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

3.6.25 · D3 · Physics › Spacecraft Structures & Systems Engineering › Link budget — path loss, EIRP, G - T, Eb - N0

Yeh link-budget topic note ka hands-on companion hai. Parent note ne chaar pillars banaye the — EIRP, path loss, $G / T$ , aur $E_{b} / N_{0}$ . Yahaan hum unhe use karte hain, aur deliberately har awkward corner dhundte hain: tiny numbers, huge numbers, "it doesn't close" failure case, aur exam trick.

Kuch bhi start karne se pehle, aayein us ek grammatical rule ko pin down karein jisme poora page bolta hai.

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

Figure 1 — Hum decibels mein kyun jeete hain. Dono bars ek hi number carry karti hain. Left par laal badi bar raw linear ratio "ek billion" (

1 0^{9}

) hai, itni oonchi ki draw karne ke liye bhi log axis chahiye. Right par chhoti neeli bar wahi ratio decibels mein hai: ek friendly

90

. Peela arrow "10 log10" labelled ek hi operation hai jo left bar ko right mein compress kar deta hai. Neeche ke har example right-hand bar par rahte hain.

Scenario matrix

Yeh hai har class of case jo ek link-budget problem aap par throw kar sakti hai. Har ek worked example jo follow karta hai use woh cell(s) tag kiya gaya hai jise woh cover karta hai, toh end mein koi bhi cell unvisited nahi rahegi.

#	Cell class	Kya tricky banata hai	Covered by
A	Basic forward pass	numbers seedha plug in, koi trap nahi	Ex 1
B	Bahut bada input (deep-space range)	huge path loss, exponent bookkeeping	Ex 2
C	Bahut chhota / paas ka input (LEO, short range)	chhota loss, kya same formula aur bhi kaam karta hai?	Ex 3
D	Zero / degenerate input	$R_{b} \to$ tiny, $r \to$ tiny, $G = 0$ dBi isotropic	Ex 4
E	Sign flip — negative margin	woh link jo fail hoti hai; minus ko interpret karo	Ex 5
F	Limiting behaviour	range double karo, frequency double karo → kitne dB?	Ex 6
G	Real-world word problem	woh data rate choose karo jo link just close kare	Ex 7
H	Exam-style twist	units mismatch (Hz vs dB-Hz), Shannon comparison	Ex 8

Constants aur ek aur symbol jo hum har jagah reuse karte hain:

Yahaan $C$ received carrier power hai, $N_{0}$ noise power per hertz hai (equal to $k T$ ), $C / N_{0}$ unka ratio hai (units dB-Hz), $R_{b}$ bit rate bits per second mein hai, aur $E_{b} / N_{0}$ energy-per-bit to noise-density ratio hai jo error rate decide karta hai.

Example 1 — Basic forward pass (Cell A)

Forecast: Computing se pehle guess karo — kya $C$ dBW mein positive hoga ya negative? (Ek negative dBW ka matlab sirf "ek watt se kam" hai.)

Power ko dBW mein convert karo. $P_{t} = 10 lo g_{10} (20) = 13.01 dBW$ . Yeh step kyun? Poora budget ek sum hai, isliye pehle har quantity ko ek dB number banana zaroori hai.
EIRP transmit power plus transmit gain hai. $EIRP = 13.01 + 15 = 28.01 dBW$ . Yeh step kyun? Parent note se, $EIRP = P_{t} \cdot G_{t}$ linear units mein, jo dB mein addition hai.
Received carrier $= EIRP - L_{p} + G_{r} = 28.01 - 180 + 40 = - 111.99 dBW$ . Yeh step kyun? Path loss yahaan akela minus hai; receive antenna gain kuch wapas laata hai.

Verify: $- 111.99$ dBW $= 1 0^{- 11.199} W \approx 6.3 \times 1 0^{- 12} W$ — sub-picowatt, exactly woh "faint whisper" scale jo hum ek real radio link ke liye expect karte hain. Sign negative hai → 1 W se far below, jaise forecast kiya tha. ✓

Example 2 — Bahut bada input: deep-space path loss (Cell B)

Forecast: Constant $32.45$ fixed hai; kya range term ya frequency term total mein dominate karega?

