Why is the sine term eliminated when computing the Fourier transform of the rectangular pulse?

Conditions

• The integration interval is $[-1, 1]$.
• The function being integrated includes $\sin(\omega t)$.
• $\omega$ is a constant with respect to the integration variable $t$.

Reasoning, step by step

1. Expand the complex exponential using Euler's formula: $e^{-i\omega t} = \cos(\omega t) - i\sin(\omega t)$.
2. Identify the parity of the cosine term: $\cos(\omega(-t)) = \cos(\omega t)$, so it is even.
3. Identify the parity of the sine term: $\sin(\omega(-t)) = -\sin(\omega t)$, so it is odd.
4. Apply the property of integrals over symmetric intervals: $\int_{-a}^{a} f(t) dt = 0$ if $f(t)$ is odd.
5. Conclude that $\int_{-1}^{1} \sin(\omega t) dt = 0$.
6. Retain only the cosine term: $\int_{-1}^{1} \cos(\omega t) dt$.
7. Simplify further using the even property of cosine: $2 \int_{0}^{1} \cos(\omega t) dt$.

Example

The video states: 'Because this is an even function... this is an odd function... so the odd function vanishes.' The whiteboard shows the transition from $\int_{-1}^{1} (\cos \omega t - i \sin \omega t) dt$ to $2 \int_{0}^{1} \cos \omega t dt$.

Common misconceptions

• Thinking the sine term is zero because $\sin(0)=0$; it is zero because of the symmetry of the integral.
• Assuming the cosine term also vanishes; it does not because it is even.
• Confusing the variable $t$ with $\omega$ when determining parity; parity is determined with respect to the integration variable $t$.

Watch the explanation

Explore next

• What remains after the sine term is eliminated?
• How is the cosine integral evaluated?
• Does this symmetry argument hold for non-symmetric intervals?