## Q. 2.15

For a QPSK system and the given parameters, determine
a.  Carrier power in dBm.
b.  Noise power in dBm.
c.  Noise power density in dBm.
d.  Energy per bit in dBJ.
e.  Carrier-to-noise power ratio in dB.
f.  $E_{b}/N_{0}$ ratio.

C = $10^{-12}$ W                    $f_{b}$ = 60 kbps

N = 1.2 × $10^{-14}$ W           B = 120 kHz

## Verified Solution

a.  The carrier power in dBm is determined by substituting into Equation 28:

$C_{(dBm)}=10 \log\frac{C_(watts)}{0.001}$        (28)

$C=10 \log\!\frac{10^{-12}}{0.001}=-90\,dBm$

b.  The noise power in dBm is determined by substituting into Equation 30:

$N_{(dBm)}=10 \log\frac{KTB}{0.001}$        (30)

$N=10 \log\!\frac{1.2×10^{-14}}{0.001}=-109.2\,dBm$

c.  The noise power density is determined by substituting into Equation 40:

$N_{0(dBm)}=N_{(dBm)}- 10 \log B$        (40)

$N_{0}$ = -109.2 dBm – 10 log 120 KHz = -160 dBm

d.  The energy per bit is determined by substituting into Equation 36:

$E_{b\,(dBJ)}=10 \log\frac{C}{f_{b}}$        (36)

$E_{b}=10 \log\!\frac{10^{-12}}{60\,kbps}=-167.8\, dBJ$

e.  The carrier-to-noise power ratio is determined by substituting into Equation 34:

$E_{b\,(dBJ)}=10 \log E_{b}$        (34)

$\frac{C}{N}=10 \log\frac{10^{-12}}{1.2\,×\,10^{-14}}=19.2\, dB$

f.  The energy per bit-to-noise density ratio is determined by substituting into Equation 45:

$\frac{E_{b}}{N_{0}}(dB)=10 \log\frac{C}{N}+10\log\,\frac{B}{f_{b}}$           (45)

$\frac{E_{b}}{N_{0}}=19.2 +10 \log\!\frac{120\,kHz}{60\,kbps}=22.2\,dB$