Question

Q.7. Assume a radar receiver with gain of 100, has an IF bandwidth of 4GHz, and a noise figure [10M] of 6.3, transmitting 31.62KW of power at a frequency of 8 GHz, for a target cross section of 8m². (Assume TO = 290 degrees kelvin., and k= 1.38 x10-23) A. Convert radar equation into decibel e radar equation, applying decimal log on both sides and find the new equation for range, [4M] B. Convert all the values into decibel and use it in the above expression to find maximum range. [3M) C. Calculate the Maximum range without decibel rule, (with Absolute equation) and compare the value with Answer of part-B. Justify the reasons for change in value. [3M]

Answer 1

1 Q7J/ANS! By assurning a radar receiver With gains of too, has If bandwidth figure of 6:3, transmittin 3162 kW of power at frequency of 8 ghz- torget croffection of 8m? of u ghe f noise for a C Assume To = agok f k = 188x152: A) As we know that raday ra range equation without Noise effect is given. 1114 Lчта Р as: Rmaxe- TPE her Rmax maam antenna of rece inge bw transmitting receining fto transmitting power by trans antenna. -multing antenna. Ae effective apperture area of both transmilting & receiving antenna - Radar Cross sectional area. Aanned Zvith mavelength of transmitted signal Panserger Power received by an antenna CS

N.P CF-1) * By takeling Nose effect on radar range equation : Куди Геке un x ² Por Pr= Fil figure of messe F SNRi was e Nobic figure ENRO NIP & Noise power Kg Boltsman constant to tempr. im ok. AW B IF, Bandwidth \ Ne=КТІВ V PRE KTOB (F-1) . * Rmax - Potet 114 Rmax يلى I 20 my? KTIBLF-1) Podela UDC 1? KTB CF-4 Re delt un A² KIT B (F-1) 2olog lly Roman de DB = Scanned with CamScanner 20 4 log CS

Rmax = Petelt Slog dB ung? KTB (F-1) Romax = 5log (ft Ae$} – loy [uay?KTBCF- Romax dB 4 Romex dB = 4 res 2o long (Ptfe$t) 20log (urad Ponede-CuepalkT805-0) Romanes=45 A - B] de {A= P6 Acer B= 40X KIB (F-1)} Zolog (PoAcht) = 20log (3162 AdB= 129 ) Aest m. Ae=A? un g sa = 1 = 3x68 3 8809 Ae= gana un 26400 2560 1 Scannea With Pt Ae² = 31.624639 CamScanner g=10 g & m? 2x 8 = 31.688632g 256 n 32T

* Ae= g for A= Potely 256n А= 21.62x5x Pix 3{: 6773 (256 и 12 Ado= о!g — Qola C31-67241) = о 0 AdB=30 dB 2 = Чил“ҚТВ (F-1) считех •3g xus 25 х 24р oxчx юx (61-) ВЧОО — Я xi:38 xиx 2 X S3Xe% = 14492x612 Ч (Ч 0 9 28.10) = -36 че е В = - 236 чR Jp Ч B=2log (l % Клод = ЧГА-В) 4 (3v-(-234) 244 eat g = 66 67 Promock = 66.6206 – 266. ЧR 22, CS Scalinen damScanner

*Rmax=6662dB = 66-62 dB without decibel rule 7 © Maxmi range Rmanitas WALKTI(F-U 'Ae= 9 256ū =lomax Ae = 0.0111g Re: 3162X603x(@oing 28 x6400 un x 9 x 138 46 234 298 X Y ALO (6-3-1) max Rmax 3162* 1.2588864860t 367181138x280X4853 vol4 114 Contamos con el restauro's) m (0:0210899XL0 1539, (21.0p9g ub'23" Rmax man = 2 142.9825 mande et doloy (2142-9825) = 66.6208 cs game has of part f. - 2.1429 km