty easy that's just number nine
with omega equal to two so it's just cos two t and don't forget the two on the outside so I've got
that one I did this one earlier on you know just one over s minus four so it's number five with alpha
equal to minus four so it's e to the minus minus four t it's just e to the four t this one
this one here I need to do something with this one well it's s squared plus four so we knew from
earlier this was going to be number eight or number nine again I examined the numerator now to decide
which one of those two candidates I should use well we know number nine has still a plus variable
in the top line and this one doesn't this one only has a number a constant so it's definitely number
eight but we have to be careful the omega down below here is two because it's two squared it's
omega squared so two squared is four so omega is equal to two so I must have a two in the top line
so there's the two there so you know I could write this I could have my fraction outside here
eight the thing that I want to divide by what the table want me to have which was two so that would
give me the four so I kind of jumped a step there so but that's the four there so I can then easily
invert this to give me sine two t because omega is equal to two and that's become four sine two t
and there is the investor class transform of that function that I started out with which looked
pretty hopeless really at the beginning because there was nothing in the table that looked
remotely like that but partial fractions allowed me to write this expression in a form that I could
use the tables with to invert okay so that's fine now in all these cases and most these
kids were able to factorize the denominator if we needed to there are some cases though
where we're unable to factorize the denominator easily without complex numbers or whatever we
really don't want to do that so the way we go about it about these is we complete the square
now we could very easily we could just as easily have complete the square in these
when we're able to factorize we can either factorize or complete the square either method will work
but if they don't factorize easily that avenue isn't available for us so all we can do is complete
the square so let's have a look at completing the square so there's our expression here so
you know how do I know this thing won't break up into two brackets s plus or minus something times
s plus or minus something well if I try to solve this thing equal to zero I might use the quadratic
formula to find the value of s and if I did that my what's known as the discriminant the quantity
under the square root sign in the quadratic formula b squared minus 4ac as you can see in this case
it works out to be negative 64 so I'm taking the square root of a negative number so that brings in
imaginary numbers and brings in complex numbers of course so this will not factorize easily
you will not get this s plus or minus an integer times s plus or minus an integer like we saw
previously that will not work so what we do is we complete the square here so we've got s squared
minus 6s plus 25 to complete the square who we do is we take the s here we take half of the
coefficient which is three from minus s we copy the minus down we have that all squared we subtract
the term we've got in here subtract the three square it don't forget the plus 25 now I'm sure
everyone had seen this before so I won't go into too much detail really on because it's in some of
the earlier maths courses so when I simplify this I get s minus three or squared plus 16 okay now one
thing to be careful here of here is this minus this minus three squared this is this up here shows
this is complete these two things minus three squared like we've got is minus three times three
which is minus nine it's completely different from minus three all squared which is minus three times
minus three which gives positive nine so make sure you know the distinction between these two things
this is subtract three squared in other words minus three times three and that's why
that would be minus nine plus 25 and you get 16 here so that's us complete the square so this
original expression here we can factorize into this expression like this we can then go to our tables
and you know we would look down and this certainly you know here you can think of this as s plus alpha
plus omega squared with alpha equal to minus three and omega equal to four so it's going to be one of
those two the first one though number 14 won't work because that's just a constant so this one
just similar to what we saw earlier where the cosine has a constant in the top line the sign
just has a constant it's going to be the cosine so it's going to be the cosine with alpha equal to
minus three see i've got s plus alpha top and bottom these must match these things must match
each other and they do i've got s minus three in both places here so alpha is equal to minus three
and omega which omega squared four square so omega is equal to four so i can easily write this as
e to the minus minus three t or e to the three t cause omega t becomes cos four t so that's the
initial plus transform of this expression here and again you can look at the table and you will
not find something that looks like this so you've got you've got no choice but to complete the square
the denominator does not factorize easily if it did we could use partial fractions but it doesn't
so we complete the square in the denominator move on to another example
maybe i'll write out the complete the square here i know i've done it on you can see it on screen there
but let's just see if i can maybe write it out let's get the pen going so i've got s squared plus
two s plus five so the rule for complete the square is i have my s there i have a square there
i'll subtract something and i've got plus five so what goes in here well that's a plus so i'll
