Communication Systems • Topic 3 of 3

Modulation

A message (audio) signal has a low frequency and small bandwidth. It cannot be transmitted directly as an EM wave over long distances. Modulation is the process of superimposing the low-frequency message (the modulating signal) onto a high-frequency carrier wave by varying one of the carrier's properties. The reverse process at the receiver, recovering the message from the modulated carrier, is called demodulation or detection.

Why modulation is necessary.

  • Antenna size. For efficient radiation, an antenna must be at least about one-quarter of the wavelength: $L \approx \dfrac{\lambda}{4}$. An audio signal of $15\,\text{kHz}$ has $\lambda=c/f=3\times10^{8}/15\times10^{3}=20\,\text{km}$, needing a $5\,\text{km}$ antenna — impossible. A $1\,\text{MHz}$ carrier has $\lambda=300\,\text{m}$, requiring only a $75\,\text{m}$ antenna, which is practical.
  • Effective power radiated. The power radiated by an antenna of fixed length is proportional to $\left(\dfrac{L}{\lambda}\right)^{2}=\left(\dfrac{Lf}{c}\right)^{2}$. A higher carrier frequency means more efficient radiation for the same antenna size.
  • Multiplexing. If every station sent baseband audio, all signals (in the same low band) would overlap and interfere. Assigning each a different carrier frequency lets many stations share free space without mixing; the receiver tunes to the chosen carrier.

Amplitude modulation (AM). A carrier $c(t)=A_c\sin\omega_c t$ has its amplitude varied in step with the message $m(t)=A_m\sin\omega_m t$. The modulated wave is $c_m(t)=(A_c+A_m\sin\omega_m t)\sin\omega_c t$. The amplitude of the carrier now follows the shape of the message — this is the AM envelope.

Modulation index. The depth of modulation is measured by the modulation index $\mu=\dfrac{A_m}{A_c}$. To avoid distortion the index must satisfy $\mu \le 1$ ($0\%$ to $100\%$). If $\mu>1$ (over-modulation) the envelope is clipped and the message is distorted. In terms of the modulated wave's extremes, $\mu=\dfrac{A_{\max}-A_{\min}}{A_{\max}+A_{\min}}$.

Sidebands and bandwidth of AM. Expanding the AM wave using the product-to-sum identity gives three frequency components:

  • the carrier at $\omega_c$ with amplitude $A_c$,
  • the upper side band at $\omega_c+\omega_m$, and
  • the lower side band at $\omega_c-\omega_m$,

each side band having amplitude $\dfrac{\mu A_c}{2}$. The two side frequencies are $\omega_c\pm\omega_m$. The total bandwidth of the AM signal is the spread between the two side bands: $\text{BW}=(\omega_c+\omega_m)-(\omega_c-\omega_m)=2\omega_m$, i.e. twice the highest message frequency.

Production of AM. The message and carrier are added and then passed through a non-linear (square-law) device, whose output contains the carrier and the two side bands; a band-pass filter selects the AM components, which a power amplifier then boosts before the antenna radiates them.

Detection of AM. At the receiver the AM wave is rectified (a diode passes only one half), giving a pulsating signal whose envelope is the message. An envelope detector — a diode followed by an $RC$ network — then smooths out the high-frequency carrier and recovers the original audio.

Frequency modulation (FM) — a brief look. In FM the carrier's frequency (not amplitude) is varied in proportion to the message amplitude, while the carrier amplitude stays constant. Because noise mainly disturbs amplitude, FM is far less affected by noise and gives higher-fidelity sound than AM. FM, however, needs a larger bandwidth, which is why FM radio occupies the VHF band above $88\,\text{MHz}$.

Amplitude-modulated wave: high-frequency carrier enveloped by the low-frequency messageEnvelope = messageModulated carrierAmplitude-Modulated Wave
1
Worked Example
A carrier of amplitude $A_c=10\,\text{V}$ is amplitude-modulated by a signal of amplitude $A_m=6\,\text{V}$. Find the modulation index.
Solution
  1. Modulation index $\mu=\dfrac{A_m}{A_c}$.
  2. $\mu=\dfrac{6}{10}=0.6$.
  3. As a percentage, $\mu=60\%$, which is within the safe limit $\mu\le 1$.

Answer: The modulation index is $\mu=0.6$ (i.e. $60\%$ modulation).

2
Worked Example
An AM wave has maximum amplitude $A_{\max}=16\,\text{V}$ and minimum amplitude $A_{\min}=4\,\text{V}$. Find the modulation index and the carrier amplitude.
Solution
  1. $\mu=\dfrac{A_{\max}-A_{\min}}{A_{\max}+A_{\min}}=\dfrac{16-4}{16+4}=\dfrac{12}{20}=0.6$.
  2. Carrier amplitude $A_c=\dfrac{A_{\max}+A_{\min}}{2}=\dfrac{16+4}{2}=10\,\text{V}$.
  3. Message amplitude $A_m=\dfrac{A_{\max}-A_{\min}}{2}=\dfrac{12}{2}=6\,\text{V}$ (check: $\mu=6/10=0.6$).

