Physical layer — encoding (NRZ, Manchester), bandwidth, Nyquist, Shannon-Hartley: Visual walkthrough

Step 1 — What a "signal" actually is

WHAT. Before "capacity" means anything, we must agree what travels on the wire. It is not "a number" — it is a shape: voltage plotted against time.

WHY. Everything ahead (bandwidth, levels, noise) is a statement about what shapes the wire can carry. So we need the shape-vs-time picture first.

PICTURE. Look at the curve below. The receiver at the far end sees only this — one wavy line. It must reconstruct the bits from the wiggles alone.

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

Figure s01 — A single signal drawn as voltage $v (t)$ (vertical) against time $t$ (horizontal). This wavy magenta curve is the ONLY thing the receiver ever sees; every bit must be recovered from it.

$t$ ::: time, seconds, running left→right
$v (t)$ ::: the voltage the wire holds at each instant

Step 2 — "Bandwidth" = how fast the wire is allowed to wiggle

WHAT. We compare two signals: a slow one and a fast one, sent into the same wire.

WHY this tool. We use frequency (wiggles per second) because a wire behaves like a gate: it passes slow shapes and blocks fast ones. The cut-off wiggle-speed is exactly $B$ . This is the physical fact every capacity formula rests on.

PICTURE. In the figure, the violet slow wave passes cleanly. The orange fast wave (above $B$ ) comes out crushed to almost a flat line — that lost wiggle is lost information.

Figure s02 — Two waves sent into the same wire. The violet wave (frequency below $B$ ) passes cleanly; the orange wave (frequency above $B$ ) exits flattened. The wire blocks fast wiggles.

Step 3 — The fastest countable rate of symbols is $2 B$

WHAT. We ask: given the wiggle speed-limit $B$ , how many separate symbols per second can the receiver still tell apart?

WHY $2 B$ and not $B$ . To see a symbol you need the signal to go up and come back — that is two edges per full wiggle. A wiggle of frequency $B$ therefore delivers $2 B$ distinguishable turning-points each second. This is the Sampling Theorem fact in disguise: a shape whose fastest wiggle is $B$ is fully pinned down by $2 B$ samples per second. Push past $2 B$ and neighbouring symbols smear into one — intersymbol interference.

PICTURE. The figure marks each turning-point of a bandwidth- $B$ wave with a dot: count them — two per cycle, giving $2 B$ per second.

Figure s03 — One bandwidth- $B$ wave with a violet dot on every turning-point. There are two dots per cycle (one up-edge, one down-edge), so a frequency- $B$ wave yields $2 B$ readable symbols each second.

Rate ceiling if

B = 3000

Hz?

2 \times 3000 = 6000

symbols/sec

Step 4 — Each symbol can carry more than one bit: the $lo g_{2} M$ factor

WHAT. Instead of only high/low, let each slot pick one of $M$ heights. How many bits does one such slot reveal?

WHY $lo g_{2}$ . With $n$ bits you can write $2^{n}$ different patterns. If a symbol must name one of $M$ possibilities, then $M = 2^{n}$ , so $n = lo g_{2} M$ . The logarithm is the tool that answers "how many yes/no questions ( $=$ bits) does it take to pin down one of $M$ choices?" — that is literally what $lo g_{2}$ counts.

PICTURE. The figure stacks $M = 4$ voltage bands; each band is labelled with its 2-bit code (00 01 10 11). One symbol landing in a band = 2 bits delivered.

Figure s04 — Four stacked voltage bands ( $M = 4$ ), each tagged with its 2-bit code. A symbol landing in one band delivers $lo g_{2} 4 = 2$ bits at once.

Bits per symbol for

M = 8

lo g_{2} 8 = 3

bits

Step 5 — Multiply the two: the Nyquist law

WHAT. We now own two independent facts:

Step 3: at most $2 B$ symbols each second.
Step 4: each symbol is worth $lo g_{2} M$ bits.

Multiply symbols-per-second by bits-per-symbol → bits per second.

WHY multiply. Rate and payload are independent knobs. (symbols/sec) $\times$ (bits/symbol) $=$ bits/sec, exactly like (steps/sec) $\times$ (metres/step) $=$ metres/sec. This is the Nyquist capacity — the ceiling for a noiseless channel.

