When we provide a sinsusoidal input, the evaluation of the solution of the differential equation head-on becomes horrendous. So we give an imaginary, complex exponential input and solve the differential equation and finally take the imaginary part as the solution (superposition theorem). So once we find the complex amplitude we can just multiply it with \$e^{iωt}\$, take imaginary part and arrive at the solution.

We can find the sinusoidal steady-state response for the given system, with this neat trick. When we use Laplace transform, as far as I can understand, it also gives us a complex amplitude, not only for the steady-state but also includes the transient response, for any input !. The idea behind the precious trick of modelling a sinusoidal input with complex numbers was perfectly intuitive. But I've got no idea how Laplace transform does all this, I don't even understand, why it demands the use complex numbers? rather how does it find this alternative model for any given input, as we did in the previous "trick" only for sinusoidal input.

  • \$\begingroup\$ Does this answer your question electronics.stackexchange.com/questions/208458/… ? \$\endgroup\$
    – jDAQ
    Oct 31, 2019 at 6:03
  • \$\begingroup\$ @jDAQ thank you, but the answer explains everything, assuming that we understand why Laplace transform includes transient response, right ? \$\endgroup\$ Oct 31, 2019 at 6:29
  • 2
    \$\begingroup\$ Isn't that because you have to include your beginning- and end-state to the differential equation? So since you have the signal from t=0 to t=infinity, the transient is included. \$\endgroup\$
    – Swedgin
    Oct 31, 2019 at 8:00

3 Answers 3


My Laplace transform math is pretty rusty, but the neat trick behind a Laplace transform is that the test waveform that the transform compares your input function to changes amplitude over time.

Examine the Fourier transform:

\$\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t)\ e^{-i \omega t}\,dx,\$

Essentially, it takes a test sine wave (\$e^{-i \omega }\$) with frequency \$\omega\$ and determines how similar that sine wave is to your input function 1. It determines this similarity, or "cross-correlation", by multiplying that sine wave with your input function across all time and integrates. If there is little similarity between this frequency \$\omega\$ and your input, then this integration will be equal to zero.

As you've observed, this test signal never changes amplitude and can only be used to examine the steady state behavior of your input.

Now compare to the Laplace transform:

\$F(s) =\int_0^\infty f(t)e^{-st} \, dt\$

Very similar, except now \$s\$ is complex and replaces the real (not complex) \$\omega\$. In addition, the integration is now not for all time, but only the future.

The neat thing about raising to a complex power \$s\$ is that there is both a real and an imaginary component, so that if \$s= a + i b\$, then \$e^{-st} = e^{-at}e^{-ibt}\$. So now we have a sine wave that either grows exponentially, shrinks exponentially, or maintains a constant amplitude depending on the value of \$a\$.

This test signal that can change over time, plus the fact that the integration now only starts at time 0 means that we can now extract the transient response from the input signal.

\$a\$ and \$b\$ together determine a point on the complex plane (known here as the s-plane), whereas \$\omega\$ will always be on the real line. This is the extra degree of freedom that @hotpaw2 mentioned. This point defines the test signal used by the Laplace transform, and you can see the effects of moving the point around the complex s-plane in this image:

s-plane for Laplace Transform signals

Diagrams depicting the Fourier series decomposition of a signal are common (like the one here: https://en.wikipedia.org/wiki/Fourier_series#/media/File:Fourier_series_and_transform.gif)

Less common are diagrams showing a Laplace transform decomposition, but I found this example analyzing an RLC circuit: Transient response of RLC circuit

On the top left plot you can see the inductor current over time, with a large decaying transient from 0 to 5 seconds. On the bottom left plot you can see the end of the initial transient and the final steady state behavior (note the large change in plot scale!).

And the corresponding Laplace decomposition: Laplace decomposition of RLC circuit

Here you can see that the total response in the first plot is the sum of the two terms plotted here, and that the impact of the initial transient decays to zero over time.

See this page for much more information:


1: In reality, \$e^{-i \omega } = \cos x + i \sin x\$, so it's actually using both a cosine and a sine wave simultaneously (a rotating complex point) as the test function. This is how the Fourier transform gets phase information. That's also why the image above shows curly 3D spirals, since the Laplace transform also has a rotating complex point. But it's easier to just think of it as a single sine wave.

