Fourier Transform of Unit Impulse Function

The question asks us to find the Fourier transform of the unit impulse function, denoted as $\delta(t)$. The Fourier transform is a mathematical tool that converts a function from the time domain to the frequency domain, revealing the frequency components present in a signal. The unit impulse function, also known as the Dirac delta function, is a special function that is zero everywhere except at $t=0$, where its value is infinitely large, such that its integral over all time is equal to 1.

Defining the Fourier Transform

The continuous-time Fourier transform of a function $x(t)$ is defined by the integral:

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

Here, $X(\omega)$ represents the function in the frequency domain, and $\omega$ is the angular frequency.

Properties of the Unit Impulse Function

The unit impulse function $\delta(t)$ has a crucial property known as the sifting property (or sampling property). This property states that for any continuous function $f(t)$:

$$\int_{-\infty}^{\infty} f(t) \delta(t-t_0) dt = f(t_0)$$

When the impulse is located at $t_0 = 0$, the property simplifies to:

$$\int_{-\infty}^{\infty} f(t) \delta(t) dt = f(0)$$

This means that integrating a function multiplied by a unit impulse at a specific point will "sift out" or "sample" the value of the function at that specific point.

Calculating the Fourier Transform of δ(t)

To find the Fourier transform of the unit impulse function $\delta(t)$, we substitute $x(t) = \delta(t)$ into the Fourier transform definition:

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

Now, we apply the sifting property of the delta function. In this integral, the function $f(t)$ is $e^{-j\omega t}$, and the impulse is at $t_0 = 0$. Therefore, according to the sifting property, the integral evaluates to the value of $f(t)$ at $t=0$:

$$X(\omega) = \left. e^{-j\omega t} \right|_{t=0}$$

Substitute $t=0$ into the expression:

$$X(\omega) = e^{-j\omega (0)}$$

$$X(\omega) = e^{0}$$

Since any non-zero number raised to the power of zero is 1:

$$X(\omega) = 1$$

Conclusion

The Fourier transform of the unit impulse function $\delta(t)$ is 1. This means that the unit impulse in the time domain contains all frequencies equally, with a magnitude of 1, across the entire frequency domain.

Comparing this result with the given options:

• Option 1: $\pi$
• Option 2: 1
• Option 3: 0
• Option 4: $\delta(\omega)$

The calculated Fourier transform matches Option 2.