Fourier Transform of Unit Impulse Function, Constant Amplitude and Complex Exponential Function

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Fourier Transform

The Fourier transform of a continuous-time function $x(t)$ can be defined as,

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

Fourier Transform of Unit Impulse Function

The unit impulse function is defined as,

$$\mathrm{\delta(t)\:=\:\begin{cases}1 \:\: for\:t\:=\:0 \\\\0 \:\: for\:t\: \neq\: 0 \end{cases}}$$

If it is given that

$$\mathrm{x(t)\:=\:\delta(t)}$$

Then, from the definition of Fourier transform, we have,

$$\mathrm{X(\omega)\:=\:\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\:=\:\int_{-\infty}^{\infty}\delta(t)e^{-j\omega t}dt}$$

As the impulse function exists only at t= 0. Thus,

$$\mathrm{X(\omega)\:=\:\int_{-\infty}^{\infty}\delta(t) e^{-j\omega t}dt\:=\:\int_{-\infty}^{\infty}1 \:\cdot\: e^{-j\omega t}dt \:=\: e^{-j\omega t}|_{t \:=\:0}\:=\:1}$$

$$\mathrm{\therefore\:F[\delta(t)]\:=\:1\:\:or\:\:\delta(t) \overset{FT}{\leftrightarrow}1}$$

That is, the Fourier transform of a unit impulse function is unity.

The magnitude and phase representation of the Fourier transform of unit impulse function are as follows −

$$\mathrm{Magnitude,\:|X(\omega)|\:=\:1;\:\:for\:all\:\omega}$$

$$\mathrm{Phase,\:\angle X(\omega)\:=\:0;\:\:for\:all\:\omega}$$

The graphical representation of the impulse function with its magnitude and phase spectra are shown in the figure.

Fourier Transform of Constant Amplitude

If the function is given as

$$\mathrm{x(t)\:=\:1}$$

Then, the function $X(t)$ is a constant function and it is not absolutely integrable, hence its Fourier transform cannot be found directly. Therefore, the Fourier transform of $X(t)\:=\:1$ is determined through inverse Fourier transform of impulse function $[\delta(\omega)]$.

From the definition of inverse Fourier transform, we have,

$$\mathrm{x(t)\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega}$$

Let

$$\mathrm{X(\omega)\:=\:\delta(\omega)}$$

Where,

$$\mathrm{\delta(\omega)\:=\:\begin{cases}1 \:\: for\:\omega\:=\:0 \\\\0 \:\: for\:\omega\: \neq\: 0\end{cases}}$$

$$\mathrm{\therefore\:x(t)\:=\:F^{-1}[X(\omega)]\:=\:F^{-1}[\delta(\omega)]}$$

$$\mathrm{\Rightarrow\:x(t)\:=\:x(t)\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega)e^{j\omega t}d\omega\:=\:\frac{1}{2\pi}\:\cdot\:(1)\:=\:\frac{1}{2\pi}}$$

$$\mathrm{\therefore\:F^{-1}[\delta(\omega)]\:=\:\frac{1}{2\pi}}$$

$$\mathrm{\Rightarrow\:F^{-1}[2\pi\delta(\omega)]\:=\:1}$$

Hence, the Fourier transform of a constant function is,

$$\mathrm{F[1]\:=\:2\pi\delta(\omega)\:or\:\:1\overset{FT}{\leftrightarrow}2\pi\delta(\omega)}$$

When the amplitude of the constant function is A, then the Fourier transform of the function becomes

$$\mathrm{A\overset{FT}{\leftrightarrow}2\pi A\delta(\omega)}$$

Fourier Transform of Complex Exponential Function

Consider the complex exponential function as,

$$\mathrm{x(t)\:=\:e^{j\omega_{0}t}}$$

The Fourier transform of a complex exponential function cannot be found directly. In order to find the Fourier transform of complex exponential function $x(t)$, consider finding the inverse Fourier transform of shifted impulse function in frequency domain $[\delta(\omega\:-\:\omega_{0})]$.

Let

$$\mathrm{X(\omega)\:=\:\delta(\omega\:-\:\omega_{0})}$$

Then, from the definition of inverse Fourier transform, we have,

$$\mathrm{x(t)\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d{\omega}}$$

$$\mathrm{\Rightarrow\:x(t)\:=\:F^{-1}[X(\omega)]\:=\:F^{-1}[\delta(\omega\:-\:\omega_{0})]}$$

$$\mathrm{\Rightarrow\:x(t)\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega\:-\:\omega_{0})e^{j\omega t}d{\omega}\:=\:\frac{1}{2\pi}e^{j\omega_{0} t}}$$

Therefore, the inverse Fourier transform of $\delta(\omega\:-\:\omega_{0})$ is,

$$\mathrm{F^{-1}[\delta(\omega\:-\:\omega_{0})]\:=\:\frac{1}{2\pi}e^{j\omega_{0} t}}$$

$$\mathrm{\Rightarrow\:F^{-1}[2\pi\delta(\omega\:-\:\omega_{0})]\:=\:e^{j\omega_{0} t}}$$

Hence, the Fourier transform of the complex exponential function is given by,

$$\mathrm{[e^{j\omega_{0} t}]\:=\:2\pi\delta(\omega\:-\:\omega_{0})}$$

Or, it can also be represented as,

$$\mathrm{e^{j\omega_{0} t}\overset{FT}{\leftrightarrow}2\pi\delta(\omega\:-\:\omega_{0})}$$