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Multilevel modulation significance; QPSK

Short answer

Multilevel modulation reduces bandwidth by sending log2(M) bits per symbol. QPSK uses four phases each representing a dibit. The transmitter splits bits into I and Q channels and sums after multiplication. The receiver coherently demodulates with two correlators and parallel-to-serial conversion.

What are the significances of multilevel modulation? Explain QPSK with its transmitter as well as a receiver block diagram.

Significance of multilevel modulation

In multilevel (M-ary) modulation each symbol carries n=log⁡2Mn=\log_2 M bits, instead of one bit as in binary schemes.

  • Bandwidth saving: symbol rate is Rb/log⁡2MR_b/\log_2 M, so bandwidth falls by the factor log⁡2M\log_2 M (QPSK needs half the bandwidth of BPSK).
  • Higher data rate on a band-limited channel (telephone line, radio channel) for the same bandwidth.
  • Higher spectral efficiency (bits/s/Hz); used in modems, Wi-Fi, LTE (16/64-QAM).
  • Cost: for a fixed power, points get closer as MM rises, so error probability increases or more power is needed; the receiver is more complex.

QPSK

Quadrature PSK sends two bits (a dibit) per symbol using four carrier phases:

si(t)=2ETcos⁡[2πfct+(2i−1)π4],i=1,2,3,4, 0≤t≤Ts_i(t)=\sqrt{\frac{2E}{T}}\cos\left[2\pi f_ct+(2i-1)\frac{\pi}{4}\right],\quad i=1,2,3,4,\ 0\le t\le T

with T=2TbT=2T_b and E=2EbE=2E_b. Expanding, si(t)=Ecos⁡[(2i−1)π4]ϕ1(t)−Esin⁡[(2i−1)π4]ϕ2(t)s_i(t)=\sqrt{E}\cos[(2i-1)\tfrac{\pi}{4}]\phi_1(t)-\sqrt{E}\sin[(2i-1)\tfrac{\pi}{4}]\phi_2(t), so QPSK is two BPSK signals on quadrature carriers.

Dibit (Gray)PhasePoint (si1,si2)(s_{i1},s_{i2})
10π/4\pi/4(+E/2,−E/2)(+\sqrt{E/2},-\sqrt{E/2})
003π/43\pi/4(−E/2,−E/2)(-\sqrt{E/2},-\sqrt{E/2})
015π/45\pi/4(−E/2,+E/2)(-\sqrt{E/2},+\sqrt{E/2})
117π/47\pi/4(+E/2,+E/2)(+\sqrt{E/2},+\sqrt{E/2})

Transmitter

 bits -> [NRZ level encoder: 1->+1, 0->-1]
               |
            [Demux S/P]
           /           \
     a1(t) odd      a2(t) even
        |               |
       (X)<-phi1(t)    (X)<-phi2(t)
        |               |
        +----->(+)<-----+
                |
              QPSK

ϕ1(t)=2/Tcos⁡2πfct\phi_1(t)=\sqrt{2/T}\cos 2\pi f_ct, ϕ2(t)=2/Tsin⁡2πfct\phi_2(t)=\sqrt{2/T}\sin 2\pi f_ct. Each branch is a BPSK signal at half the bit rate.

Receiver

         +->(X)->[int 0..T]->[x1>0 ? 1:0]-+
         |   ^ phi1                       |
 x(t) ---+                              [Mux P/S]-> bits
         |   v phi2                       |
         +->(X)->[int 0..T]->[x2>0 ? 1:0]-+

Each correlator output is compared with a zero threshold; the two decisions are multiplexed back into one stream.

  • Bandwidth: half that of BPSK for the same bit rate.
  • Bit error probability: Pe=12erfcEb/N0P_e=\tfrac12\text{erfc}\sqrt{E_b/N_0}, the same as BPSK.

✓ Answer checked by Claude Opus 5.5 (high effort)

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