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Chapter 6 โ€ข Theory & Derivations

Modulation, Demodulation & Superheterodyne Radio Communication

Comprehensive mathematical theory of wireless telecommunications: baseband transmission constraints and antenna dimensions; full spectrum, power, and efficiency derivations for Amplitude Modulation (AM); diode envelope detection; Frequency Modulation (FM) and Carson's rule; complete architectural engineering of the Superheterodyne Radio Receiver.

ยง6.1 Need for Modulation & Baseband Transmission Constraints

1. Physical Limitations of Direct Baseband Transmission

In telecommunications, information signals (speech: $300\,\text{Hz} - 3.4\,\text{kHz}$; high-fidelity audio: $20\,\text{Hz} - 20\,\text{kHz}$; video: $0 - 5\,\text{MHz}$) are low-frequency electrical signals collectively termed baseband signals. Directly transmitting these signals electromagnetically over free space is practically impossible due to fundamental physical laws:

  1. Antenna Dimensions: For efficient radiation of electromagnetic energy without severe impedance reflection, an antenna must have physical dimensions comparable to at least a quarter wavelength $(\lambda / 4)$ of the transmitted wave. For an audio wave at $f = 15\,\text{kHz}$:
    $$\lambda = \frac{c}{f} = \frac{3 \times 10^8\,\text{m/s}}{15 \times 10^3\,\text{s}^{-1}} = 20,000\,\text{m} = 20\,\text{km}$$
    $$\frac{\lambda}{4} = \frac{20\,\text{km}}{4} = 5\,\text{km} \quad (\text{a 5-kilometer tall antenna is mechanically impossible})$$
    By modulating onto a high-frequency carrier (e.g., $f_c = 100\,\text{MHz}$), $\lambda = 3\,\text{m}$ and $\lambda / 4 = 75\,\text{cm}$, which is easily fabricated on portable handheld devices.
  2. Radiated Power: The power radiated by a linear antenna of length $l$ scales inversely with the fourth power of wavelength:
    $$P_{rad} \propto \left( \frac{l}{\lambda} \right)^2 \propto l^2 f^2$$
    Low frequencies radiate virtually zero power into the surrounding ether.
  3. Signal Multiplexing: If all radio stations broadcast unmodulated audio in the $20\,\text{Hz} - 20\,\text{kHz}$ band, all signals would overlap simultaneously, creating an unintelligible din. Modulation shifts each signal to a unique carrier frequency band (Frequency Division Multiplexing - FDM).

ยง6.2 Amplitude Modulation (AM): Spectrum, Power Distribution & Efficiency

1. Time-Domain Representation and Modulation Index

In Amplitude Modulation (AM), the instantaneous amplitude of a high-frequency sinusoidal carrier $v_c(t) = V_c \cos(\omega_c t)$ is varied in direct linear proportion to the instantaneous value of the modulating message signal $v_m(t) = V_m \cos(\omega_m t)$:

$$v_{AM}(t) = [V_c + v_m(t)] \cos(\omega_c t) = [V_c + V_m \cos(\omega_m t)] \cos(\omega_c t)$$

Factoring out carrier amplitude $V_c$:

$$v_{AM}(t) = V_c [1 + m_a \cos(\omega_m t)] \cos(\omega_c t)$$

where the Modulation Index $m_a$ is:

$$m_a = \frac{V_m}{V_c} = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$$
  • Under-modulation ($m_a < 1$): Carrier envelope never reaches zero; signal is cleanly recovered by simple diode envelope detectors.
  • Critical modulation ($m_a = 1$): Envelope touches zero at troughs without clipping. Maximum unclipped transmission efficiency.
  • Over-modulation ($m_a > 1$): Envelope crosses zero and inverts phase, causing severe harmonic envelope distortion and adjacent-channel splatter.

2. Frequency Spectrum & Transmission Bandwidth

Expanding $v_{AM}(t)$ trigonometrically using $\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$v_{AM}(t) = V_c \cos(\omega_c t) + \frac{m_a V_c}{2} \cos((\omega_c + \omega_m) t) + \frac{m_a V_c}{2} \cos((\omega_c - \omega_m) t)$$

The AM signal consists of exactly three distinct spectral components:

  1. Carrier Component: Frequency $f_c$, peak amplitude $V_c$.
  2. Upper Sideband (USB): Frequency $f_c + f_m$, peak amplitude $m_a V_c / 2$.
  3. Lower Sideband (LSB): Frequency $f_c - f_m$, peak amplitude $m_a V_c / 2$.

