Physical layer — encoding (NRZ, Manchester), bandwidth, Nyquist, Shannon-Hartley: Question bank

True or false — justify

A channel with more bandwidth always gives a higher bit rate.

False — bandwidth sets a ceiling (

2 B lo g_{2} M

), but noise (Shannon) or the number of levels can hold the real rate below it; the achievable rate is the smaller of Nyquist and Shannon.

Manchester encoding is more efficient than NRZ.

False — "efficient" usually means bits per Hz, and Manchester needs about double the bandwidth for the same bit rate. Its win is self-clocking and no DC, not efficiency.

Baud rate and bit rate are the same thing.

True only when

M = 2

(binary), because bit rate

=

baud

\times lo g_{2} M

and

lo g_{2} 2 = 1

. For

M > 2

the bit rate is a multiple of the baud.

Shannon's formula lets you reach any bit rate by cranking up the SNR.

True in principle — capacity grows without bound as

S / N \to \infty

— but only logarithmically, so each extra bit/sec costs an exponentially bigger

S / N

. Real hardware and physics cap it long before.

NRZ has no clock recovery problem for the data 10101010.

True — that pattern transitions every bit, so the receiver gets an edge each bit and stays synced. NRZ only fails on long runs of the same bit, not on alternating data.

Nyquist's

C = 2 B lo g_{2} M

applies to real telephone lines.

False — Nyquist assumes a noiseless channel. Real lines have noise, so the true ceiling is Shannon; using Nyquist alone gives an over-optimistic number you can't achieve.

Adding more voltage levels

M

always increases capacity.

False — on a noisy channel, more levels sit closer together and noise flips them into each other. Beyond

M \approx 1 + S / N

you gain nothing; Shannon already caps the useful

M

Manchester removes the DC component of the signal.

True — every bit is half-high and half-low, so the average voltage over each bit is zero, which lets it pass through AC/transformer-coupled media that block DC.

A 0 dB SNR means there is no signal.

False — 0 dB means

S / N = 1 0^{0} = 1

, i.e. signal power equals noise power. It's still a usable (if poor) channel:

C = B lo g_{2} (2) = B

Spot the error

"For a 30 dB line,

S / N = 30

, so

C = B lo g_{2} (31)

The error is using dB directly. Convert first:

S / N = 1 0^{30/10} = 1000

, so

C = B lo g_{2} (1001)

. dB is logarithmic; Shannon needs the linear power ratio.

"Manchester sends 2 bits per bit-period because it has 2 transitions."

Wrong — the two transitions encode one bit; the mid-bit edge carries the data and the extra edges are just the price of the scheme. Manchester is one bit per bit-period, using roughly double the bandwidth.

"With 8 levels, bit rate

=

baud

\times 8

Wrong — it's baud

\times lo g_{2} 8 =

baud

\times 3

. Levels multiply the number of distinguishable symbols, and each symbol is worth

lo g_{2} M

bits, not

M

bits.

"Nyquist gives

C = 6

Mbps and Shannon gives

C = 4

Mbps, so the channel does

6 + 4 = 10

Mbps."

Wrong on two counts — you never add the two limits, and the achievable rate is the smaller one (4 Mbps). Nyquist is a symbol-rate ceiling; Shannon is the noise ceiling.

"NRZ has a DC problem, so we just lower the voltage to fix it."

Wrong — the DC component comes from the average level being non-zero over long runs, not from the voltage magnitude. Lowering voltage shrinks everything but the average is still non-zero; you need a balanced code like Manchester.

S / N

of 1000 corresponds to 1000 dB."

Wrong — dB

= 10 lo g_{10} (1000) = 30

dB. The linear ratio and the dB value are very different numbers; confusing them is the classic Shannon trap.

Why questions

Why does NRZ specifically lose sync on long identical runs but not on mixed data?

A flat voltage gives the receiver no edges to lock its clock onto, so tiny clock-speed differences accumulate and it miscounts how many identical bits passed. Mixed data supplies frequent edges that re-align the clock.

