Wireless Communications Lecturer: Eng. MTELASTAL
Chapter3: Homework
2011-2012
Q1. Consider an RX that consists (in this sequence) of the following components: (i) an
antenna connector and feed line with an attenuation of 1.5 dB; (ii) a low-noise amplifier
with a noise figure of 4 dB and a gain of 10 dB, and a unit gain mixer with a noise figure
of 1 dB. What is the noise figure of the RX?
Atten. 1.5dB=>m=0.708 / F2=4dB=2.5 / G2=10dB=10 / F3=1.26 / G3=1
𝐹𝑒𝑞 =1
0.708 +2.5 −1
1+1.26 −1
1∗10 = 2.93 = 4.7𝑑𝐵
2. Consider a system with 0.1-mW transmit power, unit gain for the transmit and receive
antennas, operating at 50-MHz carrier frequency with 100-kHz bandwidth. What is the
receive SNR at a 100-m distance, assuming free-space propagation? How does the SNR
change when changing the carrier frequency to 500MHz and 5 GHz? Why does the 5-GHz
system show a significantly lower SNR (assume the RX noise figure is 5 dB independent
of frequency)?
SNR=Ptx+Gtx+Grx-[losses]tx+rx-[pathloss]-Pnoise
𝜆=𝑐
𝑓= 6𝑚
− pathloss =10log10 λ
4πd 2
=10log10 6
4π∗ 100 2
=−46.42dB
Pnoise=-174+10log10 (BW)+F=-119dBm (ignoring man-made noise)
SNR=-10dBm+0dB+0dB -46.42+119=62.58dB (ignoring man-made noise)
Do it for 500MHz and for 5GHz.