Exercises — SpaceWire — high-speed serial link standard for spacecraft

The two signals are Data (D) and Strobe (S). There is no separate clock wire because the clock is recovered on the receiver as $\text{RecoveredClock}(n)=\text{Data}(n)\oplus \text{Strobe}(n)$ where $\oplus$ means XOR ("output 1 only when the two inputs differ"). The encoding guarantees that on every bit period exactly one of the two lines toggles, so their combined transition activity is a clock. A physical clock wire would drift against the data across long cables (thermal, radiation damage) — this drift is called skew, and DS encoding sidesteps it entirely. See LVDS Signaling for how each of D and S is itself a differential pair.

Answer ::: EOP = End of Packet (normal, clean end). EEP = Error End of Packet (the packet was cut off by a detected error). FCT = Flow Control Token (a control character the receiver sends back to grant the sender credit for 8 more data characters of buffer space).

0x5C in binary is $01011100$, i.e. $D_7\dots D_0=0,1,0,1,1,1,0,0$. Count the ones: $0+1+0+1+1+1+0+0=4$ → an even number of ones. Equivalently the XOR of all eight bits is $0$. Odd parity means "make the total number of ones odd." We have 4 ones (even), so we must add a one: $P=1.$ (Check: 4 data-ones + 1 parity-one = 5 ones total, which is odd. ✓) Cross-connect: this single parity bit is the tripwire for a bit flipped by a cosmic ray.

Rule: Data always equals the current bit. Strobe toggles only when the bit is the same as the previous bit (so that if Data stays flat, Strobe still gives one edge).

Exactly one line toggles each period, so RecoveredClock = D XOR S ticks four times. Look at figure s02: three flat runs on Data are "rescued" by Strobe edges, and the falling bit is carried by Data itself.

First the packet's transmission time (time to shove all bits onto the wire once): $L_{\text{packet}}=\frac{512\times 8\text{bits}}{100\times 10^6\text{bits/s}}=\frac{4096}{10^8}\text{s}=40.96\mu s.$

(a) Store-and-forward: each switch must receive the entire packet before sending it on, so every hop costs one full $L_{\text{packet}}$: $L_{\text{SF}}=4\times 40.96=163.84\mu s.$

(b) Wormhole: the header carves the path (4 hops × 1 μs = 4 μs), and then the body streams straight through all reserved switches — you pay $L_{\text{packet}}$ only once because the tail is already flowing while the head reaches the far end: $L_{\text{WH}}=4\times 1+40.96=44.96\mu s.$

Speed-up $=163.84/44.96\approx 3.64\times$. The core idea from Wormhole Routing vs Store-and-Forward: wormhole overlaps transmission across all hops; store-and-forward serialises it.

(a) One-way delay $=20/(2\times 10^8)=1\times 10^{-7}$ s $=100$ ns. RTT $=2\times 100=200$ ns $=0.2\mu s$.

(b) Bits sent in one RTT $=200\times 10^6\times 0.2\times 10^{-6}=40$ bits. As characters: $40/8=5$ data characters in flight.

(c) Each FCT grants 8 characters, so credits needed $=\lceil 5/8\rceil =1$ FCT. One outstanding FCT (8 characters of granted credit) already exceeds the 5 characters the RTT can hide — so this short, fast link is comfortably covered. This is the deep reason from the parent note: RTT — not line rate — sets how much buffered credit you need. Longer cables ⇒ bigger RTT ⇒ more outstanding FCTs. Ties into Real-Time Determinism: the buffer bound is analytically known in advance.

(a) Packet transmission time: $L_{\text{packet}}=\frac{4096\times 8}{200\times 10^6}=\frac{32768}{2\times 10^8}\text{s}=163.84\mu s.$ Latency (wormhole) $=6\times 1\mu s+163.84\mu s=169.84\mu s$.

(b) Duty $=163.84\mu s/2000\mu s=0.08192=8.192\%$ of the frame period spent transmitting.

(c) Yes: $163.84\mu s\ll 2000\mu s$, and even the full delivery latency $169.84\mu s$ fits inside one 2 ms period with $>10\times$ margin. The link is not the bottleneck; there is headroom for RMAP Protocol read/write overhead and retransmission after a single-event upset.

Step 1 — parity: usable fraction after parity $=8/9\approx 0.8889$. Step 2 — control overhead: multiply by $(1-0.05)=0.95$. $\text{Payload rate}=400\times \frac{8}{9}\times 0.95\approx 400\times 0.84444\approx 337.8\text{Mbps}.$ So a 400 Mbps SpaceWire link delivers ≈ 338 Mbps of actual instrument bytes. The gap is the price of on-wire error detection plus flow control — cheap insurance against radiation, and the reason SpaceWire beats a heavier legacy bus like MIL-STD-1553.

(a) The deadline budget splits into header reservation + body streaming: $L_{\text{header}}+L_{\text{packet}}\le 250\mu s,L_{\text{header}}=5\times 1=5\mu s.$ So the body must fit in $250-5=245\mu s$: $\frac{8192\times 8}{R}\le 245\times 10^{-6}\Rightarrow R\ge \frac{65536}{245\times 10^{-6}}\approx 2.675\times 10^8\text{bps}\approx 267.5\text{Mbps}.$ Minimum line rate ≈ 267.5 Mbps.

(b) The smallest standard rate above 267.5 Mbps is 400 Mbps. Achieved latency: $L=5\mu s+\frac{65536}{400\times 10^6}\text{s}=5\mu s+163.84\mu s=168.84\mu s\le 250\mu s.✓$ Margin $=250-168.84=81.16\mu s$ — room to absorb one EEP-triggered retransmission before the deadline is threatened.

Answer ::: A shared clock wire and the data wire travel slightly different physical paths; thermal expansion and radiation damage make them arrive at different times — this timing mismatch is skew, and once skew grows past half a bit period the receiver samples the wrong bit. DS carries no clock wire at all: because exactly one of {Data, Strobe} toggles per bit, the receiver rebuilds the clock locally as $D\oplus S$ (XOR). Since both D and S ride the same twisted-pair harness and are compared against each other rather than against a distant clock, there is no separate reference to skew against — the failure mode is designed out.