A QPSK modulated signal from an additive white Gaussian noise (AWGN) channel is received with $E_b/N_0=8.4\text{dB}$ at the input of a coherent QPSK demodulator. A maximum-likelihood reception method is used in the demodulator. Assume the complimentary error function $\mathrm{er\; fc}\left(u\right)\approx \frac{1}{u\sqrt{\pi }}{\exp }^{(-u^2)}$. Which is the nearest bit error rate (BER) at the output of the demodulator?

The correct option is 10⁻⁴.

Here is the step-by-step derivation to find the bit error rate (BER) of a coherent QPSK demodulator:

Step 1: Understand the formula for QPSK Bit Error Rate (BER)
For a coherent QPSK demodulator with maximum-likelihood detection in an AWGN channel, the BER ($P_b$) is given by the Q-function:
$P_b=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$
We can express the Q-function in terms of the complementary error function, $\mathrm{erfc}\left(u\right)$, using the relation:
$Q\left(x\right)=\frac{1}{2}\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)$
Substituting $x=\sqrt{\frac{2E_b}{N_0}}$ into the relation gives the BER as:
$P_b=\frac{1}{2}\mathrm{erfc}\left(\sqrt{\frac{E_b}{N_0}}\right)$

Step 2: Convert the energy-per-bit to noise-power-density ratio from decibels (dB) to a linear scale
We are given:
$\frac{E_b}{N_0}=8.4\text{dB}$
To convert this to a linear ratio:
$\frac{E_b}{N_0}={10}^{8.4/10}={10}^{0.84}\approx 6.9183$

Step 3: Calculate the parameter $u$
Comparing the BER expression to the complementary error function argument, we have:
$u=\sqrt{\frac{E_b}{N_0}}=\sqrt{6.9183}\approx 2.6303$

Step 4: Use the given approximation for $\mathrm{erfc}\left(u\right)$
The problem provides the approximation:
$\mathrm{erfc}\left(u\right)\approx \frac{1}{u\sqrt{\pi }}\exp (-u^2)$
Now we substitute the values:
$u^2=6.9183$
$\exp (-u^2)=e^{-6.9183}\approx 0.0009895$
$u\sqrt{\pi }=2.6303\times \sqrt{3.14159}\approx 2.6303\times 1.7725\approx 4.6622$
Substituting these values back into the approximation gives:
$\mathrm{erfc}\left(2.6303\right)\approx \frac{0.0009895}{4.6622}\approx 2.1224\times {10}^{-4}$

Step 5: Compute the Bit Error Rate (BER)
Using the relation from Step 1:
$P_b=\frac{1}{2}\mathrm{erfc}\left(u\right)\approx \frac{1}{2}\times 2.1224\times {10}^{-4}=1.0612\times {10}^{-4}$
Therefore, the nearest bit error rate (BER) at the output of the demodulator is 10⁻⁴.