Question

The Fourier transform of signal $\delta(t+1) + \delta(t-1)$ is

• A. $2/(1 + j\omega)$
• B. $2/(1 - j\omega)$
• C. $2\cos \omega$
• D. $\cos \omega$

Hint:

A symmetric pair of impulses in the time domain always transforms into a cosine function in the frequency domain.
Conversely, a pair of impulses in the frequency domain corresponds to a cosine wave in the time domain.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the Fourier transform of a sum of two shifted Dirac delta functions, $\delta(t+1) + \delta(t-1)$.

Step 2: Key Formula or Approach:
We will use the standard properties of the Fourier transform:
- Linearity: $\mathcal{F}\{a \cdot x_1(t) + b \cdot x_2(t)\} = a \cdot X_1(\omega) + b \cdot X_2(\omega)$
- Time-Shifting Property: $\mathcal{F}\{x(t-t_0)\} = X(\omega) e^{-j\omega t_0}$
- Transform of impulse: $\mathcal{F}\{\delta(t)\} = 1$
- Euler's Identity for cosine:
\[ \cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2} \implies e^{j\theta} + e^{-j\theta} = 2\cos(\theta) \]

Step 3: Detailed Explanation:

• Let the signal be $x(t) = \delta(t+1) + \delta(t-1)$.

• Applying the time-shifting property to the first term $\delta(t+1)$, where $t_0 = -1$:
\[ \mathcal{F}\{\delta(t+1)\} = 1 \cdot e^{-j\omega(-1)} = e^{j\omega} \]

• Applying the time-shifting property to the second term $\delta(t-1)$, where $t_0 = 1$:
\[ \mathcal{F}\{\delta(t-1)\} = 1 \cdot e^{-j\omega(1)} = e^{-j\omega} \]

• By the linearity property, the total Fourier transform $X(\omega)$ is:
\[ X(\omega) = e^{j\omega} + e^{-j\omega} \]

• Applying Euler's Identity:
\[ X(\omega) = 2\cos(\omega) \]

Step 4: Final Answer:
The Fourier transform of the signal is $2\cos \omega$, which corresponds to Option (C).