Why is the factor exactly

2 B

(and not

B

3 B

) in Nyquist?

A signal band-limited to

B

Hz is fully described by

2 B

samples per second (Sampling Theorem); that's the fastest rate at which you can place distinguishable symbols before they smear (intersymbol interference). Fewer wastes the channel, more can't be resolved.

Why does Shannon use

lo g_{2} (1 + S / N)

with the "+1" instead of just

lo g_{2} (S / N)

The "+1" keeps capacity sensible as noise dominates: even at

S / N = 0

you get

lo g_{2} (1) = 0

(no capacity), not a negative number. It represents that the total spread the receiver sees is signal plus noise power.

Why do we care about both Nyquist and Shannon if Shannon is the "real" limit?

Shannon tells you the top speed but not how to build for it; Nyquist tells you how many levels

M

you'd need (

2 B lo g_{2} M = C_{S hann o n}

) to actually reach that speed. Together they say "aim here, using this many levels."

Why does Manchester trade bandwidth for reliability rather than the other way around?

Guaranteeing an edge every bit means the signal changes at most twice per bit, doubling the highest frequency present, which demands more bandwidth. In return the receiver never loses the clock — reliability bought with spectrum.

Why can't you beat Shannon by inventing a cleverer encoding scheme?

Shannon's bound is information-theoretic: it counts how many distinguishable messages noise permits, regardless of representation. Any encoding is just relabeling those same distinguishable states, so none can exceed the count.

Why is

S / N

a power ratio and not a voltage or amplitude ratio?

Capacity depends on how much energy separates signal from noise, and power (energy rate) is what noise adds. Amplitude relates to power by a square, which is exactly why the "levels" estimate used

M \approx 1 + S / N

Edge cases

What happens to Shannon capacity as

S / N \to 0

(noise swamps signal)?

C = B lo g_{2} (1 + 0) = B lo g_{2} (1) = 0

bits/sec — a channel drowning in noise carries no information no matter how much bandwidth it has.

What is the Nyquist capacity when

M = 1

(only one voltage level)?

C = 2 B lo g_{2} 1 = 0

bits/sec — with a single possible level every symbol is identical, so no information is conveyed. You need at least two distinguishable states.

What does

M = 2

give you in the Nyquist formula, and why is that the NRZ case?

C = 2 B lo g_{2} 2 = 2 B

; with two levels each symbol is one bit, so bit rate equals baud rate — exactly binary NRZ where baud and bit rate coincide.

At exactly

S / N = 1

(0 dB), what is Shannon capacity?

C = B lo g_{2} (2) = B

bits/sec — a channel where signal and noise powers are equal still passes one bit per second per Hz of bandwidth.

If bandwidth

B = 0

, what is the capacity under either law?

Zero for both — Nyquist gives

2 \cdot 0 \cdot lo g_{2} M = 0

and Shannon gives

0 \cdot lo g_{2} (1 + S / N) = 0

. With no frequency range to carry signal changes, nothing gets through.

For an all-zeros data stream, does Manchester still keep the receiver synced?

Yes — Manchester puts a transition in every bit regardless of the data value, so even 00000000 produces a steady stream of clock edges, unlike NRZ which would go flat.

What happens if you push symbols faster than the Nyquist rate

2 B

Adjacent symbols overlap in time faster than a

B

-Hz channel can resolve, causing intersymbol interference — the receiver can no longer tell where one symbol ends and the next begins, so the extra "speed" corrupts data.

Connections

Bandwidth vs Bit Rate — the baud/level distinction most traps here exploit
Signal-to-Noise Ratio — the dB-vs-linear confusion lives here
Sampling Theorem — source of the $2 B$ edge case reasoning
Modulation (ASK, FSK, PSK, QAM) — where the number of levels $M$ becomes physical
Data Link Layer — consumes the reliably-clocked bits these codes produce