10.2 Amplitude Modulation (AM) Principles & Receivers

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10.2 Amplitude Modulation (AM) Principles & Receivers

Key Takeaways Explain This Topic AI

AM envelope equation v_AM(t) = V_c [1 + m cos(omega_m t)] cos(omega_c t) describes a carrier wave whose peak amplitude varies directly with the modulating signal voltage.

The modulation index m = (Vmax - Vmin) / (Vmax + Vmin) determines sideband power; values exceeding 1.0 cause severe overmodulation envelope distortion and sideband splatter.

Total power in a standard double-sideband full-carrier (DSFC) AM signal is Pt = Pc(1 + m^2 / 2), allocating a maximum of 33.3% of total transmitted power to sidebands at 100% modulation.

Superheterodyne receivers convert incoming RF signals to a fixed Intermediate Frequency (IF) of 455 kHz using a local oscillator (f_LO = f_RF + f_IF), introducing potential image frequency interference at f_img = f_RF + 2 f_IF.

Last updated: July 2026

10.2 Amplitude Modulation (AM) Principles & Receivers

Board Examination Focus: AM modulation index ( m m m ), trapezoidal pattern calculations, sideband power breakdown ( P t = P c ( 1 + m 2 / 2 ) P_t = P_c(1 + m^2/2) P t = P c ( 1 + m 2 /2 ) ), superheterodyne receiver stages, 455 kHz intermediate frequency (IF), and image frequency ( f img = f rf + 2 f if f_{\text{img}} = f_{\text{rf}} + 2f_{\text{if}} f img = f rf + 2 f if ) rejection are standard PRC licensure exam questions.

Amplitude Modulation (AM) is a baseline analog modulation scheme wherein the peak amplitude of a high-frequency radio frequency (RF) carrier wave is varied in direct proportion to the instantaneous amplitude of a lower-frequency modulating information signal. Despite the transition to digital broadcasting, AM principles remain foundational in understanding sideband spectral analysis, power distribution, and superheterodyne receiver design.

Mathematical Formulation of AM Signals

Consider an unmodulated sinusoidal carrier wave v c ( t ) v_c(t) v c ( t ) and a single-tone modulating audio signal v m ( t ) v_m(t) v m ( t ) : v c ( t ) = V c cos ⁡ ( ω c t ) v_c(t) = V_c \cos(\omega_c t) v c ( t ) = V c cos ( ω c t ) v m ( t ) = V m cos ⁡ ( ω m t ) v_m(t) = V_m \cos(\omega_m t) v m ( t ) = V m cos ( ω m t )

Where V c V_c V c and V m V_m V m represent the peak amplitude voltages, and ω c = 2 π f c \omega_c = 2\pi f_c ω c = 2 π f c and ω m = 2 π f m \omega_m = 2\pi f_m ω m = 2 π f m represent the angular frequencies ( rad/s \text{rad/s} rad/s ).

In Amplitude Modulation, the instantaneous peak envelope amplitude of the carrier wave becomes V ( t ) = V c + v m ( t ) = V c + V m cos ⁡ ( ω m t ) V(t) = V_c + v_m(t) = V_c + V_m \cos(\omega_m t) V ( t ) = V c + v m ( t ) = V c + V m cos ( ω m t ) . Substituting V ( t ) V(t) V ( t ) into the carrier equation yields the complete time-domain AM wave equation: v AM ( t ) = [ V c + V m cos ⁡ ( ω m t ) ] cos ⁡ ( ω c t ) = V c [ 1 + ( V m V c ) cos ⁡ ( ω m t ) ] cos ⁡ ( ω c t ) v_{\text{AM}}(t) = [V_c + V_m \cos(\omega_m t)] \cos(\omega_c t) = V_c \left[1 + \left(\frac{V_m}{V_c}\right) \cos(\omega_m t)\right] \cos(\omega_c t) v AM ( t ) = [ V c + V m cos ( ω m t )] cos ( ω c t ) = V c [ 1 + ( V c V m ) cos ( ω m t ) ] cos ( ω c t )

Modulation Index ( m m m ) and Percent Modulation

The modulation index ( m m m ), also called modulation factor or coefficient, is the ratio of peak modulating voltage to peak carrier voltage: m = V m V c m = \frac{V_m}{V_c} m = V c V m

The percentage of modulation ( % M \%M % M ) is simply % M = m × 100 % \%M = m \times 100\% % M = m × 100% .