PICTURE. The figure shows the assembly line: $2 B$ boxes stream out per second, each box stamped with $lo g_{2} M$ bits.

Figure s05 — An assembly line: $2 B$ boxes leave the wire each second (top arrow), and every box is stamped with $lo g_{2} M$ bits. Multiplying the two gives $C = 2 B lo g_{2} M$ .

Step 6 — The degenerate question: why can't $M$ be infinite?

WHAT (the edge case Nyquist ignores). Nyquist assumed a perfect wire. Now add noise: a random voltage fuzz added to every symbol.

WHY it kills big $M$ . If levels are packed closer than the noise fuzz is tall, the receiver can't tell which band a symbol landed in — a 01 gets read as 10. So noise sets a hard cap on how many levels are usable.

PICTURE. Left panel: 4 well-spaced levels, noise bands (shaded) don't overlap → readable. Right panel: 16 crammed levels, noise bands overlap → unreadable mush.

Figure s06 — Left: 4 levels with orange noise-fuzz bands that stay separate (readable). Right: 16 levels crammed so their noise bands overlap (a symbol is misread). Noise fixes how many levels fit.

Step 7 — How many levels does noise actually allow?

WHAT. We count the number of safely separable levels given signal power $S$ and noise power $N$ .

WHY $S + N$ is the full swing. The receiver sees signal plus noise added together. For independent random quantities, powers add: the total average power at the receiver is $S + N$ . Turning that power into a voltage height (Step-7 definition: height $\propto P$ ) gives a usable swing $\propto S + N$ . That is the whole height of the voltage ruler.

WHY $N$ is the smallest safe gap. Each received symbol is jittered up or down by the noise. Because the noise is Gaussian with average power $N$ , its typical jitter is one standard deviation $= N$ . If two levels sit closer than about this jitter, the bell curves of neighbouring levels overlap and symbols get confused. So the closest levels may be spaced is $\propto N$ — that is the ruler's smallest readable rung.

WHY the " $\approx$ " (the honest constants). Dividing swing by gap:

$M \approx \frac{S + N}{N} = 1 + \frac{S}{N} .$

The proportionality constants in "swing $\propto S + N$ " and "gap $\propto N$ " are the same constant (both are voltage-per- $power$ for the same wire), so they cancel in the ratio — that is why a clean number survives. The residual " $\approx$ " hides a small factor (how many standard deviations you demand between levels for a target error rate); tightening that factor changes $M$ by a constant multiplier, not its $1 + S / N$ shape. Shannon's full proof replaces this counting heuristic with an exact information-theory argument and lands on the same formula.

PICTURE. A voltage ruler of total height $S + N$ , chopped into rungs each $N$ tall; the count of rungs is $1 + S / N$ .

Figure s07 — A voltage ruler of height $S + N$ (magenta, total swing) sliced into rungs of height $N$ (violet, the noise jitter). The number of rungs that fit is $1 + S / N$ .

Step 7b — The edge case: noise louder than signal ( $S / N < 0$ dB)

WHAT this means for Shannon. Shannon's formula does not break here — it just returns a small positive capacity:

$C = B lo g_{2} (1 + \frac{S}{N}) > 0 even when S / N < 1,$

because $1 + S / N$ is still greater than $1$ , so its $lo g_{2}$ is still positive. The channel is slow and noisy, not dead. Only if $S / N \to 0$ (no signal at all) does $C \to B lo g_{2} (1) = 0$ .

PICTURE. The ruler collapses: the total swing $S + N$ is barely taller than the single noise rung $N$ , so under one clean level fits — but Shannon still reads out a trickle of capacity.

Figure s07b — When $N > S$ the ruler height $S + N$ is barely above one noise rung $N$ : fewer than 2 clean levels fit ( $M < 2$ ), yet the Shannon curve (violet) still returns a small positive $C$ .

Step 8 — Substitute → Shannon–Hartley

WHAT. Take Nyquist ( $C = 2 B lo g_{2} M$ ) and plug in the noise-limited $M$ from Step 7.

