Let’s use a simple harmonic oscillator as an example. When we calculate the Fourier Transform (a special case of the Laplace transform) of that system we get a function that shows which frequencies of the input (the excitation) will be boosted or attenuated by that system (the frequency response). So putting in a certain frequency for $\omega$ in this function provides us this information.

But what does the sigma (assuming the Laplace variable to be $s = \sigma + j\omega$) mean when we‘re extending to the Laplace transform? We can now look at a frequency response for every sigma we are putting in. I often read it takes account for damping but how does that work? We calculated the Laplace transform for a specific physical system with already defined damping factors from its physical components. Why do we now have some kind of variable damping factor in the Laplace transform?


2 Answers 2


The Fourier Transform represents a signal as the sum of sinusoids (and/or cosinusoids). However the Laplace Transform represents a signal as the sum of damped sinusoids (and/or cosinusoids). Sigma gives the damping factor $e^{-\sigma t}$ applied to each of the sinusoidal 'harmonics', $sin(\omega t)$.

The 'variable damping factor in the Laplace transform' is very useful, for example, in stabilizing closed loop control systems where the system damping is one of the most important performance requirements.

In general the Laplace Transform can be a more efficient way to represent wideband systems.

  • $\begingroup$ They are essentially the same thing. Subject to some caveats, $s=i \omega$ trivially converts one to the other. $\endgroup$
    – John Doty
    Oct 25, 2022 at 18:10

This is not about damping and some such but rather solving an initial value problem with elements that by themselves individually do not satisfy that. Here the initial value problem is given by the requirement that it be causal, ie., that all its solutions must be $0$ for $t<0$.

As an example, connect an inductor and a resistor in series and drive them with an ideal voltage source, an ideal battery whose internal resistance is zero and its voltage between its terminals is a constant $V_0$ by turning it "On" at the instant $t=0$ to the circuit whose initial current at the at instant is $i=0$. Defining the Heaviside $1(t)$ as $1(t)=0$ if $t<0$ and $1(t)=1$ if $t>0$ you have the differential equation for the current: $$\mathcal D i =L\frac{di}{dt}+Ri=v_0(t)=V_o \rm 1(t)$$ How to solve this? You know that the differentiating $e^{pt}$ you get $pe^{pt}$ and therefore $\mathcal D e^{pt} = Lpe^{pt}+Re^{pt} = (pL+R)e^{pt}$.

If the source voltage were a multiple of $e^{pt}$ say, $v_o(t)=He^{pt}$ for some constant in time $H$ then you would immediately see that the differential equation is solved by the substitution $i(t)=\frac{H}{pL+R}e^{pt}$ whose amplitude is independent of $t$ but with dependence on $p$.

Of course, the current $i(t)$ is a real number, so in this context we should have $\Im p=0$ although any complex $p$ is also a formal solution and because of linearity so is a formal solution formed by any linear combination thereof using various other $p$ values as long as $pL+R \ne 0$.

So if we could formally write $1(t) = \sum_k H_k e^{p_kt}$ for some $p_k$ and $H_k$ then we could also say that $i_k=\sum_k \frac{H_k}{p_kL+R}e^{p_kt}$.

In fact a simple analysis shows that $$1(t)=\frac{1}{\mathfrak j 2\pi}\int_{\mathcal C} \frac{1}{z}e^{zt}dz$$ where the contour of integration is now on the complex plane $z=x+\mathfrak j y$ for some arbitrary but positive fixed $x>0$ starting at $y=-\infty$ and running to $y=+\infty$ for any $t$ positive or negative. In other words, as long as $\Re p >0$ we can represent $1(t)$ as the continuous sum of exponentials all in the right half plane. Therefore the solution to our problem is $$i(t)=\frac{1}{\mathfrak j 2\pi}\int_{-\mathfrak j \infty +0}^{\mathfrak j \infty +0} \frac{V_0}{p}\frac{1}{pL+R}e^{pt}dp$$ Above the $+0$ just means that we are running our integral parallel with the $\Im p$ axis in the RHP.

How to calculate this complex contour integral using Cauchy's residue formulas and some such you have to study the mathematics of Laplace transformation whose inverse transform is just this integral.


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