When observing an AM envelope on an oscilloscope, m m m can be determined directly from the maximum envelope peak ( V max V_{\text{max}} V max ) and minimum envelope trough ( V min V_{\text{min}} V min ): V max = V c + V m and V min = V c − V m V_{\text{max}} = V_c + V_m \quad \text{and} \quad V_{\text{min}} = V_c - V_m V max = V c + V m and V min = V c − V m

Solving these simultaneous equations for V c V_c V c and V m V_m V m yields: V c = V max + V min 2 and V m = V max − V min 2 V_c = \frac{V_{\text{max}} + V_{\text{min}}}{2} \quad \text{and} \quad V_m = \frac{V_{\text{max}} - V_{\text{min}}}{2} V c = 2 V max + V min and V m = 2 V max − V min

Substituting back into the definition of m m m gives the standard oscilloscope measurement formula: m = V max − V min V max + V min m = \frac{V_{\text{max}} - V_{\text{min}}}{V_{\text{max}} + V_{\text{min}}} m = V max + V min V max − V min Normal Envelope (m = 0.5) 100% Modulation (m = 1.0) Overmodulation (m > 1.0)

+Vmax ------------ +Vmax ------------ +Vmax ------------
+Vmin --- +Vmin = 0 -------- Zero Crossing (Envelope Clipping)
Vmin --- -Vmin = 0 -------- Sideband Splatter / Distortion
Vmax ------------ -Vmax ------------ -Vmax ------------

Under-modulation ( m < 1.0 m < 1.0 m < 1.0 ): Safe, undistorted transmission, but RF power is underutilized.

Critical modulation ( m = 1.0 m = 1.0 m = 1.0 ): Maximum undistorted power transfer where V min = 0 V_{\text{min}} = 0 V min = 0 .

Over-modulation ( m > 1.0 m > 1.0 m > 1.0 ): The modulating signal voltage exceeds carrier voltage, causing envelope phase reversals and zero-crossing clipping. This generates severe harmonic distortion and broad sideband splatter into adjacent channels.

AM Frequency Spectrum and Power Distribution

Using the trigonometric identity cos ⁡ ( A ) cos ⁡ ( B ) = 1 2 [ cos ⁡ ( A + B ) + cos ⁡ ( A − B ) ] \cos(A)\cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)] cos ( A ) cos ( B ) = 2 1 [ cos ( A + B ) + cos ( A − B )] , expanding the time-domain AM equation yields three distinct frequency components: v AM ( t ) = V c cos ⁡ ( 2 π f c t ) + m V c 2 cos ⁡ [ 2 π ( f c + f m ) t ] + m V c 2 cos ⁡ [ 2 π ( f c − f m ) t ] v_{\text{AM}}(t) = V_c \cos(2\pi f_c t) + \frac{m V_c}{2} \cos[2\pi (f_c + f_m) t] + \frac{m V_c}{2} \cos[2\pi (f_c - f_m) t] v AM ( t ) = V c cos ( 2 π f c t ) + 2 m V c cos [ 2 π ( f c + f m ) t ] + 2 m V c cos [ 2 π ( f c − f m ) t ]

Carrier Component ( f c f_c f c ): Constant amplitude V c V_c V c , carries no information.

Upper Sideband (USB, f U S B = f c + f m f_{USB} = f_c + f_m f U SB = f c + f m ): Amplitude m V c 2 \frac{m V_c}{2} 2 m V c , contains full message spectrum.

Lower Sideband (LSB, f L S B = f c − f m f_{LSB} = f_c - f_m f L SB = f c − f m ): Amplitude m V c 2 \frac{m V_c}{2} 2 m V c , identical mirror image of message spectrum.

