The received signal level for a particular digital system is 0 151 dBW and the effective noise temperature of the receiver system is 1500 K. The value of  required for a link transmitting 2400 bps is

Concept:

Here,

Where,

P_r = received power (Watt)

R_b = data rate or bit rate (bps)

 is a dimensionless parameter and is often expressed in dB.

Calculation:

Given that,

P_r = -151 dBW ; here dBW means decibel-watt

R_b = 2400 bPS

T = effective noise temperature of the receiver system = 1500 K

To convert P_r in watt from dBW, we use the relation:

P_r (dBW) = 10 log₁₀ (P_r(W)]

-151 = 10 log₁₀ [P_r(W)]

P_r(W) = 10^-15.1 = 7.94 × 10^-16 watt.

When noise Temperature is considered in the receiver system than N_o will be:

N₀ = K.T (watt/Hz)

Where,

K = Boltzmann’s constant (J/K)

= 1.38 × 10^-23 J/K

= 15.98

= 12.03 dB

Important Points:

Power ratio in dB is given by:

Ratio’s like the voltage, current, sound, and pressure levels are calculated as the ratio of squares, i.e.

dBm means decibel-milliwatt, and is defined as:

P(dBm) = 10 log₁₀ )P(mw)]

dBW means decibel-watt

P(dBW) = 10 log₁₀ [P(watt)]

P(dBW) = P(dBm) - 30