Random Processes and Analog Communications: Noise, AM, FM and the Superheterodyne

White noise is defined by its spectrum being flat: S_n(f) = N₀/2 for every f, two-sided, where N₀ is in watts per hertz. Transform that and the autocorrelation is R_n(τ) = (N₀/2) δ(τ), an impulse at the origin and exactly zero everywhere else. Those two statements are the same statement, and reading them together is the whole content of the idea: any two distinct samples of white noise are uncorrelated, no matter how close together in time you take them. That is also why white noise cannot physically exist — the total power ∫ S_n(f) df diverges — and why nobody minds. The receiver only ever sees noise through a filter of finite bandwidth, so the flat model is a convenience that is accurate over the band that matters and wrong only where nothing is measured. When the samples are additionally Gaussian and the process is used as a channel model, the standard name is AWGN: additive white Gaussian noise. The Gaussian part is a separate assumption from the whiteness — whiteness is about correlation between samples, Gaussianity is about the distribution of one sample — and an examiner who says merely white has told you nothing about the distribution.

Now pass a WSS process through an LTI system with transfer function H(f). The output is another WSS process, and its statistics follow from two results and no integration of waveforms. The mean scales by the DC gain, m_Y = m_X H(0). The spectrum scales by the squared magnitude, S_Y(f) = |H(f)|² S_X(f) — note that the phase of H(f) disappears entirely, because a phase shift moves a waveform in time and a power spectrum does not know about time. Everything else is a consequence. Output power is E[Y²] = ∫ |H(f)|² S_X(f) df. Output autocorrelation is the inverse transform of S_Y(f), which in the time domain is the double convolution R_Y(τ) = R_X(τ) * h(τ) * h(−τ) — correct, occasionally quicker, and almost never the route to take in an exam. The cross-spectrum S_(XY)(f) = H(f) S_X(f) is the one place the phase survives, and it is what a system-identification question is really asking about.

Where does N₀ come from in a real receiver? From temperature. The available noise power spectral density at the input of a matched receiver is N₀ = kT in the one-sided convention, with k = 1.38 × 10⁻²³ J/K, so at a reference temperature of 290 K, N₀ = 4.00 × 10⁻²¹ W/Hz. In the units the RF world actually uses, 10 log₁₀(4.00 × 10⁻²¹/10⁻³) = −174 dBm/Hz, and that single number is the floor under every radio ever built. Multiply by bandwidth to get power: in 1 MHz, the thermal noise floor is −174 + 60 = −114 dBm, which is 4.0 × 10⁻¹⁵ W. A receiver's own contribution is then quoted as a noise figure in dB above that floor, so a 1 MHz receiver with a 6 dB noise figure has an effective noise power of −108 dBm at its input, and a signal must beat that by the required SNR before any of this chapter's demodulators can do their job.

Full amplitude modulation writes the transmitted signal as s(t) = A_c[1 + µ m(t)] cos 2πf_c t, with m(t) normalised so that |m(t)| ≤ 1 and µ the modulation index. For a single tone m(t) = cos 2πf_m t, expanding the product gives three terms and no more: a carrier at f_c of amplitude A_c, and two sidebands at f_c ± f_m each of amplitude µA_c/2. That is the spectrum of AM in one line, and two consequences follow immediately. The bandwidth is 2W, twice the highest message frequency, regardless of µ — a 5 kHz audio signal occupies 10 kHz of spectrum whether it is modulated at 10 per cent or 90. And the modulation index is readable straight off an envelope, because the envelope swings between A_c(1+µ) and A_c(1−µ), so µ = (A_max − A_min)/(A_max + A_min). Keeping µ ≤ 1 is not an efficiency preference but a hard requirement: at µ > 1 the term 1 + µ m(t) goes negative, the envelope no longer follows the message, and an envelope detector recovers a distorted, rectified version of it. That failure has a name — overmodulation — and it is irreversible at the receiver.

