Link budget — path loss, EIRP, G - T, Eb - N0: Exercises

3.6.25 · D4 · Physics › Spacecraft Structures & Systems Engineering › Link budget — path loss, EIRP, G - T, Eb - N0

Yahan sab kuch parent link-budget note pe build karta hai. Agar koi term unfamiliar lage, toh har section ke neeche wikilinks batate hain ki woh term kahan se aaya.

The toolkit (har woh formula jo tumhe chahiye)

Shuru karne se pehle, yeh raha equations ka complete set, jisme har symbol ko plain words mein naam diya gaya hai. Neeche koi bhi aisa symbol use nahi hua jo yahan listed na ho.

Definition Ratios ke peeche teen raw quantities

Is page pe do symbols sabse zyada kaam karte hain, toh chaliye inhe kisi bhi formula se pehle naam dete hain:

$C$ = received carrier power — woh actual signal power (watts mein, ya dBW mein) jo receiver pe pahunchti hai saare gains aur losses ke baad. Ise socho "jab signal Earth pe pahunchta hai toh tank mein kitna bacha hai."
$N_{0}$ = noise power spectral density — bandwidth ke har 1 Hz slice mein packed noise power (watts per hertz mein). Yeh poori tarah temperature se set hota hai: $N_{0} = k T_{sys}$ , jahan $k$ Boltzmann's constant hai aur $T_{sys}$ system noise temperature hai. Yeh yaad rakhne ki sabse important line hai — noise per hertz bas $k$ times temperature hai.

Neeche har ratio ( $C / N_{0}$ , $E_{b} / N_{0}$ ) inhi mein se ek ko doosre se divide karta hai. Decibels mein, "divide" ban jaata hai "subtract", isliye poora budget ek lamba sum hota hai + gains aur - losses ka.

Formula Link-budget cheat sheet (sab decibels mein)

EIRP — effective isotropic radiated power (hum sabse achhe direction mein kitna power bhejte lagte hain): $EIRP = P_{t} + G_{t}$ $P_{t}$ = transmitter power dBW mein (decibels relative to 1 watt), $G_{t}$ = transmit antenna gain dBi mein (decibels relative to an isotropic radiator).

Free-space path loss (Free-Space Path Loss Derivation): $L_{p} = 32.45 + 20 lo g_{10} (r_{km}) + 20 lo g_{10} (f_{MHz})$ $r_{km}$ = distance kilometres mein, $f_{MHz}$ = frequency megahertz mein.

Figure of merit receiver ka: $\frac{G}{T} = G_{r} - 10 lo g_{10} (T_{sys})$ $G_{r}$ = receive gain (dBi), $T_{sys}$ = system noise temperature kelvin mein (Noise Temperature and Noise Figure).

Carrier-to-noise-density ratio (dB-Hz). Kyunki $C / N_{0} = C / (k T_{sys})$ , decibels mein yeh adds aur subtracts ki ek chain mein split ho jaata hai: $\frac{C}{N _{0}} = EIRP - L_{p} + \frac{G}{T} - 10 lo g_{10} (k)$ jahan $k = 1.38 \times 1 0^{- 23}$ J/K Boltzmann's constant hai, aur $10 lo g_{10} (k) = - 228.6$ dBW/K/Hz.

Energy-per-bit to noise density: $\frac{E _{b}}{N _{0}} = \frac{C}{N _{0}} - 10 lo g_{10} (R_{b})$ $R_{b}$ = data rate bits per second mein.

Link margin (safety cushion): $M = (\frac{E _{b}}{N _{0}})_{received} - (\frac{E _{b}}{N _{0}})_{required}$

Figure — Link budget — path loss, EIRP, G - T, Eb - N0

Picture padho: left side pe, $\times 2$ ka factor $+ 3.01$ dB step ban jaata hai; $\times 10$ ban jaata hai $+ 20$ dB; $\div 100$ loss ban jaata hai $- 20$ dB drop. dB (right) axis pe straight-line addition, linear (left) axis pe repeated multiplication ke corresponding hai. Yahi poori wajah hai ki hum decibels mein kaam karte hain.

Convert P (watts)

P_{dBW} = 10 lo g_{10} (P)

Kisi dB value ko wapas ratio mein undo karo

ratio

= 1 0^{(dB /10)}

N_{0}

ek line mein kya hai?

