Exercises — MIL-STD-1553 — military avionics bus

It is a Remote Terminal (RT). In 1553 there is exactly one Bus Controller (BC) and it owns every microsecond of bus time. An RT is a slave: it transmits only in response to a command word addressed to it. Speaking first would mean two devices could talk at once — a collision — which the command/response design forbids by construction.

• RT Addr = 7 → address device number 7.
• T/R = 1 → Transmit: the RT should send data (1 = transmit, 0 = receive).
• Subaddr = 2 → from its sub-channel 2.
• Word Count = 4 → send 4 data words.

Plain English: "RT #7, send me 4 words from your subaddress 2."

It is Manchester II encoding. The problem it solves is receiver clock drift: a long run of identical bits (e.g. 100 zeros) in plain NRZ gives no edges, so the receiver loses track of where bits begin. A guaranteed mid-bit edge every $1\mu s$ lets the PLL stay locked. (See Manchester Encoding.)

$20\text{bits}\times \frac{1\mu s}{\text{bit}}=20\mu s.$ One word = $20\mu s$. This is the atom of every timing calculation below.

Count every word and gap on the timeline (red bracket = total):

Total = ==$204\mu s$==.

(The parent note used a $8\mu s$ gap to get $208\mu s$; with the $4\mu s$ minimum gap we get $204\mu s$. The formula is identical — only the gap value changed.)

Time used per second: $204\mu s\times 50=10,200\mu s=10.2\text{ms}.$ One second is $1,000,000\mu s$, so: $\frac{10,200}{1,000,000}=0.0102=1.02\%.$ The bus is about 1% busy. Plenty of room — exactly why 1 Mbit/s is enough for tiny messages.

In Manchester II every bit has a mid-bit edge, so the longest stretch with no transition is half a bit before one edge plus half a bit after the next — at most one full bit-time between forced edges, i.e. $\le 2\mu s$ of same level in worst-case adjacent bits. A sync holds one level for a full $3\mu s$, which exceeds that limit. Because data cannot produce $3\mu s$ of flat level, the receiver treats it as a guaranteed word boundary — no data pattern can ever imitate it.

• Single bus: $P=p=0.002$ (0.2% per hour).
• Dual redundant: comm is lost only if both fail: $P=p\times p=0.002^2=0.000004=4\times 10^{-6}.$ Improvement factor: $\frac{0.002}{0.000004}=500.$ Dual redundancy is 500× safer here. (Independence is the key assumption — shrapnel cutting both wires together would violate it, which is why the two buses are physically routed apart.)

$V_{\text{diff}}=(V_A+\Delta V)-(V_B+\Delta V)=(5+3)-(-5+3)=8-2=10V.$ The noise $\Delta V=+3V$ appears in both terms and subtracts away. The receiver still reads $+10V$ (a clean logic 1). This is common-mode rejection: noise that hits both wires equally cancels in the difference. (Same principle as RS-485 Protocol.)

(a) Minor frame. The fastest rate is 100 Hz, so the base period is $1/100=10\text{ms}$. In every $10\text{ms}$ frame: RT-1 runs every frame, RT-2 every 2nd frame (50 Hz), RT-3 every 4th frame (25 Hz). Minor frame = ==$10\text{ms}$==.

(b) Transaction lengths (read = cmd + gap + status + N·data; write = cmd + gap + N·data + status; both have 2 words of overhead beyond data plus one gap):

• RT-1 read, 4 words: $20+4+20+4\times 20=124\mu s$.
• RT-2 read, 16 words: $20+4+20+16\times 20=364\mu s$.
• RT-3 write, 2 words: $20+4+2\times 20+20=84\mu s$.

(c) Worst-case frame. The busiest frame is the one where all three land together (every 4th frame, since RT-3 is slowest). That frame carries: $124+364+84=572\mu s.$ Utilisation in the worst frame: $\frac{572}{10,000}=0.0572=5.72\%.$ Worst-case bus load = 5.72%. Comfortably schedulable (see Real-Time Scheduling Theory and Time-Triggered Architectures).

Worst case = all 3 attempts happen (2 fail, the 3rd may also fail before giving up): $3\times 204\mu s=612\mu s.$ Worst-case = ==$612\mu s$==. A scheduler must budget this maximum, not the nominal $204\mu s$, or a noisy burst blows the frame.

Decisive reason: raw speed is irrelevant if latency is non-deterministic. Standard Ethernet (CSMA/CD, switch buffering) cannot guarantee a bounded worst-case delay — collisions and queueing add unpredictable jitter, so you cannot prove the $500\mu s$ bound. 1553's command/response schedule makes latency provable to the microsecond. Modern high-bandwidth choice: AFDX (Avionics Full-Duplex Switched Ethernet), which adds bandwidth reservation to bound latency — used for bulk sensor data while 1553 stays on the critical control loops. Contrast also with CAN Bus and ARINC 429, each with different determinism trade-offs.

Going from one to two buses cuts comm-loss probability from $p$ to $p^2$ — a huge, cheap win. Going from two to three cuts it further to $p^3$, but with $p$ already tiny (say $10^{-3}$), $p^2=10^{-6}$ already clears Level A's requirement. The third bus improves from $10^{-6}$ to $10^{-9}$ — more reliability than the mission needs, at the cost of extra weight, wiring, transceivers, and validation effort. In avionics, weight is fuel and cost is schedule; you buy exactly the reliability the certification demands and no more. This is diminishing returns meeting a fixed requirement.

(a) RT-4 read, 32 words (recall word count 0 encodes 32, the maximum): $20+4+20+32\times 20=684\mu s.$ (b) RT-4 runs every frame, so it adds to the worst frame (4.1's $572\mu s$): $572+684=1256\mu s.$ Utilisation: $\frac{1256}{10,000}=0.1256=12.56\%.$ (c) At 12.56% the bus is still easily schedulable — 87% idle margin remains, ample room for retries and future growth. The design closes.