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CY-304 · Digital Systems/Quick Revision Short Notes

Digital Systems (CY-304) - Unit 5 Short Notes

How unit 5 is examined

This unit covers sampling, TDM, PCM, quantization error, BPSK/BFSK and Shannon capacity; BPSK/BFSK and Shannon's theorem carry the most marks, and the Shannon numerical (30 dB, 3 kHz) is a repeat pattern.

Introduction to Digital Communication: Nyquist sampling theorem

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Definition. <mark>A band-limited signal with highest frequency $f_m$ can be recovered exactly from its samples if it is sampled at a rate $f_s \ge 2f_m$.</mark>

Key points.

  1. The minimum rate $2f_m$ is the Nyquist rate; sampling below it causes aliasing, where high frequencies appear as false low ones.
  2. Signals are first passed through an anti-aliasing low-pass filter, then sampled at $f_s \ge 2f_m$.
  3. At the receiver an ideal low-pass filter of cut-off $f_m$ reconstructs the original signal from the samples.
  4. Voice (about 3.4 kHz) is sampled at 8 kHz in telephony.

Asked: [7 marks] (May 2019) What is quantization error? Explain sampling theorem. Asked: [14 marks, part] (Nov 2018) Short note: Nyquist sampling theorem (one of six, answer any four).

Time division multiplexing

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Definition. Multiplexing sends many signals over one channel to save cost. <mark>Time Division Multiplexing (TDM) gives each message its own short time slot in turn, so many signals share one channel one after another.</mark>

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u5-01" viewBox="0 0 596 252" width="596" height="252" role="img" aria-label="TDM. M = message inputs, Com = commutator (multiplexer), Ch = channel, Dec = decommutator (demultiplexer), O = outputs. Both switches rotate in sync."><style>#dsfig-u5-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u5-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u5-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u5-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u5-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u5-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u5-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u5-01 .t{fill:#16181D;font-weight:500}#dsfig-u5-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u5-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u5-01 .dot{fill:#16181D}#dsfig-u5-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u5-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u5-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u5-01 .ah{fill:#454C5A}#dsfig-u5-01 .ah.hi{fill:#2340B8}#dsfig-u5-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u5-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u5-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u5-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u5-01 .e{stroke:#B1B7C3}html.dark #dsfig-u5-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u5-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u5-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u5-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u5-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u5-01 .t{fill:#E6E8ED}html.dark #dsfig-u5-01 .t.inv{fill:#0F1115}html.dark #dsfig-u5-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u5-01 .dot{fill:#E6E8ED}html.dark #dsfig-u5-01 .ann{fill:#8FA3FF}html.dark #dsfig-u5-01 .lbl{fill:#858D9C}html.dark #dsfig-u5-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u5-01 .ah{fill:#B1B7C3}html.dark #dsfig-u5-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u5-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u5-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u5-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah27" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh27" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M55.8,50.5 L151.5,114.4" marker-end="url(#ah27)"/><path class="e" d="M59,126 L148,126" marker-end="url(#ah27)"/><path class="e" d="M55.8,201.5 L151.5,137.6" marker-end="url(#ah27)"/><path class="e" d="M188,126 L277,126" marker-end="url(#ah27)"/><path class="e" d="M317,126 L406,126" marker-end="url(#ah27)"/><path class="e" d="M442.8,115.5 L538.5,51.6" marker-end="url(#ah27)"/><path class="e" d="M446,126 L535,126" marker-end="url(#ah27)"/><path class="e" d="M442.8,136.5 L538.5,200.4" marker-end="url(#ah27)"/><circle class="n" cx="40" cy="40" r="18"/><text class="t" x="40" y="40" dy=".35em" text-anchor="middle">M1</text><circle class="n" cx="40" cy="126" r="18"/><text class="t" x="40" y="126" dy=".35em" text-anchor="middle">M2</text><circle class="n" cx="40" cy="212" r="18"/><text class="t" x="40" y="212" dy=".35em" text-anchor="middle">M3</text><circle class="n" cx="169" cy="126" r="18"/><text class="t" x="169" y="126" dy=".35em" text-anchor="middle">Com</text><circle class="n" cx="298" cy="126" r="18"/><text class="t" x="298" y="126" dy=".35em" text-anchor="middle">Ch</text><circle class="n" cx="427" cy="126" r="18"/><text class="t" x="427" y="126" dy=".35em" text-anchor="middle">Dec</text><circle class="n" cx="556" cy="40" r="18"/><text class="t" x="556" y="40" dy=".35em" text-anchor="middle">O1</text><circle class="n" cx="556" cy="126" r="18"/><text class="t" x="556" y="126" dy=".35em" text-anchor="middle">O2</text><circle class="n" cx="556" cy="212" r="18"/><text class="t" x="556" y="212" dy=".35em" text-anchor="middle">O3</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">TDM. M = message inputs, Com = commutator (multiplexer), Ch = channel, Dec = decommutator (demultiplexer), O = outputs. Both switches rotate in sync.</figcaption></figure>

Key points.