$f$ ko MHz mein daalo. $8.4 GHz = 8400 MHz$ . Yeh step kyun? $32.45$ constant kilometres aur megahertz ke liye calibrated hai; kuch aur feed karo toh number nonsense hai.
Range term. $20 lo g_{10} (2.5 \times 1 0^{8}) = 20 (8.3979) = 167.96 dB$ . Yeh step kyun? Ek bade number ka $lo g_{10}$ bas uske "digits ki sankhya" hai, toh ek soo-million-km distance sirf ~168 dB hai.
Frequency term. $20 lo g_{10} (8400) = 20 (3.9243) = 78.49 dB$ . Yeh step kyun? Higher frequency ka matlab chhota wavelength hai, aur fixed gain wale receiving antenna ki short wavelengths par smaller effective collecting area hoti hai — isliye wave ka zyada bada fraction miss ho jaata hai. Woh penalty frequency ke saath badhti hai, aur yeh isi $20 lo g_{10}$ term se formula mein enter hoti hai.
Sum. $L_{p} = 32.45 + 167.96 + 78.49 = 278.90 dB$ . Yeh step kyun? Har contribution independent hai, isliye add karte hain.

Verify: Range term (168 dB) frequency term (78 dB) ko dominate karta hai, jaise forecast kiya tha — deep space mein distance hi killer hai. Linear terms mein $1 0^{27.89}$ roughly observable universe ke volume ka ek grain of sand se ratio hai — space sachmuch itna unforgiving hai. Parent ke $278.89$ dB (rounding) se match karta hai. ✓

Figure 2 — Mars ke liye path-loss tower. Teen stacked bars total

L_{p}

build karte hain: ek fixed yellow base (

32.45

dB, unit-conversion constant), ek tall red slab (

167.96

dB range term), aur ek chhota blue slab (

78.49

dB frequency term). Laal range slab blue frequency slab se double se bhi zyada badi hai — yeh ek tasveer hai kyun distance, frequency nahi, deep-space ka dushman hai.

32.45

kahaan se aata hai yeh jaanne ke liye dekho Free-Space Path Loss Derivation.

Example 3 — Bahut chhota input: ek nearby LEO pass (Cell C)

Forecast: Mars (279 dB) ke comparison mein, kitna chhota — aadha? Ek tenth?

Range term. $20 lo g_{10} (500) = 20 (2.6990) = 53.98 dB$ . Yeh step kyun? Same formula — ismein kuch bhi assume nahi kiya ki distance badi hogi, isliye chhote $r$ ke liye bhi identically kaam karta hai.
Frequency term. $20 lo g_{10} (437) = 20 (2.6405) = 52.81 dB$ .
Sum. $L_{p} = 32.45 + 53.98 + 52.81 = 139.24 dB$ .

Verify: $139$ dB versus Mars ka $279$ dB — dB mein loss aadha hai, lekin kyunki dB logarithmic hain, linear power mein woh $1 0^{(279 - 139) /10} = 1 0^{14}$ hai: Mars signal ek soo trillion guna kamzor hai "sirf" double the dB ke bawajood. Yahi Cell C ka sabak hai: chhote dB differences mein enormous linear differences chhupe hote hain. ✓

Example 4 — Degenerate inputs: isotropic antenna, unit bit rate (Cell D)

Forecast: (a) mein, kya ek 0-dBi gain kuch add karta hai? (b) mein, kya ek 1-bit/s rate kuch subtract karta hai?

(a) Power convert karo. $5 W = 10 lo g_{10} 5 = 6.99 dBW$ . Yeh step kyun? Standard pehla move.
(a) Gain add karo. $EIRP = 6.99 + 0 = 6.99 dBW$ . Yeh step kyun? $0$ dBi ka matlab gain factor $= 1$ hai; antenna kuch bhi focus nahi karta, isliye EIRP raw power ke barabar hai. Yahi isotropic ki definition hai — woh degenerate baseline jiske against sab gains measure kiye jaate hain.
(b) Rate in dB. $10 lo g_{10} (1) = 0 dB-Hz$ . Yeh step kyun? $lo g_{10} 1 = 0$ : ek one-bit-per-second link saari power ek hi bit par har second ke liye spread karta hai, kuch subtract nahi karta.
(b) Energy per bit. $E_{b} / N_{0} = 40 - 0 = 40 dB$ .

Verify: Dono degenerate inputs apni term ko zero par collapse kar dete hain, exactly jaise $lo g_{10} (1) = 0$ demand karta hai. Ek slower link ka hamesha higher $E_{b} / N_{0}$ hota hai — yahaan sabse slow possible rate har bit ko poora $C / N_{0}$ deti hai. Consistent. ✓

Example 5 — Woh link jo fail hoti hai: negative margin (Cell E)

Forecast: Margin = available minus required. Kya woh positive hoga ya negative?