Image sources: https://www.dsprelated.com/freebooks/mdft/Comparing_Analog_Digital_Complex.html https://www.mathworks.com/help/symbolic/solve-differential-equations-using-laplace-transform.html

  • \$\begingroup\$ Yours is the only answer to partially address the OP's question, but I don't think that the added sigma is relevant (it just makes the integral converge for a large class of signals). I believe it's the integration limit placed at 0 that makes the difference. All pregress history is enclosed in the initial conditions at 0 and these, as Swedgin pointed out, are automatically included when you solve the differential equations by means of the property of differentiation. L{f'} = s F - f(0). In fact IIRC one can also define a unilateral Fourier transform, but convergence will be troublesome. \$\endgroup\$ Oct 31, 2019 at 22:46
  • \$\begingroup\$ Wow, thank you very much, I've not completely understood this, please bare with me. I'm not able get why integrating from 0 suddenly includes transients, and why Fourier transforms don't include transients. Can you please give an example, maybe a RC circuit with a 5V DC input. \$\endgroup\$ Nov 1, 2019 at 2:17
  • \$\begingroup\$ When you transform a differential equation in y you need to transform y and its derivatives y', y'',... Now, what is the (bilateral) Fourier Transform of y'? Hint: math.stackexchange.com/questions/430858/… . And what is the (unilateral) Laplace Transform of y'? Hint: mathalino.com/reviewer/advance-engineering-mathematics/… . So, which one lets you put the initial value y(0) in it and why? The (unilateral) Laplace transform does, and the reason is that one of the extremes of integration is 0 and f doesn't need 2 B 0. \$\endgroup\$ Nov 1, 2019 at 2:56
  • \$\begingroup\$ (to be zero). So when you use Laplace you can specify the initial conditions directly when you turn y(t) in Y(S), or in case of a circuit, you can model the capacitor with an initial voltage as an impedance 1/sC with a voltage source to represent the initial state. Either way, you can put the initial condition in right from the start, and avoid determining constants in the expression of the general complete solution by specifying initial conditions at the end. \$\endgroup\$ Nov 1, 2019 at 3:01
  • \$\begingroup\$ @AravindhVasu As I said at the beginning of my answer, my Laplace transform math is rusty now (much more intuitive than mathematically rigorous), so I won't venture to create my own example right now. There are many out there. I'll add a link to one in my answer. \$\endgroup\$
    – remcycles
    Nov 1, 2019 at 17:37

When you take an integral transform you are projecting the function of interest to a different coordinate system that still retains all the properties of the original function. Think of how you can write a vector in both rectangular and polar coordinate system, while still being the same vector. This is what you are doing with integral transforms, but, instead of going to the two dimensional polar space, you are projecting into an infinite dimensional Hilbert space. So, rather than having \$x\$, \$y\$ etc., you have infinitely many basis of the form \$e^{j\omega}\$.

To understand what is happening, consider a simple vector projection. This is how you obtain the components of the same vector in a different coordinate system.



To obtain the component of the vector \$\mathbf{a}\$ in the \$\mathbf{b}\$ direction you simply take the dot product

$$\mathbf{a} \cdot \mathbf{c}= \sum_i a_i c_i$$

with the unit vector \$\mathbf{c}\$, parallel to \$\mathbf{b}\$. If the vector \$\mathbf{c}\$ is complex, the dot product is with its complex conjugate. Hence

$$\mathbf{a} \cdot \mathbf{c}^* = \sum_i a_i c_i^*.$$

When you are projecting a function you have infinitely many components \$a(t)\$. With an integral transform, such as the Fourier transform, you have infinitely many basis \$e^{j\omega}\$. Each basis has a component \$e^{j\omega t}\$. Thus the sum over \$i\$ becomes the integral over \$t\$

$$\int_{-\infty}^\infty a(t) \left( e^{j\omega t} \right)^*\ \mathrm{d}t = \int_{-\infty}^\infty a(t) e^{-j\omega t} \ \mathrm{d}t.$$

So when you take an integral transform you are left with the same function but in a different coordinate system. Since it is the same function, it contains the transient response in both coordinate systems.

