Q2. A plot of dB Power (10log10Pr) vs Log Distance (log10d)

%Two-Ray Model %Plotting dB Power vs log distance

clear all; f=9*10^8; c=3*10^8; R=-1; ht=50; hr=2; Gl=1; Gr=[1,.316,.1,.01]; d=1:100000; db=10.*log10(d);

counter=1; figure(1);

lambda = c/f; l=sqrt((ht-hr)^2.+d.^2); r=sqrt((ht+hr)^2.+d.^2); dc=4*ht*hr/lambda; dcrt=dc:1:100000; ang=2.*pi.*(r-l)./lambda; s1=(lambda/(4*pi))^2;

for counter=1:1:4 GR=Gr(counter);

Pr=s1.*(abs((sqrt(Gl))./l)+(R.*(sqrt(GR))./r.*exp((sqrt(-1)).*ang))).^2;

subplot(2,2,counter);

plot(db,10*log10(Pr)-10*log10(Pr(1)));

pd1=1:1:dc; pd2=dc:1:100000;

hold on;

plot(10*log10(pd1),-20*log10(pd1),'r');

plot(10*log10(pd2),-40*log10(pd2),'g'); end hold off

Q2. Find the median path loss under the Hata model assuming fc = 900 MHz, ht = 20m, hr =

5m and d = 100m for a large urban city, a small urban city, a suburb, and a rural area.

Explain qualitatively the path loss differences for these 4 environments.

Ans: fc = 900 MHz, ht = 20m, hr = 5m d = 100m

𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = 69.55 + 26.16 log10(𝑓𝑐) − 13.82 log10(ℎ𝑡𝑒) − 𝑎(ℎ𝑟𝑒) + (44.9

− 6.55 log10(ℎ𝑡𝑒)) log10(𝑑)

i. Large Urban City

𝑎(ℎ𝑟) = 3.2(log10(11.75ℎ𝑟))2 − 4.97𝑑𝐵

𝑎(ℎ𝑟) = 3.2(log10(11.75 × 5))2 − 4.97𝑑𝐵

𝑎(ℎ𝑟) = 6.35𝑑𝐵

𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = 69.55 + 26.16 log10(9 × 108) − 13.82 log10(20) − 6.35𝑑𝐵 + (44.9− 6.55 log10(20)) log10(100)

𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = (69.55 + 234.24 − 17.98 − 6.35 + 72.76)𝑑𝐵

𝑳𝟓𝟎, 𝒖𝒓𝒃𝒂𝒏(𝒅𝑩) = 𝟑𝟓𝟐. 𝟐𝟐𝒅𝑩

ii. Small Urban City

𝑎(ℎ𝑟) = (1.1 log10(9 × 108) − 0.7)(5) − (1.56 log10(9 × 108) − 0.8)𝑑𝐵 𝑎(ℎ𝑟) = (45.75 − 13.17)𝑑𝐵

𝑎(ℎ𝑟) = 32.58𝑑𝐵

𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = 69.55 + 26.16 log10(9 × 108) − 13.82 log10(20) − 32.58𝑑𝐵 + (44.9− 6.55 log10(20)) log10(100)

𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = (69.55 + 234.24 − 17.98 − 32.58 + 72.76)𝑑𝐵

𝑳𝟓𝟎, 𝒖𝒓𝒃𝒂𝒏(𝒅𝑩) = 𝟑𝟐𝟓. 𝟗𝟗𝒅𝑩

iii. Suburban

𝐿50, 𝑠𝑢𝑏𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = 𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) − 2 [log10 (𝑓𝑐

28)]

2

− 5.4

𝐿50, 𝑠𝑢𝑏𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = 325.99𝑑𝐵 − 2 [log10 (9 × 108

28)]

2

− 5.4

𝐿50, 𝑠𝑢𝑏𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) = (325.99 − 112.71 − 5.4)𝑑𝐵

𝑳𝟓𝟎, 𝒔𝒖𝒃𝒖𝒓𝒃𝒂𝒏(𝒅𝑩) = 𝟐𝟎𝟕. 𝟖𝟕𝒅𝑩

iv. Rural

𝐿50, 𝑟𝑢𝑟𝑎𝑙(𝑑𝐵) = 𝐿50, 𝑢𝑟𝑏𝑎𝑛(𝑑𝐵) − 4.78[log10 𝑓𝑐]2 + 18.33 log10 𝑓𝑐 − 𝐾

𝐿50, 𝑟𝑢𝑟𝑎𝑙(𝑑𝐵) = 325.99𝑑𝐵 − 4.78[log10 𝑓𝑐]2 + 18.33 log10 𝑓𝑐 − 𝐾

𝐾 = 35.94

𝐿50, 𝑟𝑢𝑟𝑎𝑙(𝑑𝐵) = 325.99𝑑𝐵 − 4.78[log10(9 × 108)]2 + 18.33 log10(9 × 108) − 35.94

𝐿50, 𝑟𝑢𝑟𝑎𝑙(𝑑𝐵) = 325.99𝑑𝐵 − 383.25 + 164.13 − 35.94 𝑳𝟓𝟎, 𝒓𝒖𝒓𝒂𝒍(𝒅𝑩) = 𝟕𝟎. 𝟗𝟑𝒅𝑩

Q3: The following table lists a set of empirical path loss measurements.