N_{0} = k T_{sys}

— har 1 Hz slice mein noise power

C

kya represent karta hai?

antenna tak pahunchne wala received carrier (signal) power

Level 1 — Recognition

L1.1 — Power to dBW

Ek spacecraft transmitter 40 W output karta hai. Ise dBW mein express karo.

Recall Solution

KYA: ek linear power ko decibels mein convert karo. KYUN: poora budget dB mein kiya jaata hai taaki hum add kar sakein. $P_{t} = 10 lo g_{10} (40) = 10 \times 1.602 = 16.02 dBW$ Answer: $\approx 16.02$ dBW.

L1.2 — EIRP by addition

Wahi 40 W transmitter 18 dBi gain wale antenna ko feed karta hai. EIRP nikalo.

Recall Solution

dB mein, EIRP bas ek sum hai: $EIRP = P_{t} + G_{t} = 16.02 + 18 = 34.02 dBW$ Answer: $\approx 34.02$ dBW. (Koi multiplication nahi — yahi decibels ka poora point hai.)

L1.3 — Reading G/T

Ek receiver ka gain $G_{r} = 55$ dBi aur system noise temperature $T_{sys} = 100$ K hai. $G / T$ compute karo.

Recall Solution

$\frac{G}{T} = 55 - 10 lo g_{10} (100) = 55 - 10 \times 2 = 55 - 20 = 35 dB/K$ Answer: 35 dB/K.

Level 2 — Application

L2.1 — Path loss at the Moon

Ek lunar orbiter ke liye link ka free-space path loss compute karo: distance $r = 3.84 \times 1 0^{5}$ km, frequency $f = 2.2$ GHz (S-band).

Recall Solution

Step 1 — units. KYUN: constant 32.45 is assumption ke saath derive kiya gaya ki $r$ km mein hai aur $f$ MHz mein; kuch aur feed karo toh constant galat ho jaata hai. $r = 3.84 \times 1 0^{5}$ km (already km mein); $f = 2.2$ GHz $= 2200$ MHz. Step 2 — plug in. KYUN har term: $20 lo g_{10} (r)$ term inverse-square spreading loss hai dB mein likha hua (signal $4 π r^{2}$ area ke sphere pe spread hota hai, figure s02 dekho); $20 lo g_{10} (f)$ term isliye hai kyunki fixed dish ke liye higher-frequency wave ka catching aperture smaller hota hai. $L_{p} = 32.45 + 20 lo g_{10} (3.84 \times 1 0^{5}) + 20 lo g_{10} (2200)$

$20 lo g_{10} (3.84 \times 1 0^{5}) = 20 \times 5.584 = 111.69$ dB
$20 lo g_{10} (2200) = 20 \times 3.342 = 66.85$ dB $L_{p} = 32.45 + 111.69 + 66.85 = 210.99 dB$ Answer: $\approx 211.0$ dB.

Neeche ki picture dikhati hai kyun distance tumhe dB cost karti hai: wahi power ek bade sphere pe spread hone ka matlab hai kam power per square metre.

L2.2 — Same distance, higher frequency

L2.1 repeat karo lekin X-band pe, $f = 8.4$ GHz. Path loss kitne dB se badh jaata hai?

Recall Solution

Sirf frequency term badalta hai. 2200 MHz se 8400 MHz jaana $8400/2200 = 3.818$ ka factor hai. $Δ L_{p} = 20 lo g_{10} (3.818) = 20 \times 0.5818 = 11.64 dB$ Toh $L_{p} = 210.99 + 11.64 = 222.63$ dB. Answer: path loss $\approx 11.6$ dB badhti hai ( $\approx 222.6$ dB tak). KYUN badhti hai: higher frequency = shorter wavelength = fixed physical dish ke liye smaller effective aperture, isliye free-space loss (jo $f^{2}$ ke saath scale karta hai) badhta hai.

L2.3 — Eb/N0 from C/N0

Ek link $C / N_{0} = 68$ dB-Hz deliver karta hai aur $R_{b} = 1$ Mbps pe chalta hai. $E_{b} / N_{0}$ nikalo.

Recall Solution

KYUN rate subtract karte hain: signal ke har second mein $R_{b}$ bits hote hain, toh received power $R_{b}$ bits per second mein share hoti hai — zyada bits matlab har bit ke liye kam energy, isliye $10 lo g_{10} (R_{b})$ subtract karte hain. $R_{b} = 1 \times 1 0^{6}$ toh $10 lo g_{10} (1 0^{6}) = 60$ dB-Hz. $\frac{E _{b}}{N _{0}} = 68 - 60 = 8 dB$ Answer: 8 dB.

