Phase Modulation (PM)

Phase Modulation is an angle modulation technique where the instantaneous phase of the carrier is varied linearly with the message signal. While PM and FM are closely related mathematically, PM has distinct characteristics that make it fundamental to digital communication systems, where phase changes represent digital symbols.

Mathematical Representation

PM Signal Expression

s(t) = Ac · cos[2πfc·t + kp · m(t)]

Where:

• kp = phase sensitivity (radians/volt)
• m(t) = message signal
• kp · m(t) = instantaneous phase deviation

For Single-Tone Modulation

m(t) = Am · cos(2πfm·t)

s(t) = Ac · cos[2πfc·t + βp · cos(2πfm·t)]

Where βp = kp · Am is the phase modulation index (peak phase deviation in radians).

Instantaneous Frequency of PM

fi(t) = fc + (1/2π) · d[kp·m(t)]/dt = fc + (kp/2π) · m'(t)

For single tone: fi(t) = fc - (kp·Am·fm) · sin(2πfm·t)

Maximum frequency deviation in PM: Δf = kp · Am · fm = βp · fm

Relationship Between FM and PM

Comparison: PM vs FM

PM Waveform

Spectrum of PM

The PM spectrum has the same mathematical form as FM (Bessel function expansion):

s(t) = Ac Σ Jn(βp) · cos[2π(fc + n·fm)·t + nπ/2]

PM Bandwidth

Using Carson's Rule:

BT = 2(Δf + fm) = 2(βp·fm + fm) = 2fm(βp + 1)

Key difference from FM: In PM, if message amplitude doubles:

• βp doubles → Δf doubles → BW approximately doubles
• In FM, if amplitude doubles: Δf doubles but βf also doubles → BW changes similarly

But if message frequency doubles (same amplitude):

• PM: βp unchanged, fm doubled → BW = 2·(2fm)·(βp+1) → BW roughly doubles
• FM: βf halved, fm doubled → Δf unchanged → BW changes less

Narrowband Phase Modulation (NBPM)

When βp << 1 radian (typically < 0.5 rad):

s(t) ≈ Ac·cos(2πfc·t) - Ac·βp·m(t)·sin(2πfc·t)

This approximation shows NBPM looks like AM but with a 90° phase-shifted sideband pattern:

PM Generation Methods

Direct Method: Varactor Diode Modulator

Indirect Method: FM Modulator with Differentiator

Since FM phase ∝ ∫m(t)dt, using dm/dt as input gives phase ∝ ∫dm/dt·dt = m(t) → PM!

PM Demodulation

PLL-based PM Demodulator

Differentiate-then-Envelope Detect

Solved Example 1

Problem: A phase modulator has kp = 5 rad/V. The message signal is m(t) = 2cos(2π×1000t). Find (a) peak phase deviation, (b) frequency deviation, (c) bandwidth.

Solution:

(a) Peak phase deviation: Δφ = kp × Am = 5 × 2 = 10 radians

(b) Frequency deviation: Δf = kp × Am × fm = 5 × 2 × 1000 = 10,000 Hz = 10 kHz Alternatively: Δf = βp × fm = 10 × 1000 = 10 kHz ✓

(c) Bandwidth (Carson's rule): BT = 2(Δf + fm) = 2(10,000 + 1000) = 22 kHz Or: BT = 2fm(βp + 1) = 2(1000)(11) = 22 kHz ✓

Solved Example 2

Problem: Compare PM and FM when the message frequency changes from 1 kHz to 2 kHz with the same amplitude Am = 3V. Given kf = 10 kHz/V and kp = 5 rad/V.

Solution:

For FM:

• Δf = kf × Am = 10,000 × 3 = 30 kHz (unchanged with frequency)
• At fm = 1 kHz: β = 30, BW = 2(30k + 1k) = 62 kHz
• At fm = 2 kHz: β = 15, BW = 2(30k + 2k) = 64 kHz (slight increase)

For PM:

• Δφ = kp × Am = 5 × 3 = 15 rad (unchanged)
• At fm = 1 kHz: Δf = 15 × 1k = 15 kHz, BW = 2(15k + 1k) = 32 kHz
• At fm = 2 kHz: Δf = 15 × 2k = 30 kHz, BW = 2(30k + 2k) = 64 kHz (doubles!)

Key insight: PM frequency deviation (and bandwidth) increases with message frequency, while FM deviation stays constant.

Applications of PM

1. Digital Communications: PSK, QPSK, 8-PSK are all forms of PM with discrete phase states
2. Satellite Links: Phase modulation used extensively for its constant envelope
3. Indirect FM Generation: Armstrong method uses PM to generate FM
4. Radar Systems: Phase-coded pulse compression
5. Optical Communications: Phase modulation in coherent optical systems

Interview Questions

Q1: What is the fundamental difference between FM and PM in terms of frequency deviation?

In FM, the frequency deviation Δf = kf·Am is proportional only to message amplitude and independent of message frequency. In PM, the frequency deviation Δf = kp·Am·fm depends on both amplitude AND frequency. This means PM emphasizes higher message frequencies (acts as a natural pre-emphasis), while FM treats all frequencies equally. This is why FM needs artificial pre-emphasis for noise improvement, while PM inherently provides it.

Q2: Why is PM more relevant to digital communications than FM?

In digital communication, information is represented by discrete phase states (0°, 90°, 180°, 270° in QPSK). PM directly maps digital symbols to carrier phase values, making it natural for PSK/QPSK/8-PSK systems. The constant envelope property of PM also means power amplifiers can operate in saturation (maximum efficiency) without distortion, which is critical for satellite and mobile transmitters where power efficiency is paramount.

Q3: How can you convert between FM and PM using simple circuits?

PM from FM: Differentiate the message signal before applying it to an FM modulator. Since FM phase = ∫m(t)dt, using dm/dt as input gives phase = ∫(dm/dt)dt = m(t), which is PM. Conversely, FM from PM: Integrate the message before applying to a PM modulator. This relationship is used in the Armstrong indirect FM transmitter.

Q4: Why does PM not need pre-emphasis unlike FM?

FM demodulated noise has a parabolic (f²) power spectral density, meaning higher message frequencies suffer more noise. PM demodulated noise is uniform across the message bandwidth because the PM demodulator output is proportional to phase (not frequency). Since PM frequency deviation naturally increases with message frequency (Δf ∝ fm), it provides a built-in emphasis of high frequencies that counteracts any noise advantage at lower frequencies.