# Unilateral Laplace Transform vs Bilateral Fourier Transform

I would like to know why when we find the Laplace transform we use the one-sided (unilateral) version (all Laplace transform tables I can find are one-sided, like this one http://people.seas.harvard.edu/~jones/es154/Laplace/Table_pairs.html)

$$F(s)=\int^\infty_0e^{-st}f(t)dt$$

but when we find the Continuous Time Fourier Transform (CTFT) we use the two-sided (bilateral) version

$$F(\omega)=\int^\infty_{-\infty}e^{-j\omega t}f(t)dt$$

(a typical table http://www.mechmat.ethz.ch/Lectures/tables.pdf). What is the intuition or reasoning behind this? Since the analogy is often made that the CTFT is simply the Laplace transform with the real part of $s$ set to $0$, it would be natural to conclude that the limits of integration should not change.

• The second link is busted. Jul 14 '19 at 4:16

The reason behind this is not a mathematical reason but rather is an attempt to give an application to the Laplace transform to the analysis of physical systems. This is because any real physical system must necessarily be a casual system.

1. The Laplace transform is used because it is more generic and provide more information than the Fourier transform. Furthermore, it is easier to manipulate.
2. The use of unilateral or bilateral transform should be done with extreme care, depending on the type of causality of the system being analyzed:

• No-Causal System (Anticipative System): Bilateral definition (complete definition): \begin{align} \mathcal{L}\left\{ f(t) \right\} &\triangleq \int_{-\infty}^{+\infty} e^{-st} f(t)dt = F(s) &&& \mathcal{F}\left\{ f(t) \right\} &\triangleq \int_{-\infty}^{+\infty} e^{-j\omega t} f(t)dt = F(\omega) \end{align}
• Causal System (Non-anticipative System): Unilateral definition: \begin{align} \mathcal{L}\left\{ f(t) \right\} &\triangleq \int_{0}^{+\infty} e^{-st} f(t)dt = F(s) &&& \mathcal{F}\left\{ f(t) \right\} &\triangleq \int_{0}^{+\infty} e^{-j\omega t} f(t)dt = F(\omega) \end{align}
3. The choice of using the Fourier transform instead of the Laplace transform, is fully valid. But remember three key things:

• Fourier provides less information than Laplace.
• Fourier is more complex than Laplace.
• Use a bilateral or unilateral Fourier definition, according to the causality of the system.

If you do not know or fully understand what I'm talking about, let me explain...

The Black Box Model

Suppose you have a black box in which a signal $x(t)$ is inputted. This signal is processed in the black box and produces an output signal $y(t)$. Then it says that this black box processes the input signal and it produces the output signal, both in the time domain. Note that the black box model can be used to model any type of system. There are no limitations on it. This black box can be modeled mathematically with a function $h(t)$ (also in the time domain) so that you can set the output signal $y(t)$ by convolution process of the system function $h(t)$ and input signal $x(t)$. Algebraically: $$y(t) = h(t) \ast x(t)$$

How the Black Box Model is related to the Laplace and Fourier transforms?

The relationship between the Laplace/Fourier transform and convolution is the following property: $$y(t) = h(t) \ast x(t) \quad\xrightarrow{\mathcal{F}}\quad Y(\omega) = H(\omega) X(\omega)$$ Is interpreted as the transform of a convolution between $h(t)$ and $x(t)$ is multiplication of $H(\omega)$ and $X(\omega)$. So, the convolution in the time domain becomes a multiplication in the frequency domain. This is very useful since compute a multiplication is much simpler than computing a convolution.

Note that there is no restriction on using the Fourier transform or the Laplace transform to do this. However, the definition of the Laplace transform is more generic with respect to the Fourier transform, since $s=\sigma + \omega i$. \begin{align} \mathcal{L}\left\{ f(t) \right\} &\triangleq \int_{-\infty}^{+\infty} e^{-st} f(t)dt = F(s) &&& \mathcal{F}\left\{ f(t) \right\} &\triangleq \int_{-\infty}^{+\infty} e^{-j\omega t} f(t)dt = F(\omega) \end{align} $$\Longrightarrow\quad \begin{matrix} f(t) & \quad\xrightarrow{\mathcal{L}}\quad & F(s) \\ f(t) & \quad\xrightarrow{\mathcal{F}}\quad & F(\omega) \end{matrix} \quad\therefore\quad \left. F(s) \right|_{\sigma = 0} = F(\omega)$$ For this reason, the Laplace transform is used. In addition, the variable $s$ provides additional information because:

• Variable $\omega$ is related to the permanent system response; and
• Variable $\sigma$ is related to the inherent attenuation of the system.

$$y(t) = h(t) \ast x(t) \quad\xrightarrow{\mathcal{L}}\quad Y(s) = H(s) X(s)$$ Therefore, the Laplace transform is a powerful tool to analyze a general system, and even allows obtain the function $h(t)$ and/or $y(t)$ using the inverse transform with the following procedure:

How does all this is related with unilateral or bilateral definition of the Laplace transform?