The transmission bandwidth required for double-sideband full-carrier AM is:

$$BW = f_{USB} - f_{LSB} = (f_c + f_m) - (f_c - f_m) = 2 f_m$$

3. Power Derivation and Transmission Efficiency

Total average power $P_t$ delivered to an antenna resistance $R$ is the sum of powers in each spectral component:

$$P_c = \frac{V_c^2 / 2}{R} = \frac{V_c^2}{2 R}$$
$$P_{USB} = P_{LSB} = \frac{(m_a V_c / 2)^2 / 2}{R} = \frac{m_a^2}{4} \left( \frac{V_c^2}{2 R} \right) = \frac{m_a^2}{4} P_c$$
$$P_{SB} = P_{USB} + P_{LSB} = \frac{m_a^2}{2} P_c$$
$$P_t = P_c + P_{SB} = P_c \left( 1 + \frac{m_a^2}{2} \right)$$

The Transmission Efficiency $\eta$ is the ratio of useful information power (sideband power) to total transmitted power:

$$\eta = \frac{P_{SB}}{P_t} = \frac{\frac{m_a^2}{2} P_c}{P_c (1 + \frac{m_a^2}{2})} = \frac{m_a^2}{2 + m_a^2}$$

At maximum 100% modulation ($m_a = 1$):

$$\eta_{max} = \frac{1^2}{2 + 1^2} = \frac{1}{3} \approx 33.33\%$$

At least $66.7\%$ of total broadcast power is permanently consumed by the unmodulated carrier, which transmits zero actual message information but enables low-cost diode detection at receivers.

ยง6.3 AM Demodulation (Diode Envelope Detector) & Frequency Modulation (FM)

1. Diode Envelope Detector and Distortion Limits

An envelope detector comprises a silicon diode in series with a parallel $RC$ filter load. On positive carrier peaks, the diode conducts and charges capacitor $C$ to peak envelope voltage. When carrier voltage drops, the diode becomes reverse-biased, and $C$ discharges through $R$.

To follow the audio modulation envelope faithfully without carrier ripple:

  1. $RC \gg 1/f_c$ (Capacitor discharges slowly between carrier RF cycles, filtering RF ripple).
  2. $RC \le \frac{1}{\omega_m} \frac{\sqrt{1 - m_a^2}}{m_a}$ (To prevent Diagonal Peak Clipping). If $RC$ is too large, the capacitor voltage cannot decay as fast as the modulation envelope, clipping the audio waveform.

2. Frequency Modulation (FM): Theory and Carson's Bandwidth Rule

In Frequency Modulation, the instantaneous frequency $f_i(t)$ of the carrier varies linearly with message voltage $v_m(t)$ while amplitude $V_c$ remains constant:

$$f_i(t) = f_c + k_f v_m(t) = f_c + \Delta f \cos(\omega_m t)$$

where $\Delta f = k_f V_m$ is the Peak Frequency Deviation.

The FM Modulation Index $\beta$ is:

$$\beta = \frac{\Delta f}{f_m}$$

In time domain, the FM wave is expressed using Bessel functions of the first kind $J_n(\beta)$:

$$v_{FM}(t) = V_c \sum_{n=-\infty}^{\infty} J_n(\beta) \cos((\omega_c + n \omega_m) t)$$

Theoretically, FM contains infinite sideband pairs. However, sidebands with $|n| > (\beta + 1)$ have negligible amplitude (< 1%). The total occupied bandwidth is given by Carson's Bandwidth Rule:

$$BW_{FM} \approx 2 (\Delta f + f_m) = 2 f_m (1 + \beta)$$

For standard commercial FM broadcast ($\Delta f = 75\,\text{kHz}$, $f_{m,max} = 15\,\text{kHz}$):

$$\beta = \frac{75}{15} = 5 \implies BW_{FM} = 2 (75 + 15) = 180\,\text{kHz}$$

Adding a $20\,\text{kHz}$ guard band establishes the standard $200\,\text{kHz}$ channel allocation per FM broadcast station.

ยง6.4 Superheterodyne Radio Receiver Architecture & Image Frequency

1. Architecture of the Superheterodyne Receiver

Direct tuned radio frequency (TRF) receivers suffer from two fatal flaws: bandwidth varies across the tuning dial ($BW = f_0 / Q$), and multiple cascading tuned circuits cannot track simultaneously without instability. Edwin Armstrong solved this by inventing the Superheterodyne Receiver, which downconverts all incoming RF carriers to a fixed Intermediate Frequency (IF).

