Laplace domain is also known as "s domain".

Is there any difference between "s domain" and "frequency domain"? Can we use both terms interchangeably?

If we want to convert a time domain signal to frequency domain, can we use Laplace transform?


2 Answers 2


The s domain is synonymous with the "complex frequency domain", where time domain functions are transformed into a complex surface (over the s-plane where it converges, the "Region of Convergence") showing the decomposition of the time domain function into decaying and growing exponentials of the form $e^{st}$ where $s$ is a complex variable. In terms of its real and imaginary components, $s=\sigma + j\omega$. Thus we have $e^{st}=e^{(\sigma + j\omega)t} = e^{\sigma t}e^{j\omega t}$.

$e^{\sigma t}$ for real $\sigma$ is simply a decaying or growing exponential. $e^{j\omega t}$ for real $\omega$ is a phasor spinning at a constant rate in time with magnitude 1 (Magnitude 1, angle $\omega t$.) For a more complete intuition on the $e^{j\omega t}$ representation of what a single frequency tone is (that is used throughout DSP), please refer to this link.

The more commonly used unilateral Laplace Transform assumes the time domain is causal and therefore 0 for all time less than zero. If we restrict $s$ to be the $j\omega$ axis, then we would get the Fourier Transform with the same causality assumption (if the unilateral Laplace Transform was used).

enter image description here

It is convenient and sufficient to just show on the s-plane the locations of the extreme singularities; the poles and zeros (where poles are where the surface goes to infinity and zeros are where the surface goes to zero) as every other point on the surface as a complex value with magnitude and phase is uniquely determined by the poles and zeros. Below is an example showing a pole at the origin. If the function is causal then the Region of Convergence (ROC) is the right half plane to the right of the pole. I show a plot below of the magnitude of $1/s$ for all values of $s$, showing the surface I mention (not shown, but the surface would also have a phase at each point on the surface). Due to the ROC, this is only valid on half the plane.


Another example showing both the utility of the Laplace Transform and its relation to the Fourier Transform is shown below. Here we have the Laplace Transform of the impulse response for a filter. The Fourier Transform of a filter's impulse response is its frequency response, but with the Laplace Transform and specifically the locations of the poles and zeros on the s-plane we have much more information beyond what we can get out of Fourier alone. This includes stability, response and settling time, and insights into further tuning and adjustment. If we slice the given surface along the $j\omega$ axis, we would get the Fourier Transform from the Laplace Transform, as shown in the upper right hand corner.

Filter Example

In the plot I use the term "correlation" here somewhat loosely, as it provides great intuition. We can think of the Laplace Transform as correlating our function $x(t)$ to all possible values of $e^{st}$ and showing us in the complex surface, the relative magnitude (strength) and phase for all possible values of $e^{st}$ in $x(t)$. What is actually occurring is better termed as a mapping from one space to another, but as I detail further here, the operation is indeed very similar to correlation as an integration of complex conjugate products, with the distinction here that both the real and imaginary terms of the exponent for the basis function $e^{st}$ are negated in the Laplace Transform given as:

$$X(s) = \int_0^\infty x(t)e^{-st}dt$$

For the curious, the above filter is a 2nd order filter, with two complex poles in the left half plane (so stable) at $s=-0.2\pm j0.5$. The impulse response is:

$$x(t) = 2e^{-0.2t}\sin(0.5t), t\ge 0$$

Which results in the Laplace Transform as given in the plot.

The two poles at $s=-0.2\pm j0.5$ in the resulting Laplace Transform suggest that the impulse response $x(t)$ has two fundamental components consisting of decaying exponentials given by $e^{st}$ for $s$ at the given pole locations. (The surface of the Laplace Transform is a result of a form of correlation, and will be maximum (poles) where the components correlate the strongest):

$$e^{st}=e^{(-0.2\pm j0.5)t} = e^{0.2t}e^{j0.5t}, \text{ and } e^{0.2t}e^{-j0.5t}$$

Indeed we see the factor $e^{0.2t}$ directly in the expression for $x(t)$, and note from Euler's identity that:

$$\sin(0.5t) = \frac{e^{j0.5t}-e^{-j0.5t}}{2j}$$

Showing how both rotating phasor components given by $e^{\pm j 0.5t}$ do exist in $x(t)$. Do not be thrown off by the $j$ in the denominator and the subtraction for $e^{-j0.5t}$; the surface as depicted above is the magnitude plot, but each sample on that surface is complex values with a magnitude and phase.

The resulting Frequency Response magnitude, as the Fourier Transform of the impulse response, extracted from Laplace by restricting $s$ to be the $j\omega$ axis, is shown below. This is plotted on a log log plot with just the positive frequencies as we would typically view the magnitude response for a continuous time filter:

low pass filter


Laplace transform, Fourier transform, z transform, ... are all transforms from the time domain to the frequency domain.

See @GrapefruitIsAwesome's link, and/or: relation and difference between fourier laplace and z transforms


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