Laplace Tranform Notes Kerala University Semester 6
9 views 0 purchase
Course
MM1
Institution
UNIVERSITY OF KERALA
This Note covers all the topics on Laplace transform Sem 6 kerala university .Examples also provided with their solution to get a hands on experience on the topics . i think this would be helpful for students who are currently pursuing bsc maths .
Sumesh S. S.*
Assistant Professor
Department of Mathematics
Mar Ivanios College, Thiruvananthapuram.
Contents
1 Laplace transform 1
2 Laplace transform of some elementary functions 3
3 Transforms of Derivatives and Integrals 14
4 Unit Step Function (Heaviside Function) 25
5 Unit impulse function 30
6 Periodic Functions 31
7 Convolution 33
8 Differentiation and Integration of Transforms 38
9 Inverse Laplace Transforms 46
1 LAPLACE TRANSFORM
Rb
A relation of the form F(s) = a k(s,t) f (t) dt which transforms a given function f (t) into
another function F(s), is called an integral transform. Here K(s,t) is called the kernal of the
* +919895050982
1
, 2
transform and F(s) the transform of f (t). The most common integral transforms are
(i) a = 0, b = ∞, K(s,t) = e−st , (Laplaces)
1
(ii) a = −∞, b = ∞, K(s,t) = √ eist (Fourier)
2π
The idea behind any transform is that given problem can be solved more easily in the trans-
formed domain. Laplace transform reduces the problem of solving a differential equation to an
algebraic problem. It is widely used in problems where the
Definition 1.1
Let f (t) be a function defined for all t ≥ 0. The Laplace transform of f (t) is defined as
Z ∞
L [ f (t)] = e−st f (t) dt = F(s), (1)
0
provided the integral exists. Here s is a parameter real or complex. The function f (t)
whose transform is F(s) is said to be the inverse transform of F(s) and is denoted by
L −1 [F(s)]. Thus if L [ f (t)] = F(s), then f (t) = L −1 [F(s)].
EXISTENCE OF LAPLACE TRANSFORM
A function f (t) is said to be of exponential order or satisfies growth restriction if there exist
constants M and a such that | f (t)| ≤ M eat for all positive t.
Theorem 1.1 (Existence Theorem for Laplace Transforms)
If f (t) is defined and piecewise continuous on every finite interval on the semi-axis t ≥ 0
and is of exponential order for all for all t ≥ 0 and some constants M and a, then the
Laplace transform of f (t) exists for all s > a.
Proof. Since f (t) is piecewise continuous, e−st f (t) is integrable over any finite interval on the
Sumesh S S sumeshsmath@gmail.com
, 3
t-axis. Also f (t) is of exponential order so that for s > a we get,
Z ∞
|L [ f (t)]| = e−st f (t) dt
0
Z ∞
≤ |e−st f (t)| dt
Z0∞
= e−st | f (t)| dt
Z0 ∞
≤ e−st Meat dt
0
Z k
= M lim e−(s−a)t dt
k→∞ 0
" #k
e−(s−a)t
= M lim
k→∞ −(s − a)
0
" #
e−(s−a)k 1
= M lim −
k→∞ −(s − a) −(s − a)
M
= .
s−a
Hence L [ f (t)] exists for s > a.
I Note that Z ∞ Z k
f (x) dx = lim f (x) dx
a k→∞ a
provided, the limit on the right side exist.
LINEARITY OF THE LAPLACE TRANSFORM
The Laplace transform is a linear operation. i.e. for any function f (t) and g(t) whose Laplace
transform exist and any constants a and b,
L [a f (t) + bg(t)] = aL [ f (t)] + bL [g(t)].
Proof. By definition,
Z ∞
L [a f (t) + bg(t)] = e−st [a f (t) + bg(t)] dt
0Z
∞ Z ∞
−st
=a e f (t) dt + b e−st g(t) dt
0 0
= aL [ f (t)] + bL [g(t)].
2 LAPLACE TRANSFORM OF SOME ELEMENTARY
FUNCTIONS
Sumesh S S sumeshsmath@gmail.com
, 4
I Note that x→∞
lim e−x = 0 and lim e−ax = 0 for any a > 0.
x→∞
1
(i) L (1) = , s > 0.
s
Proof. By definition,
Z ∞
L (1) = e−st 1 dt
0
Z k
= lim e−st 1 dt
k→∞ 0
−st k
e
= lim
k→∞ −s 0
−sk
e 1
= lim +
k→∞ −s s
1
= 0+
s
1
= , s > 0.
s
1
(ii) L (eat ) = , s > a.
s−a
Proof.
Z ∞
L [e ] =
at
e−st eat dt
Z0 ∞
= e−(s−a)t dt
0
" #k
e−(s−a)t
= lim
k→∞ −(s − a)
0
" #
e−(s−a)k 1
= lim +
k→∞ −(s − a) s−a
1
= if s > a.
s−a
n!
(iii) L (t n ) = where n is a positive integer.
sn+1
1
Proof. We prove this formula by induction. For n = 0, t n = t 0 = 1 and L (1) = . Thus
s
the result is true for n = 0. We now make the induction hypothesis that it holds for any
n!
positive integer n. i.e. L (t n ) = n+1 .
s
Sumesh S S sumeshsmath@gmail.com
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller ajaybabu. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.99. You're not tied to anything after your purchase.
