I have become a bit confused about these topics. They've all started looking the same to me. They seem to have the same properties such as linearity, shifting and scaling associated with them. I can't seem to put them separately and identify the purpose of each transform. Also, which one of these is used for frequency analysis?

I couldn't find (with Google) a complete answer that addresses this specific issue. I wish to see them compared on the same page so that I can have some clarity.


The Laplace and Fourier transforms are continuous (integral) transforms of continuous functions.

The Laplace transform maps a function \$f(t)\$ to a function \$F(s)\$ of the complex variable s, where \$s = \sigma + j\omega\$.

Since the derivative \$\dot f(t) = \frac{df(t)}{dt} \$ maps to \$sF(s)\$, the Laplace transform of a linear differential equation is an algebraic equation. Thus, the Laplace transform is useful for, among other things, solving linear differential equations.

If we set the real part of the complex variable s to zero, \$ \sigma = 0\$, the result is the Fourier transform \$F(j\omega)\$ which is essentially the frequency domain representation of \$f(t)\$ (note that this is true only if for that value of \$ \sigma\$ the formula to obtain the Laplace transform of \$f(t)\$ exists, i.e., it does not go to infinity).

The Z transform is essentially a discrete version of the Laplace transform and, thus, can be useful in solving difference equations, the discrete version of differential equations. The Z transform maps a sequence \$f[n]\$ to a continuous function \$F(z)\$ of the complex variable \$z = re^{j\Omega}\$.

If we set the magnitude of z to unity, \$r = 1\$, the result is the Discrete Time Fourier Transform (DTFT) \$ F(j\Omega)\$ which is essentially the frequency domain representation of \$f[n]\$.

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    \$\begingroup\$ The s in the Laplace Transform is a complex number, say a+j\$\omega\$, so its a more general transform than the completely imaginary Fourier. In fact, so long as you're in the Region of Convergence, it's fair game to go back and forth between the two just by replacing j\$\omega\$ with s and vice versa \$\endgroup\$ – Scott Seidman Oct 25 '13 at 16:19
  • \$\begingroup\$ I find it useful to think of the Fourier transform as something you apply to periodic signals, and the Laplace transform as something you apply to time-varying signals. (This is a consequence of what @ScottSeidman explained above.) \$\endgroup\$ – Li-aung Yip Oct 26 '13 at 6:23
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    \$\begingroup\$ @Alfred: You haven't actually addressed which one of these is used for frequency analysis - for completeness it is probably worth mentioning that that most people use the FFT for frequency analysis, and how the FFT fits in with the things already listed. \$\endgroup\$ – Li-aung Yip Oct 26 '13 at 6:32
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    \$\begingroup\$ @Li-aungYip, I think you may be conflating the Fourier series and the Fourier transform. The Fourier series is for periodic functions; the Fourier transform can be thought of as the Fourier series in the limit as the period goes to infinity. So, the Fourier transform is for aperiodic signals. Also, since periodic signals are necessarily time-varying signals, I don't "get" the distinction you're drawing. \$\endgroup\$ – Alfred Centauri Oct 26 '13 at 11:51
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    \$\begingroup\$ @Li-aungYip Also, FFT is used to compute DFT which is not DTFT. DFT is like taking samples in the frequency domain after having a DTFT (which is continuous for aperiodic signals). It is just a tool used in computers for fast computations (okay, we can use it manually too). But FFT comes after you are past DTFT and CTFT. \$\endgroup\$ – Anshul Oct 30 '13 at 11:07

Laplace transforms may be considered to be a super-set for CTFT (Continuous-Time Fourier Transforms). You see, on a ROC (Region of Convergence) if the roots of the transfer function lie on the imaginary axis, i.e. for s=σ+jω, σ = 0, as mentioned in previous comments, the problem of Laplace transforms gets reduced to Continuous Time Fourier Transform. To rewind back a little, it would be good to know why Laplace transforms evolved in the first place when we had Fourier Transforms. You see, convergence of the function (signal) is a compulsory condition for a Fourier Transform to exist (absolutely summable), but there are also signals in the physical world where it is not possible to have such convergent signals. But, since analysing them is necessary, we make them converge, by multiplying a monotonously decreasing exponential e^σ to it, which makes them converge by its very nature. This new σ+jω is given a new name 's', which we often substitute as 'jω' for sinusoidal signals response of causal LTI (Linear Time-Invariant) systems. In the s-plane, if the ROC of a Laplace transform covers the imaginary axis, then it's Fourier Transform will always exist, since the signal will converge. It is these signals on the imaginary axis which comprise of periodic signals e^jω = cos ωt + j sin ωt (By Euler's).

Much in the same way, z-transform is an extension to DTFT (Discrete-Time Fourier Transforms) to, first, make them converge, second, to make our lives a lot easier. It's easy to deal with a z than with a e^jω (setting r, radius of circle ROC as untiy).

Also, you are more likely to use a Fourier Transform than Laplace for signals which are non-causal, because Laplace transforms make lives much easier when used as Unilateral (One sided) transforms. You could use them on both sides too, the result will work out to be the same with some mathematical variation.

  • \$\begingroup\$ Your answer is saviour .... thumbs up for so precise and great explanation .. \$\endgroup\$ – pravin poudel Sep 5 '18 at 4:39

Fourier transforms are for converting/representing a time-varying function in the frequency domain.

A laplace transform are for converting/representing a time-varying function in the "integral domain"

Z-transforms are very similar to laplace but are discrete time-interval conversions, closer for digital implementations.

They all appear the same because the methods used to convert are very similar.


I will try to explain the difference between Laplace and Fourier transformation with an example based on electric circuits. So, assume we have a system that is described with a known differential equation, let say for example that we have a common RLC circuit. Also assume that a common switch is used to switch ON or OFF the circuit. Now if we want to study the circuit in the sinusoid steady state we have to use Fourier transform. Otherwise, if our analysis include the switch ON or switch OFF the circuit we have to implement the Laplace transformation for the differential equations.

In other words the Laplace transformation is used to study the transient evolution of the system´s response from the initial state to the final sinusoid steady state. It Includes not only the transient phenomenon from the initial state of the system but also the final sinusoid steady state.


Different tools for different jobs. Back at the end of the sixteenth century astronomers were starting to do nasty calculations. Logarithms were first calculated to transform multiplication and division into easier addition and subtraction. Likewise, Laplace and Z transforms turn nasty differential equations into algebraic equations that you have a chance of solving. Fourier series were originally invented to solve for heat flow in bricks and other partial differential equations. Application to vibrating strings, organ pipes, and time series analysis came later.


In any LTI system for calculating transfer function we use only laplace transform instead of fourier or z transform because in fourier we get the bounded output ;it doesn't go to infinity. And z transform is used for discrete signals but the LTI systems are continous signals so we cannot use z transform .. Therefore by using laplace transform we can calculate transfer function of any LTI system.


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