AM Frequency Spectrum

When an amplitude-modulated signal is analyzed in the frequency domain, it reveals a distinct spectral structure that directly determines the bandwidth requirement and the information content of the transmitted signal. Understanding the AM frequency spectrum is essential for designing transmitters, allocating channel bandwidth, and analyzing power distribution in analog communication systems.

Core Concept Explanation

When the AM signal equation s(t) = Ac [1 + mu*cos(2*pi*fm*t)] * cos(2*pi*fc*t) is expanded using the trigonometric identity cos(A)cos(B) = 0.5[cos(A+B) + cos(A-B)], three distinct frequency components emerge. The first is the carrier component at frequency fc with amplitude Ac. The second is the upper sideband (USB) at frequency fc + fm with amplitude mu*Ac/2. The third is the lower sideband (LSB) at frequency fc - fm with amplitude mu*Ac/2.

In the frequency domain, the AM signal is represented as the Fourier transform S(f), which shows three impulses (for single-tone modulation) at fc - fm, fc, and fc + fm. The carrier impulse has strength proportional to Ac while each sideband has strength proportional to mu*Ac/2. The critical observation is that both sidebands carry identical information since they are mirror images about the carrier frequency. This redundancy is what SSB modulation later exploits.

The bandwidth of a single-tone AM signal is the frequency span occupied by the spectrum, which equals 2*fm. For a general message signal with maximum frequency W Hz (bandwidth W), the AM bandwidth is 2*W. This is the minimum channel bandwidth required to pass the AM signal without distortion. Each sideband individually occupies a bandwidth equal to W.

Mathematical Expression

Starting from the time-domain AM equation: s(t) = Ac*cos(2*pi*fc*t) + (mu*Ac/2)*cos(2*pi*(fc+fm)*t) + (mu*Ac/2)*cos(2*pi*(fc-fm)*t). Taking the Fourier transform of each cosine term using the identity that the FT of A*cos(2*pi*f0*t) = (A/2)[delta(f - f0) + delta(f + f0)], the one-sided spectrum at positive frequencies shows three spectral lines. The ratio of sideband amplitude to carrier amplitude is always mu/2.

For a practical message with spectrum M(f) band-limited to W Hz, the AM spectrum is: S(f) = (Ac/2)[delta(f - fc) + delta(f + fc)] + (mu*Ac/2)[M(f - fc) + M(f + fc)]. This shows that the AM spectrum is the message spectrum translated (shifted) to be centered at fc, appearing on both sides of the carrier. The positive frequency portion spans from fc - W to fc + W, confirming BW = 2W.

Practical Understanding

In AM broadcasting, each station is allocated a channel bandwidth of 10 kHz (in the medium wave band). Since audio bandwidth is limited to approximately 5 kHz (for voice) to match this allocation, the LSB occupies fc - 5 kHz to fc and the USB occupies fc to fc + 5 kHz. Adjacent channels are separated by 10 kHz to prevent adjacent channel interference, which occurs when one station's sideband energy overlaps with the neighboring station's channel.

The fact that both sidebands carry identical information is a fundamental inefficiency of conventional AM. In double sideband suppressed carrier (DSB-SC) modulation, the carrier is removed to improve power efficiency. In single sideband (SSB), one sideband is also removed, halving the bandwidth requirement. Understanding the AM spectrum is the foundation for analyzing all these variants.

Given:
Message signal bandwidth W = 4 kHz, carrier frequency fc = 900 kHz, modulation index mu = 0.8

Why this formula applies:
AM bandwidth = 2*W for a message with maximum frequency W. Sideband frequencies = fc +/- fm.

Formula:
BW = 2 * W
LSB spans from: fc - W to fc
USB spans from: fc to fc + W

Substitution:
W = 4 kHz, fc = 900 kHz

Calculation:
BW = 2 * 4 = 8 kHz
LSB: 900 - 4 = 896 kHz to 900 kHz
USB: 900 kHz to 900 + 4 = 904 kHz
Sideband amplitude (for single tone peak): mu*Ac/2 = 0.8*Ac/2 = 0.4*Ac

Final Answer:
AM signal bandwidth = 8 kHz, occupying 896 kHz to 904 kHz. Each sideband is 4 kHz wide. Sideband peak amplitude = 0.4*Ac.

Exam Tip: GATE often shows an AM spectrum and asks to identify bandwidth or modulation index. Remember BW = 2*fm, and mu = 2*(sideband amplitude)/(carrier amplitude) from the spectrum. Also note that in DSB-SC the carrier line disappears but sidebands remain at the same frequencies.

Quick Revision

• AM spectrum has three components: carrier at fc (amplitude Ac), LSB at fc-fm (amplitude mu*Ac/2), USB at fc+fm (amplitude mu*Ac/2).
• Bandwidth of AM signal: BW = 2*fm for single tone, BW = 2*W for message with bandwidth W.
• Both sidebands (USB and LSB) carry identical information. LSB and USB are mirror images about fc.
• Ratio of sideband amplitude to carrier amplitude in spectrum = mu/2.
• For mu = 1 (100% modulation), sideband amplitude = Ac/2. Total sideband power = Ac^2/4 (each sideband = Ac^2/8).
• Adjacent channel interference occurs if AM channel spacing is less than 2*W.
• Trap: DSB-SC removes the carrier line from the spectrum but sidebands remain at the same positions fc+/-fm.