Why Can't Standard AM Put More Than a Third of Its Power in the Sidebands?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•6 min read

Because the carrier always carries the same power while the sidebands only get μ²/2 of it, and the modulation index μ cannot go above 1 without the envelope crossing zero and distorting. The share of power in the sidebands is μ² ÷ (2 + μ²), which at μ = 1 is 1 ÷ 3. With a 10 V carrier and μ = 0.5, for example, the carrier carries 50 W and each sideband just 3.125 W, so only 11.1% of the 56.25 W total carries the message.

What you'll learn

  • Where the AM power formula Pt = Pc(1 + μ²/2) comes from
  • Why the sideband share peaks at exactly one third
  • A worked example checked against a simulated waveform and its spectrum
  • How the formula extends to several modulating tones
  • Why broadcast AM accepts the waste, and what DSB-SC and SSB do instead

Conventions used here

Standard AM (double sideband with full carrier) modulated by a single tone: s(t) = Ac[1 + μ cos(2πfmt)] cos(2πfct), with carrier amplitude Ac, carrier frequency fc, message frequency fm much smaller than fc, and modulation index 0 ≤ μ ≤ 1. Powers are average powers into a 1 Ω reference resistor, so a sinusoid of amplitude A has power A²/2.

Where does Pt = Pc(1 + μ²/2) come from?

Multiply out the product with the identity cos A cos B = ½[cos(A − B) + cos(A + B)]: s(t) = Ac cos(2πfct) + (μAc/2) cos(2π(fc + fm)t) + (μAc/2) cos(2π(fc − fm)t). That is three sinusoids at different frequencies: the carrier and an upper and a lower sideband. Over a long enough time sinusoids at different frequencies average to zero when multiplied together, so their powers simply add. The carrier contributes Pc = Ac²/2, and each sideband (μAc/2)²/2 = μ²Pc/4. Together the two sidebands carry μ²Pc/2, so Pt = Pc(1 + μ²/2).

Line spectrum of the AM signal with modulation index 0.5Three spectral lines. The carrier at fc has amplitude 10 volts and carries 50 watts into 1 ohm. The lower and upper sidebands at fc minus fm and fc plus fm each have amplitude 2.5 volts, mu Ac over 2, and carry 3.125 watts each. Total power is 56.25 watts, of which 11.1 percent is in the sidebands.02.557.510amplitude (V)2.5 Vfc − fm10 Vfc2.5 Vfc + fmμ = 0.5: carrier 50 W, each sideband 3.125 W
The spectrum for Ac = 10 V and μ = 0.5: a 10 V carrier at fc and two 2.5 V sidebands at fc ± fm. Only the sidebands depend on the message.

Why does the share stop at one third?

Divide the sideband power by the total: η = (μ²Pc/2) ÷ [Pc(1 + μ²/2)] = μ² ÷ (2 + μ²). This rises with μ, so the best you can do is the largest allowed μ. That limit is μ = 1, because the envelope Ac(1 + μ cos 2πfmt) must stay positive for a simple envelope detector to recover the message. At μ = 1.2 the envelope's minimum would be 1 − 1.2 = −0.2 times Ac, the envelope folds over, and the detector output is distorted. So the maximum is η = 1 ÷ (2 + 1) = 1/3: even fully modulated, two thirds of the transmitted power sits in a carrier that carries no information.

Share of AM power in the sidebands against modulation indexComputed efficiency eta equals mu squared over 2 plus mu squared, in percent, for mu from 0 to 1. It rises from 0 to 11.1 percent at mu equals 0.5 and reaches its maximum of 33.3 percent at mu equals 1, the largest index standard AM allows without distortion.00.250.50.751010203040modulation index μsideband share η (%)μ = 1: 33.3%μ = 0.5: 11.1%
The sideband share η = μ²/(2 + μ²), computed for μ from 0 to 1. It is 11.1% at μ = 0.5 and reaches 33.3% at μ = 1.

Worked example: Ac = 10 V, μ = 0.5

  • Carrier power: Pc = 10² ÷ 2 = 50 W.
  • Each sideband: amplitude μAc/2 = 2.5 V, so power 2.5² ÷ 2 = 3.125 W. Both together: 6.25 W.
  • Total: Pt = 50 × (1 + 0.25/2) = 56.25 W.
  • Sideband share: η = 0.25 ÷ 2.25 = 11.1%.
  • Check by simulation: generating the waveform with fc = 100 kHz and fm = 1 kHz over 10 ms and taking its mean square gives 56.25 W, and its FFT shows lines of exactly 10 V, 2.5 V and 2.5 V.

What if several tones modulate the carrier?

Each tone adds its own pair of sidebands, and their powers add, so the formulas keep the same form with an effective index μt = √(μ1² + μ2² + ...). Two tones with μ1 = 0.6 and μ2 = 0.8 give μt = √(0.36 + 0.64) = 1, and again a sideband share of 33.3%.

How does the antenna current change with modulation?

Power into a fixed antenna resistance goes as the square of the current, so the same ratio applies to current squared: It = Ic √(1 + μ²/2), where Ic is the unmodulated carrier current. At μ = 0.5 the current rises by only 6.1% (a factor of 1.0607), and even at μ = 1 by 22.5% (1.2247). Questions often run this backwards: if an ammeter in the antenna lead reads 8 A unmodulated and 8.93 A with modulation, then μ = √(2[(8.93/8)² − 1]) = 0.701. Substituting 0.701 back into the forward formula gives 8 × √(1 + 0.2457) = 8.93 A, as it should.

Why does broadcast AM put up with it?

Because the carrier is what makes the receiver cheap. With a full carrier present, the message is simply the envelope of the received signal, and a diode, a resistor and a capacitor can recover it. Removing the carrier, as double-sideband suppressed-carrier (DSB-SC) does, puts all the transmitted power into the sidebands but forces the receiver to regenerate a carrier of exactly the right frequency and phase. Single sideband (SSB) goes further, dropping one sideband as well to halve the bandwidth, at the cost of an even more demanding receiver.

Common mistakes

  • Writing Pt = Pc(1 + μ²) or Pc(1 + μ/2). The sidebands together carry μ²Pc/2.
  • Giving one sideband's power when the question asks for both, or the other way round.
  • Using the sideband amplitude μAc/2 directly as a power without squaring and halving.
  • Assuming η can be pushed to 50% by raising μ. Above μ = 1 the envelope distorts.
  • Applying the one-third limit to DSB-SC or SSB, which have no carrier to waste power on.

Check your understanding

  • Q1. What share of the power is in the sidebands at μ = 0.8? Answer: 0.64 ÷ 2.64 = 24.2%.
  • Q2. A transmitter's carrier power is 400 W and it is modulated with μ = 1. What are the total and sideband powers? Answer: Pt = 400 × 1.5 = 600 W, of which 200 W is in the sidebands.
  • Q3. An AM transmitter radiates 1200 W in total at μ = 0.6. What is the carrier power? Answer: Pc = 1200 ÷ (1 + 0.18) = 1016.9 W.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists amplitude modulation and demodulation and the spectra of AM and FM under Section 7, Communications (checked against the official PDF from IIT Madras on 10 October 2026). Every power figure was computed from the formulas and checked against a simulated waveform's mean square and FFT in code.

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