Avionics Data Buses Questions and Answers

Answer: A

Dividing gives \( 32/100\,000 = 3.2 \times 10^{-4} \) second. Word gaps of at least four bit times separate consecutive words, so the achievable word rate is somewhat below the raw figure suggests. Gaps consume part of the capacity.

Answer: B

Correct answer: 2.56 milliseconds

Transmission time is the number of bits divided by the rate:

\[ t = \frac{32}{12\,500} = 2.56 \times 10^{-3}\ \mathrm{s} = 2.56\ \mathrm{ms} \]

At the high-speed rate of 100 kbit/s the same word takes 320 microseconds, eight times less. The low-speed rate is retained for parameters that change slowly - fuel quantity, configuration discretes - where the update interval matters more than the latency, and it uses less bandwidth per source.

Answer: C

Correct answer: 256

An eight-bit field has

\[ 2^{8} = 256 \]

distinct values, so 256 labels are available. They are conventionally written in octal, from 001 to 377.

The space is not as generous as it looks, since labels are assigned by the standard so that equipment from different manufacturers agrees on their meaning. It is extended in practice by the source destination identifier bits and by equipment identifiers, which allow the same label to carry different parameters in different systems.

Answer: D

Correct answer: 99999

In binary coded decimal each digit is stored separately and can take only the values 0 to 9, so five digits give a maximum of 99999 whatever the number of bits used to hold them.

The distractor 524287 is \( 2^{19} - 1 \), which is what pure binary in the same 19 bits would give - considerably more. BCD wastes that capacity deliberately: it maps directly onto decimal displays and instrument readouts, so no conversion is needed at either end.

Answer: A

Dividing gives \( 20/10^6 = 2 \times 10^{-5} \) second. A message of a command word, thirty two data words and a status word therefore occupies about 680 microseconds plus the response gaps. Short messages bound the occupancy.

Answer: B

Correct answer: 34

Simply add the words:

\[ 1 + 32 + 1 = 34 \]

Thirty-two is the maximum number of data words a single MIL-STD-1553 message may carry, so this is the largest controller-to-terminal transfer the standard allows. The five-bit word count field is why: it encodes 1 to 31 directly and uses zero to mean 32. A transfer of more data must be split across several messages, each with its own command and status words.

Answer: C

Correct answer: 60 per cent

Bus loading is traffic over capacity:

\[ \frac{60}{100} \times 100 = 60\% \]

The distractor 40 per cent is the margin remaining rather than the loading. Sixty per cent is high for an avionics bus: designers commonly aim nearer 50 to leave room for later functions and to keep worst-case latency comfortably bounded. A bus already at this level constrains what can be added to the aircraft afterwards.

Answer: D

Correct answer: 250 kilobits per second

The bandwidth allocation gap is the minimum interval between successive frames on a virtual link, so the maximum rate is one maximum-size frame per gap:

\[ \frac{1000\ \mathrm{bits}}{4 \times 10^{-3}\ \mathrm{s}} = 250\,000\ \mathrm{bit/s} = 250\ \mathrm{kbit/s} \]

This is the mechanism that makes an Ethernet network deterministic. Every virtual link's traffic is bounded in advance by its own gap and frame size, so the total offered to any switch port can be computed and the worst-case delay demonstrated.

Answer: A

Correct answer: 0.64 per cent

High-speed ARINC 429 runs at 100 kbit/s. One 32-bit word twenty times a second is

\[ 32 \times 20 = 640\ \mathrm{bit/s} \]

\[ \frac{640}{100\,000} \times 100 = 0.64\% \]

A single parameter therefore costs very little, which is why one bus can carry many. The arithmetic also shows where loading comes from: it is the number of parameters and their update rates together, not any one of them, and a fast-updating attitude parameter costs several times this.

Answer: B

Correct answer: 1 megahertz

Manchester encoding places a transition at the middle of every bit, and a run of identical bits also produces transitions at the bit boundaries. The highest fundamental frequency therefore equals the bit rate: 1 MHz for 1 Mbit/s.

That is the cost of the scheme - twice the bandwidth of a simple non-return-to-zero code carrying the same data. It buys clock recovery from the data itself and a signal with no DC component, which is what allows the transformer coupling MIL-STD-1553 depends on.