Physical layer — encoding (NRZ, Manchester), bandwidth, Nyquist, Shannon-Hartley: Worked examples

Before we compute anything, let's re-anchor the symbols so no notation is used before it is earned.

The two laws (both from the parent):

Before the examples, one bridge idea we'll reuse in the "beat Shannon?" twist:

The scenario matrix

Every capacity question is one (or a blend) of these cells. The worked examples below each carry a tag like [Cell A2] so you can see the coverage.

#	Cell	What makes it special	Which law
A1	Noiseless, given $M$	plug straight in	Nyquist
A2	Noiseless, solve for $M$	invert the log	Nyquist
B1	Noisy, SNR linear	plug straight into Shannon	Shannon
B2	Noisy, SNR in dB	convert dB→linear first	Shannon
B3	Noisy, $S / N < 1$ (negative dB)	$1 + S / N$ between 1 and 2	Shannon
C	Both limits given	take the smaller; find $M$ to reach ceiling	Both
D1	Degenerate: $S / N = 0$ (no signal)	capacity collapses	Shannon
D2	Degenerate: $B = 0$ (no pipe)	capacity collapses	either
E	Limiting: $S / N \to \infty$ or huge dB	log grows slowly	Shannon
F	Real-world word problem	translate English → symbols	Both
G	Exam twist	dB trap / units trap / "beat Shannon?"	Both

The examples: A1, A2, B1, B2, B3, C, D (both), E, F, G — together they touch every cell.

Worked examples

Example 1 — [Cell A1] Noiseless, levels given

Example 2 — [Cell A2] Noiseless, solve for the number of levels

Example 3 — [Cell B1] Noisy channel, SNR given as a plain linear ratio

Worked example Statement

A noisy channel: $B = 4000$ Hz, and the signal-to-noise ratio is given directly as $S / N = 15$ (a plain number, already linear — no decibels in sight). Find $C$ .

Forecast: guess whether $C$ is closer to $4000$ or to $16000$ bit/s.

Choose the law. Noise present ⇒ Shannon. Why? $M$ isn't ours to choose; noise sets the ceiling.
Check the units of the SNR. It's already a linear power ratio (no "dB" written), so no conversion is needed — we drop it straight in. Why check? Half of all mistakes are converting a number that was already linear.
Plug in. $C = B lo g_{2} (1 + S / N) = 4000 lo g_{2} (1 + 15) = 4000 lo g_{2} 16$ .
Evaluate. $lo g_{2} 16 = 4$ (since $16 = 2^{4}$ ), so $C = 4000 \times 4 = 16000$ bit/s.

Verify: $1 + 15 = 16 = 2^{4}$ makes the log land exactly on 4. ✓ Forecast: $16000$ bit/s. This is the cleanest cell — nothing to convert, nothing to combine.

Example 4 — [Cell B2] Noisy channel, SNR given in decibels

Example 5 — [Cell B3] Noisy, but noise stronger than signal ( $S / N < 1$ )

Worked example Statement

A weak link: $B = 2000$ Hz, SNR $= - 3$ dB (yes, negative decibels — the signal is below the noise). Find $C$ .

Forecast: guess whether $C$ is zero, near $B$ , or above $2 B$ .

Convert dB → linear. $\frac{S}{N} = 1 0^{- 3/10} = 1 0^{- 0.3} \approx 0.501$ . Why? Negative dB just means the power ratio is less than 1 — signal quieter than noise, but not silent.
Watch what happens inside the log. $1 + S / N = 1 + 0.501 = 1.501$ , which sits between 1 and 2. Why does this matter? Shannon's formula still applies; the " $1 +$ " guarantees the argument stays above 1, so $lo g_{2} (1.501) > 0$ — capacity is small but not zero.
Evaluate. $lo g_{2} (1.501) \approx 0.586$ , so $C = 2000 \times 0.586 \approx 1172$ bit/s.

Verify: $2^{0.586} \approx 1.501$ ✓, and $0 < 1172 < 2000 = B$ , exactly what "a channel worse than 0 dB but still alive" should give. The " $+ 1$ " inside the log is what keeps a below-noise channel from collapsing to zero. ✓

Example 6 — [Cell C] Both limits: find the achievable rate AND the levels

Example 7 — [Cells D1 & D2] Degenerate inputs

Example 8 — [Cell E] Limiting behaviour: cranking SNR huge

The figure below draws exactly this: capacity vs SNR, with the two dots you just computed.

Figure — Physical layer — encoding (NRZ, Manchester), bandwidth, Nyquist, Shannon-Hartley

Example 9 — [Cell F] Real-world word problem

Worked example Statement

A modem vendor claims 56 kbit/s over a phone line with $B = 3400$ Hz. What SNR (in dB) would the pure Shannon limit demand — and is the claim physically comfortable?

