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Okay Good morning friends, we shall continue our discussions on Laplace transform. Yesterday,
we derived a transforms of some standard functions will continue with that before we go to the
set of other properties I just like to mention there are function f (t) is Laplace transformable
if f(t) mod e to the power minus sigma t dt, 0 to infinity is a finite quantity that is
it is less than infinity if you have a function say f (t) equal to say t squared or t cubed
will this property hold good for this f (t).
Let us see, f (t) in to e to the power minus sigma t as t10 ds to infinity, how much is
this product t cubed e to the power minus sigma t okay limit t10 in to infinity, so
what will be the limit. I can always write this as e to the power sigma t, I can write
as 1 by e to the power sigma t so 1 plus sigma t sorry, 1 minus 1 plus sigma t plus sigma
squared t squared by factorial 2 and so on and the t cube on the numerator can be brought
back here, it can be brought down here. So this will be 1 by limit t10 in to infinity
1 by t cubed plus sigma by t squared plus sigma squared by factorial 2 t all right like
that plus sigma cubed by factorial 3 t cube will get cancelled and after that you will
get sigma 4 by factorial 4 t and so on okay. Now how much is this coming to 1 by t cube
it will be 0, this will be 0, this will be 0, this will be finite. So it would be equal
to 1 by sigma cubed by factorial 3 plus this will be infinite terms multiplied by t so
it will be equal to 0 this is may be t 10 to infinity, so this will be 0.
so a function now when t tends to 0, when tends to 0 this is a infinity, this is a infinity,
this is a infinity, these are all 0's so 1 by infinity that is again 0. So what will
this function look like t cubed is a function which is going up like this to be parallel
alright and t cubed in to e to the power minus sigma t, e to the power minus sigma t is a
function going like this. So the product will be starting from 0 ending at 0 so somewhere
it may have a peak and then it will go like this.
So the area will be finite this is going to 0, so area is finite. So I have just taken
an example with t cube it can be show on that for any t to the power n it will be always
converging to finite value and hence the integration from 0 to infinity will also give you some
finite sum it will be less then infinity, so it is transformable. Functions of this
kind, so e to the power minus or plus a t squared if f (t) is of this kind then you
will find again by taking the limits this will not tend to a finite value. So this is
not transformable but any other functions set e to the power n are functions of this
kind they will be transformable.
So for any transformable functions you have to check whether this property holds good
or not, f (t) may go to infinity but f (t) in to e to the power minus sigma t this integration
this entire product if it goes to a finite value then and the integration if it gives
you a finite value then it is transformable. Next will take up some properties of Laplace
transform, Laplace transform of shifted function. Last time we discussed about some shifted
functions how to represent them suppose f (t) is a function like this then f (t) minus
tau in to u t minus tau is basically this same function starting after the interval
tau okay.
So what will be the Laplace transform of this in terms of Laplace transform of this function
okay, what will be the Laplace transform of this function in terms of the Laplace transform
of this function, you would like to find out that. Now Laplace transform of f (t) minus
tau u t minus tau will be equal to integration 0 to infinity f (t) minus tau u t minus tau
e to the power minus s t dt okay, by definition. So that gives me equal to 0 to infinity now
between 0 and infinity if I take a function now the function starts after tau that means
it is 0 here. So the integration from 0 to infinity means tau to infinity so I can break
it up in to 0 to tau f t minus tau, u t minus tau e to the power minus s (t) dt plus tau
to infinity f (t) minus tau, u t minus tau e to the power minus s (t) dt and this part
is 0, 0 to tau is equal to 0 plus tau to infinity t minus tau u t minus tau, e to the power
minus s t dt.