An envelope detector is a diode into a parallel RC. The diode charges the capacitor on each carrier peak, and between peaks the capacitor discharges through R, so the output tracks the envelope. The whole design is one inequality with two ends: 1/f_c ≪ RC ≪ 1/W. The lower bound says the capacitor must not discharge appreciably between carrier peaks, or the output carries carrier ripple. The upper bound says it must discharge fast enough to follow the message down, or the output cuts corners on falling edges — a defect with the descriptive name diagonal clipping. The precise upper bound for a tone is RC ≤ √(1 − µ²)/(µ ω_m), which is worth reading rather than memorising: the deeper the modulation, the faster the envelope can fall, and so the tighter the ceiling on RC. At µ → 1 the permitted RC goes to zero, which is the analytical statement of why 100 per cent modulation is hard on a simple detector even though it is legal.

Coherent (synchronous) detection is the alternative, and the only option for DSB-SC and SSB because those have no envelope to follow. Multiply the received signal by a locally generated cos(2πf_c t + θ) and low-pass the product. For DSB-SC the recovered output is proportional to m(t) cos θ, so a phase error θ costs amplitude by cos θ — at θ = 60° you keep half the signal, and at θ = 90° you keep none at all. That last case is quadrature null, and it is the mechanism behind both the failure and its useful cousin: two independent messages can share one carrier in quadrature, which is exactly QAM in the next chapter. A frequency error Δf in the local oscillator is worse than a phase error because it makes θ a ramp in time, so the output beats in and out at Δf. For SSB the same frequency error does something qualitatively different — it shifts every message component by the same absolute Δf rather than scaling it, so speech becomes unintelligible at surprisingly small errors, which is why SSB voice links are specified to tens of hertz.

In angle modulation the amplitude is constant and the message rides in the argument of the cosine: s(t) = A_c cos[2πf_c t + φ(t)]. Two choices of φ(t) give the two schemes. Phase modulation sets φ(t) = k_p m(t), so the phase follows the message. Frequency modulation sets the instantaneous frequency f_i(t) = f_c + k_f m(t), which means φ(t) = 2πk_f ∫ m(t) dt — the phase follows the integral. That single difference is the whole relationship between them: FM is PM of the integrated message, and PM is FM of the differentiated message. For a tone m(t) = cos 2πf_m t, FM gives s(t) = A_c cos(2πf_c t + β sin 2πf_m t) with β = Δf/f_m, the modulation index, where Δf = k_f A_m is the peak frequency deviation. Notice what β depends on in each scheme, because this is where the marks are. In FM, Δf is set by the message amplitude alone, so doubling f_m at constant amplitude halves β. In PM, β_p = k_p A_m does not involve f_m at all, so the equivalent deviation Δf = β_p f_m doubles when f_m doubles. The consequence for bandwidth is opposite in the two cases and is a standard question.

The spectrum of FM is not three lines. Expanding cos(2πf_c t + β sin 2πf_m t) in Bessel functions gives an infinite set of components at f_c ± n f_m, n = 0, 1, 2, …, with amplitudes A_c J_n(β). Three structural facts fall out. The spectrum is infinitely wide in principle, so any FM bandwidth is a truncation and you must be told the criterion. The total power is A_c²/2 whatever β is, because Σ J_n²(β) = 1 — angle modulation redistributes power among sidebands but never changes the total, which is why an FM transmitter runs its output stage in saturation and an AM transmitter cannot. And the carrier component A_c J_0(β) vanishes entirely at β = 2.405, the first zero of J_0, which is a real and slightly startling fact: a correctly set FM transmitter can have no energy at its own nominal carrier frequency. Carson's rule is the standard truncation: B ≈ 2(Δf + f_m) = 2(β + 1) f_m, and its two limits are worth naming. For β ≪ 1, narrowband FM, this gives 2f_m — the same bandwidth as AM, which is the whole reason NBFM exists. For β ≫ 1, wideband FM, it gives 2Δf — bandwidth set by the deviation and hardly at all by the message.