  • \$\begingroup\$ How does this answer the OP's question? The input signal does not necessarily carry information about the initial state of the system (unless they are all zero). Nor does the transfer function. Where do these conditions fit in your picture? For example with an RLC circuit where in one instance C has 5V, L has 1A at t=0, and another instance where Vc(0)=-1V and IL(0)=.2 mA... Where are those conditions, necessary to compute the transient, stored? Also, you are considering integration from minus infinity to plus infinity. Where do you place initial conditions? \$\endgroup\$ Oct 31, 2019 at 23:10
  • \$\begingroup\$ @SredniVashtar There are many ways to set up the RLC system in your example. We need to set up the system in such a way that it is not stable at t = 0. Otherwise it would have already converged to its steady state. One way to achieve this is to hold the system at the initial conditions for t < 0. This means that there is always some type of windowing effect, e.g. we set up the current through the capacitor to be zero for t < 0, introducing a step window. My answer simply states that the integral transform preserves all the information, so I did not explain the specific techniques used. \$\endgroup\$
    – user110971
    Oct 31, 2019 at 23:32
  • \$\begingroup\$ My objection is that such information is not there at all, in the first place. When you solve the ODE in the time domain, you get a solution in term of arbitrary constants whose value is determined by imposing the initial conditions. When you solve it with Laplace, you input the initial condition when you transform the ODE in AE. The initial conditions (either in the form of previous state of charge/current in the dynamical components or as initial values imposed by the excitations) determine the transient but they are usually not included in the transfer function (nor in the input signal). \$\endgroup\$ Nov 1, 2019 at 0:01
  • \$\begingroup\$ Otherwise, one should specify a transfer function for every possible set of ICs. Instead, if we do not directly transform the differential equation, the ICs are used to determine the coefficients of the generic natural response associated with the poles of the transfer function. Still, they have to be added in, even if at a later stage of solving. \$\endgroup\$ Nov 1, 2019 at 0:05
  • \$\begingroup\$ @SredniVashtar You do get different transfer function depending on the initial conditions, but it only affects the DC frequency in terms of Dirac deltas. The Dirac delta gets introduced when you have integration in the time domain. See here: thefouriertransform.com/transform/fourierintegralproperty.gif. \$\endgroup\$
    – user110971
    Nov 1, 2019 at 0:22

Steady state Fourier analysis is handy if you have a single infinitely long single frequency pure sinusoid. You can readily compute the response of an LTI system to this infinitely long sinusoid because it maps to a single point in the Fourier domain.

The problem is that in the real world, signals have a finite length (an envelope), which maps to a infinite batch (integral) of frequencies in the Fourier domain. And computing the response to an infinite batch of frequencies might require an infinite amount of computation, which could take a very long time.

The Laplace transform adds another degree of freedom, by including time in its integral kernel. It thus can represent an infinite batch of frequencies in a single complex (or 2D) point, thus allowing computing a system response to these infinite batches of infinite sinusoids with less than an infinite amount of chalkboard. An transient signals can be decomposed into batches of these infinite batches.

So consider it (Laplace transform) to be a mathematical trick to do an infinite amount of single frequency steady state (Fourier transform) analysis in finite time (and chalkboard), by adding another degree of freedom.

Added: Note that the Fourier transform of a decaying exponential is infinitely long. Infinite in the Fourier space, but a single (complex) point in Laplace space.

  • \$\begingroup\$ Thank you for answering, but I don't understand, "The Laplace transform adds another degree of freedom, by including time in its integral kernel. It thus can represent an infinite batch of " this part. What do you mean by it can represent an infinite batch of frequencies? \$\endgroup\$ Oct 31, 2019 at 17:12
  • \$\begingroup\$ It represents an infinite integral, and not just over a single frequency of sinusoid. Whereas a point in the Fourier domain is for just one frequency. \$\endgroup\$
    – hotpaw2
    Oct 31, 2019 at 17:19

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