Level 3 — Analysis

L3.1 — Hum range kitni double kar sakte hain?

Ek link abhi exactly 3 dB of margin ke saath close (kaam kar raha) hai. Pointing aur atmosphere ko ignore karte hue, hum kitna extra path loss absorb kar sakte hain, aur woh range multiplier kya correspond karta hai?

Recall Solution

KYA: margin woh spare loss hai jo hum afford kar sakte hain. KYUN: $L_{p}$ ke alawa har term fixed hai, toh 3 dB margin = link tootne se pehle 3 dB aur $L_{p}$ . Path loss $20 lo g_{10} (r)$ ke saath scale karta hai. Range multiplier $k_{r}$ ke liye $20 lo g_{10} (k_{r}) = 3$ dB set karte hain: $k_{r} = 1 0^{3/20} = 1 0^{0.15} = 1.413$ Answer: hum margin khatam hone se pehle $\approx 1.41 \times$ zyada door ja sakte hain. ( $2 \times$ nahi! Range double karne ke liye $20 lo g_{10} (2) = 6.02$ dB chahiye.)

L3.2 — Data rate se margin trade karna

Ek downlink mein $R_{b} = 4$ Mbps pe $E_{b} / N_{0} = 6$ dB hai, lekin modem ko reliable BPSK decoding ke liye $E_{b} / N_{0} = 9.6$ dB chahiye. Hum power ya antennas nahi badal sakte. Exactly 9.6 dB hit karne ke liye data rate kitni girani padegi?

Recall Solution

Key idea: $C / N_{0}$ fixed hai (power, loss, G/T unchanged). Sirf $R_{b}$ hi $E_{b} / N_{0}$ ko move karta hai, $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b})$ ke through. Hum $9.6 - 6 = 3.6$ dB short hain. Rate giraa ke exactly utna recover karte hain: $10 lo g_{10} (\frac{4 Mbps}{R _{b}}) = 3.6 \Rightarrow \frac{4 Mbps}{R _{b}} = 1 0^{0.36} = 2.291$ $R_{b} = \frac{4}{2.291} = 1.746 Mbps$ Answer: $\approx 1.75$ Mbps. Slow karna margin khareedata hai kyunki har bit ko wahi received power ka bada hissa milta hai.

L3.3 — Kaunsa upgrade jeetega?

Tumhara link 2 dB short hai. Tum ya toh (A) receiver cool kar sakte ho, $T_{sys}$ 120 K se 60 K karo, ya (B) 3 dB transmit power amplifier add karo. Kaun zyada gain deta hai, aur winner need se kitna zyada hai?

Recall Solution

Option A $G / T$ ko $- 10 lo g_{10} (T_{sys})$ term se badalta hai. $T$ ko 120 K se 60 K pe halving karna: $Δ (G / T) = - 10 lo g_{10} (60) + 10 lo g_{10} (120) = 10 lo g_{10} (120/60) = 10 lo g_{10} (2) = 3.01 dB$ Option B EIRP mein exactly 3.00 dB add karta hai. Dono $C / N_{0}$ (aur isliye $E_{b} / N_{0}$ ) mein one-for-one flow hote hain. A thoda sa jeetat hai: 3.01 dB vs 3.00 dB. Dono 2 dB shortfall clear karte hain — A mein $3.01 - 2 = 1.01$ dB ka naya margin bacha. KYUN temperature halving ≈ 3 dB: noise power $T_{sys}$ ke proportional hai, aur power ratio halving hamesha $10 lo g_{10} (2) = 3.01$ dB hoti hai. Yeh wahi 3 dB hai jo is subject mein har jagah milta hai.

Level 4 — Synthesis

L4.1 — Poora Mars downlink budget banao

Given:

$P_{t} = 12$ W
$G_{t} = 25$ dBi
$r = 2.0 \times 1 0^{8}$ km, $f = 8.4$ GHz
Ground station $G / T = 50$ dB/K (ek DSN 34-m dish)
$R_{b} = 500$ kbps

EIRP, $L_{p}$ , $C / N_{0}$ , aur $E_{b} / N_{0}$ nikalo. Agar FEC ke saath BPSK ko $E_{b} / N_{0} = 2.5$ dB chahiye, toh margin kya hai?

