3
$\begingroup$

This video (no need to actually watch it) makes a great point. If we interpret the $f$ in $f(x)$ as a function of time, then the fourier transform of $f$ takes the representation of this function in terms of time to a new representation of $f$ simply in terms of frequency, & this makes sense intuitively since a function can be represented, via Fourier series, as a sum of sine functions, & because any sine function is completely determined by it's amplitude, frequency & phase we should be able to completely characterize the function represented in terms of time by a new function in terms of frequency - if we know the amplitude & phase at every frequency in our sum of sines (& this is found by integrating over all time, across the entire spectrum, in the time representation). From this perspective the utility of complex numbers leaps out since amplitude & phase can be represented in a single complex number...

In other words, take a function, represent it as it's fourier series, decompose the coefficients into of the sines & cosines into their integral representations, do the algebra to reduce this to the Fourier transform representation, then if you wish represent this in terms of complex numbers - this decomposes a function represented in terms of time to a new representation in terms of frequency. This whole process can also be achieved by "pre-multiplying" the function by a complex exponential with a purely imaginary argument and integrating.

From this perspective the Laplace transform is merely just the Fourier transform where the complex exponential also has a real argument.

But I just do not see how that extra step falls out of the development I've written above, I don't see where the real argument rears it's ugly headm or what it means.

$\endgroup$
1
$\begingroup$

Ok, now I think that we have a point.

The Laplace Transform $$ X(s)=\int_{0}^{\infty} {f(t)e^{-s t} dt} $$ with: $s=\sigma + j\omega$ , needs to converge to zero to be defined. This doesn't happen with Fourier Transform that is defined independently by the convergence of the integral. Both Transforms represent the function as sum of sinusoid, so the only way to converge the function to zero for $t \to \infty $ is multiply it for an exponential elevated to a value that is greater than its abscissa of convergence (a definite sigma that allows the exponential to go to zero faster that $f(t)$ goes to infinity).

This means that we have the Fourier Transform of $f(t)e^{-\sigma t} $, so: $$X(s)=\mathscr{F}[f(t)e^{-\sigma t}] $$

Infact, making this, we have the function represented by sinusoid that converge to zero for $t\to \infty$, that is the definition of Laplace Transform.

$\endgroup$
1
$\begingroup$

We have Fourier Transform $$ X(\omega)=\int_{-\infty}^{\infty} {f(t)e^{-j\omega t} dt} $$

And the LaplaceTransform $$ X(s)=\int_{-\infty}^{\infty} {f(t)e^{-s t} dt} $$ With: $s=\sigma + j\omega$

They are absolutely the same if $\sigma=0$ $$X(s)|_{s=j\omega} = X(j\omega) $$

In general, the Laplace Transform is simply: $$X(s)=\int_{-\infty}^{+\infty} f(t)e^{-(\sigma+j\omega)t}dt=\int_{-\infty}^{+\infty} [f(t)e^{-\sigma t}] e^{-j\omega t}dt $$ This means that we have the Fourier Transform of $f(t)e^{-\sigma t} $, so: $$X(s)=\mathscr{F}[f(t)e^{-\sigma t}] $$

Infact, the F.T. represents a function as composed by harmonics. They, on the complex plane, are points just on the imaginary axe ($\sigma=0$).

$\endgroup$
  • 1
    $\begingroup$ If the Laplace transform is just the Fourier transform of $f(t)e^{-\sigma t}$, why in the world would one want to take the Fourier transform of $f(t)e^{-\sigma t}$? I think I can see where the Fourier transform itself comes from (I've explained what I think it is in the beginning of my question), but I don't see how taking the Fourier transform of $f$ after it's been pre-multiplied by $e^{-\sigma t}$ means anything, why one wants to do this, or how it springs out of any theory or anything - any ideas? $\endgroup$ – bolbteppa Oct 1 '13 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.