Fourier transform has different types like continuous Fourier transform (CFT), Discrete time Fourier transform (DTFT) and Discrete Fourier transform ( DFT).

CFT can be used in case of continuous aperiodic signals while DFT for discrete aperiodic signals .

On the other hand, Laplace transform can be used in case of continuous signals and Z transform for discrete signals.

So I want to ask that can use of Fourier transform be minimized completely with the help of Laplace and Z transform?

  • $\begingroup$ Why would you do that? Laplace is the general equation which integrate over the whole complex plane. Fourrier is just about the imaginary axis. $\endgroup$
    – MathieuL
    Jun 9, 2016 at 18:10
  • $\begingroup$ @MathieuL if you can compute everything that can be obtained by Fourier transform with the help of Laplace and Z-transform , why to go for Fourier ? $\endgroup$
    – devraj
    Jun 9, 2016 at 18:18
  • $\begingroup$ because fourrier is easier to compute the inverse fourrier transform than the Laplace inverse transform. Laplace take into the account the transient part of a signal as the opposite the Fourrier TF only consider the permanent regime. $\endgroup$
    – MathieuL
    Jun 9, 2016 at 18:36
  • $\begingroup$ Laplace inverse transform a real pain to compute, it is really hard, we talk high level math here. $\endgroup$
    – MathieuL
    Jun 9, 2016 at 18:37
  • 2
    $\begingroup$ "... while DFT for discrete aperiodic signals." uhm no, the DFT maps a discrete and periodic sequence (with period $N$) from one domain to another discrete and periodic sequence (of the same period $N$) in the reciprocal domain. (uniform sampling in one domain always causes periodic extension in the reciprocal domain.) $\endgroup$ Jun 9, 2016 at 23:59

3 Answers 3


The answer to your last question is definitely 'no'. The point hotpaw2 makes in his answer is very relevant: the FFT is an efficient implementation of the DFT, and there are no equivalently efficient implementations for the numerical computation of the $\mathcal{Z}$-transform or the Laplace transform.

But that's not the only reason. There are important functions (or sequences) for which the Laplace transform or the $\mathcal{Z}$-transform don't even exist, whereas the Fourier transform does. E.g., take a sinusoid or a complex exponential extending from $-\infty$ to $\infty$. These functions (or sequences) don't have a Laplace transform or a $\mathcal{Z}$-transform. Other important examples are impulse responses of ideal frequency selective filters such as low pass or high pass. They can be represented in terms of the sinc function, which can only be transformed using the Fourier transform, but not using the (bilateral) Laplace transform or - in discrete time - the (bilateral) $\mathcal{Z}$-transform.

So even if formally it looks like the Fourier transform is a special case of the Laplace transform or the $\mathcal{Z}$-transform, that's generally not case. One reason for that is the incorporation of the theory of distributions in the theory of the Fourier transform (i.e., the use the Dirac delta impulse), which makes it possible to compute the transform of functions like $\sin(\omega_0t)$ or $e^{j\omega t}$. The latter is not possible using the (bilateral) Laplace transform (or, in discrete time, the bilateral $\mathcal{Z}$-transform).

When people see the definitions of the (bilateral) Laplace transform and of the Fourier transform

$$X_L(s)=\int_{-\infty}^{\infty}x(t)e^{-st}dt\\ X_F(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

it may seem obvious to them that both transforms become identical by substituting $s=j\omega$. This, however, is generally not true. The pitfall here is the fact that the substitution does not take into account the convergence of the improper integrals. Depending on $x(t)$, the Laplace integral might not converge for $s=j\omega$, so $X_F(j\omega)$ might not even exist. The substitution is only valid if the region of convergence (ROC) of the Laplace integral includes the $j\omega$-axis. A completely analogous argument is true for the $\mathcal{Z}$-transform and the DTFT. In that case the substitution $z=e^{j\omega}$ is only valid if the ROC includes the unit circle.

The last paragraph may seem to imply that the Laplace transform and the $\mathcal{Z}$-transform are simply more general than the respective versions of the Fourier transform. However, this is also not true, as already mentioned above, because there are functions (sequences) that can only be treated by the Fourier transform, but not by the Laplace transform ($\mathcal{Z}$-transform).