WHY. Nyquist gives the rate structure; Step 7 gives the largest $M$ the noise permits. Combining them yields the true ceiling — the one you cannot beat by cleverness. Here we treat $M$ as continuous (integer rounding is a later, rate-lowering step, so the result stays an upper bound).

C = 2 B lo g_{2} 1 + \frac{S}{N} = 2 B \cdot \frac{1}{2} lo g_{2} (1 + \frac{S}{N}) = B lo g_{2} (1 + \frac{S}{N})

The became a factor of $\frac{1}{2}$ because $lo g_{2} x = \frac{1}{2} lo g_{2} x$ (a square root is a power of $\frac{1}{2}$ ), and that $\frac{1}{2}$ cancels the $2$ in $2 B$ .

PICTURE. The figure lines up the algebra visually: $2 B$ meets $\frac{1}{2}$ , they cancel to $B$ ; the under the log flattens to $(1 + S / N)$ .

Figure s08 — The algebra collapse: $2 B lo g_{2} 1 + S / N$ , the square root pulls out a $\frac{1}{2}$ , the $\frac{1}{2}$ cancels the $2$ , leaving $C = B lo g_{2} (1 + S / N)$ .

Step 9 — Using both together: which ceiling wins?

WHAT. Given a real channel, Shannon says the maximum bits/sec; Nyquist says how many levels you'd need to reach it. You take the smaller achievable rate.

WHY. Shannon is a wall you cannot climb over even with infinite $M$ . Nyquist then tells you the practical (whole-number) $M$ that just touches that wall.

PICTURE. A bar chart: Shannon ceiling as a solid wall, Nyquist rising with more levels until it just reaches the wall at the right integer $M$ .

Figure s09 — Violet bars: Nyquist rate for $M = 2, 4, 8, 16$ . Magenta line: the Shannon wall (6 Mbps). The orange dot marks $M = 8$ , the smallest integer level count that reaches the wall.

The one-picture summary

Figure s10 — The whole chain in one frame: wire speed-limit $B$ → $2 B$ symbols/sec → each worth $lo g_{2} M$ bits → multiply = Nyquist → Gaussian noise caps $M$ at $1 + S / N$ → substitute = Shannon.

The whole chain in one frame: wire wiggle-cap $B$ → $2 B$ symbols/sec → each symbol worth $lo g_{2} M$ bits → multiply for Nyquist → noise caps $M$ at $1 + S / N$ → substitute → Shannon.

Recall Feynman retelling (cover and explain it back)

Imagine a conveyor belt of little boxes. The wire can only push out $2 B$ boxes every second — push faster and the boxes crash into each other (that's the bandwidth limit, and the "2" is because each box needs an up-edge and a down-edge to be seen). Into each box you can put a height — and if you allow $M$ different heights, each box carries $lo g_{2} M$ bits, because it takes $lo g_{2} M$ yes/no questions to name one of $M$ heights. Multiply boxes-per-second by bits-per-box and you get Nyquist: $2 B lo g_{2} M$ . Now the room gets foggy — noise. Fog blurs the heights, so if you use too many heights they blur into each other. The number of heights you can still tell apart is about $1 + S / N$ : the total loudness you can swing is $S + N$ (signal and fog powers add), and the smallest safe step is one fog-wobble $N$ , and because those two heights use the same volts-per-loudness constant it cancels, leaving $1 + S / N$ . Since heights must be whole levels you round that down — never below 2. Slot that $M$ back into Nyquist, the square root turns the $2 B$ into just $B$ , and out pops Shannon: $C = B lo g_{2} (1 + S / N)$ — the true top speed no trick can beat, even when the fog is thicker than the signal (then you just get a slow trickle, not zero).

Connections

Parent topic
Sampling Theorem — source of the $2 B$ in Step 3
Signal-to-Noise Ratio — the $S / N$ used in Steps 7–8
Bandwidth vs Bit Rate — the baud vs bits relation of Step 4
Modulation (ASK, FSK, PSK, QAM) — how the $M$ levels are physically made
Data Link Layer — error-correcting codes that rescue the $S / N < 1$ case