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Although this topic has already been addressed in multiple popular questions of SE but i have few confusions in this regard

Number 1)

Link of question https://electronics.stackexchange.com/questions/86489/relation-and-difference-between-fourier-laplace-and-z-transforms

In the top voted answer there was a sentence that i highlighted

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 $f(t)$

So what does that imply? If Fourier transform is essentially the frequency domain representation of $f(t)$ then does that mean Laplace transform is not frequency domain representation

Number 2)

Link of question

What are the advantages of Laplace Transform vs Fourier Transform in signal theory?

In top voted answer, there was a sentence

Laplace transforms can capture the transient behaviors of systems. Fourier transforms only capture the steady state behavior.

What does that imply? Fourier transform cannot be used for studying the transient behavior and Laplace transform cannot be used for studying steady state behaviour?

One last confusion: Which transform is more commonly used in practial applications ?Laplace transform or Fourier transform?

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3 Answers 3

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Concerning your first question, both, the Laplace and the Fourier transform, are frequency domain representations of a function or signal. In the Fourier transform we deal with a real-valued frequency variable $\omega$, whereas in the Laplace transform we have a generally complex-valued independent variable (usually $s$), the imaginary part of which equals frequency: $s=\sigma+j\omega$.

Your second question can be answered in a very simple way: the quoted sentence is wrong (which I also mentioned in a comment to that answer).

As for "practical application", I would say that when you talk about causal systems implemented with lumped elements, then the (unilateral) Laplace transform is probably used more often. It is also more straightforward to take initial conditions into account when using the unilateral Laplace transform.

The Fourier transform is more suited to idealized systems such as ideal frequency-selective filters (lowpass, highpass, etc.). Note that the latter cannot be treated by the Laplace transform. I point this out because another common misconception is that the Laplace transform is more general than the Fourier transform. It is not, both transforms have their merits for solving certain problems.

You should search for more questions and answers on Fourier and Laplace transforms on this site, many things have been said already.

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  • $\begingroup$ I wouldn't call it a misconception but an interpretation of what we would call "more general": For instance, the Fourier Transform is the Unilateral Laplace Transform of a causal function evaluated on the $j \omega$ axis. In this case the Fourier Transform is by mathematical statement a selection or restricted portion of the Laplace Transform- so in that sense what does "more general" mean? $\endgroup$ Dec 13, 2021 at 13:00
  • $\begingroup$ (Even with that said I see and agree with your point that they each have their purposes- but I also share that "misconception" in observing the relationship between the two so wanted to get to the bottom of it really being a misconception or not) $\endgroup$ Dec 13, 2021 at 13:24
  • $\begingroup$ In a comment to this answer, you wrote: "It's not true that the Fourier transform can only capture the steady state behavior. The frequency response of an LTI system, which is the Fourier transform of its impulse response, completely characterizes its behavior, and hence it covers transients as well as the steady state response." What did you mean by this? My understanding is that $e^{st}$ can be decomposed into a product of a decaying/increasing exponential and a sinusoidal term, which capture... $\endgroup$
    – mhdadk
    Dec 13, 2021 at 14:21
  • $\begingroup$ ...the steady-state (as time goes to infinity) and transient behavior respectively. The Fourier transform only consists of the sinusoidal term and not the decaying/increasing exponential, so how can it capture steady state behavior? $\endgroup$
    – mhdadk
    Dec 13, 2021 at 14:21
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    $\begingroup$ @DanBoschen: The bilateral Laplace transform converges on a strip in the complex plane, and that strip must have a non-zero width. You can't compute the (bilateral) Laplace transform of $\sin(\omega_0t)$, or of $e^{j\omega_0t}$, or of $\textrm{sinc}(t)$. Yet, all of these functions have a Fourier transform. $\endgroup$
    – Matt L.
    Dec 13, 2021 at 17:07
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So what does that imply? If Fourier transform is essentially the frequency domain representation of $f(t)$ then does that mean Laplace transform is not frequency domain representation

There's probably some variation in terminology, but most authors consider both of them (and the z transform) to be frequency-domain techniques.

What does that imply? Fourier transform cannot be used for studying the transient behavior and Laplace transform cannot be used for studying steady state behaviour?