To demodulate FM you must turn frequency into amplitude, and there are three standard ways. A slope detector puts the signal on the linear flank of a tuned circuit's response, so a frequency swing becomes an amplitude swing that an ordinary envelope detector can read; simple, and non-linear away from the centre. A balanced discriminator uses two such circuits detuned either side of f_c and subtracts, which cancels the curvature and doubles the slope. A phase-locked loop is the modern answer: the loop forces a VCO to track the incoming instantaneous frequency, and the control voltage driving the VCO is the demodulated message. The PLL wins because it works down to low signal levels where a discriminator has already collapsed. Whichever is used, the front end must be preceded by a limiter, because the whole premise of angle modulation is that amplitude carries no information — so amplitude variations are, by definition, noise, and clipping them away costs nothing and removes a great deal. This is the property that makes FM immune to fading and impulsive interference in a way AM can never be.

One more standard piece of FM practice follows from the noise mechanism just described. Because the discriminator differentiates, its output noise spectral density rises as f² — the recovered noise is not flat but tilted, so the highest audio frequencies come out worst. The cure is pre-emphasis and de-emphasis: boost the high audio frequencies before modulating and cut them by the exactly complementary amount after demodulating. The message is restored unchanged, the noise picked up in the channel is cut without being boosted first, and the improvement is free apart from a slight increase in peak deviation. Broadcast FM uses a 75 µs time constant for this, corresponding to a corner at 1/(2π × 75 × 10⁻⁶) = 2.12 kHz, which is why an FM receiver's audio chain has a fixed, standardised roll-off built into it.

A receiver has to do two things that fight each other: tune across a band, and select one channel out of it sharply. Building a filter that is both tunable and sharp is hard, because its bandwidth would have to stay constant while its centre frequency moved over a two-to-one range. The superheterodyne solves this by refusing to try. A local oscillator mixes the incoming signal down to a fixed intermediate frequency, and all the selectivity lives in a fixed IF filter that never has to move. What tunes is only the local oscillator, and an oscillator is easy to tune. The chain is therefore: RF amplifier with broad tuning, mixer, LO, fixed IF amplifier with the real selectivity and most of the gain, detector, audio. The mixer produces both f_LO − f_s and f_LO + f_s; the IF filter picks one. Choosing f_LO above the signal — high-side injection — is standard, and the reason is a range calculation rather than a preference.

The image frequency is the price of mixing. The IF filter passes whatever comes out of the mixer at the IF, and two input frequencies do that: the wanted f_s = f_LO − IF and the unwanted f_img = f_LO + IF. Subtract and you get the relation to carry: f_img = f_s + 2 × IF. There is nothing the IF filter can do about it, because by the time the image reaches the IF filter it is at the IF and indistinguishable from the signal. The only defence is before the mixer, in the tuned RF stage, and this is the reason a broadcast receiver has an RF selectivity requirement at all. The quality of that defence is quoted as the image rejection ratio, and for a single tuned circuit of quality factor Q it is IRR = √(1 + Q²ρ²) with ρ = f_img/f_s − f_s/f_img.

1. Compute the LO first, not the image. f_LO = f_s + IF for high-side injection; every other quantity in the problem follows from it, and writing it down prevents the commonest slip, which is adding the IF once instead of twice.
2. The image is 2 × IF away from the signal, on the far side of the LO. So it is above the signal for high-side injection and below it for low-side, and questions do vary the injection side.
3. All the selectivity is in the IF; all the image rejection is in the RF. Confusing these is the conceptual error the question is built on: no amount of IF filtering removes an image, and no amount of RF filtering separates adjacent channels.
4. A second mixing stage is the industrial answer. A double-conversion receiver uses a high first IF to push the image out of band and a low second IF to get the selectivity, taking both benefits at the cost of one more oscillator.

The EC paper is 65 questions for 100 marks in 180 minutes, of which General Aptitude is 10 questions for 15 marks, leaving the technical subject to carry the rest. The marking asymmetry should shape how you attack this material: a wrong MCQ costs one third of the question's value, so a one-mark MCQ down to two plausible options is worth answering and a blind guess among four is worth exactly nothing on average, while MSQ and NAT questions carry no negative marking at all — which means every numerical-answer question in this section should be attempted, even on a rough estimate, because there is no downside. That matters more here than in most sections, because so many of the results in this chapter are single numbers you can bound quickly: a modulation efficiency cannot exceed one third for full AM, a bandwidth cannot be less than 2W for AM, an image is always 2 × IF away, and a noise-equivalent bandwidth for one pole is always π/2 times the 3 dB figure.