Question

The energy of a signal \(A\delta[n]\) is _______.

• A. \(A^2\)
• B. \(\dfrac{A^2}{2}\)
• C. \(\dfrac{A^2}{4}\)
• D. 0

Hint:

The energy of \(A\delta[n]\) is always \(A^2\), because \(\delta[n]\) is non-zero only at \(n = 0\). Only one term contributes to the summation.

Answer 1

The Correct Option is A

Understanding the Signal:
\( \delta[n] \) is the unit impulse (or Dirac delta function in discrete time). Its defining property is:
\[ \delta[n] = \begin{cases} 1, & \text{if } n = 0 \\ 0, & \text{otherwise} \end{cases} \]

Given signal: \( x[n] = A\delta[n] \)

Energy of a discrete-time signal:
\[ E = \sum_{n=-\infty}^{\infty} |x[n]|^2 \]

Now,
\[ x[n] = A\delta[n] \Rightarrow |x[n]|^2 = A^2 \delta[n] \]
\[ E = \sum_{n=-\infty}^{\infty} A^2 \delta[n] = A^2 \sum_{n=-\infty}^{\infty} \delta[n] = A^2 (1) = A^2 \]