Formula sheet

r

Bits carried per signal element, dimensionless

Data elements

Count of bits

Signal elements

Count of signal elements used to carry them

r = 1 is one bit per pulse. r below 1 means it takes more than one signal element to carry a bit — the cost the biphase line codes in the next topic pay for a guaranteed transition every interval.

Bit rate

Data rate, bits per second (bps)

Signal rate

Baud rate, signal elements per second (baud)

Follows directly from the two rate definitions on this page — bits per second equals (bits per signal element) times (signal elements per second). Nyquist's log2(L) term, in the next topic, is this same relationship for the special case where every signal element carries an equal share of L levels.

C

Channel capacity, bits per second

B

Bandwidth, hertz

P

Received signal power

N

Noise power

P/N is a POWER ratio, not decibels. If a question gives you dB, convert first with SNR = 10^(dB/10).

B

Bandwidth, hertz

L

Number of distinct signal levels

Assumes a noiseless channel. L must be at least 2 — one level carries no information.

dB

Signal-to-noise ratio in decibels

Ten, not twenty. The 20 log10 form is for amplitude or voltage ratios, not power.

Bit Rate

bps

Sampling Rate

samples per second

bits per sample

count, fixed by the precision chosen

r

Bits carried per signal element

L

Number of distinguishable signal levels — amplitudes, frequencies or phases, depending on the scheme

f_{\max}

Highest frequency present in the original analog signal, hertz

This is the condition PCM's sampling step must meet to make accurate reconstruction possible. Telephone voice is taken as topping out at 4000 Hz, which is where the 8000 samples/s figure below comes from.

n

Number of channels multiplexed

B_{\text{ch}}

Bandwidth of one channel, hertz

B_{\text{guard}}

Bandwidth of one guard band, hertz

For n channels there are n-1 guard bands between them — one fewer than the number of channels, because a guard band only sits between two neighbours.

T_{\text{slot}}

Duration of one input slot, seconds

R_{\text{in}}

Input data rate, bps

n

Number of input slots per frame

T_{\text{out slot}}

Duration of one output slot, seconds

k

Message length in bits

n

Codeword length in bits, message plus redundancy

Lower rate means better protection and lower throughput. It is a direct trade.

n-k

Number of redundancy bits added

M(x)

The message as a polynomial

x^{n-k}

Shift left by the number of redundancy bits, appending zeros

R(x)

The remainder after dividing by the generator g(x)

The generator has length (n-k)+1 and is known to both sender and receiver.

d_min

The minimum Hamming distance of the code — the smallest number of bit positions in which any two valid codewords differ

s

Number of errors the code is guaranteed to detect

t

Number of errors the code is guaranteed to correct

Standard result in coding theory. Lecture 3 does not derive this — it is included because Hamming-distance questions are still fair game in the exam.

T_{fr}

Frame transmission time, seconds

T_p

Maximum one-way propagation time across the collision domain, seconds

This is the whole reason Ethernet has a minimum frame size. A frame shorter than 2Tp could finish transmitting before a collision at the far end gets back to tell the sender about it.

m

Number of collisions this frame has suffered so far

K

Randomly chosen backoff multiplier, redrawn after every collision

The range doubles with each collision until it caps at m = 10 (K up to 1023), and the station gives up after 16 attempts. Both cap numbers are standard IEEE 802.3, not lecture content — see the note on the page.

d

Maximum distance between the two furthest stations, metres

v

Propagation speed in the medium, metres per second

T_p

Maximum one-way propagation time, seconds

Propagation-delay-only figure. The real limit for 10 Mbps Ethernet is under half of this — see the collision domain section.

HLEN

Header Length field, counted in 4-byte (32-bit) words

HLEN = 5, the minimum with no options, gives 5 x 4 = 20 bytes, exactly the fixed part.

Lecture 5's own figure: about 4295 million addresses total, of which about 3706 million are described as usable.

\wedge

Bitwise logical AND, applied bit by bit across all 32 bits

This single operation is what a router performs on every packet it forwards.

x

Number of host bits, that is 32 minus the prefix length

The minus 2 removes the network address and the broadcast address. It does NOT apply to /31 or /32.

b

Number of bits borrowed from the host portion

Fragmentation Offset

Value stored in the 13-bit offset field, counted in 8-byte units

The field never stores a byte count directly. Always multiply the field's value by 8 to get the real position in the original datagram.

Compare against IPv4's 2^32. The jump is not "a bit more room" — it is enough that address exhaustion stops being a design constraint at all.