Margin definition. $Margin = (E_{b} / N_{0})_{available} - (E_{b} / N_{0})_{required} = 2.3 - 4.0 = - 1.7 dB$ . Yeh step kyun? Budget tabhi "closes" hota hai jab jo aapke paas hai woh jo chahiye use exceed kare; difference safety cushion hai.
Sign interpret karo. Negative margin ka matlab hai signal $1 0^{1.7/10} = 1.48 \times$ zyada weak hai — aap zaroori power se 48% short hain. Yeh step kyun? dB deficit ko linear mein convert karna exactly dikhata hai ki aap kitna miss kiya.
Fix options. $1.7$ dB recover karne ke liye aap (i) bit rate ko $1 0^{0.17} = 1.48$ factor se drop kar sakte ho, ya (ii) EIRP $1.7$ dB raise kar sakte ho, ya (iii) $G / T$ $1.7$ dB improve kar sakte ho. Yeh step kyun? Kyunki sab kuch dB mein additive hai, kahiin ka bhi deficit kahiin bhi usi number of dB se repair kiya ja sakta hai.

Verify: $- 1.7$ dB margin ka matlab BER itna high hai ki lock nahi hoga — receiver noise sunega. $R_{b}$ ko, maano, $2$ Mbps se $2/1.48 = 1.35$ Mbps tak reduce karo aur term $- 10 lo g_{10} (R_{b})$ $1.7$ dB badh jaati hai, margin ko $0$ par push karti hai. Signs aur remedies consistent. ✓ Forward Error Correction dekho ki kaise coding required threshold ko neeche laata hai.

Example 6 — Limiting behaviour: range aur frequency double karna (Cell F)

Forecast: Aage padhne se pehle har ek ke liye "3 dB" ya "6 dB" guess karo.

Doubling ka rule. Kisi bhi $20 lo g_{10}$ term ke liye, argument double karne se $20 lo g_{10} (2) = 20 (0.3010) = 6.02 dB$ add hota hai. Yeh step kyun? Kyunki $lo g_{10} (2 x) = lo g_{10} x + lo g_{10} 2$ ; constant $lo g_{10} 2 \approx 0.301$ yaad rakhne layak hai.
(a) Range double karo. $L_{p}$ $6.02 dB$ badhta hai. Yeh step kyun? $r$ ek $20 lo g_{10}$ ke andar hai, isliye uski doubling ek "6-dB brick" cost karti hai.
(b) Frequency double karo. $L_{p}$ bhi $6.02 dB$ badhta hai. Yeh step kyun? $f$ bhi ek $20 lo g_{10}$ ke andar hai — identical geometry.

Verify: Example 3 vs ek hypothetical $1000$ -km pass check karo: range term hogi $20 lo g_{10} (1000) = 60.00$ dB, $53.98$ dB se upar — $6.02$ dB ki rise. Exactly ek brick. ✓

Figure 3 — Path loss har doubling mein 6 dB badhta hai. Neeli curve

L_{p}

ko distance ke against logarithmic horizontal axis par plot karta hai, jo inverse-square law ko ek seedhi rising line mein badal deta hai. Do peele dots

500

km aur

1000

km par hain; unke beech step karna — ek doubling — line ko exactly

6.02

dB (peela arrow) oopar uthata hai. Akaila laal dot kaafi door right mein Mars hai

\sim 279

dB par. Yeh Fris Transmission Equation ek tasveer ke roop mein hai.

Example 7 — Real-world word problem: data rate choose karo (Cell G)

Forecast: Deep-space rates famously slow hoti hain — kilobits? Megabits?

EIRP. $P_{t} = 10 lo g_{10} (15) = 11.76 dBW$ , toh $EIRP = 11.76 + 22 = 33.76 dBW$ . Yeh step kyun? Same opening move; transmit side ek number mein collapse ho jaati hai.
Carrier-to-noise density. $\frac{C}{N _{0}} = 33.76 - 278.89 + 49.02 + 228.6 = 32.49 dB-Hz .$ Yeh step kyun? Yeh master equation hai — woh sab kuch jo receiver ke paas hai isse pehle ki hum decide karein kitni fast spend karni hai. ( $+ 228.6$ woh $- 10 lo g_{10} k$ term hai; negative subtract karna add karta hai.)
Rate ke liye solve karo. $E_{b} / N_{0} = C / N_{0} - 10 lo g_{10} (R_{b})$ . $E_{b} / N_{0} = 4.0$ set karke: $10 lo g_{10} (R_{b}) = 32.49 - 4.0 = 28.49 dB-Hz .$ Yeh step kyun? Maximum rate par hum exactly threshold energy per bit spend karte hain — margin zero.
Log undo karo. $R_{b} = 1 0^{2.849} = 706 bit/s$ . Yeh step kyun? $1 0^{(\cdot) /10}$ $10 lo g_{10}$ ka inverse hai, dB-Hz ko wapas bits per second mein badalta hai.

