What is the modulation index of AM and what happens when it exceeds 1?

The modulation index μ = k_a·A_m is the ratio of the peak message-induced amplitude change to the carrier amplitude; from the envelope it is (A_max − A_min)/(A_max + A_min). For μ ≤ 1 the envelope never crosses zero and faithfully follows the message. For μ > 1 the envelope reaches zero and the carrier reverses phase, so an envelope detector outputs |1 + μ·m(t)|, which is distorted, and the extra harmonics spread the spectrum beyond 2W.

Why is conventional AM power-inefficient, and how do DSB-SC and SSB improve on it?

Total power is P_c(1 + μ²/2), and the carrier term P_c carries no information, so even at μ = 1 only one third of the power is in the sidebands. DSB-SC removes the carrier so all transmitted power carries the message, and SSB additionally removes one redundant sideband, halving the bandwidth to W. The price is that both need a coherent receiver with carrier recovery, whereas conventional AM can use a cheap diode envelope detector.

How do you choose the RC time constant of a diode envelope detector?

RC must be much larger than the carrier period 1/f_c so the capacitor holds the peak between carrier cycles and the ripple is small. It must also be much smaller than 1/W so the output can follow the fastest message variation; for a tone the strict condition against diagonal clipping is RC ≤ √(1 − μ²)/(2πf_m·μ). So 1/f_c ≪ RC ≪ 1/W, and the upper limit tightens at high modulation depth.

What happens in coherent detection of DSB-SC if the local oscillator has a phase or frequency error?

Multiplying A_c·m(t)·cos(2πf_c·t) by cos(2πf_c·t + φ) and low-pass filtering gives an output proportional to m(t)·cos φ. A fixed phase error therefore attenuates the output, and at φ = 90° the output vanishes (quadrature null effect). A frequency error Δf gives m(t)·cos(2πΔf·t), a beating that makes the output wax and wane, so a carrier-recovery loop such as a Costas loop or squaring loop is needed.

Where are AM principles used in instrumentation rather than broadcasting?

Many AC-excited sensors produce AM directly: an LVDT or strain-gauge bridge excited by a carrier gives an output whose amplitude is the measurand, and a phase-sensitive (synchronous) demodulator recovers magnitude and sign. Lock-in amplifiers modulate a weak signal to a reference frequency and coherently detect it, rejecting 1/f noise and drift. Chopper-stabilised amplifiers use the same modulate–amplify–demodulate idea to avoid DC offset drift.

An AM transmitter's antenna current rises from 8 A to 8.93 A on modulation. What is the modulation index?

Antenna current goes as the square root of power, so I_t/I_c = √(1 + μ²/2). Then μ² = 2[(8.93/8)² − 1] = 2 × 0.246 = 0.492, giving μ ≈ 0.70. The power increase is only about 24.6%, which is why AM transmitters spend most of their power on the carrier.

Amplitude modulation and demodulation

AM puts the message on the carrier envelope: modulation index, power and efficiency, DSB-SC/SSB/VSB, envelope and coherent detection.

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Why it matters

Amplitude modulation (AM) is the simplest way to put a low-frequency signal onto a high-frequency carrier, and it is the reference against which every other modulation scheme is compared. Medium-wave broadcasting, aircraft VHF radio, carrier-type instrumentation amplifiers and lock-in detection all use AM ideas. The same mathematics (mixing, sidebands, envelope detection) reappears in superheterodyne receivers and in sensor signal conditioning such as LVDT demodulators.

Key ideas

Why modulate at all. Antennas must be a sizeable fraction of a wavelength, and a 3 kHz audio wave has a wavelength of 100 km. Moving the message up to a carrier frequency f_c makes antennas practical, lets many users share the spectrum by frequency division and lets us choose a band with good propagation.

Conventional AM (DSB with full carrier, DSB-FC). The carrier amplitude follows the message: s(t) = A_c[1 + k_a·m(t)]·cos(2πf_c·t). For a tone m(t) = A_m·cos 2πf_m·t, the modulation index is μ = k_a·A_m (equivalently A_m/A_c when k_a = 1/A_c). Expanding gives a carrier at f_c and two side frequencies at f_c ± f_m, each of amplitude μA_c/2. The bandwidth is B = 2f_m (or 2W for a message band-limited to W).

Modulation index from the envelope. On an oscilloscope, with maximum envelope A_max and minimum A_min, μ = (A_max − A_min)/(A_max + A_min). If μ > 1 (over-modulation) the envelope crosses zero, phase reversals appear and an envelope detector distorts the output.

