Definitions for the BER (bit error rate) parameter table

Telecommunications Notes for Electrical, Electronic and Telecommunications Engineering, made by me.

Definitions for the BER (bit error rate) parameter table

Q(z) = 1/sqrt(2*Pi) int(z..infinity) exp(-u^2/2) du

If there is noise at the receiver input (assumed to be additive white Gaussian noise), the following is sampled at the output:

output = r_o = s_o + n_o

where s_o is a constant (s_o1 for a sent 1, s_o2 for a sent 0) and n_o is a zero-mean Gaussian random variable (the noise component). The constants s_o1 and s_o2 are associated with known input signaling waveforms s_1(t) and s_2(t), for a given type of receiver.

N_o/2

is the power spectral density (PSD) of the noise at the receiver input. Calling:

E_d

to the energy of the difference signal s_d(t)=s_1(t)-s_2(t) at the input of the receiver:

E_d = int(0..T) {s_1(t)-s_2(t)}^(1/2) dt

then the average energy per bit E_b is defined as a certain function of E_d that depends on the type of signaling.

Example.

For bit reception via QPSK-type bandpass signaling, according to the table (see textbooks), the minimum TX bandwidth required is R/2 where R is the bit rate and the BER is:

Q[sqrt(2*(E_b/N_o))]

requiring coherent detection.

Example.

For OOK bandpass signaling bit reception, the minimum TX bandwidth required is the bit rate R and the BER is:

Q[sqrt(E_b/N_o)]

for coherent detection, and:

1/2*exp[-1/2*(E_b/N_o)]

for non-coherent detection, and it must be true that (E_b/N_o) > 1/4.

Example.

There is a reception of the unipolar signaling type with white Gaussian noise at the input of a receiver filter, where E_b = A^2/(2R), the bit rate (data rate) is R = 9600 bps, the PSD (density power spectral) noise is 3*10^-5. Calculate (E_b/N_o) at the input of the filter, the minimum bandwidth and the corresponding BER.

Solution.

(E_b/N_o)_dB = 10 log_10 (E_b/N_o) = 10 log_10 ([A^2/(2R)]/[6*10^-5])

since PSD = N_o/2 = 3*10^-5, then N_o = 2*PSD = 6*10^-5. I mean:

= 10 log_10 (A^2/[2R*6*10^-5]) = 10 log_10 (A^2/[12*10^-5*9600]) = 10 log_10 (A^2/1.152) = 10 [2 log_10 A - log_10 1.152] = 10[2 log_10 A - 0.06145] dB

The minimum transmission bandwidth required is:

R/2 = 9600/2 = 4800bps

The BER is:

Q[sqrt(E_b/N_o)] = Q[sqrt([A^2/(2R)]/[6*10^-5])] = Q[sqrt(A^2/[12*10^-5 *9600])] = Q[sqrt(A^2/1.152)] = Q[A/sqrt(1.152)] = Q[A/1.0733]

For example, if parameter A = 5, then:

(E_b/N_o)_dB = 10[2 log_10 A - 0.06145] = 10[2 log_10 5 - 0.06145] = 13.4 dB

and also:

BER = Q[A/1.0733] ​​= Q[5/1.0733] ​​= Q[4.6585] = 1/sqrt(2*Pi) int(4.6585..infinity) exp(-u^2 /2) du = 0.000001593 = 1.6*10^-6

Sources:

1. Couch, "Digital and Analog Communications Systems."

2. Hsu, "Schaum's Outlines of Theory and Problems of Analog and Digital Communications", Schaum's Outlines series.

I highly recommend these two texts to learn Telecommunications; start with Schaum's first, then study the other!