  1. A commutator (rotary switch) samples each input in turn and the decommutator, rotating in synchronism, routes each pulse to its output.
  2. One pass through all channels is a frame, and each channel's slot repeats once per frame.
  3. Frame synchronisation bits mark the start of each frame so the receiver stays aligned.
  4. Synchronous TDM gives every channel a fixed slot whether or not it has data; asynchronous (statistical) TDM assigns slots only to active channels, with an address tag on each.
  5. Advantages over FDM: no crosstalk between channels, the full bandwidth is used by one signal at a time, circuits are digital and cheap, and no filters are needed for separation.
  6. Its drawbacks are the need for accurate synchronisation and a bandwidth that grows with the number of channels.
Basis Synchronous TDM Asynchronous TDM
Slot Fixed, even if idle Given only to active input
Efficiency Low when inputs are idle High
Address Position gives it Address tag needed

Answer frame. Open with the multiplexing definition; draw the commutator-channel-decommutator figure; develop points 1-4 then 5; close with the synchronous vs asynchronous table. For the TDM vs FDM question add point 5 and define synchronous TDM.

Asked: [7 marks] (Jun 2023, Dec 2025) What is multiplexing? Explain the working of TDM with a diagram; differentiate synchronous and asynchronous TDM. Asked: [7 marks] (Dec 2020) Advantages of TDM over FDM; define synchronous TDM. Asked: [14 marks, part] (May 2019) Explain any two: TDM is one option.

PCM

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Definition. <mark>Pulse Code Modulation (PCM) converts an analog signal into a digital one by sampling, quantizing and encoding each sample into a binary code word.</mark>

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u5-02" viewBox="0 0 596 80" width="596" height="80" role="img" aria-label="PCM. Smp = sampler, Qnt = quantizer, Enc = encoder, Ch = channel with regenerative repeaters, Dec = decoder, LPF = low-pass reconstruction filter."><style>#dsfig-u5-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u5-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u5-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u5-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u5-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u5-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u5-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u5-02 .t{fill:#16181D;font-weight:500}#dsfig-u5-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u5-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u5-02 .dot{fill:#16181D}#dsfig-u5-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u5-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u5-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u5-02 .ah{fill:#454C5A}#dsfig-u5-02 .ah.hi{fill:#2340B8}#dsfig-u5-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u5-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u5-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u5-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u5-02 .e{stroke:#B1B7C3}html.dark #dsfig-u5-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u5-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u5-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u5-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u5-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u5-02 .t{fill:#E6E8ED}html.dark #dsfig-u5-02 .t.inv{fill:#0F1115}html.dark #dsfig-u5-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u5-02 .dot{fill:#E6E8ED}html.dark #dsfig-u5-02 .ann{fill:#8FA3FF}html.dark #dsfig-u5-02 .lbl{fill:#858D9C}html.dark #dsfig-u5-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u5-02 .ah{fill:#B1B7C3}html.dark #dsfig-u5-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u5-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u5-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u5-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah28" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh28" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,40 L105,40" marker-end="url(#ah28)"/><path class="e" d="M145,40 L191,40" marker-end="url(#ah28)"/><path class="e" d="M231,40 L277,40" marker-end="url(#ah28)"/><path class="e" d="M317,40 L363,40" marker-end="url(#ah28)"/><path class="e" d="M403,40 L449,40" marker-end="url(#ah28)"/><path class="e" d="M489,40 L535,40" marker-end="url(#ah28)"/><circle class="n" cx="40" cy="40" r="18"/><text class="t" x="40" y="40" dy=".35em" text-anchor="middle">In</text><circle class="n" cx="126" cy="40" r="18"/><text class="t" x="126" y="40" dy=".35em" text-anchor="middle">Smp</text><circle class="n" cx="212" cy="40" r="18"/><text class="t" x="212" y="40" dy=".35em" text-anchor="middle">Qnt</text><circle class="n" cx="298" cy="40" r="18"/><text class="t" x="298" y="40" dy=".35em" text-anchor="middle">Enc</text><circle class="n" cx="384" cy="40" r="18"/><text class="t" x="384" y="40" dy=".35em" text-anchor="middle">Ch</text><circle class="n" cx="470" cy="40" r="18"/><text class="t" x="470" y="40" dy=".35em" text-anchor="middle">Dec</text><circle class="n" cx="556" cy="40" r="18"/><text class="t" x="556" y="40" dy=".35em" text-anchor="middle">LPF</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">PCM. Smp = sampler, Qnt = quantizer, Enc = encoder, Ch = channel with regenerative repeaters, Dec = decoder, LPF = low-pass reconstruction filter.</figcaption></figure>