Answer: $\mu=0.6$ and the carrier amplitude $A_c=10\,\text{V}$.

3
Worked Example
A $1.0\,\text{MHz}$ carrier is amplitude-modulated by a $5\,\text{kHz}$ tone. Find the frequencies of the two side bands and the bandwidth of the AM signal.
Solution
  1. Side bands are at $f_c\pm f_m$.
  2. Upper side band $=1000\,\text{kHz}+5\,\text{kHz}=1005\,\text{kHz}$.
  3. Lower side band $=1000\,\text{kHz}-5\,\text{kHz}=995\,\text{kHz}$.
  4. Bandwidth $=2f_m=2\times5=10\,\text{kHz}$.

Answer: Side bands at $995\,\text{kHz}$ and $1005\,\text{kHz}$; bandwidth $=10\,\text{kHz}$.

4
Worked Example
A message signal of frequency $10\,\text{kHz}$ and peak voltage $10\,\text{V}$ modulates a carrier of frequency $1\,\text{MHz}$ and peak voltage $20\,\text{V}$. Write the modulation index and the amplitude of each side band.
Solution
  1. $\mu=\dfrac{A_m}{A_c}=\dfrac{10}{20}=0.5$.
  2. Each side band has amplitude $\dfrac{\mu A_c}{2}=\dfrac{0.5\times20}{2}=5\,\text{V}$.
  3. Side bands lie at $1\,\text{MHz}\pm 10\,\text{kHz}=0.99\,\text{MHz}$ and $1.01\,\text{MHz}$.

Answer: $\mu=0.5$; each side band has amplitude $5\,\text{V}$ at $0.99\,\text{MHz}$ and $1.01\,\text{MHz}$.

5
Worked Example
An audio signal of $15\,\text{kHz}$ is to be radiated directly. Find the wavelength and the minimum antenna length ($\lambda/4$). Comment on feasibility.
Solution
  1. Wavelength $\lambda=\dfrac{c}{f}=\dfrac{3\times10^{8}}{15\times10^{3}}=2\times10^{4}\,\text{m}=20\,\text{km}$.
  2. Minimum antenna length $L=\dfrac{\lambda}{4}=\dfrac{20}{4}=5\,\text{km}$.
  3. A $5\,\text{km}$ antenna is impractical, so the audio must be modulated onto a high-frequency carrier first.

Answer: $\lambda=20\,\text{km}$ and $L=5\,\text{km}$ — far too large, which is exactly why modulation is needed.

6
Worked Example
A carrier wave is represented by $c_m(t)=(15+9\sin\omega_m t)\sin\omega_c t$ volts. Identify $A_c$, $A_m$, the modulation index and whether over-modulation occurs.
Solution
  1. Comparing with $(A_c+A_m\sin\omega_m t)\sin\omega_c t$: $A_c=15\,\text{V}$ and $A_m=9\,\text{V}$.
  2. Modulation index $\mu=\dfrac{A_m}{A_c}=\dfrac{9}{15}=0.6$.
  3. Since $\mu=0.6<1$, the wave is properly modulated; no over-modulation or distortion.

Answer: $A_c=15\,\text{V}$, $A_m=9\,\text{V}$, $\mu=0.6$; modulation is within limits (no over-modulation).

Key Points

  • Modulation superimposes a low-frequency message on a high-frequency carrier; demodulation recovers it at the receiver.
  • Modulation is needed for a practical antenna size ($L\approx\lambda/4$), efficient power radiation, and multiplexing many stations.
  • In AM the carrier amplitude follows the message: $c_m(t)=(A_c+A_m\sin\omega_m t)\sin\omega_c t$, with modulation index $\mu=\dfrac{A_m}{A_c}\le 1$.
  • AM produces a carrier plus two side bands at $\omega_c\pm\omega_m$ (each of amplitude $\mu A_c/2$); bandwidth $=2\omega_m$.
  • AM is generated with a non-linear device and detected by an envelope (diode + RC) detector; FM varies frequency instead and resists noise better.
Tap an option to check your answer0 / 4
Q1.In amplitude modulation the modulation index is defined as:
Explanation: The modulation index $\mu=\dfrac{A_m}{A_c}$ measures how deeply the carrier is modulated; it must not exceed 1.
Q2.An AM wave consists of which frequency components?
Explanation: AM produces the carrier at $\omega_c$ plus an upper side band at $\omega_c+\omega_m$ and a lower side band at $\omega_c-\omega_m$.
Q3.If a $600\,\text{kHz}$ carrier is modulated by a $4\,\text{kHz}$ signal, the bandwidth of the AM wave is:
Explanation: Bandwidth $=2f_m=2\times4\,\text{kHz}=8\,\text{kHz}$ (the spread between the two side bands).
Q4.Compared with AM, frequency modulation (FM) is preferred for high-fidelity sound mainly because:
Explanation: Noise corrupts amplitude; since FM carries information in frequency (constant amplitude), it is far more noise-immune, giving better fidelity at the cost of more bandwidth.