Verify: $\approx 706$ bit/s — sach mein high-gain antennas aur coding se pehle early Mars downlinks ka ballpark. Wapas plug karo: $10 lo g_{10} (706) = 28.49$ , toh $E_{b} / N_{0} = 32.49 - 28.49 = 4.0$ dB exactly threshold par. ✓ Faster rates ke liye better Modulation Schemes (BPSK, QPSK) ya Deep Space Network (DSN) ke 70-m dishes chahiye.

Example 8 — Exam-style twist: units aur Shannon ceiling (Cell H)

Forecast: Kya true bit rate Shannon capacity se kabhi exceed kar sakti hai? (Nahi — yahi toh baat hai.)

(a) Trap pakdo. Student ne ek raw hertz count ( $1000$ ) ko ek decibel quantity ( $32.49$ dB-Hz) se subtract kiya. Aap sirf dB ko dB se subtract kar sakte ho, isliye bandwidth ko pehle decibels mein convert karna zaroori hai. Yeh step kyun? $C / N_{0}$ ek ratio-per-hertz hai dB mein; jis noise se ise divide karna hai woh $N = N_{0} B$ hai, aur dB mein divide karne ka matlab hai $B$ ki dB value subtract karna — kabhi bhi raw Hz number nahi.
(a) Bandwidth convert karo. $B = 1000 Hz \to 10 lo g_{10} (1000) = 30 dB-Hz$ . Yeh step kyun? Ab dono quantities ek hi dB ruler par hain aur legally subtract kiye ja sakte hain.
(a) Correct SNR. $SNR = C / N_{0} - B_{dB} = 32.49 - 30 = 2.49 dB$ . Yeh step kyun? SNR $C / N$ hai jahaan $N = N_{0} B$ ; dB mein woh exactly $C / N_{0}$ minus $B$ -in-dB hai. (Student ka $- 967.5$ ka answer nonsense tha; real SNR ek modest $+ 2.49$ dB hai.)
(b) Linear SNR. $SNR_{lin} = 1 0^{2.49/10} = 1.774$ . Yeh step kyun? Shannon-Hartley Theorem apne logarithm ke andar ek linear ratio use karta hai, isliye humein pehle dB chhodni padegi.
(b) Capacity. $C_{shan} = B lo g_{2} (1 + SNR_{lin}) = 1000 lo g_{2} (2.774) = 1000 (1.472) = 1472 bit/s$ . Yeh step kyun? Shannon ka law SNR aur bandwidth given hone par error-free bits ki absolute ceiling set karta hai.

Verify: $706 bit/s < 1472 bit/s$ — hamaari practical rate Shannon ceiling se comfortably neeche hai, jaise physics demand karta hai. $\approx 2 \times$ gap exactly woh "coding gap" hai jise close karne ke liye Forward Error Correction exist karta hai. ✓

Recall Self-test: kya tum bottom line tak pahunch sakte ho?

Ek dB kisi ratio ke kaun se function ka das guna hai? ::: $lo g_{10}$ of the ratio. Range ya frequency mein se kisi ko bhi double karne se kitne dB ka path loss add hota hai? ::: About $6.02$ dB. $- 1.7$ dB ka margin matlab signal kitni baar zyada weak hai (linear)? ::: $1 0^{0.17} \approx 1.48 \times$ too weak. $C / N_{0}$ (dB-Hz) se $E_{b} / N_{0}$ (dB) jaane ke liye aap kya subtract karte ho? ::: $10 lo g_{10} (R_{b})$ , the bit rate in dB. $C / N_{0}$ se subtract karne se pehle bandwidth ko Hz se dB-Hz mein kyun convert karna zaroori hai? ::: Sirf dB ko dB ke saath add/subtract kar sakte hain — raw Hz aur dB mix karna meaningless hai. $0$ -dBi antenna EIRP mein kitna contribute karta hai? ::: Kuch nahi — yeh isotropic baseline hai, gain factor 1. $G / T$ mein $T$ kya stand karta hai, aur kya bada better hai? ::: System noise temperature in kelvins; smaller is better (kam receiver hiss).