# Relation between Laplace Transform and Fourier Transform

## Fourier Transform

The Fourier transform is a transformation technique which is used to transform the signals from continuous-time domain to the corresponding frequency domain.

Mathematically, if $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a continuous-time domain function, then its Fourier transform is given by,

$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{j\omega t}}\:\mathit{dt}} \:\:\:\:\:\:...(1)}$$

## Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a time domain function, then its Laplace transform is defined as −

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}\:\:\:\:\:\:...(2)}$$

Where, s is a complex variable and it is given by,

$$\mathrm{\mathit{s}\:\mathrm{=}\:\sigma \:\mathrm{+}\:\mathit{j\omega}}$$

## Relation between Laplace and Fourier Transforms

From the definition of Fourier transform, we have the Fourier transform of a time-domain function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a continuous sum of exponential functions of the form $\mathit{e^{j\omega t}}$, which means it uses addition of waves of positive and negative frequencies. The Fourier transform is used for solving the differential equations that relate the input and output of a system.

In order to solve these differential equations, first the differential equations are to be converted into the algebraic equations and then the algebraic equations are manipulated to obtain the Fourier transform of the output $\mathit{Y}\mathrm{\left(\mathit{\omega}\right)}$ as the function of frequency response $\mathit{H}\mathrm{\left(\mathit{\omega}\right)}$ and the Fourier transform of input $\mathit{X}\mathrm{\left(\mathit{\omega}\right)}$ of the system. Finally, the output as the function of time is then obtained by taking the inverse Fourier transform of the output $\mathit{Y}\mathrm{\left(\mathit{\omega}\right)}$.

But the Fourier transform does not exist for many signals such as $\left|\mathit{x}\mathrm{\left(\mathit{t}\right)} \right|$ because it is not absolutely integrable. Also, to analyse the unstable systems the Fourier transform cannot be used.

Therefore, the Laplace transform is used where the Fourier transform cannot be used. The Laplace transform redefines the transform and includes an exponential convergence factor σ along with jω. Therefore, using the Laplace transform the time-domain signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ can be represented as a sum of complex exponential functions of the form $\mathit{e^{st}}$.

Due to the inclusion of the exponential convergence factor (σ), the function $\left|\mathit{x}\mathrm{\left(\mathit{t}\right)} \right|$ becomes absolutely integrable. Therefore, the Laplace transform exists for such functions for which the Fourier transform does not exist.

Now, the Fourier transform of a continuous-time signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is defined as −

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{j\omega t}}\:\mathit{dt}}\:\:\:\:\:\:...(3)}$$

And the inverse Fourier transform is given by,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t }\right)}\:\mathrm{=}\:\frac{1}{2\pi}\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{\omega}\right)}\mathit{e^{\mathit{j\omega t}}\:\mathit{d\omega }}\:\:\:\:\:\:...(4)}$$

Replacing ω by $\sigma \:\mathrm{+}\:\mathit{j\omega}$ in the definition of Fourier transform, we get,

$$\mathrm{\mathit{X}\left(\sigma \:\mathrm{+}\:\mathit{j\omega}\right)\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathrm{\left(\sigma \:\mathrm{+}\:\mathit{j\omega} \right )}t}\:\mathit{dt}}\:\:\:\:\:\:...(5)}$$

Then, the inverse Fourier transform, i.e., $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is given by,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t }\right)}\:\mathrm{=}\:\frac{1}{2\pi}\int_{-\infty }^{\infty}\mathit{X}\mathrm{\left(\sigma \:\mathrm{+}\:\mathit{j\omega} \right )}\mathit{e^{\mathrm{\left(\sigma \:\mathrm{+}\:\mathit{j\omega} \right )}t}\:\mathit{d\omega }}\:\:\:\:\:\:...(6)}$$

The term $\sigma \:\mathrm{+}\:\mathit{j\omega}$ is denoted by s. Then,

$$\mathrm{\mathit{j}\:\mathrm{=}\:\frac{\mathit{ds}}{\mathit{d\omega}}\:\mathrm{or}\:\mathit{d\omega}\:\mathrm{=}\:\frac{\mathit{ds}}{\mathit{j}}}$$

On putting these values in equations (5) & (6), we get,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{s }\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}\:\:\:\:\:\:...(7)}$$

And

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t }\right)}\:\mathrm{=}\:\frac{1}{2\pi\mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty } \right)}}^{\mathrm{\left(\sigma \mathrm{+}\mathit{j\infty} \right)}}\mathit{X}\mathrm{\left(\mathit{s}\right)}\mathit{e^{\mathit{st}}\:\mathit{d\omega }}\:\:\:\:\:\:...(8)}$$

The equations (7) and (8) constitutes the bilateral Laplace transform pair or the complex Fourier transform pair. Therefore, the Laplace transform is just the complex Fourier transform of a signal. Hence, the Fourier transform is equivalent to the Laplace transform evaluated along the imaginary axis of the s-plane, i.e.,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}|_{\mathit{s=j\omega}}}$$

In other words, we can say the Laplace transform of a function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is equal to the Fourier transform of $\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{\sigma t}}}$, i.e.,

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)} \right]}\:\mathrm{=}\:\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{\sigma t}}} \right ]}}$$