Question

The Fourier transform of the following signal is \[ x(t) = \begin{cases} 1+\cos \pi t & |t|<1 \\ 0 & |t|>1 \end{cases} \]

• A. \( 2\text{sinc}(\omega) + \text{sinc}(\omega-\pi) - \text{sinc}(\pi+\omega) \)
• B. \( 2\sin(\omega) + \sin(\pi-\omega) - \sin(\pi+\omega) \)
• C. \( \text{sinc}(\pi-\omega) - \text{sinc}(\pi+\omega) \)
• D. \( \cos(\omega) + \cos(\pi-\omega) \) (Assuming \(\text{sinc}(x) = \sin(x)/x\))

Hint:

\(\mathcal{F}\{\text{rect}(t/T)\} = T \text{sinc}(\omega T/2)\) where \(\text{sinc}(x)=\sin(x)/x\). For width 2 (T=2), \(T/2=1\), so \(2\text{sinc}(\omega)\).
Modulation property: \(g(t)\cos(\omega_0 t) \leftrightarrow \frac{1}{2}[G(\omega-\omega_0) + G(\omega+\omega_0)]\).
\(\text{sinc}(x)\) is an even function.

Answer 1

The Correct Option is A

The signal \(x(t) = (1 + \cos \pi t) \cdot p_2(t)\), where \(p_2(t)\) is a rectangular pulse of width 2, non-zero from \(t = -1\) to \(t = 1\).

\(\mathcal{F}\{p_2(t)\} = 2 \frac{\sin(\omega \cdot 1)}{\omega \cdot 1} = 2\text{sinc}(\omega)\) (using \(\text{sinc}(x) = \frac{\sin(x)}{x}\)).

Let \(x(t) = p_2(t) + p_2(t) \cos(\pi t)\). 

\(\mathcal{F}\{p_2(t)\} = 2\text{sinc}(\omega)\).

For the term \(p_2(t) \cos(\pi t)\), we use the modulation property: If \(g(t) \leftrightarrow G(\omega)\), then \(g(t) \cos(\omega_0 t) \leftrightarrow \frac{1}{2}[G(\omega-\omega_0) + G(\omega+\omega_0)]\).

Here \(g(t) = p_2(t)\), \(G(\omega) = 2\text{sinc}(\omega)\), and \(\omega_0 = \pi\).

So, \(\mathcal{F}\{p_2(t) \cos(\pi t)\} = \frac{1}{2}[2\text{sinc}(\omega - \pi) + 2\text{sinc}(\omega + \pi)] = \text{sinc}(\omega - \pi) + \text{sinc}(\omega + \pi)\).

Therefore, \(X(\omega) = 2\text{sinc}(\omega) + \text{sinc}(\omega - \pi) + \text{sinc}(\omega + \pi)\).

Option (a) is \( 2\text{sinc}(\omega) + \text{sinc}(\omega - \pi) - \text{sinc}(\pi + \omega) \).

Since \(\text{sinc}(x)\) is an even function (\(\frac{\sin(x)}{x} = \frac{\sin(-x)}{-x}\)), \(\text{sinc}(\pi + \omega) = \text{sinc}(-(\pi + \omega)) = \text{sinc}(\omega + \pi)\).