Key points.

  1. The sampler samples the analog input at $f_s \ge 2f_m$ (after an anti-aliasing filter) to give discrete-time pulses.
  2. The quantizer rounds each sample to the nearest of $L = 2^n$ levels, which makes the amplitude discrete but adds quantization error.
  3. The encoder turns each level into an $n$-bit binary word, giving a bit rate of $n f_s$ bits per second.
  4. The channel carries the bit stream, and repeaters regenerate it, so noise does not build up.
  5. The decoder converts code words back to quantized pulses and the low-pass filter smooths them into the analog output.
  6. PCM is noise-immune and easy to multiplex, but needs more bandwidth and suffers quantization noise.

Answer frame. Open with the PCM definition; draw the seven-block figure; develop transmitter points 1-3 then receiver point 5; close with points 4 and 6 (advantages, bandwidth cost).

Asked: [7 marks] (Nov 2019, Dec 2020, Dec 2024) Draw the block diagram of PCM and explain it; explain PCM with a neat diagram; short note on PCM.

Quantization error

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Definition. Sampling is discretization in time, quantization is discretization in amplitude to a finite set of levels, and <mark>quantization error is the difference between the actual sample value and its quantized value.</mark>

Key points.

  1. Sampling picks the signal value at instants $T_s = 1/f_s$ apart.
  2. Quantization maps each sample to one of $L = 2^n$ levels separated by the step size $\Delta = (V_{max}-V_{min})/L$.
  3. The error is $e = x - x_q$ and lies within $\pm\Delta/2$ for a uniform quantizer.
  4. The noise power is $\Delta^2/12$, so a smaller step means less noise.
  5. Signal-to-quantization-noise ratio for a full-scale sine is $\text{SNR} \approx 6.02n + 1.76$ dB, so each extra bit gives about 6 dB.
  6. Noise is reduced by using more bits (a smaller step), or by non-uniform quantization with companding, which gives small signals finer steps.

Answer frame. Open with the three definitions in order (sampling, quantization, error); show $e = x - x_q$ and $\Delta^2/12$; close with the noise-reduction methods in point 6.

Asked: [7 marks] (Nov 2019) Explain the terms sampling, quantization and quantization error. Asked: [7 marks] (Dec 2025) Explain quantization and quantization error in PCM; how can quantization noise be reduced? Asked: [14 marks, part] (Nov 2018) Short note: quantization error. Asked: [7 marks] (May 2019) What is quantization error? (with sampling theorem, see Nyquist section)

Introduction to BPSK and BFSK modulation schemes

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Definition. BPSK (Binary Phase Shift Keying) switches the carrier phase between $0^\circ$ and $180^\circ$ by the data bit; <mark>BFSK (Binary Frequency Shift Keying) switches the carrier frequency between $f_1$ for bit 1 and $f_2$ for bit 0.</mark>