Power. The carrier carries no information but takes most of the power: P_t = P_c(1 + μ²/2). Even at μ = 1 only one third of the power is in the sidebands, so the transmission (power) efficiency η = μ²/(2 + μ²) is at most 33.3%. For several tones, the effective index is μ_t = √(μ_1² + μ_2² + …). Because antenna current is proportional to the square root of power, I_t = I_c·√(1 + μ²/2).

Suppressed-carrier variants.

DSB-SC: s(t) = A_c·m(t)·cos 2πf_c·t. All power is in the sidebands, bandwidth is still 2W, but detection needs a coherent (synchronous) local carrier.
SSB: only one sideband, bandwidth W, half the DSB power for the same sideband level; generated by filtering or by the phase-shift (Hilbert) method.
VSB: one sideband plus a vestige of the other; used where the message has energy near DC, such as analog TV video.

Generation. Low-level: a square-law or switching modulator followed by a band-pass filter, or an analog multiplier. High-level: a class-C amplifier whose collector supply is varied by the message. DSB-SC is made with a balanced or ring modulator, which cancels the carrier.

Demodulation.

Envelope detector: a diode, a capacitor C and a load R. It works only for conventional AM with μ ≤ 1. The time constant must satisfy 1/f_c ≪ RC ≪ 1/W so that the capacitor holds between carrier peaks but still follows the envelope. If RC is too large the output cannot follow a falling envelope (diagonal clipping); for a tone, avoid it with RC ≤ √(1 − μ²)/(2πf_m·μ).
Square-law detector: works at low levels but produces second-harmonic distortion.
Coherent detector: multiply by a local carrier cos(2πf_c·t + φ) and low-pass filter. Output is proportional to m(t)·cos φ, so a phase error attenuates DSB-SC (and nulls it at φ = 90°); a frequency error produces beating. A Costas loop recovers the carrier for DSB-SC.

Formulas

s(t) = A_c[1 + μ·cos(2πf_m·t)]·cos(2πf_c·t) A_c carrier amplitude (V), μ modulation index (dimensionless, 0 to 1 for distortion-free envelope detection), f_m tone frequency (Hz), f_c carrier frequency (Hz).

μ = (A_max − A_min)/(A_max + A_min) A_max, A_min maximum and minimum envelope amplitude (V).

P_c = A_c²/(2R), P_t = P_c(1 + μ²/2), P_SB = P_c·μ²/2 P_c carrier power, P_t total power, P_SB power in both sidebands (W); R load resistance (Ω).

η = μ²/(2 + μ²) η transmission efficiency (fraction of power in sidebands).

I_t = I_c·√(1 + μ²/2) I_t, I_c RMS antenna current with and without modulation (A).

μ_t = √(μ_1² + μ_2² + …) μ_t effective index for multi-tone modulation.

B_AM = B_DSB = 2W, B_SSB = W B transmission bandwidth (Hz), W highest message frequency (Hz).

RC ≤ √(1 − μ²)/(2πf_m·μ) Envelope-detector condition to avoid diagonal clipping; R (Ω), C (F).

Worked examples

Example 1 (standard). A carrier of amplitude 10 V is amplitude modulated by a 5 kHz tone to μ = 0.5 and fed to a 50 Ω antenna. Find the bandwidth, carrier power, total power and efficiency.

B = 2f_m = 2 × 5 kHz = 10 kHz.
P_c = A_c²/(2R) = 10²/(2 × 50) = 1 W.
P_t = P_c(1 + μ²/2) = 1 × (1 + 0.25/2) = 1.125 W.
η = μ²/(2 + μ²) = 0.25/2.25 = 0.111. Answer: B = 10 kHz, P_c = 1 W, P_t = 1.125 W, η ≈ 11.1%.

Example 2 (GATE level). The unmodulated antenna current of an AM transmitter is 8 A. With a single tone it rises to 8.93 A. (a) Find μ. (b) A second tone with μ_2 = 0.4 is added. Find the new antenna current.

I_t/I_c = √(1 + μ²/2) gives μ² = 2[(I_t/I_c)² − 1] = 2[(8.93/8)² − 1] = 2 × 0.2460 = 0.4920.
μ = √0.4920 = 0.70.
Effective index: μ_t = √(0.70² + 0.4²) = √(0.49 + 0.16) = 0.806.
I_t = 8 × √(1 + 0.806²/2) = 8 × √1.325 = 9.21 A. Answer: (a) μ ≈ 0.70; (b) I_t ≈ 9.21 A.

Example 3 (detector design). An envelope detector with C = 10 nF must demodulate AM with μ = 0.5 and f_m up to 5 kHz. Find the largest R that avoids diagonal clipping.

