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

Do signals hain Data (D) aur Strobe (S). Koi alag clock wire nahi hai kyunki clock ko receiver pe recover kiya jaata hai: $\text{RecoveredClock}(n)=\text{Data}(n)\oplus \text{Strobe}(n)$ jahan $\oplus$ ka matlab hai XOR ("output 1 tabhi jab dono inputs alag hon"). Encoding yeh guarantee karta hai ki har bit period mein dono lines mein se exactly ek toggle karti hai, toh unki combined transition activity hi ek clock ban jaati hai. Ek physical clock wire lambi cables pe data ke against drift karti — thermal, radiation damage ki wajah se — is drift ko skew kehte hain, aur DS encoding isse poori tarah bypass kar deta hai. LVDS Signaling dekho yeh samajhne ke liye ki D aur S mein se har ek khud ek differential pair hai.

Answer ::: EOP = End of Packet (normal, clean end). EEP = Error End of Packet (packet ek detected error ki wajah se beech mein cut off ho gaya). FCT = Flow Control Token (ek control character jo receiver wapas bhejta hai taaki sender ko 8 aur data characters ke buffer space ka credit mile).

0x5C binary mein $01011100$ hai, yaani $D_7\dots D_0=0,1,0,1,1,1,0,0$. Ones count karo: $0+1+0+1+1+1+0+0=4$ → even number of ones. Equivalently saaton bits ka XOR $0$ hai. Odd parity ka matlab hai "ones ki total count odd banao." Hamare paas 4 ones hain (even), toh humein ek aur one add karna hoga: $P=1.$ (Check: 4 data-ones + 1 parity-one = 5 ones total, jo odd hai. ✓) Cross-connect: yeh single parity bit ek cosmic ray se flip hue bit ka tripwire hai.

Rule: Data hamesha current bit ke barabar hoti hai. Strobe sirf tab toggle karta hai jab bit pichle bit ke barabar ho (taaki agar Data flat rahe, Strobe phir bhi ek edge de).

Har period mein exactly ek line toggle karti hai, isliye RecoveredClock = D XOR S chaar baar tick karta hai. Figure s02 dekho: Data pe teen flat runs ko Strobe edges "rescue" karti hain, aur falling bit Data khud carry karti hai.

Pehle packet ka transmission time (sabhi bits ek baar wire pe daalne ka time): $L_{\text{packet}}=\frac{512\times 8\text{bits}}{100\times 10^6\text{bits/s}}=\frac{4096}{10^8}\text{s}=40.96\mu s.$

(a) Store-and-forward: har switch ko aage bhejna shuru karne se pehle poora packet receive karna padta hai, toh har hop pe ek full $L_{\text{packet}}$ cost hoti hai: $L_{\text{SF}}=4\times 40.96=163.84\mu s.$

(b) Wormhole: header path carve karta hai (4 hops × 1 μs = 4 μs), aur phir body seedha stream through karti hai saare reserved switches mein — tum $L_{\text{packet}}$ sirf ek baar pay karte ho kyunki tail already flow kar raha hota hai jab head doosri end tak pahunchta hai: $L_{\text{WH}}=4\times 1+40.96=44.96\mu s.$

Speed-up $=163.84/44.96\approx 3.64\times$. Core idea Wormhole Routing vs Store-and-Forward se: wormhole transmission ko saare hops pe overlap karta hai; store-and-forward isse serialize karta hai.

(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) Ek RTT mein bheje gaye bits $=200\times 10^6\times 0.2\times 10^{-6}=40$ bits. Characters mein: $40/8=5$ data characters in flight.

(c) Har FCT 8 characters grant karta hai, toh credits needed $=\lceil 5/8\rceil =1$ FCT. Ek outstanding FCT (8 characters ka granted credit) RTT ke 5 characters se pehle se zyada hai — toh yeh short, fast link aaram se cover ho jaati hai. Yahi deep reason hai parent note se: RTT — line rate nahi — decide karta hai ki kitna buffered credit chahiye. Lambi cables ⇒ bada RTT ⇒ zyada outstanding FCTs. Real-Time Determinism se connected: buffer bound pehle se analytically known hota hai.

(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) Haan: $163.84\mu s\ll 2000\mu s$, aur poori delivery latency $169.84\mu s$ bhi ek 2 ms period ke andar fit ho jaati hai $>10\times$ margin ke saath. Link bottleneck nahi hai; RMAP Protocol read/write overhead aur single-event upset ke baad retransmission ke liye headroom hai.

Step 1 — parity: parity ke baad usable fraction $=8/9\approx 0.8889$. Step 2 — control overhead: $(1-0.05)=0.95$ se multiply karo. $\text{Payload rate}=400\times \frac{8}{9}\times 0.95\approx 400\times 0.84444\approx 337.8\text{Mbps}.$ Toh ek 400 Mbps SpaceWire link ≈ 338 Mbps actual instrument bytes deliver karta hai. Yeh gap on-wire error detection aur flow control ki price hai — radiation ke against sasti insurance, aur yehi wajah hai ki SpaceWire ek heavier legacy bus jaise MIL-STD-1553 se behtar hai.

(a) Deadline budget header reservation aur body streaming mein split hoti hai: $L_{\text{header}}+L_{\text{packet}}\le 250\mu s,L_{\text{header}}=5\times 1=5\mu s.$ Toh body $250-5=245\mu s$ mein fit honi chahiye: $\frac{8192\times 8}{R}\le 245\times 10^{-6}\Rightarrow R\ge \frac{65536}{245\times 10^{-6}}\approx 2.675\times 10^8\text{bps}\approx 267.5\text{Mbps}.$ Minimum line rate ≈ 267.5 Mbps.

(b) 267.5 Mbps se upar ka sabse chhota standard rate 400 Mbps hai. 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 $=250-168.84=81.16\mu s$ — ek EEP-triggered retransmission absorb karne ki jagah hai deadline threaten hone se pehle.

Answer ::: Ek shared clock wire aur data wire thode alag physical paths se travel karti hain; thermal expansion aur radiation damage unhe alag times pe pahunchata hai — yeh timing mismatch skew hai, aur jab skew half a bit period se zyada badh jaaye toh receiver galat bit sample karta hai. DS mein koi clock wire hoti hi nahi: kyunki {Data, Strobe} mein se exactly ek har bit mein toggle karta hai, receiver clock ko locally $D\oplus S$ (XOR) ke roop mein rebuild karta hai. Kyunki D aur S dono same twisted-pair harness se chalta hain aur ek doosre ke against compare hote hain na ki ek door clock ke against, koi alag reference nahi hai jiske against skew ho sake — failure mode ko design out kar diya gaya hai.