The received signal level for a particular digital system is 0 15

SOLUTION

Concept:

\(\frac{{{E_b}}}{{{N_0}}} = \frac{{Energy\;received\;per\;information\;bit}}{{noise\;power\;spectral\;density}}\) 

Here,

\({E_b} = {P_r} \times {T_b} = \frac{{{P_r}}}{{{R_b}}}\) 

Where,

P_r = received power (Watt)

R_b = data rate or bit rate (bps)

\(\frac{{{E_b}}}{{{N_0}}}\) is a dimensionless parameter and is often expressed in dB.

\(\therefore {\left( {\frac{{{E_b}}}{{{N_0}}}} \right)_{dB}} = 10{\log _{10}}\left( {\frac{{{E_b}}}{{{N_0}}}} \right)\) 

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

\(\therefore \frac{{{E_b}}}{{{N_0}}} = \frac{{{P_r}}}{{{R_b} \times \left( {KT} \right)}} = \frac{{7.94 \times {{10}^{ - 16}}}}{{2400 \times 1.38 \times {{10}^{ - 23}} \times 1500}}\) 

\(= \frac{{1.598 \times {{10}^{ - 22}}}}{{{{10}^{ - 23}}}}\)

\(= \frac{{1.598}}{{{{10}^{ - 1}}}}\)

= 15.98

\(\therefore {\left( {\frac{{{E_b}}}{{{N_0}}}} \right)_{dB}} = 10{\log _{10}}\left( {\frac{{{E_b}}}{{{N_0}}}} \right) = 10{\log _{10}}\left( {15.98} \right)\)

= 12.03 dB

Important Points:

Power ratio in dB is given by:

\(dB = 10{\log _{10}}\left( {\frac{{{P_1}}}{{{P_2}}}} \right)\) 

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

\(\therefore dB = 20{\log _{10}}\left( {\frac{{{V_1}}}{{{V_2}}}} \right)\) 

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