If a random communication signal has PDF given by P(X) = 2ae-b|x| for all x, -∞ < x < +∞ then :

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  1. b = 2a
  2. b = 4a
  3. b = a/2
  4. b = 0

Answer (Detailed Solution Below)

Option 2 : b = 4a
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Detailed Solution

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Concept:

For any valid probability density function (PDF), the total area under the curve must be 1:

\( \int_{-\infty}^{\infty} P(x)\, dx = 1 \)

Given:

\(P(x) = 2ae^{-b|x|} \quad \text{for all } x \in (-\infty, \infty)\)

Calculation:

Since the function is even (depends on |x|), integrate over the positive side and double it:\[ \int_{-\infty}^{\infty} 2ae^{-b|x|} dx = 2 \cdot \int_0^{\infty} 2ae^{-bx} dx = 4a \cdot \int_0^{\infty} e^{-bx} dx \]
\(= 4a \cdot \left[ \frac{1}{b} \right] = \frac{4a}{b}\)

 

Set equal to 1 (normalization condition):

\[ \frac{4a}{b} = 1 \Rightarrow b = 4a \]

Hence, the correct option is 2

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