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u5-03" viewBox="0 0 467 166" width="467" height="166" role="img" aria-label="BPSK transmitter. NRZ = NRZ level encoder (bit 1 to +1, bit 0 to -1), Mod = balanced (product) modulator, Osc = carrier oscillator."><style>#dsfig-u5-03 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u5-03 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u5-03 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u5-03 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u5-03 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u5-03 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u5-03 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u5-03 .t{fill:#16181D;font-weight:500}#dsfig-u5-03 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u5-03 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u5-03 .dot{fill:#16181D}#dsfig-u5-03 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u5-03 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u5-03 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u5-03 .ah{fill:#454C5A}#dsfig-u5-03 .ah.hi{fill:#2340B8}#dsfig-u5-03 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u5-03 .wl .t{font-size:12px;font-weight:700}#dsfig-u5-03 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u5-03 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u5-03 .e{stroke:#B1B7C3}html.dark #dsfig-u5-03 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u5-03 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u5-03 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u5-03 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u5-03 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u5-03 .t{fill:#E6E8ED}html.dark #dsfig-u5-03 .t.inv{fill:#0F1115}html.dark #dsfig-u5-03 .kd{stroke:#E6E8ED}html.dark #dsfig-u5-03 .dot{fill:#E6E8ED}html.dark #dsfig-u5-03 .ann{fill:#8FA3FF}html.dark #dsfig-u5-03 .lbl{fill:#858D9C}html.dark #dsfig-u5-03 .ptr{fill:#8FA3FF}html.dark #dsfig-u5-03 .ah{fill:#B1B7C3}html.dark #dsfig-u5-03 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u5-03 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u5-03 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u5-03 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah29" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh29" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,126 L148,126" marker-end="url(#ah29)"/><path class="e" d="M188,126 L277,126" marker-end="url(#ah29)"/><path class="e" d="M298,59 L298,105" marker-end="url(#ah29)"/><path class="e" d="M317,126 L406,126" marker-end="url(#ah29)"/><circle class="n" cx="40" cy="126" r="18"/><text class="t" x="40" y="126" dy=".35em" text-anchor="middle">Bit</text><circle class="n" cx="169" cy="126" r="18"/><text class="t" x="169" y="126" dy=".35em" text-anchor="middle">NRZ</text><circle class="n" cx="298" cy="126" r="18"/><text class="t" x="298" y="126" dy=".35em" text-anchor="middle">Mod</text><circle class="n" cx="298" cy="40" r="18"/><text class="t" x="298" y="40" dy=".35em" text-anchor="middle">Osc</text><circle class="n" cx="427" cy="126" r="18"/><text class="t" x="427" y="126" dy=".35em" text-anchor="middle">Out</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">BPSK transmitter. NRZ = NRZ level encoder (bit 1 to +1, bit 0 to -1), Mod = balanced (product) modulator, Osc = carrier oscillator.</figcaption></figure>

Key points.

  1. BPSK signal: $s(t) = A\cos(2\pi f_c t)$ for bit 1 and $A\cos(2\pi f_c t + \pi) = -A\cos(2\pi f_c t)$ for bit 0.
  2. The NRZ encoder gives $\pm1$, and the balanced modulator multiplies it with the carrier, so the phase flips by $180^\circ$.
  3. BPSK has one basis function and two antipodal points at $\pm\sqrt{E_b}$, with $P_e = Q\left(\sqrt{2E_b/N_0}\right)$, and needs coherent detection.
  4. BFSK: $s_i(t) = \sqrt{2E_b/T_b}\cos(2\pi f_i t)$, $i = 1, 2$, generated by two oscillators and a switch, or a VCO driven by the data.
  5. BFSK detects coherently (two correlators, compare) or non-coherently (two band-pass filters and envelope detectors).
  6. Orthogonal BFSK has $\rho = 0$ (spacing $\Delta f = k/2T_b$) with $P_e = Q\left(\sqrt{E_b/N_0}\right)$.
  7. Non-orthogonal BFSK: $\phi_1 = s_1/\sqrt{E_b}$ by Gram-Schmidt, $s_1 = \sqrt{E_b}\,\phi_1$, $s_2 = \rho\sqrt{E_b}\,\phi_1 + \sqrt{E_b(1-\rho^2)}\,\phi_2$, where $\rho = \frac{1}{E_b}\int_0^{T_b} s_1 s_2\,dt$.
  8. The distance is $d^2 = 2E_b(1-\rho)$, so $P_e = Q\left(\sqrt{E_b(1-\rho)/N_0}\right)$; non-zero positive $\rho$ shrinks $d$ and raises the error.

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Basis BPSK BFSK
Parameter varied Phase ($0^\circ$, $180^\circ$) Frequency ($f_1$, $f_2$)
Bandwidth About $2/T_b$ (less) Larger (about $2/T_b + \Delta f$)
Error rate at same $E_b/N_0$ Lower (3 dB better) Higher
Noise immunity Best Good
Detection Coherent only Coherent or non-coherent
Complexity Simple but needs carrier recovery Simple non-coherent receiver

Answer frame. BPSK question: define; draw the block diagram; give the equations, waveform and constellation; close with $P_e$. BFSK question: define; equations; transmitter and detector blocks; close with the orthogonality condition. Non-orthogonal: define, Gram-Schmidt, vector figure, error effect. Compare: table, then conclude BPSK for reliability, BFSK for simple or fading links.

