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25-Comp-B5 Computer Communications · December 2018

Question 7 of 9: QPSK Modulation of a Digital Sequence

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: Stallings, Data and Computer Communications, 10th ed. — sampling and aliasing (Ch.5, Q1), cascaded gains/losses and decibels (Ch.3, Q2), Shannon–Hartley channel capacity (Ch.3, Q3), AM/FM analog modulation (Ch.5, Q4), LAN/network topologies (Ch.16, Q6), QPSK digital modulation (Ch.5, Q7), IP addressing and subnetting (Ch.18, Q8), and physical/link/network-layer terminology (Ch.3, 9, 11, 17, Q9); Kurose & Ross, Computer Networking: A Top-Down Approach, 8th ed. — error detection via CRC (Ch.5, Q5), IP addressing (Ch.4, Q8), and TCP/IP terminology (Ch.1, Q9).

This is a choose-any-5-of-9 exam; all nine questions are answered below.

Question 7: QPSK Modulation of a Digital Sequence (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Dibit → phase map11→$\pi/4$, 10→$3\pi/4$, 00→$5\pi/4$, 01→$7\pi/4$
Bit sequence (14 bits)0 0 1 0 0 1 1 1 1 0 0 0 0 1

Find. The QPSK-modulated waveform for the given bit sequence, i.e. the sequence of carrier phases and the resulting sketch.

Approach. QPSK sends 2 bits per symbol, so first split the 14-bit stream into 7 consecutive dibits, look each up in the given phase table, then sketch one constant-amplitude carrier segment per symbol period at the assigned phase — the waveform is continuous in amplitude but its PHASE jumps discontinuously at every symbol boundary.

  1. Split the sequence into dibits. Grouping the 14 bits two at a time in the order given: $$\underbrace{00}_{1}\ \underbrace{10}_{2}\ \underbrace{01}_{3}\ \underbrace{11}_{4}\ \underbrace{10}_{5}\ \underbrace{00}_{6}\ \underbrace{01}_{7}$$ which is exactly 7 symbols (14 bits ÷ 2 bits/symbol), consistent with QPSK's 2-bits-per-symbol rate.
  2. Map each dibit to its carrier phase using the table given in the question:
    Symbol1234567
    Dibit00100111100001
    Phase$5\pi/4$$3\pi/4$$7\pi/4$$\pi/4$$3\pi/4$$5\pi/4$$7\pi/4$
  3. Draw the waveform. Each symbol interval carries $s(t)=\cos(2\pi f_c t+\theta_k)$ at its own phase $\theta_k$ for the symbol's duration, with the SAME constant amplitude and carrier frequency $f_c$ throughout (only the phase changes symbol-to-symbol) — visible below as a phase "kink" at every dashed symbol boundary rather than any change in amplitude or envelope.
  4. Explain the sketch. Five of the six symbol boundaries (1→2, 3→4, 4→5, 5→6, 6→7) are quarter-turn ($\pm\pi/2$) phase steps, the smallest jump this constellation allows; boundary 2→3 (dibit 10→01, $3\pi/4\to7\pi/4$) is a full $\pi$ (180°) reversal, the largest possible jump, since $10$ and $01$ sit at diametrically opposite phases in the constellation. No two consecutive symbols repeat the same dibit in this particular sequence, so every one of the six boundaries shows a visible phase discontinuity in the sketch — there is no "flat" (same-phase) stretch anywhere in this example.
QPSK signal for 00 10 01 11 10 00 01 — phase jumps at each symbol boundary 00(5π/4)10(3π/4)01(7π/4)11(π/4)10(3π/4)00(5π/4)01(7π/4) t
Fig. Q7 — QPSK waveform for the 7-symbol sequence; each vertical divider is a symbol boundary where only the carrier's PHASE changes (amplitude and frequency stay constant).
Symbol #1234567
Dibit00100111100001
Phase$5\pi/4$$3\pi/4$$7\pi/4$$\pi/4$$3\pi/4$$5\pi/4$$7\pi/4$