When you're being mathematically strict with yourself, Fourier transform analysis is easier to use with steady-state behavior, and Laplace transform analysis is easier for transient behavior.

Also, Fourier analysis is easier by far for systems that involve heterodyning (i.e., multiplication of a signal with a sine wave).

One last confusion: Which transform is more commonly used in practial applications ?Laplace transform or Fourier transform?

I use both.

When I have my signal processing hat on, I use the Fourier transform up to the point where I need to actually implement a filter, at which point I go to the z domain or Laplace domain.

When I have my control systems hat on, I just start in Laplace or z.

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The answer to your second question - whether the Fourier transform can capture transient or just steady-state behavior - is rather subtle.

As long as the transient behavior has a Fourier transform, then the Fourier transform will capture it. In this sense, the Fourier transform can capture transient behavior. The subtlety is that the transient behavior may or may not have a Fourier transform.

For example, consider a lightly damped, driven oscillator $x(t)$ with the equation of motion $$\frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2 x = f(t), \tag{1}$$ with $\gamma < \omega_0/2$ and $f(t)$ a specified external driving function. Fourier transforming the equation of motion gives the solution $$x(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{\tilde{f}(\omega)}{-\omega^2 + i \gamma \omega + \omega_0^2} e^{i \omega t} d\omega, \tag{2}$$ where $$\tilde{f}(\omega) := \int_{-\infty}^\infty f(t) e^{-i \omega t} dt$$ is the Fourier transform of the driving function $f(t)$.

Now consider the undriven case where $f(t) \equiv 0$. The general solution to the equation of motion (1) is the decaying exponential $$f(t) = A e^{-\frac{1}{2} \gamma t} \cos \left( \sqrt{\omega_0^2 - \frac{1}{4} \gamma^2}\ t + \delta \right), \tag{3}$$ where $A$ and $\delta$ are arbitrary real constants determined by the initial conditions. But in this case, $\tilde{f}(\omega) \equiv 0$, and so equation (2) simply gives the trivial solution $x(t) \equiv 0$. Similarly, if the driving force is sinusoidal at a given frequency $\omega_d$, then equation (2) above will only return the sinusoidal steady-state solution at frequency $\omega_d$ and not the decaying transients. This is origin of the claim that "The Fourier transform can only capture steady-state behavior."

What's going on here? Why isn't the Fourier transform capturing the transient homogeneous term (3)?

The answer is that the transient term (3) grows exponentially as $t \to -\infty$, so it is not a tempered distribution and it does not have a Fourier transform. So the Fourier transform method cannot capture it. You need the Laplace transform in order to capture it.

But there's another twist to the story. The transient solution (3) is a formal solution to the equation of motion, but it isn't very physical, because it extends back to negative infinite time. A more realistic setup is to "start the clock" at some initial time $t_0$ with some initial displacement $x(t_0) = x_0$ and initial velocity $\frac{dx}{dt}|_{t_0} = v_0$, and only consider the solution to the initial-value problem for times $t > t_0$. This setup is usually solved using the Laplace transform if you care about the decaying transients, but the Fourier transform can often handle it as well. The trick is to encode the initial conditions into an effective driving term via Dirac delta functions. In the case of the equation of motion (1), we can encode the intial conditions into the Dirac delta function driving terms $$f(t) = x_0\ \delta'(t - t_0) + v_0\ \delta(t - t_0)$$ (plus any "usual", continuous driving terms). These effective driving terms will generate a nonzero Fourier transform $\tilde{f}(\omega)$ - even in the undriven case - and equation (2) will indeed return the correct answer, including the decaying transients. In this case, the signal x(t) will be identically 0 for $t < t_0$ and will have a jump discontinuity to the appropriate initial conditions at $t = t_0$ (very similar to the setup for the Laplace transform method).

However, if the "transients" are exponentially growing rather than decaying, then this trick won't work, and the Fourier transform method will still fail.

To summarize: you can still use the Fourier transform method to solve initial-value problems and capture the behavior of transients that decay, or grow at most polynomially fast - but it requires a little more setup than the Laplace transform method does. But the Fourier transform method can't to capture "transients" that grow exponentially fast, although the Laplace transform method can.

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