# Is the Laplace transform a special case of Fourier transform? (Not the other way around)

Always had a thought about why Laplace transform reveals the transient properties of the system? My doubt is based on the following fact, Fourier transform is given as

$$\begin{equation} \mathscr{F}\left\lbrace f(t)\right\rbrace = \int_{-\infty}^\infty f(t) e^{ -j \omega t} dt \end{equation}$$

Where Mathematically and intuitively we believe that the angular frequency $$\omega$$ takes only real value.

What if, instead of taking real angular frequencies, if the variable $$\omega$$ assumes a complex angular frequency in the form $$\beta - j \alpha$$ , then,

$$j \omega t = j (\beta - j \alpha) t = (\alpha + j \beta ) t = s t$$

While taking Fourier transform w.r.t $$\omega$$, the quantity  $$\beta$$ will be real angular frequency in radians per second and $$\alpha$$ will be the $$\textbf{imaginary angular }$$ frequency in radians per second.

$$\begin{equation} \int_{-\infty}^\infty f(t) e^{ -j \omega t} dt = \int_{-\infty}^\infty f(t) e^{ - s t} dt = \mathscr{L}\left\lbrace f(t)\right\rbrace \end{equation}$$

Hence is it mathematically correct to consider bilateral Laplace transform as a special case of Fourier transform (not the other way around) when $$\omega$$ takes a complex angular form $$\beta - j \alpha$$ ?  I believe the fact that $$\omega$$ can take complex values is the reason why we get transient properties of the system when using Laplace transform.

The Fourier Transform is the Laplace Transform with the complex variable s restricted to be the imaginary axis on the s plane. For this reason the Fourier Transform only exists when the imaginary axis is within the region of convergence. The variable s is called a "complex frequency" as it is the frequency variable that can take on real ($$\sigma$$) and imaginary ($$\omega$$) components. That said, I would view the Fourier Transform as a subset of the Laplace Transform, or the Laplace Transform as an expansion on the Fourier Transform that provides a lot more functionality and can exist when the Fourier Transform can't.

This is also the reason that the frequency response for a system with a general transfer function $$H(s)$$ is given as $$H(j\omega)$$.

When a system is restricted to $$s= j\omega$$ as the input, then the input is restricted to be only sinusoids or signals given by $$e^{st}$$ with $$s = j\omega$$ which maintain a constant magnitude with time. By allowing s to have real and imaginary components as in $$s = \sigma + j\omega$$ then we also allow the input to grow or shrink with time, depending on which point in the s plane is used as the input to the system.

• what is the physical meaning of $\sigma$ ( say $\alpha$ in my question ) – abhilash Mar 16 at 15:37
• Consider the function $e^{-\sigma t}$--- it is simply a decaying exponential for positive t and real $\sigma$. So $\sigma$ provides a decay to the envelope of an otherwise sinusoidal function when you have both $\sigma$ and $\omega$ terms. Thus we see that under the integral how we can have a "region of convergence" for all values of $\omega$ that are greater than the right most pole (for causal functions). Try solving for the integral otherwise and you will see that the solution explodes--- so you can consider $\sigma$ to be a convergence factor for when the Fourier Transform would not exist – Dan Boschen Mar 16 at 16:14
• Agreed, I concur, it’s the decaying exponential, but still, I am unclear about the factor ‘$\sigma$’, let me explain, 1. Suppose we want to find out:- what part of a function $f(t)$ varies along a circle at a particular speed –or-- what component of $f(t)$ is having a periodic variation that moves with a specific time period. What we need to do is to correlate complex function $f(t)$ with a unit circle $e^{ j \omega t}$. \begin{equation} f(t) \otimes e^{ j \omega t} = \int_{-\infty}^\infty f(t) (e^{ j \omega t})^* dt \hspace{1.5cm} (with zero lag, \tau =0) \end{equation} – abhilash Mar 17 at 19:22
• Which is nothing but our famous Fourier transform. Here $'\omega'$ (speed of variation) is a factor we are throwing into the correlation process to get the magnitude of the variation (amplitude as output). 2. By utilizing a similar thought process, what is the significance of $'\sigma'$, -or- what is the actual meaning of the factor $'\sigma'$ which we are feeding into the transform process; to pull out the amplitude variation. – abhilash Mar 17 at 19:23
• @abhilash it is a decay or growth factor we can apply to the function auch that we can correlate to it and have that converge where it otherwise wouldn’t — and importantly it helps us identify the points of extrema (zeros and poles) that define all dynamics of the system. If you didn’t have the ability to vary the real part of the exponent, you would not find these solutions. – Dan Boschen Mar 17 at 19:29

I think this question is based on a wrong premise: "[...] why Laplace transform reveals the transient properties of the system". It's not true that the transients can only be obtained from the Laplace transform. The Fourier transform can do the same, assuming it exists. What is true is that it is more convenient to use the unilateral Laplace transform for taking into account non-zero initial conditions. But note that even this can be done with the Fourier transform if the initial conditions are modeled as separate sources.

So, to answer your question, no, the Laplace transform is not a special case of the Fourier transform. They are different tools with partially different (but overlapping) applications. For analyzing causal systems with possibly non-zero initial conditions the unilateral Laplace transform is a very practical tool. The Fourier transform is better suited for analyzing ideal systems (such as ideal frequency selective filters) or systems with idealized input signals (such as pure sinusoids). Note that there are signals for which the Fourier transform exists but the (bilateral) Laplace transform doesn't (e.g., sinusoids, complex exponentials, or impulse responses of ideal brick wall filters).

• L wouldn't it be "for which the Fourier Transform exists but the bilateral Laplace transform doesn't"? As I have often seen it go under the assumption that when not mentioned explicitly the unilateral transform is to be assumed. The Laplace transform for a sinusoid for t>=0 and 0 for t<0 does exist. – Dan Boschen Mar 16 at 16:21
• @DanBoschen: Well, that's a matter of definition, but for me it doesn't make sense to talk about the transform of a signal extending from $-\infty$ to $+\infty$ that only takes into account the interval $(0,\infty)$. So in that sense, the Laplace transform of $x(t)=\sin(\omega_0t)$ doesn't exist. What does exist is the Laplace transform of $x(t)=\sin(\omega_0t)u(t)$, but that's not what I was referring to. – Matt L. Mar 16 at 17:19
• @DanBoschen: Anyway, thanks for pointing out a possible misunderstanding, I'll edit my answer. – Matt L. Mar 16 at 17:19
• Right, just a matter of convention really: since the term “Laplace Transform” alone is so ubiquitous- maybe we shouldn’t all be so quick to assume it is the unilateral when not mentioned otherwise; but I think that is typically the case so will avoid misunderstanding indeed. – Dan Boschen Mar 16 at 17:23
• @DanBoschen: I'd say it has to do with the type of convergence we expect from different transforms. With the Fourier transform we're happy with less stringent definitions of convergence. But that would make a great question anyway, even though it might be a better fit at math.SE. – Matt L. Mar 16 at 18:54