The required channel bandwidth for standard AM is twice the maximum modulating frequency: B W = 2 f m , max BW = 2 f_{m,\text{max}} B W = 2 f m , max

AM Power Calculations

Assuming an antenna load resistance R R R , total average power ( P t P_t P t ) is the sum of carrier power ( P c P_c P c ) and sideband powers ( P U S B + P L S B P_{USB} + P_{LSB} P U SB + P L SB ): P c = V c 2 2 R and P U S B = P L S B = ( m V c / 2 ) 2 2 R = m 2 V c 2 8 R = m 2 4 P c P_c = \frac{V_c^2}{2 R} \quad \text{and} \quad P_{USB} = P_{LSB} = \frac{(m V_c / 2)^2}{2 R} = \frac{m^2 V_c^2}{8 R} = \frac{m^2}{4} P_c P c = 2 R V c 2 and P U SB = P L SB = 2 R ( m V c /2 ) 2 = 8 R m 2 V c 2 = 4 m 2 P c

Summing all components yields the total transmitted power formula: P t = P c + P U S B + P L S B = P c + m 2 4 P c + m 2 4 P c = P c ( 1 + m 2 2 ) P_t = P_c + P_{USB} + P_{LSB} = P_c + \frac{m^2}{4} P_c + \frac{m^2}{4} P_c = P_c \left(1 + \frac{m^2}{2}\right) P t = P c + P U SB + P L SB = P c + 4 m 2 P c + 4 m 2 P c = P c ( 1 + 2 m 2 )

Total antenna current ( I t I_t I t ) in terms of unmodulated carrier current ( I c I_c I c ) is: I t = I c 1 + m 2 2 I_t = I_c \sqrt{1 + \frac{m^2}{2}} I t = I c 1 + 2 m 2

AM Transmission Mode Carrier Present? Sidebands Transmitted Power Efficiency at 100 % 100\% 100% Mod. Application

DSFC (Double Sideband Full Carrier) Yes USB + LSB 33.3 % 33.3\% 33.3% in sidebands ( 66.7 % 66.7\% 66.7% wasted in carrier) Commercial AM Broadcast ( 535 − 1705 kHz 535-1705\text{ kHz} 535 − 1705 kHz )

DSBSC (Double Sideband Suppressed Carrier) No USB + LSB 100 % 100\% 100% in sidebands Analog subcarriers, stereo multiplexing

SSBSC (Single Sideband Suppressed Carrier) No USB or LSB only 100 % 100\% 100% in single sideband HF Military, Aviation, Amateur Radio

Superheterodyne Receiver Architecture

Invented by Edwin Armstrong in 1918, the superheterodyne receiver converts all incoming RF signals to a fixed lower frequency called the Intermediate Frequency (IF) before demodulation. This design resolves the severe selectivity and gain instability limitations of Tuned Radio Frequency (TRF) receivers.

Block Diagram Breakdown Antenna --> [ RF Amp ] --> [ Mixer / Converter ] --> [ IF Amp (455 kHz) ] --> [ Diode Detector ] --> [ Audio Amp ] --> Speaker

^
|
[ Local Osc (f_LO) ] <------------------------------ AGC Feedback

RF Amplifier: Low-noise preamplifier providing initial selectivity and suppressing image frequency responses.

Local Oscillator (LO) & Mixer: Heterodynes incoming f RF f_{\text{RF}} f RF with f LO f_{\text{LO}} f LO to create sum ( f LO + f RF f_{\text{LO}} + f_{\text{RF}} f LO + f RF ) and difference ( f LO − f RF f_{\text{LO}} - f_{\text{RF}} f LO − f RF ) frequencies. High-side injection is standard ( f LO = f RF + f IF f_{\text{LO}} = f_{\text{RF}} + f_{\text{IF}} f LO = f RF + f IF ).

IF Amplifier: Fixed-frequency tuned amplifier tuned to f IF = 455 kHz f_{\text{IF}} = 455\text{ kHz} f IF = 455 kHz (for standard AM broadcast). Provides the bulk ( 60 − 80 dB 60-80\text{ dB} 60 − 80 dB ) of receiver gain and overall adjacent-channel selectivity.

Diode Peak Detector: Rectifies and filters the 455 kHz 455\text{ kHz} 455 kHz IF signal to recover the original audio envelope.

Automatic Gain Control (AGC): Feeds a negative DC voltage proportional to received carrier strength back to RF/IF stages to maintain uniform audio output level across weak and strong stations.

Image Frequency and Selectivity Analysis

An image frequency ( f img f_{\text{img}} f img ) is an unwanted RF signal located on the opposite side of the local oscillator frequency from the desired station ( f RF f_{\text{RF}} f RF ) by an amount equal to 2 f IF 2 f_{\text{IF}} 2 f IF .

