ELEC3506

Physical LayerLecture 224 min

Modulation and PCM — Riding Bits on a Carrier, and Back

Four ways to carry digital data on an analog carrier, and the three-step process that turns an analog signal into digital in the first place.

By the end of this page you should be able to

  • Describe ASK, FSK, PSK and QAM, and say which carrier property each one varies
  • Rank ASK, FSK, PSK and QAM by susceptibility to noise
  • State the three steps of PCM and compute bits per sample and the resulting bit rate
  • Apply the Nyquist sampling rule to a signal of known bandwidth

The idea

The previous topic split a signal into amplitude, frequency and phase, and said amplitude and phase set the number of usable levels while frequency sets how many signal elements fit into a second. This topic is that split put to work in two directions.

Digital-to-analog: you have bits, and the channel — a phone line, a radio link — only carries an analog carrier wave. You pick one of the carrier’s three properties and vary it to represent your bits. That gives you ASK, FSK and PSK, one scheme per property, plus QAM which varies two at once.

Analog-to-digital: you have an analog signal — a voice — and you need it as bits to put on a digital network. That is what PCM does, in three mechanical steps.

Digital-to-analog: ASK, FSK, PSK

How it works

One property changes, two stay fixed

ASK — Amplitude Shift Keying. The carrier’s amplitude is varied to represent binary 1 or 0; frequency and phase stay constant. BASK is ASK implemented with just two amplitude levels, the normal case.

FSK — Frequency Shift Keying. The carrier’s frequency is varied; amplitude and phase stay constant. The simplest form is BFSK, Binary FSK.

PSK — Phase Shift Keying. The carrier’s phase is varied; amplitude and frequency stay constant. The simplest form is Binary PSK; a common step up is QPSK, Quadrature PSK.

Amplitude, frequency and phase shift keying, each drawn as a carrier wave carrying the bit sequence 1, 0, 1: ASK varies the wave height, FSK varies how many cycles fit in each bit interval, and PSK keeps height and cycle rate constant but flips the wave upside down when the bit changes. Below them, a four-point QAM constellation diagram plots each 2-bit symbol as a point on in-phase and quadrature axes.ASK — Amplitude Shift Keyingheight carries the bit101FSK — Frequency Shift Keyingcycle rate carries the bit101PSK — Phase Shift Keyingphase reversal carries the bit101QAM constellation — M = 4, two bits per symbolIQ00011110
ASK is the one most wrecked by noise, because noise directly adds or subtracts amplitude — the exact quantity carrying the bit. PSK and FSK survive noise better because the bit lives in a shape (phase, frequency) that additive noise distorts less directly.

Where marks get lost

ASK is the one noise wrecks

The lecture states this outright: ASK is highly susceptible to noise. Noise is additive, and it acts directly on amplitude — the exact property ASK uses to carry the bit. A noise spike and a genuine amplitude change look the same to the receiver.

FSK and PSK are not immune to noise, but the bit lives in frequency or phase rather than amplitude, which additive noise disturbs less directly. If a question asks which scheme is worst for a noisy channel, the answer is ASK, and the reason is that the noise and the data are the same kind of quantity.

QAM — using two properties at once

How it works

ASK and PSK combined

QAM — Quadrature Amplitude Modulation — also called M-QAM — combines ASK and PSK. It uses two separate PSK modulators driving in-phase and quadrature (out-of-phase) components: two composite signals of the same frequency but different phases, added together.

With 2 bits carried per signal element, the resultant wave has 4 possible phases: 45°, −45°, 135°, −135° — one phase per 2-bit combination.

Aside

The diagram’s constellation labels its four points using the standard Gray-coded 4-QAM mapping, where adjacent points differ by only one bit. The lecture’s own slide gives the same four phase angles, but its text describing which 2-bit value sits at which phase did not extract cleanly, so this page does not present a specific bit-to-phase mapping as the lecture’s answer — only the four phases themselves and the 2-bits-per-symbol count, both of which the lecture states plainly.

SchemeProperty variedHeld constantNoise sensitivity
ASKAmplitudeFrequency, phaseHighest — noise acts directly on amplitude
FSKFrequencyAmplitude, phaseLower than ASK
PSKPhaseAmplitude, frequencyLower than ASK
QAMAmplitude and phase togetherFrequencyCarries the most bits per symbol, 2+ per baud
This is the shape of Q7 and Q8 in tutorial 2: name what changes, and rank the noise sensitivity.

Worked example

Bits per baud, three schemes

The formula is r = log2(L), where L is the number of distinguishable levels the scheme uses — amplitudes for ASK, frequencies for FSK, phases for PSK.

  1. ASK, 4 amplitude levels. r = log2(4) = 2 bits per signal element.

  2. FSK, 8 frequencies. r = log2(8) = 3 bits per signal element.

  3. PSK, 4 phases. r = log2(4) = 2 bits per signal element — the same as 4-amplitude ASK, because both use 4 distinguishable levels, just of different carrier properties.