Now t minus tau I can put another variable t1 so dt is equal to d (t1) and t is equal
to t1 plus tau, so replace t by t1 plus tau, so when t is tau when t is tau t1 is 0, so
this limit become 0 to infinity f (t1) u t1 actually f (t1) u (t1) from 0 it does not
mean much here I can break it up in to e to the power minus s (t1), e to the power minus s tau,
e to the power minus s tau can be taken out because the integration is with respect to
t1, so thank you very much 0 to infinity f (t1) u (t1), e to the power minus s (t1) d
(t1) and what is this f of s. So in terms of the original transform the shifted function
has a transform that is e to the power minus s tau in to f (s) okay. So now let us derived
some of those the Laplace transform of some of the shifted functions, we discussed yesterday.
Let us take a simple example, suppose this is A, T this we wrote as equal to a function
like this plus function like this minus A and plus A, is it not? So what will be the
Laplace transform of the first function A by s this one A by s shifted by T, so multiplied
by e to the power minus s t just now we have seen that any function if it is shifted by
tau so the original Laplace transform multiplied by e to the power minus s tau, this will be
the result. So this is A by s in to 1 minus e to the power minus s t okay. Now suppose
we take T very very small T very very small 10 in to 0 and A is 10 in to infinity such
that A in to T is 1 we get a unit impulse, is it not? When T tends to 0 and A tends to
infinity but the product is finite and in this case if it is 1 then what will be this
this will tend to for this f (t) is nothing but delta t okay.
So this value are f (t) so Laplace transform of delta t we want to compute will substitute
that condition here A by s in to 1 minus what is e to the power minus s t, 1 minus s t plus
s squared T squared by factorial 2 and so on okay. Now let me simplify here itself 1
will go A by s in to s in to T then will find 1 s will go s t squared and so on okay, A
by s, s t minus s squared T squared by factorial 2 and so on. So that gives me equal to AT
minus sorry As t squared by factorial 2 plus A s squared T cube by factorial 3 and so on
and this AT is 1 and this is AT in to T, T tends to 0 means this may be all 0 okay T
tends to 0. So Laplace transform of delta T is unity okay what would be the Laplace
transform of a periodic function. So let us take periodic function like this and so on
so this is of magnitude 10 this is 1, 2, 3, 4 and so on.
So let us take first of all the the first block which is repeated after every period,
so what is the Laplace transform of this first block, could you please tell me? It will be
10 by s in to 1 minus e to the power minus 1 s is it not tau is 1. Let us call it F1
s then what would be f (s) the same thing being repeated, so it would be F1 (s) plus
you F1 (s) is repeated after 2 seconds so F1 (s) in to e to the power minus 2 s plus
F1 s in to e to the power minus 4 s and so on up to infinity okay. So F1 s if I take
common will be 1 plus e to the power minus 2 s plus e to the power minus 4 s and so on
which means F1 (s) in to its a geometric series common ratio is e to the power minus 2 s.
So 1 minus e to the power minus 2s, now F1 (s) is already obtained substitute that what
you get F1 (s) equal to already known so just repeating it here therefore f s will be 10
by s 1 minus e to the power minus s, 1 minus e to the power minus 2 s okay e to the power
1 minus e s goes out, is that all right.
Let us take another example, periodic function once again like this this is 10 this is minus
10 okay 1, 2, 3, 4 and so on what would be the Laplace transform of this function? Can
you write in 1 row, what is this? F1 (s) is see if it is a same block being repeated with
alternate signs then there is no problem I can take this as the primary block, first
initial block whose Laplace transform is already known then what will be f (s) will be F1 (s)
in to 1 then minus e to the power it is shifted by 1 second and then its sign is changed so
e to the power minus s plus e to the power minus 2 s minus e to the power minus 3 s plus
e to the power minus 4 s and so on, is that all right. So how much is this common ratio
is now minus e to the power minus s so it will be F1 (s) by 1 plus e to the power minus
s, so that is 10 by s1 minus e to the power minus s by 1 plus e to the power minus s very
familiar form know what is it?
Can I can I not take e to the power s by 2 multiplied throughout by e to the power s
by 2, so it will become e to the power s by 2 minus e to the power minus s by 2 divided
by e to the power s by 2 plus e to the power minus s by 2. So this will be not 2 j why
it j I am not putting j omega and all that this is simply cos this is sin or sin and
cos, so it will be 10 so it will be 10 by s tan hyperbolic s by 2.