Recall Solution

Step 1 — EIRP. $P_{t} = 10 lo g_{10} (12) = 10.79$ dBW. $EIRP = 10.79 + 25 = 35.79 dBW$ Step 2 — Path loss. $f = 8400$ MHz, $r = 2.0 \times 1 0^{8}$ km.

$20 lo g_{10} (2.0 \times 1 0^{8}) = 20 \times 8.301 = 166.02$ dB
$20 lo g_{10} (8400) = 20 \times 3.924 = 78.49$ dB $L_{p} = 32.45 + 166.02 + 78.49 = 276.96 dB$ Step 3 — C/N0. KYUN $+ 228.6$ : term hai $- 10 lo g_{10} (k) = - (- 228.6) = + 228.6$ — hum tiny noise density $N_{0} = k T_{sys}$ se divide kar rahe hain, aur ek tiny number se divide karna ratio ko huge banata hai, isliye yeh add hota hai. $\frac{C}{N _{0}} = 35.79 - 276.96 + 50 - (- 228.6) = 37.43 dB-Hz$ Step 4 — Eb/N0. $R_{b} = 5 \times 1 0^{5}$ , toh $10 lo g_{10} (5 \times 1 0^{5}) = 56.99$ dB-Hz. $\frac{E _{b}}{N _{0}} = 37.43 - 56.99 = - 19.56 dB$ Step 5 — Margin. $M = - 19.56 - 2.5 = - 22.06$ dB. Link FAIL ho gaya 22 dB se. Interpretation: 500 kbps pe yeh budget hopeless hai. Fix ke liye L4.2 dekho.

L4.2 — Failing link ko close karo

L4.1 ka system lo. Baki sab fixed rakhte hue, highest data rate kya hai jo exactly required $E_{b} / N_{0} = 2.5$ dB deti hai?

Recall Solution

KYUN yeh kaam karta hai: $C / N_{0}$ hardware aur geometry ki ek fixed property hai — yeh data rate ke baare mein nahi jaanta. Rate sirf $- 10 lo g_{10} (R_{b})$ term se enter hoti hai, toh hum bas poochhte hain "kaun si rate $E_{b} / N_{0}$ ko exactly requirement pe laati hai?" $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b}) = 2.5$ rearrange karo: $10 lo g_{10} (R_{b}) = 37.43 - 2.5 = 34.93 dB-Hz$ dB undo karo (10 ko dB/10 power pe raise karo, kyunki woh $10 lo g_{10}$ ko invert karta hai): $R_{b} = 1 0^{34.93/10} = 1 0^{3.493} = 3113 bits/s \approx 3.1 kbps$ Answer: $\approx 3.1$ kbps. Deep-space links sach mein itni slow hote hain — isliye missions heavy FEC aur small data rates use karte hain.

Level 5 — Mastery

L5.1 — Shannon sanity check

Hamare L4 link ka $C / N_{0} = 37.43$ dB-Hz hai. Maano channel bandwidth $B = 1$ MHz hai. Shannon-Hartley Theorem $C_{cap} = B lo g_{2} (1 + SNR)$ use karke, kya L4.2 ki 3.1 kbps rate theoretical ceiling ke paas bhi hai?

Recall Solution

Step 1 — bandwidth mein SNR (linear) nikalo. KYUN $B$ se divide karte hain: $C / N_{0}$ signal-to-noise hai per hertz; ordinary SNR paane ke liye us noise ko un sab $B$ hertz pe spread karna hoga jo channel occupy karta hai, yaani $SNR = (C / N_{0}) / B$ . dB mein: $37.43 - 10 lo g_{10} (1 0^{6}) = 37.43 - 60 = - 22.57$ dB. Ratio ke roop mein: $SNR = 1 0^{- 22.57/10} = 1 0^{- 2.257} = 5.53 \times 1 0^{- 3}$ . Step 2 — Shannon capacity. KYUN approximation: jab $SNR ≪ 1$ (yahan yeh $\sim 0.006$ hai, 1 se bahut neeche), curve $lo g_{2} (1 + x)$ apni tangent $lo g_{2} (1 + x) \approx x / ln 2$ ke kaafi paas hoti hai. Ise use karne se calculator ki zaroorat nahi aur physics dikhti hai — jab signal weak hota hai toh capacity SNR mein linear hoti hai. $C_{cap} = 1 0^{6} \times lo g_{2} (1 + 5.53 \times 1 0^{- 3}) = 1 0^{6} \times 7.96 \times 1 0^{- 3} = 7960 bits/s$ Step 3 — compare. Hamara practical 3113 bps $3113/7960 = 39%$ Shannon limit pe baitha hai. Answer: haan, hum ceiling ke $\approx 2.6$ factor ke andar hain — respectable, aur baaki gap wahi hai jo better FEC aur modulation band karne ki koshish karte hain.

