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COE 434 SOLUTION for version 2 Quiz 1 distancetop to

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Student Name: Student ID:
↑
Question 1 [4 points]: State whether the following statements are True or False.
1. The wavelength of a radio signal is the distance covered by the signal in one second.⑤ #False False
2. InfraRed (IR) signal cannot penetrate through a solid wall.② =>
True true
3. =>
The gain of an isotropic antenna is 1 dBi. False false it's o

4. The Doppler frequency shift at the receiver will be zero if the transmitter and the receiver are
not moving relative to each other.&True true

Question 2 [16 points]: Pr f

A base station is transmitting +20 dBm power at a frequency of 1.2 GHz. A mobile unit is present
d
Seh
at a distance of 1 Km from the base station. The mobile unit receiver sensitivity is -90 dBm.
- Gr
The
6
= 0

base station uses isotropic antenna, while the mobile unit uses an antenna with a gain of 2 dBi.
- -

Ignore cable and connector losses.

(a) Calculate the power received (in Watts and dBm) at the mobile unit.
-

(b) Find the fade margin (in dBs), if any, present in the system.
-

Note: c2 ex Sen-Pr
Pr = P Gt.Gr margin =

out in
t
(4fd ) 2

f= frequency (Hz); Pt = Transmitted power (Watts); Pr = Received power (Watts);

Gt= Transmitter Antenna Gain; Gr= Receiver Antenna Gain; d = distance between antennas (m);
c = speed of light (= 3 * 108 m/s)

Pt = 20 dBm = 10.log10(Pt (in W) / 10-3) → Pt (in W) = 102 * 10-3 W= 0.1 W ~

Gt= 1 (isotropic antenna) ~
Gr (in dBi)= 2 dBi = 10. log10(Gr ) → Gr = 100.2 = 1.585 ~
Plugging in the values in the above Friis formula, and calculating..

Received Power (in Watts) = Pr = 6.273 *10-11 Watts -

Received Power (in dBms) = Pr (in dBms)= 10.log10(Pr (in W)/10-3) = -72.03dBm

Fade Margin = Received Power (dBms) – Rx Sensitivity (dBms)

Fade Margin = -72.03 –(-90) = 17.97 dBs ~

log(s) 1
w = room w
0
20 = 10 => Pe = .

( Gt) = 1
log G =
0 = 10

10
log(6r) => Gr 1 585
(0x12x109)
2 2= = .

Po =
0 .
1
x 1 985 x
.

1
W
-

2
= 62 .
73x10
-
o
Pr(dBm) = 10
log() = -

72 025 .
dBm

tade 72 013 17 98. dB

margin
=
90
- .

(Sem-Pr)
COE 434 SOLUTION for version 1 Quiz 1
Student Name: Student ID:
Question 1 [4 points]: State whether the following statements are True or False.
1. The wavelength of a radio signal is the distance covered by the signal in one second.⑤ False
2. InfraRed (IR) signal cannot penetrate through a solid wall.= True
3. The gain of an isotropic antenna is 1 dBi. =
False
4. The Doppler frequency shift at the receiver will be zero if the transmitter and the receiver are
not moving relative to each other.-True

Question 2 [16 points]: -find Po

-x
109

W
A base station is transmitting +25 dBm power at a frequency of 1.2 GHz. A mobile unit is present
Cry
at a distance of 1 Km from the base station. The mobile unit receiver sensitivity is -90 dBm. The
C
aBitO
base station uses isotropic antenna, while the mobile unit uses an antenna with a gain ofO5 dBi.
Ignore cable and connector losses.
(a) Calculate the power received (in Watts and dBm) at the mobile unit.
(b) Find the fade margin (in dBs), if any, present in the system.
Note: c2 othis
D
Pr = Pt
(4fd ) 2
Gt.Gr
-
f= frequency (Hz); Pt = Transmitted power (Watts); Pr = Received power (Watts);
Gt= Transmitter Antenna Gain; Gr= Receiver Antenna Gain; d = distance between antennas (m);
c = speed of light (= 3 * 108 m/s)

Pt = 25 dBm = 10.log10(Pt (in W) / 10-3) → Pt (in W) = 102.5 * 10-3 W= 316.23* 10-3 W

a)
Gt= 1 (isotropic antenna)
find
Gr (in dBi)= 5 dBi = 10. log10(GPR
(calculate
r ) → Gr = 10
Pr3.163
0.5 = 6+
,
62)
,

Plugging in the values in above Friis formula, and calculating..

Received Power (in Watts) = Pr = 3.96 *10-10 Watts

b)
Received Power (in dBms) = Pr (in dBms)=
90 - Pr
This is fade
-3 margin
C 10.log10(Pr (in W)/10 ) = -64.02dBm
=

Fade Margin = Received Power (dBms) – Rx Sensitivity (dBms)

Fade Margin = -64.02 –(-90) = 25.98 dBs