For high-side local oscillator injection ( f LO = f RF + f IF f_{\text{LO}} = f_{\text{RF}} + f_{\text{IF}} f LO = f RF + f IF ): f img = f RF + 2 f IF f_{\text{img}} = f_{\text{RF}} + 2 f_{\text{IF}} f img = f RF + 2 f IF

When f img f_{\text{img}} f img enters the mixer alongside f RF f_{\text{RF}} f RF , heterodyning produces an identical difference frequency: f img − f LO = ( f RF + 2 f IF ) − ( f RF + f IF ) = f IF f_{\text{img}} - f_{\text{LO}} = (f_{\text{RF}} + 2 f_{\text{IF}}) - (f_{\text{RF}} + f_{\text{IF}}) = f_{\text{IF}} f img − f LO = ( f RF + 2 f IF ) − ( f RF + f IF ) = f IF

Both stations pass directly into the IF amplifier, causing severe unintelligible interference.

Image Rejection Ratio (IRR)

The ability of the preselector RF stage to attenuate the image frequency is quantified by the Image Rejection Ratio: IRR = 1 + Q 2 ρ 2 \text{IRR} = \sqrt{1 + Q^2 \rho^2} IRR = 1 + Q 2 ρ 2

Where Q Q Q is the quality factor of the RF tuned circuit, and ρ \rho ρ is the frequency ratio factor: ρ = f img f RF − f RF f img \rho = \frac{f_{\text{img}}}{f_{\text{RF}}} - \frac{f_{\text{RF}}}{f_{\text{img}}} ρ = f RF f img − f img f RF

Worked Example: Image Frequency and Power Calculation

Problem: An AM superheterodyne receiver with an IF of 455 kHz 455\text{ kHz} 455 kHz is tuned to a station at f RF = 1000 kHz f_{\text{RF}} = 1000\text{ kHz} f RF = 1000 kHz . The transmitter has an unmodulated carrier power of 10 kW 10\text{ kW} 10 kW .

(a) Determine the local oscillator frequency ( f LO f_{\text{LO}} f LO ) and image frequency ( f img f_{\text{img}} f img ).

(b) If the transmitter is modulated at 80 % 80\% 80% ( m = 0.8 m = 0.8 m = 0.8 ), calculate total transmitted power ( P t P_t P t ) and power in each sideband.

Solution:

Calculate local oscillator frequency ( f LO f_{\text{LO}} f LO ): f LO = f RF + f IF = 1000 kHz + 455 kHz = 1455 kHz f_{\text{LO}} = f_{\text{RF}} + f_{\text{IF}} = 1000\text{ kHz} + 455\text{ kHz} = 1455\text{ kHz} f LO = f RF + f IF = 1000 kHz + 455 kHz = 1455 kHz

Calculate image frequency ( f img f_{\text{img}} f img ): f img = f RF + 2 f IF = 1000 kHz + 2 ( 455 kHz ) = 1910 kHz f_{\text{img}} = f_{\text{RF}} + 2 f_{\text{IF}} = 1000\text{ kHz} + 2(455\text{ kHz}) = 1910\text{ kHz} f img = f RF + 2 f IF = 1000 kHz + 2 ( 455 kHz ) = 1910 kHz

Calculate total transmitted power ( P t P_t P t ): P t = P c ( 1 + m 2 2 ) = 10 kW ( 1 + 0.8 2 2 ) = 10 kW ( 1 + 0.32 ) = 13.2 kW P_t = P_c \left(1 + \frac{m^2}{2}\right) = 10\text{ kW} \left(1 + \frac{0.8^2}{2}\right) = 10\text{ kW} (1 + 0.32) = 13.2\text{ kW} P t = P c ( 1 + 2 m 2 ) = 10 kW ( 1 + 2 0. 8 2 ) = 10 kW ( 1 + 0.32 ) = 13.2 kW

Calculate power in each sideband ( P U S B = P L S B P_{USB} = P_{LSB} P U SB = P L SB ): P U S B = P L S B = m 2 4 P c = 0.64 4 × 10 kW = 1.6 kW P_{USB} = P_{LSB} = \frac{m^2}{4} P_c = \frac{0.64}{4} \times 10\text{ kW} = 1.6\text{ kW} P U SB = P L SB = 4 m 2 P c = 4 0.64 × 10 kW = 1.6 kW