0
\$\begingroup\$

After seeing a question on one of the posts I got this confusion in my mind. Here I'm not much interested in just mathematical explanation but I would like to know how a transformation actually turns a time domain signal to frequency domain. And what frequency domain representation actually means.Since the signal say a unit step has 0 frequency. Is it just a plot of magnitude of signal with frequency parameter or there is more to it?

\$\endgroup\$
  • 2
    \$\begingroup\$ After seeing a question on one of the posts I got this confusion in my mind. Which question was it? \$\endgroup\$ – K. Rmth Dec 30 '15 at 11:40
  • \$\begingroup\$ Laplace transform and the idea of frequency domain analysis \$\endgroup\$ – manav.tix Dec 30 '15 at 13:36
  • \$\begingroup\$ For info,the question OP is referring to is here. \$\endgroup\$ – K. Rmth Dec 30 '15 at 14:38
4
\$\begingroup\$

I can only explain why they work mathematically, b/c the way time-domain functions get turned into s-domain functions is mathematically-based.

Fourier and Laplace Transforms are based on a mathematical concept known as an integral kernel. An integral kernel is defined to be the \$f(x,t)\$ term in an integral of the form: $$ F(x) = \int_{t_0}^{t_1}f(x,t)g(t)dt $$

\$f(x,t)\$ defines a family of functions parameterized by x, such as \$e^{-jxt}\$, for example. There are an infinite number of functions that take this form one for each (uncountably!) infinite number of values of \$x\$ from, for instance, \$-\infty\$ to \$\infty\$.

Let's assume we're given a value of \$x\$, call it \$x_0\$, and a function \$g(t)\$ that we want to analyze. Let's assume the domain is \$\pm\infty\$. The above integral can be written as: $$ A = \int_{-\infty}^{\infty}f(x_0,t)g(t)dt $$

The above integral is known as an inner product, analogous to the concept for vectors in linear algebra. Informally, an inner product produces a number \$A\$ which measures "how similar in shape is \$f(t)\$ to \$g(t)\$?".

It turns out by summing every single function of the family \$f(x,t)\$, each with a constant multiplier \$A\$ obtained by evaluating the inner product with \$g(t)\$ at each \$x\$, one can get the original \$g(t)\$ as a result.

Now, to answer your question, Fourier Transform decomposes a function \$g(t)\$ into a set of complex exponentials \$f(x,t)\$ of the form \$e^{-jxt}\$. These are very closely related to sinusoids of an angular velocity \$x_0\$ by taking the magnitude of the resulting complex number.

Laplace transforms are similar, but of the form \$e^{-xt}\$, where \$x\$ is a complex number, and \$t\$ is limited from 0 to \$\infty\$ (\$x\$ is normally called \$s\$ in this transform. I am using \$x\$ for consistency). They don't have as nice an intuitive interpretation that I'm aware of. Wikipedia says the Laplace Transform decomposes functions into moments. Analyzing the values of the Laplace transform for values of \$x\$ of the form \$jx_0\$ gets you back to the sinusoidal interpretation as above.

As Lorenzo also points out, there are convenient properties of integral transforms that appear in the frequency domain, such as convolution of two functions \$f(x_0, t)\$ and \$g(t)\$ become a multiplication in the s-domain.

\$\endgroup\$
  • \$\begingroup\$ Thank for support, now I understand use of transforms, and their basic theory but I still have some confusion on what frequency domain representation implies in physical terms. \$\endgroup\$ – manav.tix Dec 30 '15 at 14:59
  • \$\begingroup\$ If you integrate over t, there won't be any t remaining in the result. \$\endgroup\$ – The Photon Dec 30 '15 at 16:26
  • \$\begingroup\$ @ThePhoton Fixed my integrals. Not sure what I was thinking there. \$\endgroup\$ – cr1901 Dec 30 '15 at 16:43
  • 1
    \$\begingroup\$ @manav.tix The frequency domain can be thought of as sums of sinusoids that extend in time from \$-\infty\$ to \$\infty.\$ We trade information about time (since the sinusoids don't have a well-defined beginning and end) for information about sinusoidal content. Obviously, we can't create sinusoids like that in reality, but the model is still very useful. Real signals are generated in the time domain, as a function of time, not as a function of infinitely-long sinusoids. \$\endgroup\$ – cr1901 Dec 30 '15 at 16:50
  • \$\begingroup\$ @cr1901 Plz correct me if my assumption is flawed but then transformation changes entire signal and its properties but energy remains conserved thus the information stored too is saved. And we analyze the signal as a sum of sinusoids for various frequency value to see its amplitude variations. \$\endgroup\$ – manav.tix Dec 30 '15 at 18:54
1
\$\begingroup\$

Well, from what you say it is not clear what you mean by "how the transform works", since you don't want a mathematical explanation. How transforms work is something mathematical.

Anyway, I guess you are interested in the rationale behind their application in the circuit engineering fields and the like.

In that respect you can view transforms as a way to represent a circuit or a signal in a different way, but a way that preserves (apart from pathological cases) the "information" of the original circuit/signal.

For example, when you apply Laplace transform to a circuit to perform transient analysis in the s-domain you are exploiting the math properties of the L-transform to simplify your work. And when you use it to compute the transfer function of a linear time-invariant system you are using the math properties of the L-transform to represent that system in a different, equivalent way. This equivalent representation can be advantageous, for instance, if you are interested in the stability of the system, since this property is related to the position of the poles of its transfer function.

To summarize, it is all about choosing the right representation that helps you analyze or design a circuit/system/signal in the easiest possible way. As an analogy, think of a civil engineer designing a building. He can choose different representations for the building according to what he is going to design: he must use technical drawing to represent the exact measurements for walls and other structures, but he may resort to 3D modeling (either real scaled-down models or CAD simulations) to see the impact of the building on the surrounding environment.

\$\endgroup\$
  • \$\begingroup\$ dear friend I from the line "I'm not much interested in just mathematical explanation" wanted that I would like its maths to be explained along with its physical applications. If you can make your answer more informative with maths then please go ahead. It would be great. \$\endgroup\$ – manav.tix Dec 30 '15 at 13:40
0
\$\begingroup\$

A Fourier transform is essentially the infinite period time (\$T\to\infty\$) limit of the Fourier Series. This means applying the transform will transform a time-dependent signal to it's frequency spectrum. Because \$T\to\infty\$ , the difference between the harmonic frequencies approaches zero, and the spectrum becomes continuous. By adding all the frequencies in the spectrum mathematically together again, you also get the inverse transform.

\$\endgroup\$

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.