There is a property of physical systems called causality, where the output of the black box depends on past and current input but not future inputs. This idea is intuitive. A clear example is the following: the weather tomorrow will depend of present and past conditions, but can never depend on the weather the day after tomorrow. In the Wikipedia link, this idea is better explained.

And what influences a system is causal or not? That it influences the definition of the system function $h(t)$: $$h(t) \mbox{ is a casual system} \quad\Leftrightarrow\quad \forall t\in\mathbb{R},\, t < 0:\quad h(t) = 0$$

Therefore, when computing the Laplace transform of a causal system, you get the unilateral definition of the Laplace transform: $$\mathcal{L}\left\{ h(t) \right\} \triangleq \underbrace{\int_{-\infty}^{+\infty} e^{-st} h(t)dt}_{\mbox{Bilateral Def.}} = \underbrace{\int_{0}^{+\infty}e^{-st} h(t)dt }_{\mbox{Unilateral Def.}}$$ Note that it is extremely important force analysis of a physical system to a causal system, especially since the system function $h(t)$ can be mathematically defined for $t <0$. For this reason, it applied directly the unilateral Laplace transform for real physical systems (or more generally speaking, for any casual system).

There are also no-causal systems (such as software that digitally processes an image) where the system state may depend on future states of the system ("future" pixels). In these cases, you must necessarily apply the bilateral transform.

• +1, perfectly structured, formatted and nice pictures. 10x. Jan 13 '17 at 9:50
• Great answer! Why is the Fourier transform more commonly used in the bilateral form then, as opposed to the Laplace transform?
– hddh
Aug 27 '18 at 7:47
• @hddh Rigorously, in all cases you should use bilateral notation because it's the formal definition. Remember that the unilateral form is a consequence of $h(t) = 0$ for all $t<0$ to force the condition of causality in physical systems. Considering this margin note, as Fourier transform is usually used to analyze periodic functions (sine and cosine), therefore $h(t) \neq 0$ for all $t < 0$ which implies that it's not possible to reduce the bilateral form to the unilateral one.
– Noir
Sep 5 '18 at 17:18
• @Noir I understand arriving at the Laplace transform via operator calculus, but I don't understand, the process of " Applying Laplace transform on both sides" I made some interpretation, but then it lead to the question, why it's more efficient? but I'm not sure, can you please take a look at math.stackexchange.com/questions/3431062/… Nov 12 '19 at 1:02

TL;DR

In short, If the system is not IAR (iff LTI) you can only use the unilateral transform. Otherwise, it is possible to, also, use the bilateral transform. If the system is not BIBO stable, you can only use the Laplace transform. Otherwise, you can use the Fourier transform as well.

With respect to the excellent answer by @Noir, I would like to add a few details and remarks.

First, it is important to understand that both Fourier and Laplace bilateral transforms are tools aimed at analyzing systems easily. None of them can be used without some assumptions. While Noir's assumption of a black-box model can describe any system, a system can only be defined as a convolution system if it is both linear and time-invariant (LTI). Otherwise, it has no impulse response kernel $$h(t)$$, which can describe it as a convolutional system. In such a case, if there is no $$h(t)$$ there is no bilateral $$H^L(s)$$ or $$H^F(\omega)$$ to speak of. An example of such a system is simply the system $$y(t)=\Psi \left\{ x \right\} (t)=x^2(t)$$ which is obviously not linear or $$y(t)=\Psi \left\{ x \right\} (t)=tx(t)$$ which is clearly not time-invariant.

The simplest and most common way to define a system (if possible) is by an ordinary diffraction equation (ODE) and it is an LTI system if it is initially at rest (IAR). This condition is met if the output of the system is $$0$$ as long as the input is $$0$$. In a mathematical representation: $$x(t)=0\forall t IAR is "stronger" than the initial conditions of an ODE in the sense that we don't get a value for a specific time point, but a set of values prior to a specific time. If the system is not an ODE the discussion of an LTI system is by the specific linearity and time-invariancy definitions and everything formerly discussed does not apply including the existence of transforms.

The IAR LTI state is an assumption on the system which conditions the use of bilateral Laplace and\or Fourier transforms. The unilateral Laplace transform, on the other hand, considers the initial conditions of the signal in the form: $$\mathcal{L}_+\left\{ x' \right\} (t)=sX_+^L(s)-x(0^-)$$ $$\mathcal{L}_+\left\{ x^{(n)} \right\} (t)=s^nX_+^L(s)-\sum_{k=1}^n x^{(k-1)}(0^-)s^{n-k}$$ In that sense, the unilateral transform can analyze non-LTI (none IAR) systems and is, therefore, very useful.

A second remark on the assumptions is in the difference between Fourier and Laplace transform. The Fourier transform is a private case of the Laplace transform given in the projection $$H^F(\omega)=H^L(s=j\omega)$$. This assumption implies that the imaginary axis $$s=j\omega$$ is included in the region of conversion (ROC) of the Laplace transform. If we analyze a system, the inclusion of the $$j\omega$$ axis is an iff for the bounded input bounded output (BIBO) stability of the system. If the system is not BIBO stable the axis is not in the ROC and only the Laplace transform is defined. There is no Fourier transform for such a case.