Distance from Transmitter Pr/Pt

5 m -60 dB

25 m -80 dB

65 m -105 dB

110 m -115 dB

400 m -135 dB

1000 m -150 dB

(a) Find the parameters of a simplified path loss model plus log normal shadowing that

best fit this data.

(b) Find the path loss at 2 Km based on this model.

(c) Find the outage probability at a distance d assuming the received power at d due to

path loss alone is 10 dB above the required power for nonoutage.

Ans:

a). Finding the path loss component 𝜸

𝐹(𝛾, 𝐾) = ∑[𝑀𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑(𝑑𝑖) − 𝑀𝑚𝑜𝑑𝑒𝑙(𝑑𝑖)]2

6

𝑖=1

𝑀𝑚𝑜𝑑𝑒𝑙(𝑑𝑖) = 𝐾 − 10𝛾 log10 𝑑 𝐹(𝛾, 𝐾) = (−60 − 𝐾 + 10𝛾 log10 5) 2 + (−80 − 𝐾 + 10𝛾 log10 25) 2

+ (−105 − 𝐾 + 10𝛾 log10 65) 2 + (−115 − 𝐾 + 10𝛾 log10 110) 2 + (−135 − 𝐾 + 10𝛾 log10 400) 2 + (−150 − 𝐾 + 10𝛾 log10 1000) 2

𝐹(𝛾, 𝐾) = (−60 − 𝐾 + 6.9897𝛾)2 + (−80 − 𝐾 + 13.9794𝛾)2 + (−105 − 𝐾 + 18.1291𝛾)2

+ (−115 − 𝐾 + 20.4139𝛾)2 + (−135 − 𝐾 + 26.0206𝛾)2 + (−150 − 𝐾 + 30𝛾)2 Section A of the above equation (−60 − 𝐾 + 6.9897𝛾)2

= 3600 + 60𝐾 − 419.382𝛾 + 60𝐾 + 𝐾2 − 6.9897𝛾𝐾 − 419.382𝛾 − 6.9897𝛾𝐾+ 48.8559𝛾2

(−60 − 𝐾 + 6.9897𝛾)2 = 3600 + 120𝐾 − 838.764𝛾 + 𝐾2 − 13.9794𝛾𝐾 + 48.8559𝛾2 Section B

(−80 − 𝐾 + 13.9794𝛾)2

= 6400 + 80𝐾 − 1118.352𝛾 + 80𝐾 + 𝐾2 − 13.9794𝛾𝐾 − 1118.352𝛾 − 13.9794𝛾𝐾+ 195.4236𝛾2

(−80 − 𝐾 + 13.9794𝛾)2 = 6400 + 160𝐾 − 2236.704𝛾 + 𝐾2 − 27.9588𝛾𝐾 + 195.4236𝛾2 Section C (−105 − 𝐾 + 18.1291𝛾)2

= 11025 + 105𝐾 − 1903.5555𝛾 + 105𝐾 + 𝐾2 − 18.1291𝛾𝐾 − 1903.5555𝛾− 18.1291𝛾𝐾 + 328.6643𝛾2

(−105 − 𝐾 + 18.1291𝛾)2 = 11025 + 210𝐾 − 3807.111𝛾 + 𝐾2 − 36.2582𝛾𝐾 + 328.6643𝛾2 Section D (−115 − 𝐾 + 20.4139𝛾)2

= 13225 + 115𝐾 − 2347.5985𝛾 + 115𝐾 + 𝐾2 − 20.4139𝛾𝐾 − 2347.5985𝛾− 20.4139𝛾𝐾 + 416.7273𝛾2

(−115 − 𝐾 + 20.4139𝛾)2 = 13225 + 230𝐾 − 4695.197𝛾 + 𝐾2 − 40.8278𝛾𝐾 + 416.7273𝛾2 Section E (−135 − 𝐾 + 26.0206𝛾)2

= 18225 + 135𝐾 − 3512.781𝛾 + 135𝐾 + 𝐾2 − 26.0206𝛾𝐾 − 3512.781𝛾− 26.0206𝛾𝐾 + 677.0716𝛾2

(−135 − 𝐾 + 26.0206𝛾)2 = 18225 + 270𝐾 − 7025.562𝛾 + 𝐾2 − 52.0412𝛾𝐾 + 677.0716𝛾2

Section F

(−150 − 𝐾 + 30𝛾)2 = 22500 + 150𝐾 − 4500𝛾 + 150𝐾 + 𝐾2 − 30𝛾𝐾 − 4500𝛾 − 30𝛾𝐾 + 900𝛾2

(−150 − 𝐾 + 30𝛾)2 = 22500 + 300𝐾 − 9000𝛾 + 𝐾2 − 60𝛾𝐾 + 900𝛾2

Summation 𝐹(𝛾, 𝐾) = 74975 + 1290𝐾 − 27603.338𝛾 − 231.0654𝛾𝐾 + 6𝐾2 + 2566.7427𝛾2

On Differentiating

𝜕(𝐹(𝛾, 𝐾))

