Question

A QPSK modulated signal from an additive white Gaussian noise (AWGN) channel is received with an \(E_b/N_o=8.4\) dB at the input of a coherent QPSK demodulator. A maximum likelihood reception method is used in the demodulator.
Assume the complementary error function \[ erfc(u)\cong\left[1/(u\sqrt{\pi})\right]\exp(-u^2). \]
Which is the nearest bit error rate (BER) at the output of the demodulator?

• A. \(10^{-3}\)
• B. \(10^{-4}\)
• C. \(10^{-5}\)
• D. \(10^{-6}\)

Hint:

QPSK behaves like two independent BPSK links, so \(P_b=\tfrac{1}{2}erfc(\sqrt{E_b/N_o})\).

Answer 1

The Correct Option is B

Step 1: Convert Eb/No from dB into a plain ratio.
\[ \frac{E_b}{N_o}=10^{8.4/10}=10^{0.84} \] Since $10^{0.8}\approx6.31$ and $10^{0.04}\approx1.096$, \[ 10^{0.84}\approx6.31\times1.096\approx6.92 \]

Step 2: Recall the bit error rate formula for coherent QPSK.
With Gray coded, maximum likelihood coherent detection, QPSK behaves like two independent BPSK channels on the in-phase and quadrature axes, and the bit error probability is \[ P_b=\frac{1}{2}\,erfc\left(\sqrt{\frac{E_b}{N_o}}\right) \]

Step 3: Compute the argument of erfc.
\[ u=\sqrt{6.92}\approx2.63 \]

Step 4: Apply the given approximation for erfc(u).
\[ erfc(u)\approx\frac{\exp(-u^2)}{u\sqrt{\pi}} \] Here $u^2\approx6.92$ which is just $E_b/N_o$ itself, so \[ \exp(-6.92)\approx9.9\times10^{-4} \] and \[ u\sqrt{\pi}\approx2.63\times1.772\approx4.66 \] So \[ erfc(u)\approx\frac{9.9\times10^{-4}}{4.66}\approx2.13\times10^{-4} \]

Step 5: Get the bit error rate.
\[ P_b=\frac{1}{2}\times2.13\times10^{-4}\approx1.06\times10^{-4} \]

Step 6: Round to the nearest power of ten.
$1.06\times10^{-4}$ is closest to $10^{-4}$.

Step 7: Check the other options.
$10^{-3}$ would result from forgetting the square root, that is, using $u=E_b/N_o$ directly inside erfc. $10^{-5}$ and $10^{-6}$ would need a much larger $E_b/N_o$ or a mistaken doubling of the exponent inside the exponential term.

Step 8: Final conclusion.
\[ \boxed{10^{-4}} \]