Example Question Fe

Example Question

1. Noise:

A receiver with a 75Ω input resistance operates at a temperature of 31°C. The received signal is at 89MHz with a bandwidth of 6MHz. The received signal voltage of 8.3µV is applied to an amplifier with a noise figure of 3 dB.

1. Calculate PNOISE and PSIGNAL                                        [3 marks]

1. If an increase of output performance of at least 25% is desired, calculate the noise temperature of the amplifier. [7 marks]

Answer:

Tc = 273+31 = 304K

Vn = Sqrt(4kTBR) = sqrt(4 x1.38x10E-23 x 304 x 6x 10E6 x 75) = 2.75μV        

Pnoise = Vn^2/R  = 0.1pW        

Psignal = Vs^2/R = 0.918 pW        

SNRIN = Psignal/Pnoise = 9.18        

SNRdB = 10 log 9.18 = 9.63 dB        

NF = 10 log F

3 dB = 10 log F

F = 2                

SNROUT = SNRIN / F

= 9.18/2 = 4.59        

And increase of 25% performance

SNROUT increase by 25% or noise factor decrease by 25%                

Choose noise factor decrease by 25%

2 – (25%) = 2 x 0.75 = 1.5                

To find the temperature of amplifier substitute into TN = 290(F – 1)

So TN = 290(1.5 – 1) = 145 K or -128°C                

1. TDM/FDM:

Table 1 shows GSM frequency bands, while Figure 1 illustrates GSM timeslots and frame structure.

1. What is the multiplexing technique used in GSM? Justify your answer.                                                                                                         [2 marks]
2. Determine average data rate.                                                        [2 marks]

1.  Particularly for the P-GSM 900 system, what is the bandwidth allocation for each user?                                                                         [1 mark]

Table 1