Forecast: guess whether the required SNR is around 20 dB, 40 dB, or 50 dB.

Translate. Need $C = 56000$ bit/s, $B = 3400$ Hz. Set Shannon: $56000 = 3400 lo g_{2} (1 + S / N)$ . Why Shannon? Real phone lines are noisy; this is the true ceiling.
Isolate the log. $lo g_{2} (1 + S / N) = \frac{56000}{3400} \approx 16.47$ . Why divide by $B$ ? To free the log so we can undo it.
Undo $lo g_{2}$ . $1 + S / N = 2^{16.47} \approx 90800$ , so $S / N \approx 90800$ . Why $2^{(\cdot)}$ ? Inverse of $lo g_{2}$ .
Back to dB. $(S / N)_{d B} = 10 lo g_{10} (90800) \approx 10 \times 4.958 \approx 49.6$ dB. Why $lo g_{10}$ here? dB is defined with base-10 log, not base-2 — different log for a different job.

Verify: feed $S / N = 90800$ back: $3400 lo g_{2} (90801) \approx 3400 \times 16.47 \approx 56000$ ✓. Interpretation: ~50 dB is very clean — real lines sit near 30–40 dB, which is why raw 56k is hard downstream and needs digital tricks. Ballpark forecast (~50 dB) confirmed.

Example 10 — [Cell G] Exam twist: "beat Shannon with more levels?"

Worked example Statement

Exam claim: "A noisy channel has $B = 1$ MHz, $S / N = 63$ . Shannon says $C = 6 \times 1 0^{6}$ bit/s. If I just use $M = 64$ levels, Nyquist gives $2 (1 0^{6}) lo g_{2} 64 = 12 \times 1 0^{6}$ bit/s — so I beat Shannon!" Where is the error, and what is the real achievable rate?

Forecast: guess which number is the true ceiling.

Compute both. Shannon: $1 0^{6} lo g_{2} (1 + 63) = 1 0^{6} lo g_{2} 64 = 1 0^{6} \times 6 = 6 \times 1 0^{6}$ bit/s. Nyquist with $M = 64$ : $2 \times 1 0^{6} \times 6 = 12 \times 1 0^{6}$ bit/s. Why both? To expose the trap the twist relies on.
Spot the fallacy. Nyquist assumes you can reliably distinguish all 64 levels. But we already derived $M_{m a x} = 1 + S / N$ up top: $M_{m a x} = 1 + 63 = 64 = 8$ , not 64. Why $\cdot$ ? Because the honest Nyquist rate can't exceed Shannon — that inequality forced $M \leq 1 + S / N$ .
Honest Nyquist. With $M = 8$ : $2 \times 1 0^{6} lo g_{2} 8 = 2 \times 1 0^{6} \times 3 = 6 \times 1 0^{6}$ bit/s. Why? Now Nyquist and Shannon agree — noise already fixed the level count.
Conclusion. The real ceiling is the smaller: $6 \times 1 0^{6}$ bit/s. You cannot beat Shannon by inventing levels the noise will smear together.

Verify: $64 = 8$ ✓, Nyquist $(M = 8) = 6 \times 1 0^{6} =$ Shannon ✓. The " $12 \times 1 0^{6}$ " is a mirage from illegal levels — exactly the parent's "Using Nyquist on a noisy channel" mistake.

Recall

Recall Cover the answers

Which cell is "solve for $M$ "? ::: A2 — invert Nyquist: $lo g_{2} M = C / (2 B)$ , then $M = 2^{l o g_{2} M}$ If SNR is given as a plain number (no dB), do you convert? ::: No — it's already linear; drop it straight into Shannon (Cell B1) First move when SNR is in dB? ::: Convert to linear: $S / N = 1 0^{dB /10}$ What happens when $S / N < 1$ (negative dB)? ::: $1 + S / N$ lands between 1 and 2, so $C$ is small but still positive What is $C$ when $S / N = 0$ ? ::: Zero — $lo g_{2} (1 + 0) = lo g_{2} 1 = 0$ What is $C$ when $B = 0$ ? ::: Zero — $B$ is a factor in both laws Formula for the most levels noise allows? ::: $M_{m a x} = 1 + S / N$ , from $2 B lo g_{2} M \leq B lo g_{2} (1 + S / N)$ Doubling SNR in dB does what to $C$ ? ::: Roughly doubles it (the log tames a squared linear jump)

Connections

Parent topic
Signal-to-Noise Ratio — the dB↔linear conversions used throughout
Modulation (ASK, FSK, PSK, QAM) — where the $M$ levels physically live
Bandwidth vs Bit Rate — the $lo g_{2} M$ bits-per-symbol link
Sampling Theorem — origin of the $2 B$ factor in Nyquist