The functional block stages are:

  1. RF Amplifier: Provides low-noise amplification, improves receiver sensitivity, and provides pre-selection to reject image frequencies.
  2. Local Oscillator (LO): Generates an unmodulated RF sine wave at frequency $f_{LO}$. Standard designs use high-side (supradyne) injection: $f_{LO} = f_s + f_{IF}$.
  3. Mixer (Frequency Converter): Multiplies the incoming signal $f_s$ and local oscillator $f_{LO}$, producing sum and difference frequencies: $f_{LO} \pm f_s$.
  4. IF Amplifier: Highly selective, high-gain tuned amplifier permanently aligned to the fixed Intermediate Frequency ($f_{IF} = 455\,\text{kHz}$ for AM; $f_{IF} = 10.7\,\text{MHz}$ for FM). All adjacent-channel selectivity and main voltage gain are provided here.
  5. Demodulator / Detector: Diode envelope detector (AM) or ratio detector / PLL (FM) recovering the baseband audio.
  6. Audio Power Amplifier: Drives low-impedance speakers.
  7. Automatic Gain Control (AGC): Feeds rectified DC voltage back to the RF and IF stages, maintaining constant audio output volume regardless of incoming antenna signal strength variations.

2. Image Frequency ($f_{img}$) and Image Rejection Ratio (IRR)

Because the mixer responds to the absolute difference $|f_{LO} - f_{signal}| = f_{IF}$, an unwanted signal at the Image Frequency $f_{img}$ will also mix with $f_{LO}$ to produce the exact same IF:

$$f_{img} - f_{LO} = f_{IF} \implies f_{img} = f_{LO} + f_{IF} = (f_s + f_{IF}) + f_{IF} = f_s + 2 f_{IF}$$

Once the image frequency enters the mixer, the IF amplifier cannot distinguish it from the desired station, causing severe simultaneous heterodyne whistle and crosstalk.

Image rejection must therefore be performed entirely by the RF pre-selection filter before the mixer. The Image Rejection Ratio (IRR) is:

$$\text{IRR} = \sqrt{1 + Q^2 \rho^2}$$

where $Q$ is the quality factor of the tuned RF input circuit, and $\rho$ is the fractional frequency separation:

$$\rho = \frac{f_{img}}{f_s} - \frac{f_s}{f_{img}}$$

In decibels, $\text{IRR}(\text{dB}) = 20 \log_{10}(\text{IRR})$. High IF choices (e.g. 10.7 MHz) push the image frequency far away, making it easy for simple RF front-ends to eliminate image interference completely.

Solved Problem Example 6.1: AM Transmitter Power, Spectrum & Efficiency Calculation

An AM broadcast transmitter radiates a total power of $P_t = 11.8\text{ kW}$ when modulated by a single sinusoidal audio tone with a modulation index $m = 0.60$. (a) Calculate the unmodulated carrier power $P_c$. (b) Determine the total sideband power $P_{\text{SB}}$ and the power in each individual sideband. (c) If the modulation depth increases to 100% ($m = 1.0$), calculate the new total radiated power and the percentage power efficiency of the transmitter.

Step 1: Calculate Unmodulated Carrier Power $P_c$

Using the AM total power relation:

$$P_t = P_c \left( 1 + \frac{m_a^2}{2} \right)$$
$$1 + \frac{m_a^2}{2} = 1 + \frac{0.6^2}{2} = 1 + \frac{0.36}{2} = 1.18$$
$$P_c = \frac{P_t}{1.18} = \frac{11.8\,\text{kW}}{1.18} = 10.0\,\text{kW}$$

Step 2: Calculate Power in Sidebands

$$P_{total,SB} = P_t - P_c = 11.8\,\text{kW} - 10.0\,\text{kW} = 1.8\,\text{kW}$$
$$P_{USB} = P_{LSB} = \frac{P_{total,SB}}{2} = \frac{1.8\,\text{kW}}{2} = 0.9\,\text{kW} = 900\,\text{W}$$

Step 3: Calculate Transmission Efficiency $\eta$

$$\eta = \frac{P_{total,SB}}{P_t} \times 100\% = \frac{1.8\,\text{kW}}{11.8\,\text{kW}} \times 100\% \approx 15.25\%$$

Step 4: Spectral Frequencies Present

$$f_c = 1000\,\text{kHz}$$
$$f_{USB} = f_c + f_m = 1000\,\text{kHz} + 5\,\text{kHz} = 1005\,\text{kHz}$$
$$f_{LSB} = f_c - f_m = 1000\,\text{kHz} - 5\,\text{kHz} = 995\,\text{kHz}$$
$$BW = 2 f_m = 10\,\text{kHz}$$

Step 5: Total Radiated Power at $m_a = 0.9$

$$P_t' = P_c \left( 1 + \frac{m_a^2}{2} \right) = 10.0\,\text{kW} \times \left( 1 + \frac{0.9^2}{2} \right) = 10.0 \times (1 + 0.405) = 14.05\,\text{kW}$$
Solved Problem Example 6.2: FM Signal Analysis & Carson's Rule Bandwidth Calculation

A commercial FM signal is given by the mathematical expression: $v(t) = 12 \cos\left( 2\pi \times 10^8 t + 6 \sin(2\pi \times 10^4 t) \right)\text{ V}$. (a) Determine the carrier frequency $f_c$ in MHz and the modulating signal frequency $f_m$ in kHz. (b) Find the modulation index $\beta$ and maximum frequency deviation $\Delta f$. (c) Using Carson's rule, compute the transmission bandwidth required to pass 98% of the signal power.