Okay another example, if I have a pulse train 0, 2, 4, 6 and so on this is delta t, this
is delta t minus 2 and so on. So what would be f s for this for the first one it is 1
for the next one it is 1 in to e to the power minus 2 s plus e to the power minus 4 s and
so on so it will be 1 by 1 minus e to the power minus 2 s is that alright if I have
alternately coming say plus 1 minus 1 delta t minus delta t plus delta t minus delta t,
we shift then only change will be here okay plus, very good. Now you have another important
property suppose Laplace transform of f (t) is given as f (s) what would be the Laplace
transform of f dash t, where f dash denotes the first derivative what will it be. So by
definition this will be 0 to infinity d dt f (t) e to the power minus st dt.
So if you integrate by parts alright this you can take as integration of u dashed v
is u v minus integration u v dashed okay. So if I take this as this as v okay this can
be differentiated this can be integrated so it will be f (t) if I integrate this in to
v, e to the power minus s (t), 0 to infinity minus derivative of this will become minus
s in to e to the power minus s t and integration of this is f (t) dt so this 1 will give me
if I put infinity this becomes 0 minus if I put 0 it will be minus f (0) minus and minus
will make it plus and integration of f (t) e to the power minus s t dt is f (s). So it
is s in to f (s) where f (s) is the Laplace transform of original function f t minus f
at 0 okay.
Now we shall derive the Laplace transform of integrals of functions from here. You can
go from the basic definitions there is another way of looking at it. Let us take f (t) as
some f0 (t), df (t) by dt as df0 by dt as some f1 (t) therefore d square f (t) by dt
squared is df1 t by dt as f2 (t) okay. So f0 (t), f1 (t), f2 (t) mean they are all successive
derivatives and corresponding Laplace transforms we denote as capital F0 (s), F1 (s), F2 (s)
and so on. So just now we have derived F1 (s) is what derivative F1 (t) its Laplace
transform is F1 (s), so it will be in terms of F0 (s), f0 (s) minus f0, f0 at 0 is that
alright.
Similarly, F2 (s) will be s in to F1 (s) minus f1 at 0 if I substitute F1 (s) here it will
be s square f0 (s) minus s minus f10 that means if I take the nth derivative and its
Laplace transform will be in terms of the original Laplace transform s to the power
n sorry this n minus s to the power n minus 1 f00 minus s to the power n minus 2 f10 and
so on minus s to the power 0 when I am calculating fn then fn minus 1 0 okay is just by induction.
We have gone downward from f0 derivative is f1 next derivative is f2 etcetera if you go
up it will be integration all right. So these equations can be just a little bit of manipulation
I can write on this side F1 s is s f0, so F0 (s) is how much in terms of F1 it will
be F1 (s) by s plus f00 by s okay.
So sorry, if you are given the Laplace transform of F1 in terms of that that is f1 (t) is known
in terms of that can you compute the Laplace transform of its high order function by high
order I mean the integral lower order means derivative, so F0 s that is the integral of
f1 (t) what will be its Laplace transform. So F0 (s) will be the Laplace transform of
the original function divided by s. Now plus what is f00 the integral evaluated at t equal
to 0 what does it mean? So it is like this if you have a current expression for example,
current is for example, in a capacitor 1 by c integral I dt, 0 to t is the voltage at
any instant if I take if I have to take the Laplace transform of this given the Laplace
transform of I (t) as I (s) what will be v (s), v (s) is integral of I is it not. So
if I have to compute v (s) it will be 1 by c in to integral of this means original Laplace
transform divided by s then plus plus what is it I0 what is f0 it is integral of that
so integral of I dt okay divided by s and this is to be evaluated at 0 from where do
we start, so 0 minus to 0 plus or basically whatever has been there before the counting
of time before t equal to 0 so that gives me integral of I dt is what charge q, is it
not?
So integral I dt means this is to be evaluated at t equal to 0 means the initial charge q0
alright mind you it is not initial value of i initial value of integration of i, when
you take the integration so initial value of not f1 f0 means integration of f1 alright.
