Question:hard
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 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?
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?
Show Hint
QPSK behaves like two independent BPSK links, so \(P_b=\tfrac{1}{2}erfc(\sqrt{E_b/N_o})\).
Updated On: Jul 20, 2026
- \(10^{-3}\)
- \(10^{-4}\)
- \(10^{-5}\)
- \(10^{-6}\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Convert dB to a ratio using natural logarithms.
\[ \frac{E_b}{N_o}=10^{0.84}=e^{0.84\ln10} \] Since $\ln10\approx2.3026$, the exponent is $0.84\times2.3026\approx1.934$, so \[ \frac{E_b}{N_o}\approx e^{1.934}\approx6.92 \]
Step 2: Write the QPSK error probability using the Q function first.
For coherent QPSK, $P_b=Q(\sqrt{2E_b/N_o})$, and using $Q(x)=\tfrac{1}{2}erfc(x/\sqrt2)$ this becomes \[ P_b=\frac{1}{2}erfc\left(\sqrt{\frac{2E_b/N_o}{2}}\right)=\frac{1}{2}erfc\left(\sqrt{\frac{E_b}{N_o}}\right) \] which is the same expression used before, confirming the formula.
Step 3: Evaluate the argument.
\[ u=\sqrt{6.92}\approx2.63,\qquad u^2\approx6.92 \]
Step 4: Evaluate $\exp(-u^2)$ using the split $6.92=7-0.08$.
\[ \exp(-6.92)=\exp(-7)\exp(0.08)\approx(9.12\times10^{-4})(1.083)\approx9.88\times10^{-4} \]
Step 5: Divide by $u\sqrt{\pi}$.
\[ u\sqrt{\pi}\approx4.66,\qquad erfc(u)\approx\frac{9.88\times10^{-4}}{4.66}\approx2.12\times10^{-4} \]
Step 6: Halve to get the BER.
\[ P_b\approx1.06\times10^{-4} \] which rounds to \[ \boxed{10^{-4}} \] This matches the value obtained directly from the decibel split method.
\[ \frac{E_b}{N_o}=10^{0.84}=e^{0.84\ln10} \] Since $\ln10\approx2.3026$, the exponent is $0.84\times2.3026\approx1.934$, so \[ \frac{E_b}{N_o}\approx e^{1.934}\approx6.92 \]
Step 2: Write the QPSK error probability using the Q function first.
For coherent QPSK, $P_b=Q(\sqrt{2E_b/N_o})$, and using $Q(x)=\tfrac{1}{2}erfc(x/\sqrt2)$ this becomes \[ P_b=\frac{1}{2}erfc\left(\sqrt{\frac{2E_b/N_o}{2}}\right)=\frac{1}{2}erfc\left(\sqrt{\frac{E_b}{N_o}}\right) \] which is the same expression used before, confirming the formula.
Step 3: Evaluate the argument.
\[ u=\sqrt{6.92}\approx2.63,\qquad u^2\approx6.92 \]
Step 4: Evaluate $\exp(-u^2)$ using the split $6.92=7-0.08$.
\[ \exp(-6.92)=\exp(-7)\exp(0.08)\approx(9.12\times10^{-4})(1.083)\approx9.88\times10^{-4} \]
Step 5: Divide by $u\sqrt{\pi}$.
\[ u\sqrt{\pi}\approx4.66,\qquad erfc(u)\approx\frac{9.88\times10^{-4}}{4.66}\approx2.12\times10^{-4} \]
Step 6: Halve to get the BER.
\[ P_b\approx1.06\times10^{-4} \] which rounds to \[ \boxed{10^{-4}} \] This matches the value obtained directly from the decibel split method.
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