RC ≤ √(1 − μ²)/(2πf_m·μ) = √0.75/(2π × 5000 × 0.5) = 0.866/15 708 = 55.1 µs.
R ≤ 55.1 µs / 10 nF = 5.51 kΩ. Answer: R ≤ about 5.5 kΩ (and RC must still be much larger than 1/f_c).

Common mistakes

Using P_t = P_c(1 + μ²) instead of (1 + μ²/2); each sideband carries only P_c·μ²/4.
Adding modulation indices of two tones directly. They add as squares (powers), not linearly.
Using an envelope detector for DSB-SC or for over-modulated AM. The envelope is |m(t)|, not m(t).
Forgetting that current goes as the square root of power, so a 12.5% power rise is only a 6.1% current rise.
Choosing RC only to smooth the carrier ripple and ignoring diagonal clipping at high f_m and high μ.
Claiming SSB halves the power required for the same signal quality without stating what is held fixed; compare at the same sideband power.

For GATE IN

Power, efficiency and antenna-current numericals, including multi-tone modulation.
Reading μ from A_max and A_min, or from a given s(t) expression and its spectrum.
Bandwidths of AM, DSB-SC, SSB and VSB, and which detector works for which.
Coherent detection with phase or frequency error in the local oscillator.
Envelope-detector time-constant conditions.

Quick check

An AM envelope swings between 4 V and 16 V. What is μ?
What fraction of total power is in the sidebands at μ = 1?
Which AM variant needs only bandwidth W?
A DSB-SC receiver's local carrier has a 60° phase error. By what factor is the output scaled? Answers: 1. μ = 12/20 = 0.6. 2. One third (33.3%). 3. SSB. 4. cos 60° = 0.5.

Interview questions

All Communication and Optical Instrumentation interview questions

Try answering each one aloud before you open it.

1.What is the modulation index of AM and what happens when it exceeds 1?Concept
The modulation index μ = k_a·A_m is the ratio of the peak message-induced amplitude change to the carrier amplitude; from the envelope it is (A_max − A_min)/(A_max + A_min). For μ ≤ 1 the envelope never crosses zero and faithfully follows the message. For μ > 1 the envelope reaches zero and the carrier reverses phase, so an envelope detector outputs |1 + μ·m(t)|, which is distorted, and the extra harmonics spread the spectrum beyond 2W.
2.Why is conventional AM power-inefficient, and how do DSB-SC and SSB improve on it?Concept
Total power is P_c(1 + μ²/2), and the carrier term P_c carries no information, so even at μ = 1 only one third of the power is in the sidebands. DSB-SC removes the carrier so all transmitted power carries the message, and SSB additionally removes one redundant sideband, halving the bandwidth to W. The price is that both need a coherent receiver with carrier recovery, whereas conventional AM can use a cheap diode envelope detector.
3.How do you choose the RC time constant of a diode envelope detector?Concept
RC must be much larger than the carrier period 1/f_c so the capacitor holds the peak between carrier cycles and the ripple is small. It must also be much smaller than 1/W so the output can follow the fastest message variation; for a tone the strict condition against diagonal clipping is RC ≤ √(1 − μ²)/(2πf_m·μ). So 1/f_c ≪ RC ≪ 1/W, and the upper limit tightens at high modulation depth.
4.What happens in coherent detection of DSB-SC if the local oscillator has a phase or frequency error?Concept
Multiplying A_c·m(t)·cos(2πf_c·t) by cos(2πf_c·t + φ) and low-pass filtering gives an output proportional to m(t)·cos φ. A fixed phase error therefore attenuates the output, and at φ = 90° the output vanishes (quadrature null effect). A frequency error Δf gives m(t)·cos(2πΔf·t), a beating that makes the output wax and wane, so a carrier-recovery loop such as a Costas loop or squaring loop is needed.
5.Where are AM principles used in instrumentation rather than broadcasting?Concept
Many AC-excited sensors produce AM directly: an LVDT or strain-gauge bridge excited by a carrier gives an output whose amplitude is the measurand, and a phase-sensitive (synchronous) demodulator recovers magnitude and sign. Lock-in amplifiers modulate a weak signal to a reference frequency and coherently detect it, rejecting 1/f noise and drift. Chopper-stabilised amplifiers use the same modulate–amplify–demodulate idea to avoid DC offset drift.
6.An AM transmitter's antenna current rises from 8 A to 8.93 A on modulation. What is the modulation index?Concept
Antenna current goes as the square root of power, so I_t/I_c = √(1 + μ²/2). Then μ² = 2[(8.93/8)² − 1] = 2 × 0.246 = 0.492, giving μ ≈ 0.70. The power increase is only about 24.6%, which is why AM transmitters spend most of their power on the carrier.

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