Also take a look at the following answers to related questions: answer 1, answer 2, answer 3.


The short answer is yes, if you have the Laplace or Z-transform of a function you do not need the Fourier transform.

This is because the CFT is a special case of the Laplace transform and the DTFT is a special case of the Z transform. The Fourier transform is used to find the complex sinusoids that compose a function, whereas the Laplace transform finds all complex exponentials that compose a function.

Knowing this, it is simple to derive the Fourier transform from the Laplace transform. The Laplace transform is defined as:

$$ \int_{-\infty}^{\infty}e^{-s}f(t)\,dt\\ s = \alpha+j\omega $$

To find the Fourier transform from the Laplace transform, all you must do is assert $\alpha = 0$ for all $\omega$ and $t$. This evaluates the Laplace transform along the imaginary axis only, as any component lying directly on the imaginary axis is a complex sinusoid. This leaves us with: $$ \int_{-\infty}^{\infty}e^{-j\omega}f(t)\,dt\\ $$

Which is in fact the Fourier transform.

A similar relationship applies between the DTFT and the Z-transform. The Z-transform is defined as:

$$ \sum_{n=-\infty}^\infty x[n]Z^{-n} $$ Where $Z$ is a complex variable. To find all complex sinusoids in the discrete world we must constrain $Z$ to fall on the unit circle: $$ Z = e^{j\omega} $$

Which leaves us with

$$ \sum_{n=-\infty}^\infty x[n]e^{-j\omega n} $$ Or the discrete time Fourier transform.

The reason the Fourier transform exists is because there are a lot of applications that only need the real frequencies of a function. If only the real frequencies are needed there is no point calculating the rest of the Laplace domain. Why do more work than necessary?

For more information check out the Wikipedia articles https://en.wikipedia.org/wiki/Discrete-time_Fourier_transform#Relationship_to_the_Z-transform and https://en.wikipedia.org/wiki/Laplace_transform#Fourier_transform And I would also strongly recommend the text book Linear Systems and Signals by B.P. Lathi for a much more indepth discussion on Laplace, Z, and Fourier transforms.

  • $\begingroup$ hey welcome, Matt. $\endgroup$ Jun 9, 2016 at 23:56
  • $\begingroup$ It's generally not true that by setting $s=j\omega$ or $z=e^{j\omega}$ you obtain the Fourier transform from the Laplace transform or the Z-transform. You have to take into account the convergence of the corresponding integral and sum. It might be that the integral or the sum become meaningless (i.e., don't converge). The above substitution step is only valid if the region of convergence contains the $j\omega$-axis (Laplace transform), or the unit circle (Z-transform). Apart from that, you also have functions that can only be treated by the Fourier transform, and not by the other transforms. $\endgroup$
    – Matt L.
    Jun 10, 2016 at 8:05
  • $\begingroup$ @MattL. Since the formula is exactly the same... it must be true, no? But yes, the Fourier transform may not converge for all functions, that also means Laplace won't exist for e^0 multiplier (the horizontal line at 0 in imag/real plane). So the result is still exactly the same... $\endgroup$
    – Dole
    Jun 10, 2016 at 14:50
  • 1
    $\begingroup$ @MattL. I agree that there are exceptions and that they must converge but to deny that the Fourier transform is a special case of the Laplace transform is being very nitpicky, especially in practical analysis. $\endgroup$
    – Matt
    Jun 10, 2016 at 15:41
  • 1
    $\begingroup$ @MattL. As I said before, I agree with you, but the majority of the time it is more convenient to view the Fourier transform as a special case of the Laplace transform. It is rare that this doesn't hold true for real world systems. $\endgroup$
    – Matt
    Jun 10, 2016 at 17:01

The DFT and IDFT have a computationally efficient implementation in the form of the FFT. This efficiency can not be replaced or minimized by the LT or ZT in the general case.

  • 1
    $\begingroup$ I have been seeing many of your answers. You answers are very brief and short but contains sea of information . your method is very good but sometimes it takes months for common person to extract information from your answers :-) $\endgroup$
    – devraj
    Jun 10, 2016 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.