Communication Protocols in Embedded Systems Questions and Answers

Answer: C

A small amount of data with a bounded latency is exactly what interrupt transfers provide, the host polling at the interval the device requests. Bulk suits mass storage, isochronous suits audio, and control carries the setup traffic.

Answer: D

Correct answer: isochronous, which sacrifices retries for guaranteed bandwidth

Audio needs its data on time rather than perfectly, so isochronous transfers reserve bandwidth in every frame and do not retry a failed packet. A lost sample is a click; a retransmitted one arriving late would be a gap. Bulk transfers make the opposite trade, guaranteeing delivery but no timing, which is right for a disk and wrong for a microphone.

Answer: A

The media access controller and the physical layer chip are separate functions, and the reduced variant of the interface halves the pin count by doubling the clock. Some parts integrate the physical layer as well, at the cost of the magnetics still being external.

Answer: B

The 64-byte minimum was set so that a station could still detect a collision before it finished transmitting on the longest legal shared segment. Shorter payloads are padded, which is why a very small message costs as much airtime as a moderate one.

Answer: C

Correct answer: RAM for the buffers and the connection state

TCP needs a send and receive buffer for every connection, plus the control block holding sequence numbers, window sizes and timers, and the retransmission queue - and the buffers must be large enough to hold unacknowledged data. That memory is what a small microcontroller runs out of first. The checksum is integer arithmetic, and neither an RTOS nor a precision clock is required.

Answer: D

Three bytes of nine bit times each is twenty-seven bits, plus the start and stop conditions and the setup, which comes to roughly thirty-six bit times at ten microseconds each. The overhead of the acknowledgement bit is a ninth of the traffic and is easy to overlook.

Answer: A

The reciprocal of 360 microseconds is about 2800, and a real design would allow for the bus being shared with other devices. Whether that suffices is the question a sensor sampling requirement has to be checked against.

Answer: B

Five hundred thousand bits a second divided by a hundred and ten is roughly 4500. Bit stuffing makes the frame length depend on the data, so a worst-case figure some ten per cent lower is what a loading calculation should use.

Answer: C

Because arbitration is by fixed priority, the lowest priority frames wait for every higher one and their worst-case latency climbs steeply as the bus fills. Thirty to fifty per cent is the load usually designed for in a vehicle network.

Answer: D

Eight bits per byte at 8 MHz is one microsecond a byte, so 512 of them take about half a millisecond plus whatever gaps the driver leaves between bytes. Those gaps dominate on an interrupt-per-byte driver, which is the argument for DMA.