Asked: [7 marks] (Dec 2020) Compare the BPSK and BFSK modulation schemes. Asked: [7 marks] (Jun 2023) With a block diagram explain the working of BPSK modulation. Asked: [7 marks] (Dec 2023) Explain BFSK modulation schemes. Asked: [7 marks] (Dec 2024) Explain the geometrical representation of non-orthogonal BFSK. Asked: [14 marks, part] (May 2019, Nov 2018) BFSK as an "any two/four" option.

Shannon's theorem for channel capacity

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Definition. ==Shannon's theorem states that the capacity of a band-limited channel with Gaussian noise is $C = B\log_2(1 + S/N)$ bits/s, and error-free transmission is possible at any rate $R \le C$.==

Formula. $$C = B\log_2\left(1 + \frac{S}{N}\right)$$ $$\text{SNR}_{linear} = 10^{\text{SNR}_{dB}/10}$$

Key points.

  1. Mutual information $I(X;Y) = H(X) - H(X|Y)$ is the information the output carries about the input.
  2. Channel capacity is its maximum over input distributions, $C = \max_{p(x)} I(X;Y)$.
  3. For a Gaussian channel this maximum gives the formula above, so capacity grows with bandwidth $B$ and SNR.
  4. Capacity rises only logarithmically with SNR but linearly with bandwidth.
  5. For $R > C$ no coding can give reliable transmission; for $R \le C$ suitable coding gives error as small as wanted.
  6. As $B \to \infty$, $C \to 1.44\,S/N_0$, a finite limit.

Example. Given $\text{SNR} = 30$ dB, $B = 3$ kHz.

Step Working
Linear SNR $10^{30/10} = 1000$
Capacity $C = 3000\log_2(1001)$
Value $3000 \times 9.967$

Shannon limit = 29,901 bits/s (about 29.9 kbps).

Answer frame. Open with the theorem; define mutual information and its link to $C$; state the formula; do the dB conversion first, then substitute; close with the meaning of $R \le C$.

Asked: [7 marks] (Dec 2023) What is mutual information and how is it related to channel capacity? Find the Shannon limit for SNR 30 dB, bandwidth 3 kHz. Asked: [14 marks, part] (May 2019) Explain any two: Shannon's theorem, TDM, BFSK, Sample and Hold. Asked: [14 marks, part] (Nov 2018) Short note (any four): Shannon's theorem, BFSK, quantization error, Flash RAM, demultiplexer, Nyquist theorem.

Last-minute revision

  • Sampling theorem: $f_s \ge 2f_m$; below it causes aliasing.
  • TDM: time slots, commutator and decommutator, frame sync; no crosstalk.
  • PCM: sample, quantize, encode; bit rate $= nf_s$; $L = 2^n$.
  • Quantization error $e = x - x_q$, within $\pm\Delta/2$; noise power $\Delta^2/12$.
  • SNR of PCM $\approx 6.02n + 1.76$ dB.
  • BPSK phase $0^\circ$/$180^\circ$; $P_e = Q(\sqrt{2E_b/N_0})$.
  • BFSK frequencies $f_1$/$f_2$; orthogonal $P_e = Q(\sqrt{E_b/N_0})$.
  • Non-orthogonal BFSK: $d^2 = 2E_b(1-\rho)$.
  • Shannon: $C = B\log_2(1+\text{SNR})$; 30 dB is 1000; answer 29,901 bits/s.

Memory hooks

  • SQE for PCM: Sample, Quantize, Encode.
  • BPSK = Phase flip 180; BFSK = two Frequencies.
  • TDM = "take turns"; FDM = "share the room by band".
  • Each extra PCM bit adds 6 dB.
  • Shannon: bandwidth and SNR set the speed limit.

Coverage checklist

  • Introduction to Digital Communication: Nyquist sampling theorem: May 2019 sampling theorem, Nov 2018 short note.
  • time division multiplexing: Jun 2023, Dec 2025, Dec 2020, May 2019.
  • PCM: Nov 2019, Dec 2020, Dec 2024.
  • quantization error: Nov 2019, Dec 2025, May 2019, Nov 2018.
  • introduction to BPSK & BFSK modulation schemes: Dec 2020, Jun 2023, Dec 2023, Dec 2024.
  • Shannon's theorem for channel capacity: Dec 2023, May 2019, Nov 2018.
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