How to Find the Fourier Transform of a Sine or Cosine Wave
When we talk about the continuous-time Fourier transform (CTFT), it's a bit tricky to compute for pure sinusoids like sine and cosine because these functions extend infinitely in time and are not absolutely integrable.
To make the Fourier transform well-defined, we truncate the sinusoid over a finite interval, say from \( -T/2 \) to \( T/2 \), by multiplying it with a rectangular window.
Suppose you have a cosine signal:
\[ x(t) = \cos(2\pi f t) \]
Truncated over a finite interval using a rectangular window:
\[ x(t) = \cos(2\pi f t) \cdot \text{rect}\left( \frac{t}{T} \right) \]
where:
\[ \text{rect}\left(\frac{t}{T}\right) = \begin{cases} 1 & \text{if } |t| \leq \frac{T}{2} \\ 0 & \text{otherwise} \end{cases} \]
The rectangular function is 1 between \( -T/2 \) and \( T/2 \), and 0 elsewhere.
Frequency-domain Result: Convolution
By the Convolution Theorem:
Multiplication in time ⇨ Convolution in frequency
So, the Fourier transform of the truncated sinusoid becomes:
\[ X(f) = \text{FT} \left\{ \sin(2\pi f_0 t) \right\} * \text{FT} \left\{ \text{rect} \left( \frac{t}{T} \right) \right\} \]
- The Fourier transform of a sinusoid is a pair of delta functions at \( \pm f_0 \).
- The Fourier transform of a rectangular window is a sinc function: \( T \cdot \text{sinc}(T f) \).
Therefore, the result of the convolution is two sinc functions centered at \( \pm f_0 \). This shows how truncating a sinusoid spreads its energy in frequency.