How is the Fourier transform of the rectangular pulse function computed?

Q: How is the Fourier transform of the rectangular pulse function computed?

The Fourier transform of the rectangular pulse function $f(t)$ is computed by substituting its piecewise definition into the integral definition of the Fourier transform. Because $f(t)$ is zero outside the interval $[-1, 1]$ and equals 1 inside, the infinite integral reduces to a finite integral over $[-1, 1]$. Using Euler's formula, the complex exponential is expanded into cosine and sine terms. The sine term integrates to zero over the symmetric interval because it is an odd function, while the cosine term is an even function, allowing the integral to be simplified to twice the integral over $[0, 1]$. Evaluating this definite integral yields the final result $F(\omega) = 2 \frac{\sin \omega}{\omega}$.

The Fourier transform of the rectangular pulse function

f(t)

is computed by substituting its piecewise definition into the integral definition of the Fourier transform. Because

f(t)

is zero outside the interval

[-1, 1]

and equals 1 inside, the infinite integral reduces to a finite integral over

[-1, 1]

. Using Euler's formula, the complex exponential is expanded into cosine and sine terms. The sine term integrates to zero over the symmetric interval because it is an odd function, while the cosine term is an even function, allowing the integral to be simplified to twice the integral over

[0, 1]

. Evaluating this definite integral yields the final result

F(\omega) = 2 \frac{\sin \omega}{\omega}

Conditions

The function is defined as $f(t) = 1$ for $|t| \le 1$ and $0$ otherwise.
The Fourier transform is defined as $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$ .
$\omega$ is a real number.

Reasoning, step by step

Write the definition of the Fourier transform: $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$ .
Substitute the piecewise definition of $f(t)$ , restricting the integration limits to $[-1, 1]$ where $f(t)=1$ : $F(\omega) = \int_{-1}^{1} e^{-i\omega t} dt$ .
Apply Euler's formula to expand the integrand: $e^{-i\omega t} = \cos(\omega t) - i\sin(\omega t)$ .
Separate the integral into real and imaginary parts: $\int_{-1}^{1} \cos(\omega t) dt - i\int_{-1}^{1} \sin(\omega t) dt$ .
Evaluate the sine integral: Since $\sin(\omega t)$ is an odd function and the interval $[-1, 1]$ is symmetric about the origin, $\int_{-1}^{1} \sin(\omega t) dt = 0$ .
Evaluate the cosine integral: Since $\cos(\omega t)$ is an even function, $\int_{-1}^{1} \cos(\omega t) dt = 2\int_{0}^{1} \cos(\omega t) dt$ .
Compute the antiderivative: $2\int_{0}^{1} \cos(\omega t) dt = 2 \left[ \frac{\sin(\omega t)}{\omega} \right]_{0}^{1}$ .
Substitute the limits: $2 \left( \frac{\sin(\omega)}{\omega} - \frac{\sin(0)}{\omega} \right) = 2 \frac{\sin \omega}{\omega}$ .
Conclude that $F(\omega) = 2 \frac{\sin \omega}{\omega}$ .
Note that for $\omega=0$ , the limit is taken, yielding $F(0)=2$ .

Example

The video explicitly derives this on the whiteboard: ' $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt = \int_{-1}^{1} e^{-i\omega t} dt = \int_{-1}^{1} (\cos \omega t - i \sin \omega t) dt = 2 \int_{0}^{1} \cos \omega t dt = 2 \left[ \frac{\sin \omega t}{\omega} \right]_{0}^{1} = 2 \frac{\sin \omega}{\omega}$ '.

Common misconceptions

Assuming the integral of the sine term is non-zero; it vanishes due to the odd symmetry over a symmetric interval.
Forgetting to adjust the integration limits from $(-\infty, \infty)$ to $[-1, 1]$ based on the support of $f(t)$ .
Confusing the variable of integration ( $t$ ) with the frequency variable ( $\omega$ ) during the antiderivative step.

Watch the explanation

BilibiliFourier series, Fourier transform, Green's functions｜海浪晶灵aa

2:00 – 3:15Watch this moment ↗

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.