If the initial condition is given as 0 then there is no problem, so 1 by c s in to I (s)
plus if I call it q0 by c s, q0 by c is nothing but initial voltage across the capacitor,
so V0 by s okay. So when we evaluate for example when we evaluate the voltage across a capacitor you are energizing
by this that can be a resistance also you want to measure the voltage across this capacitor
then its initial voltage plus at any instant if current is It if you want to deal with
this problem in the Laplace domain then I (s) in to 1 by c s this total sum will be
the voltage across the capacitor, initial voltage across the capacitor plus the drop
due to this is the total voltage at any instant.
Now we will take up 1 or 2 small examples, then again will come to some other properties
what would be Laplace inverse of say 1 by s in to s squared plus omega square 1 by s
square plus omega squared what is the Laplace inverse, if you remember this is sin omega
t what is the Laplace transform of sin omega t omega by s square plus omega square, so
omega is not there so I will just write sin omega t by omega okay and just now we have
seen if the initial conditions are not given otherwise mention say it will be treated as
0. So what would be therefore Laplace inverse of 1 by s in to s squared plus omega square
1 by s in to s square plus omega square it will be integration 0 to t, sin omega t by
omega sorry dt, is it not? So that will give me 1 minus cos omega t by omega okay 1 comes
because you are putting limit on sin cosine omega t. So this there is another omega so
while integrating you will get omega squared sin omega t will give me another omega. So
1 by omega square 1 minus cosine omega t is that okay one may do it by partial fractions
I can write this as A by s plus b s by s square plus omega square. Let us see how much is A multiply by s put
s equal to 0.
Let it be 1 by omega squared how much is B multiply by s squared plus omega square divide
by s, so will be and then put s square plus omega square equal to 0. So if I am multiply
by s square plus omega squared this will go on the left hand side will be 1 by s then
this will be s square plus omega square in the numerator and here it will be b s and
then if I divide by s this will be s square this will be s square. So it will be 1 by
s square on this side on this side it will be A in to s square plus omega squared plus
B and now, I am making s square plus omega square equal to 0.
So that means 1 by minus 1 by omega square all right that will be B. So this one will
be 1 by omega squared in to 1 by s minus B means 1 by omega square. So s by s square
plus omega square. Again, if you take the Laplace inverse it get the same result, so
you can do it by either method alright making use of the property of integrating a function
or making partial fractions. Similarly, you can solve for1 by s square in to s square
plus omega square. So if you know the result of this once again divide by s means once
again an integral function, so you integrate this one again you will get the Laplace transform
partial fraction will be simpler, is it not? will be 1 by s square minus 1 by s square
plus omega square. So that will give me s square plus omega square minus minus s square.
So I will divide by omega square so corresponding inverse will be 1 by s square is Laplace transform
form of what ramp function t in to u t okay and this 1 1 by s square plus omega square
sin omega t by omega so it will be 1 by omega square t u t minus 1 by omega cube sin omega
t okay. I need not write u (t) all the time I told you if you write t so long as you understand
if it is a ramp function it is okay. Another important property will derive now so by definition
we have f (t) e to the power minus s (t) dt. The common term is s common variable is s
if I differentiate with respect to s what do you get if you differentiate with respect
s it will be t in to e to the power minus s (t) with a negative sign f (t) dt and what
is this Laplace transform of t, f (t) with a negative sign.
So Laplace transform of t in to f (t) is minus d f (s) by ds okay minus d f (s) by ds again
we take a few examples, what will be Laplace inverse of this 1 by s square plus B squared
if I take the derivative with respect to s what is it, derivative of this 1 by s square
plus beta beta squared yes, could you please tell me 2s minus 2s okay. So if I put a beta
here all right okay t into this is corresponding to sin beta t is it not t in to sin beta t,
t in to sin beta Laplace transform of t in to sin beta t is how much then t in to sin
beta t, t in to f (t) is d f (s) by ds with a negative sign alright.
