Consider a receiver with a noise figure of 6 db and a bandwidth of 2.1 GHz operating at room temperature. The input 1-dB compression point is 10.5 dbm and the detector at receiver output requires a minimum SNR of 7 db. Calculate the Minimum Detectable Signal (MDS) and input power Dynamic Range (DR). NF=6 db, B=2.1 GHz, SNR min =7dB T=290 k, P in,1db = 10.5 dbm Noise floor=10 log(ktb) +NF=-174 dbm/hz+10log(b)+nf= -74.8 dbm MDS=Noise floor+snr min =-72.8 dbm+7 db=-65.8 dbm The dynamic range can be defined as the difference between the input signal level that causes a 1-dB compression gain and the minimum input signal level that can be detected above the noise level. Note that MDS is also known as the receiver sensitivity! page 1 DR=P in,1db -MDS=10.5 dbm+65.8=76.3 db
A microwave receiver has an equivalent noise temperature of 45 K and is fed from an antenna with an input noise temperature of 35 K. The receiver operates over a 5 MHz bandwidth. (a) Calculate the noise power at the input of the receiver. (b) Calculate the minimum input signal power for good reception if the minimum SNR is 100:1. (c) If the receiver has a dynamic range of 81 db and sensitivity of 0.55 nw, what is the maximum allowable input signal? T e = 45 k, T in = 35 k, B = 5MHz (a) N in = k T e + T in B = 5.52 fw (b) SNR min =100 S in,min = N in SNR min = 552 fw (c) DR= S in,max senesitvity S in,max = 81 db + 10 log 0.55 10 9 WW 1 10 3 WW = 18.4 dbm page 2
An RF input signal at 1.9 GHz is down-converted with a mixer to an IF frequency of 80 MHz. Calculate the possible LO frequencies and the corresponding image frequencies? f RF = 1.9 GHz, f IF = 80 MHz f LO = f RF ± f IF = 1.98 GHz, 1.82 GHz Image frequency for f LO =1.98 GHz (f LO > f RF ): f IM = f LO + f IF =2.06 GHz Image frequency for f LO =1.82 GHz (f LO < f RF ): f IM = f LO f IF =1.74 GHz page 3
A receiver consists of three elements: a preamplifier, a mixer and an IF amplifier with noise figures 3 db, 6 db and 10 db, respectively. a) If the total gain of the receiver is 30 db and the IF amplifier has a gain of 10 db, what is the minimum gain of the preamplifier to achieve a total maximum noise figure of 5 db. b) b) Now, what would the overall noise figure become if the noise figure of the IF amplifier is increased to 20 db? page 4 Model thereceiver as follows: F 1 =2, F 2 =4, F 3 =10, G tot = G 1 G 2 G 3 =30dB=1000, G 3 =10dB=10 G 1 G 2 = 100 F tot = F 1 + F 2 1 + F 3 1 =5dB= 3.2 G 1 G 1 G 2 F tot = 2 + 4 1 G 1 b) Increase F 3 to 20dB = 100 F tot = 2 + 4 1 + (10 1) 100 G 1,min=2.7 =4.3 db + (100 1) 2.7 100 Stage 1 Stage 2 Stage 3 Pre-amp IF-amp F 1 LO G 1 G 3 = 4.1 = 6.1dB, which is a very small increase as the overall noise figure is dominated by the noise figures of early stages! F2 F 3 G2
A GSM cellular telephone system operates at a downlink frequency of 935 960 MHz, with a channel bandwidth of 200 khz, and a base station that transmits with an EIRP of 20 W. The mobile receiver has an antenna with a gain of 0 dbi and a noise temperature of 450 K, and the receiver has a noise figure of 8 db. Find the maximum operating range if the required minimum SNR at the output of the receiver is 10 db, and a link margin of 30 db is required to account for propagation into vehicles, buildings, and urban areas. f 0 = 947.5 MHz λ=0.32m, EIRP=P t G t = 20 W, G r = 0 dbi = 1 T in = 450 K, NF = 8 db F = 6.31, SNR o,min = 10 db, B ch = 200 khz LM = 30 db page 5 T tot = T in + T e = 450 K + 290 K F 1 =1990 K N o = kt tot B ch =-112.6 dbm LM=SNR o SNR o,min S o = LM + SNR o,min + N o =-72.6 dbm= 55 pw R = P tg t G r λ 2 4π 2 S o =15.35 km
Consider the following dual-conversion super heterodyne receiver: Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 RF Filter1 LNA RF Filter2 IF Filter1 IF amp1 IF Filter2 IF amp2 LO 1 LO 2 Gain (db) NF (db) IP3 in (dbm) -2.5 12.0-3.0-6.0-2.5 20.0 18.0-3.0 60.0 2.0 12.0 3.0 12.0 20.0 10.0 16.0 12.0 26.0 a) Calculate the overall noise figure of the receiver chain and additional noise figure contributions from different stages. b) Calculate the overall 3 rd order intercept point IP3 tot page 6
Stage 1 Stage 2 Stage 3 Stage 4 Stage 5 Stage 6 Stage 7 Stage 8 Stage 9 RF Filter1 LNA RF Filter2 IF Filter1 IF amp1 IF Filter2 IF amp2 LO 1 LO 2 Gain (db) NF (db) IP3 in (dbm) a) -2.5 12.0-3.0-6.0-2.5 20.0 18.0-3.0 60.0 2.0 12.0 3.0 12.0 20.0 10.0 16.0 12.0 26.0 F tot = F 1 + F 2 1 G 1 + F 3 1 G 1 G 2 + F 4 1 G 1 G 2 G 3 + + F 9 1 G 1 G 2 G 3 G 4 G 5 G 6 G 7 G 8 =1.78+ 1.58 1 0.56 + 2 1 0.56 15.85 + 15.84 1 0.56 15.85 0.5 + 1.78 1 0.56 15.85 0.5 0.25 + 2 1 + 15.84 1 + 2 1 + 0.56 15.85 0.5 0.25 0.56 0.56 15.85 0.5 0.25 0.56 100 0.56 15.85 0.5 0.25 0.56 100 63.1 100 1 0.56 15.85 0.5 0.25 0.56 100 63.1 0.5 page 7 = 1.78 + 1.04 + 0.11 + 3.34 + 0.7 + 1.6 + 0.24 + 0.00025 + 0.05 = 8.86 = 9.47 db
F add,1 = 1.78 = 2.5 db, F add,2 = 2.82 1.78 = 1.99 db, F add,3 = 2.93 = 0.17 db 2.82 F add,4 = 6.27 2.93 = 3.3 db, F add,5 = 6.97 6.27 = 0.46 db, F add,6 = 8.57 = 0.9 db 6.97 F add,7 = 8.81 8.57 = 0.12 db, F add,8 = 8.81 8.81 = 0 db, F add,9 = 8.86 = 0.02 db 8,81 b) IP3 in,tot = 1 1 + G 1 + G 1G 2 + + G 1G 2 G 3 G 4 G 5 G 6 G 7 G 8 IP3 in, 1 IP3 in, 2 IP3 in, 3 IP3 in, 9 IP3 in, 1 = IP3 in, 3 = IP3 in, 5 = IP3 in, 8 = since the corresponding elements are passive! IP3 in,tot = 1 =2.73 mw=4.36 dbm 0.056 +0.112+ 0.040 +0.158 G tot =-2.5dB+12.0dB-3.0dB-6.0dB-2.5dB+20.0dB+18.0dB-3.0dB+60.0dB=90 db IP3 out,tot = IP3 in,tot + G tot = 97.36 dbm page 8