L5.2 — Atmosphere strikes back

L4.1 ka link phir se kholo lekin survivable 3.0 kbps pe chalaate hue ( $E_{b} / N_{0} \approx 2.6$ dB, tiny margin). Ground station pe ek rainstorm 4 dB extra loss add karta hai aur $T_{sys}$ us value se badhata hai jo $G / T = 50$ deta tha, ek aisi value tak jo $G / T = 47$ dB/K deti hai. Kya link survive karta hai? 2.5 dB requirement ke against naya margin quantify karo.

Recall Solution

Step 1 — naya C/N0. Do penalties lagte hain: +4 dB extra path/atmospheric loss (4 subtract karo) aur $G / T$ 3 dB drop hoti hai (3 subtract karo). $\frac{C}{N _{0}}_{new} = 37.43 - 4 - 3 = 30.43 dB-Hz$ Step 2 — 3.0 kbps pe naya Eb/N0. $10 lo g_{10} (3000) = 34.77$ dB-Hz. $\frac{E _{b}}{N _{0}} = 30.43 - 34.77 = - 4.34 dB$ Step 3 — margin. $M = - 4.34 - 2.5 = - 6.84$ dB. Storm ke dauran link 6.84 dB se FAIL ho gaya. Engineer ka takeaway: isliye deep-space links ko weather margin built in ke saath design kiya jaata hai, aur isliye DSN mid-pass mein lower rate pe switch kar sakta hai. Baarish signal ko attenuate bhi karti hai aur antenna jo sky dekhta hai use warm bhi karti hai — ek double hit jo $G / T$ term capture karta hai.

L5.3 — 3 dB structurally kahan se aata hai?

Formulas se prove karo ki data rate half karna aur received carrier power double karna $E_{b} / N_{0}$ mein exactly wahi change produce karte hain. Kaun sa $C / N_{0}$ bhi badalta hai?

Recall Solution

Toolkit se yaad karo ki $C$ received carrier power hai (dBW) aur $N_{0} = k T_{sys}$ noise per hertz hai, toh $\frac{E _{b}}{N _{0}} = \frac{C}{N _{0}} - 10 lo g_{10} (R_{b}), \frac{C}{N _{0}} = C - 10 lo g_{10} (k T_{sys}) .$ Case 1 — data rate half karo ( $R_{b} \to R_{b} /2$ ). Rate sirf $- 10 lo g_{10} (R_{b})$ term mein aata hai: $- 10 lo g_{10} (\frac{R _{b}}{2}) = - 10 lo g_{10} (R_{b}) + 10 lo g_{10} (2) = - 10 lo g_{10} (R_{b}) + 3.01.$ Toh $E_{b} / N_{0}$ +3.01 dB badhta hai. Lekin $R_{b}$ $C / N_{0}$ mein nahi aata, toh $C / N_{0}$ unchanged rahta hai. Case 2 — received carrier double karo ( $C \to 2 C$ linear terms mein). dB mein, $C$ $10 lo g_{10} (2) = 3.01$ dB badhta hai. Kyunki $C$ $C / N_{0}$ ke andar baitha hai, woh ratio +3.01 dB badhta hai, aur kyunki $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b})$ , $E_{b} / N_{0}$ bhi +3.01 dB badhta hai. Conclusion: dono changes $E_{b} / N_{0}$ ko identical +3.01 dB se uthate hain. Fark yeh hai ki power double karna $C / N_{0}$ bhi raise karta hai, jabki rate half karna $C / N_{0}$ ko unchanged chhod deta hai (yeh bas wahi received power ko kam bits ke beech re-slice karta hai). Yahi structural wajah hai ki "3 dB" har jagah recur karta hai — yeh bas $10 lo g_{10} (2)$ hai, factor of two ka decibel fingerprint.