Exam text content

ELT-45106 : RF Eguipment for Wireless Networks Exam 16.12.2016
Compiler: Ari Asp Answer to all guestions!

Calculators are allowed in this exam. Remember the kaiku feedback

1 Explain shortly diversity methods

2: Explain the two reasons why 3. order intermodulation products are much more difficult to

handle than 2. order intermodulation products for receiver.

35 Using the values given in Figure 1, calculate the noise power at the receiver output, after the 2.
amplifier. If the cable loss increases 2 dB ( from 4 > 6 dB ), is it possible to compensate by
changing parameters of amplifiers? What would be new values of those parameters?

ANTENNA LNA MIXER — IFFILTER — AMPLIFIER
cable Amplifer = NP=6dB

      
  

 

cable loss
=4dB G=15dB
Te=150K

  

Filterloss — G=17dB
=4dB NF=6dB

Figure 1. Reveiver structure

4. Calculate the EIRP at each antenna for the DAS (Distributed Antenna System) given below. The

 

 

 

 

 

 

 

power level at the transmitter is 30 dBm. The manufacturer of the feeder (1/2”) promises a loss of
6.0. dB/100m at the carrier freguency. Gain of antenna 1 = 1 dBi, and for antenna 2,3 and 4 Gain =2
dBi.
1.ant 2.ant 3.ant —4.ant pyy
rytmi fyyzai [20
10 | 20m
|
4-way
45m i splitter : —30m
sn
Tx | 30d8m

 

5. Explain following figure. What kind of systems are behind those curves and what are the main
differences between those systems? Explain structure principles of those systems.
10— T T T

 

   

normalizeci output SNR, dB

 

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NOTE: Formulas are in the following page if you need them.
 

 

Some (more or less) useful eguations for ELT-45106 exam:

 

 

 

       

 

 

 

2
5 2
a-a-ka n y,= [19BR = =; v? (Varrar) at
” n Pa oho i 32aR. 4R
as. < n 10109, AV
» n» Pi P2 —5Blr-2'00s6) — Br 10
€ E jpx'cosO 71 1mW
=|41(z)—— d'= I 3 e! dz
10805 = 1080 G, +108 am 00. s W
Di a BoIä 1900
1010g, 2=— Y=t- >1 26. 26
J p f aN =1-h Joh CNR;,
2. 15 CNRout
4 R=0.62/7/4
divD(r,?)= p(r,t)
0 dBd=2.15 dBi R,=R+R = divB(r,f)=0
: oB
hh = 6.546:10* Jsec, Planck”s constant 4 curl E(r,t)=- (n)
c=299792 458 m/s, speed of light Vom 140 Z,=R,+jX, o
-23 (01 2, E
k=1.38-10 J/*K, Boltsmanns constant Vain 1-1F] D0log,y[rj as — culH(r, "= 2 Dn yle 2
E-1 E-1
F,y=K+2234% F(SSL =
di 0 SSI, 0 F(0,9)=g(0,9)-7(0.9)
Ta +814
G GG, V(z)=Wre*+V7 eh
= Fis < dP
T,=T,(F-1)&F=1+- masse == G=g,D =
e 2 ) TE P- a z U Ja
NF=10-10g(f) =
I(2)=1(0)sin 4(2-14)] AF=), Ae" y=PBdcos(0)+a
47U (0.9) 2 =
00 : = ;
= P=-2PR, AP =1,e*+1e% 417,6 4...] 07
2

Nu n=k-T-B-Fi'Gr
; , in(Nv/2 Ny/2
Nu ps =k:T-B-Fys Ap = 4,100 Sia v/2) Sv <= v)

sin(y/2 in(y/2)
Na v (v/2)
= rnln
Ppin)=" == joo 2 (7) =Pr[T<7]=Pr[maxfr,<7)]
w
0 =Pr[TuTa:-Ty S7]=(1-0"")

=Pt[PuDas-T <7]=112;(7)