copy that down half of that gets copied so there it goes and then whatever i write in here i subtract
it squared so this becomes s plus one all squared minus one plus five gives me plus
four like that which of course is equal to s plus one all squared plus two squared and i only write it
like that because i know that's the form that's needed for the tables so let's just compare them
so i look at the denominator i've got s plus one all squared plus two squared s plus one all squared
plus two squared so it's either going to be 14 or 15 let's now look at the numerator to narrow the
field down even further well the numerator is just a number so it's going to be number 14
15 won't work because 15 has the Laplace variable in there it's got the s in the numerator 14 is
perfect because it's just got a number but if you look at this carefully we don't quite have this in
the same form because the table tells me that whatever the omega here is and i must have exactly
the same omega value up here so i've got a two down here so that means i must have a two in the
numerator because look there's omega is equal to two and i'll have two squared there which is actually
what i've got so i need a two in the top line and that's all i've done here so this in here is
exactly what the table has me used so basically you know if you look at it i just clear this first
of all if you look at this this is of the form omega in the top line over s plus alpha squared
plus omega squared that's exactly in the form that i want to use number 14
so i've introduced a two in the top line to enable me to use number 14 but i must correct that because
that's not what i started out with and we know how to do that you multiply by a fraction this top
line has the thing that you want which is a one remember it's a one that we want in the top
i would divide by whatever the table forced me to introduce and that was a two i had to introduce
a two here because that omega had to match the omega down below so i multiply this by by a half
and then i can easily invert this bit because it's just number 14 with alpha equal to one
and omega equal to two so it gives me remembering the half it gives me a half e to the minus t
sine 2t it's just that one with alpha equal to one give me e to the minus one t omega equal to
sine 2t okay so another example of completing the square so what do we got so i think i'll
just write this one out as well so we've got s squared s squared minus six s plus 25 i can
break this into my brackets there's an s there there's a square there and a minus something
squared and don't forget the plus 25 there's a plus here so that's got to be a plus take half
of that coefficient so it gives me a three subtract the thing i've written in there so it's three
squared if you work that out you get s plus three all squared minus nine plus 25 so it's plus 16
or in other words s plus three all squared plus four squared there's a plus there okay
so that's exactly what i've got here okay i've just completed the square in the denominator
that's all i've done that's just this calculation going from left to right here that's a june blue
there the numerators are seven so let's look at the table check the denominator and i would hit
number 14 and 15 so it's certainly of the form s plus alpha all squared plus omega squared
doesn't matter that's negative just think of it as s plus minus three so it's s plus alpha all
squared plus omega squared so it's one of those two number 15 however though has the Laplace variable
has the s in the numerator our one doesn't it's just constant and that's what number 14 gives just
in the previous example so it's going to be number 14 but number 14 as i said with alpha equal to minus
three omega equal to four so that must mean that i need a four in the top line there to match the four
down here there's omega this is omega squared so omega's four so by number 14 i need an omega equal
to four up top there so i'll write that in and i'll multiply by my fraction to make sure that i'm not
changing things the fraction will have the thing that i want in the top line which is a seven and i'll
divide by what the table forced me to introduce which was a four because omega is equal to four
i had to write a four in the top line the table forced me to do that so the denominator here
must be a four the thing that i want on top which is seven down below the thing the table forced me
to introduce so i can then easily use this i've got seven over four times the inverse of this
this is just number 14 with alpha equal to minus three omega equal to four so i've got seven over
four e to the minus minus three t or e to the three t sine omega t omega's four so it's sine
40 so that is the answer there okay so that concludes this lecture so if we just have a quick
look at what we've done so we've introduced the inverse Laplace transform so you should now be
able to calculate the inverse Laplace transform of standard functions using the tables you should
be able to apply the first shifted theme to calculate the inverse Laplace transforms and you should
also be able to apply algebraic techniques which include partial fractions we looked at three different
types of partial fractions and also complete the square you'd be able to apply these methods to
calculate inverse Laplace transforms in the next class we're going to look at taking Laplace transforms
of derivatives and how this can be applied to solving basic ordinary differential equations
using Laplace transforms so just to have a quick look at what you should now be able to do
as far as the Mobius questions go you should be able to do questions one to nine and the tutorial
questions from the notes you should attempt question one parts one to eight and you should now be able
to do question two as well so that's the end of this lecture so we'll leave it at that for now
okay bye