𝜕𝛾= 0

𝜕(𝐹(𝛾, 𝐾))

𝜕𝛾= −27603.338 − 231.0654𝐾 + 5133.4854𝛾 = 0

𝜕(𝐹(𝛾, 𝐾))

𝜕𝐾= 1290 − 231.0654𝛾 + 12𝐾 = 0

−27603.338 − 231.0654𝐾 + 5133.4854𝛾 = 0

1290 − 231.0654𝛾 + 12𝐾 = 0 On Substitution;

𝜸 = 𝟒. 𝟎𝟒

𝑲 = −𝟐𝟗. 𝟕𝟏

(b) Find the path loss at 2 Km based on this model. 𝑑 = 2𝐾𝑀 𝑃𝑎𝑡ℎ 𝐿𝑜𝑠𝑠 = 𝑀𝑚𝑜𝑑𝑒𝑙(𝑑𝑖) = 𝐾 − 10𝛾 log10 𝑑

= −29.71 − (10 × 4.04) log10 2000

= −𝟏𝟔𝟑. 𝟎𝟕

(c) Find the outage probability at a distance d assuming the received power at d due

to path loss alone is 10 dB above the required power for nonoutage.

𝜎𝜓𝑑𝐵2 =

1

6∑[𝑀𝑀𝑒𝑎𝑠𝑢𝑟𝑒𝑑(𝑑𝑖) − 𝑀𝑀𝑜𝑑𝑒𝑙(𝑑𝑖)]2

6

𝑖=1

𝜎𝜓𝑑𝐵2 =

1

6 (74975 + 1290𝐾 − 27603.338𝛾 − 231.0654𝛾𝐾 + 6𝐾2 + 2566.7427𝛾2)

𝜎𝜓𝑑𝐵2 =

1

6 (74975 + (1290 × −29.71) − (27603.338 × 4.04) − (231.0654 × 4.04

× −29.71) + 6(−29.71)2 + 2566.7427(4.04)2)

𝜎𝜓𝑑𝐵2 = 9.246

𝜎𝜓𝑑𝐵 = 3.041

𝑝(𝑃𝑟(𝑑) ≤ 𝑃𝑚𝑖𝑛) = 1 − 𝑄 (𝑃𝑚𝑖𝑛 − (𝑃𝑚𝑖𝑛 + 10)

𝜎𝜓𝑑𝐵)

𝑝(𝑃𝑟(𝑑) ≤ 𝑃𝑚𝑖𝑛) = 1 − 𝑄 (1 − 11

3.041)

= 1 − 𝑄 (−10

3.041)

= 1 − 𝑄(−3.288)

= 3.64 𝑥10−4

Q4:

1. Data Collection Types and size

The data was collected from 5 different locations under various conditions and was categorized into 3 categories namely;

o Category A – Maximum path loss with hilly terrain and moderate-to-heavy tree densities.

o Category B – Flat terrain with moderate-to-heavy tree densities and hilly terrain with light tree densities.

o Category C – Minimum path loss with flat terrain and light tree densities.

95 cellular base stations were used with each transmitting a continuous wave signal at around 1.9GHz using omnidirectional azimuth pattern and gain of 8.14dBi. The mobile receiver antenna with omnidirectional azimuth pattern and gain of 2.5dBi was mounted at a height of 2m on a test van and Grayson receiver recorded the local mean power.

Distance covered ranged from few meters to 8km and the distance was estimated/calculated with the help of GPS.

2. Key Findings

𝑃𝐿 = 𝐴 + 10𝛾 log10 (𝑑 𝑑𝑜) + 𝑠⁄ i. The decibel path loss (A) are approximately 78 dB (free path loss) at 1.9GHz. This therefore

allows use of fixed-intercept approach. ii. Due to less blockage and better ground clearance, almost line of sight condition is

produced meaning the higher base antennas lead to smaller power-law exponent(𝛾). This shows that 𝛾 is strongly dependent on antenna height hb and the terrain category.

iii. Deviation of 𝛾 about its regression fit(∆𝛾) has a near-Gaussian distribution over the population of macrocells for each terrain category.

iv. The random variate s tend to be Gaussian within a given macrocell. This confirms that the shadow fading is log-normal.

v. 𝜎 is random from one macrocell to another.

3. Path loss exponent variations are Gaussian

Gaussian functions are widely used in statistics where they describe the normal distributions representing real-valued random variables whose distributions are not known. Usually to model real environment, the path loss and shadowing effects cannot be neglected. The variation in path loss follows a Gaussian distribution simply because once there is a change in the position of the receiver node, there can be variation in the path loss due to variation in the multi-path components. Also, the effects of diffraction, scattering, and reflection might change since by changing the position of the receiver can have an effect on the LOS component. Since the conditions for testing are not fixed and changes with the movement of the receiver in this case, various conditions generate various components of path loss with are not constant. To put this into considerations, Gaussian functions which uses mean distribution is used to get a linear estimate.