QAM with 4 points, as drawn above, is the same arithmetic again: log2(4) = 2 bits per symbol, which is why it is described as QPSK-equivalent at that size — the count of levels is what sets r, not which property they belong to.

AnswerASK with 4 amplitudes: 2 bits. FSK with 8 frequencies: 3 bits. PSK with 4 phases: 2 bits.

Analog-to-digital: PCM

How it works

Sampling, Quantization, Encoding

PCM — Pulse Code Modulation — is the most common technique for analog-to-digital conversion, and it is three steps, always in this order.

1. Sampling. Take the analog signal and measure it at equal intervals. This transforms the signal into a series of pulses — a PAM signal (Pulse Amplitude Modulation) — which is not yet a full digital signal, only a sampled one.

2. Quantization. Assign each sampled instance a value from a specific range. Telephone companies assign a sign (+/−) and a magnitude, usually a value from 0 to 127.

3. Encoding. Translate each quantized value into its 7-bit binary equivalent, with an 8th bit for the sign. The signal is now fully digital — 8 bits, exactly 1 byte, per sample.

Pulse code modulation in three stages over the same analogue waveform: sampling, where dots mark the signal value at equally spaced instants; quantisation, where each sample snaps to the nearest of eight horizontal levels; and encoding, where each level is written as a binary code beneath it.1 · Sample2 · Quantise (8 levels)3 · Encode010100110111110101011001000
Shown with 8 levels (3-bit codes) so the numbers stay readable — the telephone system in the source uses 128 levels (7 bits) plus a sign bit, 8 bits per sample, and 8000 samples a second, which multiply out to the 64 kbps figure worked in the lecture.

Check yourself

Why is sampling alone not enough to call the signal “digital”?

Sampling only measures the signal at instants — the result is a PAM signal, a series of pulses whose heights still vary continuously. It becomes digital only after quantization snaps each pulse to one of a fixed set of levels and encoding writes that level as bits.

How many bits per sample

Bits per PCM sample depends on the precision needed. Telephone companies use 7 bits for magnitude plus 1 bit for sign — 8 bits per sample, which is also exactly one byte, and Nyquist sets the sampling rate that makes the whole thing accurate.

Worked example

Telephone voice, 4000 Hz

  1. Apply Nyquist. The sampling rate must be at least twice the highest frequency present:

    Sampling Rate≥2×fmax⁡=2×4000=8000 samples/s\text{Sampling Rate} \geq 2 \times f_{\max} = 2 \times 4000 = 8000 \text{ samples/s}

  2. Bits per sample. 7 bits magnitude plus 1 bit sign = 8 bits per sample.

  3. Bit rate.

    Bit Rate=Sampling Rate×bits per sample=8000×8=64,000 bps=64 Kbps\text{Bit Rate} = \text{Sampling Rate} \times \text{bits per sample} = 8000 \times 8 = 64{,}000 \text{ bps} = 64 \text{ Kbps}

64 kbps is the number to have ready — it is the standard telephone PCM figure and it reappears in the multiplexing topic next.

AnswerSampling rate 8000 samples/s, bit rate 64,000 bps = 64 kbps

The direction that separates them

Analog-to-digitalDigital-to-analog
SchemesPCM, PAMASK, FSK, PSK, QAM
InputAn analog signalDigital data
OutputA digital signalAn analog carrier carrying the data
When you need itGetting voice or another analog source onto a digital networkGetting digital data across a channel that is fundamentally analog
The lecture draws this line explicitly. Get the direction backwards and you will misclassify every scheme on this page.

In the exam

  • Which property each scheme varies. ASK — amplitude. FSK — frequency. PSK — phase. QAM — amplitude and phase together. One-line answers, high hit rate.
  • Noise ranking. ASK is most susceptible, because noise acts directly on the amplitude that carries the data.
  • Bits per baud. r = log2(L). Practise it with small L — 2, 4, 8 — until it is automatic.
  • The three PCM steps, in order. Sampling produces a PAM signal, not yet digital. Quantization assigns sign and magnitude. Encoding writes the binary code, with a separate sign bit.
  • The telephone PCM numbers. 4000 Hz voice, 8000 samples/s, 8 bits/sample, 64 kbps. All four show up together in worked questions.
  • Analog-to-digital vs digital-to-analog. PCM and PAM are one direction; ASK, FSK, PSK and QAM are the other. State which direction a scheme performs before describing what it does.

Check yourself

  1. Which characteristic of the carrier does PSK vary?
  2. Of ASK, FSK, PSK and QAM, which is most susceptible to noise, and why?
  3. A scheme uses FSK with 8 distinguishable frequencies. How many bits does each signal element carry?
  4. Telephone voice tops out at 4000 Hz and PCM uses 8 bits per sample. What is the resulting bit rate?
  5. PCM and ASK/FSK/PSK/QAM sit on opposite sides of a classification the lecture draws. What is it?