So this is corresponding to sin beta t, so t in to sin beta t should be equal to derivative
of this with respect with an negative sign, so derivative is this. So this is nothing
but 2 beta s by s square plus beta square whole square is it not is that alright t in
to cosine beta t what will the Laplace transform cosine beta t is s by s square plus beta square
if you take the derivative of this how much is it? So d ds of with negative sign d ds
of s by s square plus beta squared how much is that d ds of s by s square by beta square
s square plus beta squared in to derivative of this minus 2 s squared divided by s square
plus beta square whole squared.
So that gives me s squared minus beta square divided by s square plus beta square whole
square. Okay the other 1 is t in to sin beta t2 beta s by s square plus beta square whole
square okay. So I can make any combination t in to cosine beta t plus minus t in to sin
beta t Laplace transform of this will be s square minus b is beta squared by s square
plus beta square whole squared plus minus 2 beta s by s square plus beta square whole
square okay.
Now you already know beta by s square plus beta square corresponds to sin beta t and
this one corresponds to cos beta t. So by making manipulations there are 4 results now
1 is this plus this this minus this so I have already got 4 such Laplace transforms by manipulating
you can find out the Laplace transform of this because I have break it up in to see,
if I make partial fractions or manipulate these I can get the inverse of this I leave
it as an exercise you do it yourself okay, not very difficult. Next we take up application
of Laplace transform in simple network problems, suppose you consider a simple RL circuit there
is a DC source of voltage V and the switch is put on at t equal to 0 initially this is
uncharged what would be the expression for the current i (t) you write V you are suddenly
switching on. So it will be v u (t), V is a magnitude of voltage you are applying a
step voltage and if I is the current when it is Ri, I am not writing i (t) it is understood
i is a time varying quantity l di by dt if I take Laplace transform on both sides it
will be V by s on this side it will be r is plus l di by dt and what is the Laplace transform
of the derivative function s times I (s) minus I (0) if I assume the initial condition to
be 0 then I0 is 0 so that will not come.
So it is R plus sL in to I (s) alright so like you are having voltage equal to impedance
in to current in simple AC circuit. Here, also the applied voltage in the transform
domain this is equal to R plus sL in to current in the time domain Laplace domain. So in the
transform domain I will write V is 0 excuse me v0 mean a specific value and v (s) is the
Laplace transform of v (t) in this case it is V0 in to u (t) okay. So v (s) by I (s)
that means in the transform domain voltage function by current function is equal to R
plus sL in the transform domain. So a resistance therefore these circuit I will write as v
s the voltage source and an impedance which is R plus sL and the current is I (s) so an
impedance is given by R plus sL for R and L in a similar manner we can show if there
is an if the there is a capacitance, if there is a voltage source here v (t) Ri plus 1 by
c integral I dt is equal to the applied voltage if I take the Laplace transform will be R
in to I s plus 1 over c in to s in to I (s) once again if I assume the capacitor to be
initially uncharged then that q0 by c (s) will be 0, so this side it will be v (s) or
v (s) by I (s) will turn out to be R plus 1 by c (s).
So if it is an inductive element it will be R plus sL if there is a capacitive element
present here then it will be equivalently represented by 1 by sc. So if you have RLC
wherever there are resistances you replace it by you return it that is resistance should
be returned as it is inductance will be replace by sL capacitance will be replace by 1 by
s c and in all network configurations next part will be using the transform domain variables
are sl1 by s c sorry if you do this then any network problem can be solved. Let us take
1 or 2 more examples, you have a source here a resistance here. Okay therefore, we shall
replace this entire circuit by an equivalent v1 (s), R1 you can show it like this or a
symbol like this can also be used for a generalized impedance z (s) it can be R it can be 1 by
sL, 1 by sc it can be sL or any combinations this is R2 plus sL2.
