I'm used to Fourier transform, i.e. I have a good unterstanding of how to interpret a Fourier transformed function. But I'm struggeling with the Laplace transform. What's the meaning of its complex parameter \$s = o + iw\$ (where \$w\$ is the 'normal' frequency)?

I also don't see the advantage of the Laplace transform when it comes to transfer functions. It seems to me that in the end one is just interested in the spectrum of the analyzed system which coincides with evaluating the transfer function just on the imaginary axis ('normal' frequencies). So why one wouldn't simlpy use the Fourier transform to define the transfer function like this:

$$G(w) = \frac{F\left(y(t)\right)(w)}{F\left(x(t)\right)(w)}$$

where \$w\$ is real, instead of:

$$G(s) = \frac{L\left(y(t)\right)(s)}{L\left(x(t)\right)(s)}$$

where \$s = o + iw\$.


3 Answers 3


The Laplace transform has some nice properties that help to get more insight into the behavior of linear systems.

A very nice property is that the Laplace transform evaluated along the jw-axis is equivalent to the Fourier transform, which is less abstract and easier to understand. This of course brings us back to question, why we didn't use the Fourier transform in the first place.

The answer is quite simple. We often have to analyze and work with systems that are unstable or where we want to determine whether they are stable or not. In such cases the Fourier transform fails (=does not exist). This is the reason the Laplace transform was introduced. It includes an additional term that helps the integral to converge and therefore the Laplace transform can be applied to a broader range of problems and applications.

  • \$\begingroup\$ Can you make a example of what information one could get from the transfer function in such a case? \$\endgroup\$ Commented Sep 14, 2016 at 18:18
  • 1
    \$\begingroup\$ The placement of the poles and zeros of a transfer function can be used to determine stability, frequency and time domain behavior. (en.wikipedia.org/wiki/Pole%E2%80%93zero_plot) \$\endgroup\$
    – Mario
    Commented Sep 14, 2016 at 18:27

What are the Fourier transforms for the step and ramp functions? As well stated by Prof. C.P. Quevedo: "The idea of saying that such functions are periodic, with infinite period, no longer applies (the function never returns to zero and do not have the opportunity to repeat, neither in the infinity". This is where enters the Laplace Transform. By introducing a real term \$\sigma\$ in \$s = \sigma + j\omega\$, it is possible to make the "Fourier Transform" integral (now a Laplace Transform) to converge.

$$F(s) = \int_{^{^0{-}}}^{\infty}f(t)e^{^{-st}}dt $$

In fact, the main goal of the Laplace transform is to convert a differential equation into an algebraic equation (like logarithms). After, to operate in the complex domain, expand the result with partial fractions return to time domain using (normally) transformer tables; not only to a sinusoidal signal inputs. Also, It should be remembered that the frequency response (sinusoidal) is just one of the transfer function applications.

Fourier series: A periodic time signal is seen as an infinite sum of sinusoids (discrete frequencies, harmonics).

Fourier transform: A not necessarily "periodic" time signal is seen as an infinite sum of infinitesimal scaled sinusoids (continuous frequencies). Sum -> Integral.

Laplace Transform: A not necessarily "periodic" time signal is seen as an infinite sum of infinitesimal exponentially scaled sinusoids (continuous frequencies).

ADDITION: \$\sigma\$ comes into play when when the concept of frequency is generalized to "complex frequency". When you write \$Ae^{kt}\$, the exponent should be dimensionless; so that \$k\$ should be \${second}^{-1}\$. Note the similarity with "Hertz", ie it is a type of frequency. For example, in \$Ae^{2t}\$, the \$2\$ is the frequency (events per second) with which \$A\$ will be multiplied by \$e\$. The unit of this "frequency" is Neper/second. Thus $$s [complex Neper/ second] = \sigma [Neper/second] + j\omega [radians/second]$$ Now an exponential function has a frequency, even though it differs from the traditional concept. In the Laplace Transform, the \$\sigma\$ is an appropriate value (but not only) for the integral to converge.

  • \$\begingroup\$ Does the sigma have any meaning? F(iw) gives me the systems spectrum but what is about the rest of the (right half) complex plane? \$\endgroup\$ Commented Sep 14, 2016 at 20:16
  • \$\begingroup\$ @user2224350: Edited in response above (too long here). \$\endgroup\$ Commented Sep 14, 2016 at 22:14

So typically I use the Fourier domain to analyze a signal and it's frequency components. In the Fourier domain you can do tricks to manipulate the signal (like multiply by a step function to do a LP or HP or BP filter). Although last time I was faced with this I just did the convolution of the signal against a sinc function (easier to write in a program and faster).

The Laplace domain is more so used for system analysis and control theory. Typically you will start out with the transformed version -> like a capacitor is Z = 1/(sC). And once you have figured out the transfer function of your whole system, you can see how it responds to inputs (often you will do a frequency sweep or a nyquist plot).

These concepts are not mutually exclusive. I believe Fourier is a specific case of the more generalized Laplace case (please correct me if I am wrong or elaborate as necessary because I really don't remember this point).

  • \$\begingroup\$ to answer the question you asked above, one quick analysis you could do is to find the roots of the denominator of the transfer function. If any of these are positive, your system is unstable. There is more to stability analysis than this. But this is some of the information you can get from a transfer function. H(s) = 1/(s-1) | s = 1 -> unstable H(s) = 1/(s+1) | s = -1 -> stable \$\endgroup\$
    – klamb
    Commented Sep 14, 2016 at 18:28

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