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In the modulation system, the energy per bit-to-noise power density ratio \(\frac{{{E_b}}}{{{N_o}}}\) is

(1) \(\frac{C}{N} \times \frac{{{f_b}}}{B}\)

(2) \(\frac{N}{C} \times \frac{B}{{{f_b}}}\)

(3) \(\frac{C}{N} \times \frac{B}{{{f_b}}}\)

(4) \(\frac{N}{C} \times \frac{{{f_b}}}{B}\)

Where:

N = Noise power of thermal (W)

B = Bandwidth (Hz)

C = Carrier power (W)

 fb = Bit rate (bps)


1. 1
2. 2
3. 3
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Correct Answer - Option 3 : 3

The bit energy is digital modulation is given by:

Eb = C × Tb

C = Carrier Power

Tb = Bit duration, which is the inverse of the bit rate, i.e.

\(T_b=\frac{1}{f_b}\)

The bit energy can now be written as:

\(E_b = \frac{C}{f_b}\)

Also, the total noise power over the complete Bandwidth (B) is defined as:

N = N0 × B

\(N_0 = \frac{N}{B}\)

where N0 = Noise power density

∴ The required ratio is given by:

\(\frac{E_b}{N_0} = \frac{C}{f_b}\times\frac{B}{N}\)

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