O.J. said:
I can't help but keep asserting how me textbook of differential equations just keeps stating new ideas without any background on WHY and HOW they were created. I've just started Laplace transform and the first thing i see is a blue DEFINITION box that just states the tranform equation asking us to memorise it. Can you please explain WHY is that specific function so important and HOW did they figure it out?
okay, OJ, i'll try to give you an answer where other folks here declined to. this is the pedagogy that electrical engineers have with the Laplace Transform, it may not be congruent to the actual history.
it starts out with Fourier series. using a sum of sinusoids to represent a periodic function:
x(t+T) = x(t) \quad \forall t
Fourier thought that sines and cosines having the same period
T might be able to add up to the same
x(t):
x(t) = \frac{a_0}{2} + \sum_{n=1}^{+\infty} a_n \cos( 2 \pi n t / T ) - b_n \sin( 2 \pi n t / T )
or
x(t) = \frac{a_0}{2} + \sum_{n=1}^{+\infty} g_n \cos( 2 \pi n t / T + \phi_n )
or
x(t) = \sum_{n=-\infty}^{+\infty} c_n e^{ i 2 \pi n t / T }
if you fiddle around with trig identities and complex numbers, you can see that the three expressions are the same (with some mapping between
an, bn and
gn or
cn). The latter expression is the most convenient; it's pretty easy to solve for
cn :
c_n = \frac{1}{T} \int_{t_0}^{t_0+T} x(t) e^{ -i 2 \pi n t / T } dt \quad \forall t_0
note that
cn is not a function of
t and indicates the strength of the frequency component at frequency of
n/T. now if you plug that
cn back into the
x(t) expression (and change the letter for the dummy variable of integration, you get:
x(t) = \sum_{n=-\infty}^{+\infty} \frac{1}{T} \int_{t_0}^{t_0+T} x(u) e^{ -i 2 \pi n u / T } du \quad e^{ i 2 \pi n t / T }
this is a self-contained expression for a periodic
x(t) of any finite period
T, so, if you want to represent a function that is not periodic, you can let
T get very large (like a year), and if you're careful with your Riemann sums, you can show that they become a Riemann integral. first define the arbitrary
t0 = -T/2,
x(t) = \sum_{n=-\infty}^{+\infty} \int_{-T/2}^{+T/2} x(u) e^{ -i 2 \pi n u / T } du \quad e^{ i 2 \pi n t / T } \frac{1}{T}
as T \to \infty, the
1/T becomes a differential amount of frequency and the summation becomes this integral:
x(t) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x(u) e^{ -i 2 \pi f u } du \quad e^{ i 2 \pi f t } df
or
x(t) = \int_{-\infty}^{+\infty} X(f) e^{ i 2 \pi f t } df
where
X(f) = \int_{-\infty}^{+\infty} x(t) e^{ -i 2 \pi f t } dt
so, instead of representing
x(t) as a sum of discrete frequencies of
n/T that are getting closer and closer to each other as T goes to infinity, we are representing it as a continuoum of frequencies that are infinitesimally close to eacy other. that is the
Fourier Integral.
there are some functions that the Fourier Integral does not do to well with (doesn't converge nicely), like the Heaviside unit step function, so we generalize the purely imaginary frequency term in the exponent i 2 \pi f by adding a little real part to it to make the integral converge.
\int_{-\infty}^{+\infty} x(t) e^{ -(\sigma + i 2 \pi f) t } dt
if you define this complex value as s = \sigma + i 2 \pi f or s = \sigma + i \omega, the above integral is the so-called two-sided Laplace Transform and it converges for some functions that the Fourier Transform does not.
X(s) = \int_{-\infty}^{+\infty} x(t) e^{ -s t } dt
That's the motivation behind the definition of the Laplace Transform.
