The Fourier transform of a rectangular pulse is
sinc function
The question asks us to identify the Fourier transform of a rectangular pulse. The Fourier transform is a mathematical tool that decomposes a function of time (or space) into its constituent frequencies.
Let's first define a rectangular pulse. A common way to represent a rectangular pulse centered at time \(t=0\) with amplitude \(A\) and duration \(\tau\) is:
\( f(t) = \begin{cases} A & \text{if } |t| \le \frac{\tau}{2} \\ 0 & \text{if } |t| > \frac{\tau}{2} \end{cases} \)
Alternatively, using the rectangular function notation:
\(f(t) = A \cdot \text{rect}\left(\frac{t}{\tau}\right)\)
The Fourier transform \(F(\omega)\) of a function \(f(t)\) is given by the integral:
\(F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt\)
Let's compute the Fourier transform of the rectangular pulse \(f(t)\) defined above:
\( F(\omega) = \int_{-\infty}^{\infty} A \cdot \text{rect}\left(\frac{t}{\tau}\right) e^{-j\omega t} dt \)
Since the rectangular function is non-zero only for \(-\frac{\tau}{2} \le t \le \frac{\tau}{2}\), the integral becomes:
\( F(\omega) = \int_{-\tau/2}^{\tau/2} A e^{-j\omega t} dt \)
We can pull the constant \(A\) out of the integral:
\( F(\omega) = A \int_{-\tau/2}^{\tau/2} e^{-j\omega t} dt \)
Now, integrate with respect to \(t\):
\( F(\omega) = A \left[ \frac{e^{-j\omega t}}{-j\omega} \right]_{-\tau/2}^{\tau/2} \)
\( F(\omega) = \frac{A}{-j\omega} \left( e^{-j\omega \tau/2} - e^{j\omega \tau/2} \right) \)
We can rewrite \((e^{-j\theta} - e^{j\theta})\) using Euler's formula, \(e^{j\theta} = \cos\theta + j\sin\theta\), which gives \(e^{-j\theta} - e^{j\theta} = (\cos\theta - j\sin\theta) - (\cos\theta + j\sin\theta) = -2j\sin\theta\). Letting \(\theta = \omega\tau/2\), we get \((e^{-j\omega \tau/2} - e^{j\omega \tau/2}) = -2j\sin(\omega\tau/2)\).
Substitute this back into the expression for \(F(\omega)\):
\( F(\omega) = \frac{A}{-j\omega} (-2j\sin(\omega\tau/2)) \)
\( F(\omega) = A \frac{2j\sin(\omega\tau/2)}{j\omega} \)
\( F(\omega) = A \frac{2\sin(\omega\tau/2)}{\omega} \)
To relate this to the sinc function, recall the definition of the normalized sinc function:
\( \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \)
And the unnormalized sinc function (often used in signal processing):
\( \text{sinc}(x) = \frac{\sin(x)}{x} \)
Our result is \(F(\omega) = A \frac{2\sin(\omega\tau/2)}{\omega}\). We can rewrite the denominator \(\omega\) as \(\frac{2(\omega\tau/2)}{\tau}\).
\( F(\omega) = A \frac{\sin(\omega\tau/2)}{(\omega\tau/2)/\tau} = A\tau \frac{\sin(\omega\tau/2)}{\omega\tau/2} \)
This matches the form of the unnormalized sinc function, \(\frac{\sin(x)}{x}\), where \(x = \omega\tau/2\). Therefore, the Fourier transform is \(F(\omega) = A\tau \text{sinc}(\omega\tau/2)\).
This result shows that the Fourier transform of a rectangular pulse is a sinc function.
Let's consider why the other options are incorrect:
Based on the derivation and the properties of Fourier transforms, the Fourier transform of a rectangular pulse is a sinc function.
| Time Domain Function \(f(t)\) | Frequency Domain Function \(F(\omega)\) |
|---|---|
| Rectangular Pulse \(A \cdot \text{rect}(t/\tau)\) | \(A\tau \cdot \text{sinc}(\omega\tau/2)\) |
| Triangular Pulse \(\Lambda(t/\tau)\) | \(\tau \cdot \text{sinc}^2(\omega\tau/2)\) |
| Impulse \(\delta(t)\) | \(1\) |
| Constant \(1\) | \(2\pi \delta(\omega)\) |
| Sinc Function \(A \cdot \text{sinc}(Wt)\) | \((A\pi/W) \cdot \text{rect}(\omega/(2W))\) |
| Concept | Description |
| Fourier Transform | Transforms a signal from the time domain to the frequency domain. |
| Rectangular Pulse | A function that is constant for a finite duration and zero elsewhere. |
| Sinc Function | Defined as \(\frac{\sin(x)}{x}\) (unnormalized) or \(\frac{\sin(\pi x)}{\pi x}\) (normalized). It has a main lobe and decaying side lobes. |
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