Step 1: Identify Carrier and Modulating Frequencies

Comparing with canonical FM equation $v(t) = V_c \cos(\omega_c t + \beta \sin(\omega_m t))$:

$$\omega_c = 2\pi \times 10^8\,\text{rad/s} \implies f_c = 10^8\,\text{Hz} = 100\,\text{MHz}$$
$$\omega_m = 2\pi \times 1.25 \times 10^4\,\text{rad/s} \implies f_m = 1.25 \times 10^4\,\text{Hz} = 12.5\,\text{kHz}$$

Step 2: Determine Modulation Index $\beta$ and Frequency Deviation $\Delta f$

$$\beta = 6$$
$$\beta = \frac{\Delta f}{f_m} \implies \Delta f = \beta \cdot f_m = 6 \times 12.5\,\text{kHz} = 75\,\text{kHz}$$

Step 3: Calculate Bandwidth via Carson's Rule

$$BW = 2 (\Delta f + f_m) = 2 (75\,\text{kHz} + 12.5\,\text{kHz}) = 2 \times 87.5\,\text{kHz} = 175\,\text{kHz}$$

Step 4: Calculate Total Transmitted Power into $50\,\Omega$ Load

Since FM amplitude is strictly constant ($V_c = 12\,\text{V}$), total power is unaffected by modulation:

$$P = \frac{V_c^2}{2 R_L} = \frac{(12\,\text{V})^2}{2 \times 50\,\Omega} = \frac{144}{100} = 1.44\,\text{W}$$
Solved Problem Example 6.3: Superheterodyne Receiver Tuning Range & Image Rejection Analysis

A standard AM broadcast superheterodyne receiver tunes across the medium wave band from $f_{s,\text{min}} = 535\text{ kHz}$ to $f_{s,\text{max}} = 1605\text{ kHz}$ using an intermediate frequency (IF) of $f_{\text{IF}} = 455\text{ kHz}$ with higher-side local oscillator injection ($f_{\text{LO}} = f_s + f_{\text{IF}}$). (a) Calculate the tuning range of the local oscillator $f_{\text{LO}}$. (b) Determine the capacitance ratio $C_{\text{max}} / C_{\text{min}}$ of the variable tuning capacitor. (c) When tuned to a station at $700\text{ kHz}$, determine the image frequency and the image rejection ratio in dB if the RF preselector circuit has a loaded quality factor $Q = 80$.

Step 1: Calculate Local Oscillator Tuning Range

With high-side injection ($f_{LO} = f_s + f_{IF}$):

$$f_{LO,min} = 535\,\text{kHz} + 455\,\text{kHz} = 990\,\text{kHz}$$
$$f_{LO,max} = 1605\,\text{kHz} + 455\,\text{kHz} = 2060\,\text{kHz}$$

Tuning ratio: $2060 / 990 \approx 2.08:1$. If low-side injection had been used, the range would be $80\,\text{kHz}$ to $1150\,\text{kHz}$ (a $14.4:1$ ratio), requiring an impractically large variable capacitor tuning ratio of $(14.4)^2 \approx 207:1$. High-side injection requires only $(2.08)^2 \approx 4.33:1$ capacitance range.

Step 2: Calculate Image Frequency for $f_s = 1000\,\text{kHz}$

$$f_{img} = f_s + 2 f_{IF} = 1000\,\text{kHz} + 2(455\,\text{kHz}) = 1000 + 910 = 1910\,\text{kHz}$$

Step 3: Calculate Image Rejection Ratio (IRR)

Calculate fractional separation parameter $\rho$:

$$\rho = \frac{f_{img}}{f_s} - \frac{f_s}{f_{img}} = \frac{1910}{1000} - \frac{1000}{1910} = 1.91 - 0.5236 = 1.3864$$

With $Q = 80$:

$$Q \rho = 80 \times 1.3864 \approx 110.91$$
$$\text{IRR} = \sqrt{1 + (Q \rho)^2} = \sqrt{1 + (110.91)^2} \approx 110.91$$

In decibels:

$$\text{IRR}(\text{dB}) = 20 \log_{10}(110.91) \approx 20 \times 2.0449 = 40.90\,\text{dB}$$

The RF pre-selector suppresses the image frequency by nearly $41\,\text{dB}$ ($111$ times in voltage) prior to the mixer.

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