QPSK Modulation

Quadrature Phase Shift Keying is the most widely deployed digital modulation scheme in modern wireless systems. QPSK doubles the spectral efficiency of BPSK by transmitting two bits per symbol while maintaining the same BER performance as BPSK when measured against Eb/N0. Understanding QPSK is essential for analyzing 3G, 4G LTE downlink, Wi-Fi, and satellite systems.

Core Concept: Two BPSK Streams in Quadrature

QPSK uses four phase states: 45°, 135°, 225°, 315°. Each phase state carries a 2-bit dibit (symbol of two bits), so one QPSK symbol transmits log₂(4) = 2 bits. The modulated signal is: s(t) = Ac cos(2πfct + φk), where φk is the phase of the k-th symbol. Alternatively, using the quadrature decomposition: s(t) = Ic cos(2πfct) - Qk sin(2πfct), where Ik and Qk are the in-phase and quadrature components of the symbol.

This decomposition reveals the key insight: QPSK is equivalent to two independent BPSK streams transmitted simultaneously on cosine and sine carriers. The in-phase channel carries the first bit of each dibit and the quadrature channel carries the second bit. Because the cosine and sine carriers are orthogonal, the two channels do not interfere with each other. The receiver separates them using two parallel correlators.

Mathematical Expression

With symbol period Ts = 2Tb (since each symbol carries 2 bits at bit duration Tb), the symbol energy is Es = Ac²Ts/2 = 2Eb where Eb = Ac²Tb/2. In signal space, the four symbols lie at the corners of a square of radius sqrt(Es). The minimum Euclidean distance between adjacent symbols is d = sqrt(2Es) = sqrt(4Eb) = 2sqrt(Eb).

The BER of QPSK is: Pb = Q(sqrt(2Eb/N0)). This is the same as BPSK. The reason is that each I and Q bit is protected by the same BER formula, because each bit is effectively in a BPSK link. The symbol error rate (SER) is slightly higher: Ps ≈ 2Q(sqrt(2Eb/N0)) for moderate to high SNR. The spectral efficiency doubles compared to BPSK because the same bandwidth now carries 2 bits per symbol.

Practical Understanding

QPSK is used extensively in 3G UMTS (downlink and uplink), DVB-S2 satellite broadcasting, and DOCSIS 3.0 cable systems. In LTE, QPSK is used in the Physical Downlink Shared Channel (PDSCH) when the UE reports a low Channel Quality Indicator (CQI), meaning the channel is noisy. Higher-order schemes like 16-QAM and 64-QAM are selected when the channel improves.

Offset QPSK (OQPSK) is a variant where the Q channel is delayed by Tb relative to the I channel. This prevents the 180-degree phase transitions that occur in regular QPSK when both bits of a dibit flip simultaneously. 180-degree transitions cause large amplitude fluctuations after bandlimiting filtering, which increases envelope variation and places demands on the power amplifier linearity. OQPSK limits transitions to 90 degrees maximum, reducing envelope variation.

Given:
QPSK system, Bit rate Rb = 10 Mbps, Channel bandwidth B = 6 MHz, Eb/N0 = 9 dB

Why this formula applies:
QPSK has the same BER as BPSK because each quadrature channel is an independent BPSK stream.

Formula:
Pb = Q(sqrt(2 Eb/N0))
Symbol rate Rs = Rb / 2 (since 2 bits per symbol)
Bandwidth efficiency η = Rb / B

Substitution:
Rs = 10 Mbps / 2 = 5 Msymbols/s
η = 10 Mbps / 6 MHz = 1.67 bits/s/Hz
Eb/N0 (linear) = 10^(9/10) = 7.94
2 Eb/N0 = 15.87
sqrt(15.87) = 3.984

Calculation:
Pb = Q(3.984) ≈ 3.4 × 10^-5
For comparison BPSK with same Rb would need 10 MHz bandwidth; QPSK saves 4 MHz.

Final Answer with units:
BER ≈ 3.4 × 10^-5, Symbol rate = 5 Msymbols/s, Bandwidth efficiency = 1.67 bits/s/Hz

Exam Tip: QPSK and BPSK have the same BER formula Q(sqrt(2Eb/N0)) when Eb/N0 is the x-axis. However, if the x-axis is Es/N0 instead, BPSK BER = Q(sqrt(2Es/N0)) while QPSK BER = Q(sqrt(Es/N0)) because Es = 2Eb for QPSK. Always check which energy ratio the problem uses.

• The input bit stream is split into odd and even bits (I and Q streams), each running at Rb/2 bps. Each stream modulates an orthogonal carrier, producing two independent BPSK signals.
• The transmitted QPSK signal is the sum of the I and Q channel signals: s(t) = Ik Ac cos(2πfct) - Qk Ac sin(2πfct).
• The demodulator uses two separate correlators, one for the cosine carrier (I channel) and one for the -sine carrier (Q channel). Each correlator independently recovers one bit of the dibit.
• Gray coding ensures adjacent constellation points differ by only one bit, so a wrong symbol decision in favorable SNR conditions causes at most one bit error per two-bit symbol, keeping BER close to SER/2.

Quick Revision

• QPSK maps 2 bits per symbol; 4 constellation points at 45, 135, 225, 315 degrees.
• BER = Q(sqrt(2Eb/N0)), identical to BPSK when plotted against Eb/N0.
• Bandwidth efficiency doubles: ideal η = 2 bits/s/Hz vs 1 bit/s/Hz for BPSK.
• QPSK = two independent BPSK streams on cos and sin carriers; I and Q channels are orthogonal.
• Symbol rate Rs = Rb/2; required bandwidth (Nyquist) = Rs = Rb/2 Hz.
• Gray coding limits bit errors to 1 per symbol error event at high SNR.
• Trap: if the question gives Es/N0, note Es = 2Eb for QPSK, so BER = Q(sqrt(Es/N0)) not Q(sqrt(2Es/N0)).