Consider a simple AWGN (Additive white Gaussian noise) Wireless Channel. The car
ID: 2249147 • Letter: C
Question
Consider a simple AWGN (Additive white Gaussian noise) Wireless Channel. The carrier frequency is fc = 2115MHz and the channel bandwidth is f = B = 1MHz. The RX antenna temperature can be assumed to be Ta = 1000K due to man-made noise; the RX antenna gain is always at least 0dB (worst case), and depolarization loss not more than 3dB. We want a bit error rate (BER, error probability) Pe 104 and we use 4PSK as provided in the class textbook Figure 5.13b.
a) What SNR Eb / N0 do we need?
b) For a bit-rate Rs= 500 kb/s, what is the required carrier-to-noise ratio C/N? (in dB)
1.0 1.0 10-I 10-I 10-2 10" 10 10-4 10 10-4 M=8 10 10 10-6 10-6 M=8 2 3 45 6 7 89 10 11 12 13 14 15 ENNo) (dB) (a) Multilevel FSK (MFSK) 10-7 10 2 34 5 6 7 89 10 11 12 13 14 15 (ENo) (dB) (b) Multilevel PSK (MPSK) Figure 5.13 Theoretical Bit Error Rate for Multilevel FSK and PSKExplanation / Answer
From the Fig 5.13 it is clear that we have to maintain eb/no >= 9db to get pe <= 10^-4 for 4 psk
where M=4
and for C/N
the epression for C/N is given by
C/N = (eb/no)*(fb/BW)
where fb is bt rate and is given by fs/n = fs/(log2M) = fs/(log24) =fs/2
fs is given 500kbps
therefore we get fb = 250kbps
and
band width is given 1MHz
C/N = eb/no in dB + 10log (fb/BW)
C/N = 9dB-6.02dB
C/N = 2.98 dB
Noise power is given in boltzmann's equation
N = KTB
K is Boltzmann's constant = 1.380650x10-23 J/K
T is the effective temperature in degree Kelvin which is given 1000 k
B is the receiver bandwidth which is given 1MHz
NOISE POWER is calculated 1.17* 10^-7
which is -69.31dB and loss due to depolarization is given -3dB
so total -72.31dB will be the total noise power
and carrier power = 2.98dB - 72.31dB = -69.33 dB
The path loss in dB is given by
PL = 22 dB + 20log(d/)
'PL' path loss in dB;
'd' distance between the transmitter and receiver; and
is the wavelength of the carrier =c/f
Let us assume the distance between transmitter and reciever be 100 mts
so PL would be 22dB + 20log(100/0.14)
PL=79.07dB
SO P = 79.07dB - 69.33dB
P = 9.74dB
so the transmitted power would be around 9 W