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QPSK modulation and demodulation

Short answer

QPSK splits the input bit stream into in-phase and quadrature channels, each modulating a carrier at 90° offset, then sums the results. Coherent demodulation uses two correlators with recovered carriers and zero thresholds to recover odd and even bits, recombined by a multiplexer. Signal points are (±√(E/2), ±√(E/2)), doubling bandwidth efficiency to Rb with BPSK-equivalent BER.

Explain the modulation and demodulation techniques used in QPSK.

Quadrature Phase Shift Keying (QPSK) is a 4-level PSK in which each symbol carries a pair of bits (a dibit) and the carrier takes one of four phases spaced 90∘90^\circ apart.

si(t)=2ETcos⁡[2πfct+(2i−1)π4],i=1,…,4, 0≤t≤Ts_i(t)=\sqrt{\frac{2E}{T}}\cos\left[2\pi f_ct+(2i-1)\frac{\pi}{4}\right],\quad i=1,\dots,4,\ 0\le t\le T

T=2TbT=2T_b is the symbol duration and E=2EbE=2E_b the symbol energy. Using cos⁡(A+B)\cos(A+B):

si(t)=Ecos⁡[(2i−1)π4]ϕ1(t)−Esin⁡[(2i−1)π4]ϕ2(t)s_i(t)=\sqrt{E}\cos\left[(2i-1)\tfrac{\pi}{4}\right]\phi_1(t)-\sqrt{E}\sin\left[(2i-1)\tfrac{\pi}{4}\right]\phi_2(t)

with ϕ1(t)=2/Tcos⁡2πfct\phi_1(t)=\sqrt{2/T}\cos 2\pi f_ct and ϕ2(t)=2/Tsin⁡2πfct\phi_2(t)=\sqrt{2/T}\sin 2\pi f_ct. So QPSK is the sum of two BPSK signals on quadrature carriers.

Signal space (Gray coded)

             phi2
   01 *       |       * 11
              |
  ------------+------------ phi1
              |
   00 *       |       * 10
 points at (+-sqrt(E/2), +-sqrt(E/2))

QPSK modulator

 bits-->[polar NRZ encoder]-->[Demux S/P]
                               |        |
                         a1(t) odd  a2(t) even
                         (rate Rb/2) (rate Rb/2)
                               |        |
                     phi1 ->(X)      (X)<- phi2
                               |        |
                               +->(+)<--+
                                   |
                                 QPSK s(t)
  1. The data is converted to polar NRZ (+1, -1).
  2. The demultiplexer splits it into odd and even bits, each lasting T=2TbT=2T_b.
  3. Odd bits modulate the in-phase carrier and even bits the quadrature carrier (two BPSK signals).
  4. The adder gives one of four phases; e.g. dibit 11 gives the point (+E/2,+E/2)(+\sqrt{E/2},+\sqrt{E/2}), phase 7π/47\pi/4.

QPSK demodulator (coherent)

          +->(X)->[int 0..T]-x1->[x1>0?1:0]--+
          |   ^ phi1                          |
 x(t) ----+                                 [Mux P/S]-> bits
          |   v phi2                          |
          +->(X)->[int 0..T]-x2->[x2>0?1:0]--+
  1. The received signal goes to two correlators fed with the recovered carriers ϕ1\phi_1 and ϕ2\phi_2.
  2. Because ϕ1⊥ϕ2\phi_1\perp\phi_2, each correlator sees only its own BPSK component: x1=±E/2x_1=\pm\sqrt{E/2}, x2=±E/2x_2=\pm\sqrt{E/2} (plus noise).
  3. Each output is compared with a zero threshold to give the odd and even bits.
  4. The multiplexer recombines them into the original serial stream.

Properties

FeatureQPSKBPSK
Bits/symbol21
BandwidthRbR_b2Rb2R_b
Bit error probability12erfcEb/N0\tfrac12\text{erfc}\sqrt{E_b/N_0}same
ComplexityHigherLower

QPSK doubles bandwidth efficiency without any loss in bit-error performance, which is why it is used in satellite links, DVB-S and CDMA systems. Offset QPSK (Q branch delayed by TbT_b) avoids 180∘180^\circ phase jumps.

✓ Answer checked by Claude Opus 5.5 (high effort)

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