Similarly, this 1 is R3 plus 1 by sc3 and this side you have got V2 (s). Now for calculating
the current through any branch or potential at any node you can apply supposition theorem,
nodal analysis, thermos theorem, any theorem you want by treating these as simple algebraic
operational impedances all right, treating s as an algebraic operator and treat this
as it is and then when you get the expression for current Is take the inverse you get the
current I (t). So if I am interested in the current through this I2, I2 (s) what will
be I2 (s), we can go for nodal analysis.
So let this potential be V okay VA, so VA minus V1 (s) by R1 plus VA by R2 plus sL2
plus VA minus V2 s by R3 plus 1 by sc3 equal to 0. Now VA, V1 (s) and V2 (s) will be given
to you it may be stay function ramp function or any other function so put those values
solve for VA, VA means VA (s) okay once you know VA (s), then VA (s) divided by R2 plus
sL2 therefore I2 (s) will be VA (s) by R2 plus sL2 okay. So far you have got the solutions
in terms of s now you make partial fractions expand it partial fractions and then take
the inverse you will get the current I2 okay while making partial fractions sometimes you
may come across multiple poles at a particular point say
You want to find out the partial fractions of say the denominator is s plus alpha 1 in
to s plus alpha 2 and so on but s plus some alpha k is having n number of poles that is
3, 4 multiple roots then this can be written as A1 by s plus alpha 1 plus A2 by s plus
alpha 2 and so on valuation of A1, A2 etcetera is very simple straight forward multiplied
by s plus alpha 1 then make s, s plus alpha 1 equal to 0 will get A1 and so on but this
1 will be some ak1 by s plus alpha kn plus ak2 by s plus alpha kn minus 1 and so on.
Evaluation of this becomes a little difficult straight forward if you apply that technique
then you will be able to get only a k1.
So for evaluation of a k1 you multiplied by s plus alpha k to the power n both sides,
so f (s) is multiplied by this and then put s equal to minus alpha k okay. This product
let me call it some q (s) evaluated minus alpha k then Ak2 will be derivative of this
q s and then evaluate at s equal to minus alpha k similarly Ak3 will be 1 by factorial
2, d square q (s) by ds square, second derivative we take and then substitute s equal to minus
alpha k and so on third derivative it will be factorial 3 and so on so you keep on taking
the derivatives of this product and then evaluate the function at s equal to minus alpha k that
will give you successive residues. Once you know that you can expand you can find out
the inverse okay so will stop here for today. Thank you very much you try some other problems
at home and will take up some numerical problems in the next class.
Okay Good afternoon friends, will continue with Laplace transform. Today will have some
tutorial session, will be solving some problems employing Laplace transform. Let us take simple
functions what be the Laplace transform of 1 by 2 a squared sin a t minus sin at, what
be the Laplace transform of this? So this 1 can be written as 1 by 2 a square what is
sin i e to the power a t minus e to the power minus a t divided by 2 minus sin a t. So it
is pretty simple 1 by 4 a square e to the power at will be 1 by s minus a minus 1 by
s plus a is that all right minus 1 by 2 a squared sin a t will give you a divided by
s square plus a square, is that okay? So if I add these 2 it will be s plus a minus s
plus a so twice a, so that will be 1 by 2 a into s squared minus a square minus here
1 a goes, so 1 by 2 a 1 by s squared plus a square.
So 1 by 2 a I can take out it will become 1 by s square minus a squared minus 1 by s
square plus a squared and that gives me s square plus s squared minus s square. So twice
s squared twice a will go, so a by s to the power 4 minus a to the power 4 okay plus 2
times y (s) is equal to 5 by s so it is s squared plus 3 s plus 2 in to y s which is
5 by s. So how much is y (s) 5 divided by s in to s square plus 3 s plus 2 can be written
as s plus 1 in to s plus 2. Okay I can write it as A by s plus B by s plus 1 plus c by
s plus 2 and taking the inverse of these 3, I will get the final y (t) as A into u t plus
B in to u t the power minus t plus c in to u t the power minus 2 t where A B C residues
can be calculated from here by standard partial fraction technique, okay. So we will stop
here for today we will take it up in the next class with further examples using Laplace
transform for network problems. Thank you very much.