Mobile Communications Communications 2015-2016 Dijlah University College Department of Computer Techniques Engineering
Tutorial of Chapter 3
Dr. Eng. Hussein Jamaluddin Khaleel Email:
[email protected]
Problem 3.2 (a) Explain the advantages and disadvantages of the 2-ray ground reflection model in the analysis of path loss. (b) In the following cases, tell whether the 2-ray model could be applied, and explain why or why not: ht = 35 m, hr = 3 m, d = 250 m ht = 30 m, hr = 1.5 m, d = 450 m
(c) What insight does the 2-ray model provide about large-scale path loss that was disregarded when cellular systems used very large cells?
Problem 3.2 – Solution (a) Advantage: The 2-ray ground reflection model considers both the direct path and a ground reflected propagation path between transmitter and receiver and therefore it provides more accurate estimation of the signal power compared to free space model.
Disadvantage: The 2-ray ground reflection model does not account for terrain, vegetation and buildings. (b) The 2-ray ground reflection model can be applied when ≫
ℎ ℎ
For ht = 35 m, hr = 3 m, d = 250 m:
ℎ ℎ = 10.25 >> 250 and therefore the model can be applied. ℎ ℎ = 6.71 >> 450 and therefore the model can be applied. (c) The 2-ray model shows that at large d , Pr decreases with d 4, and Pr and the path loss become independent of the frequency f .
Problem 3.6* Find the received power using the approximated 2-ray ground reflection model. Assume the height of the transmitter is 40 m and the height of the receiver is 3 m. The frequency is 1800 MHz, and unity gain antennas are used. Calculate the received power at distances 1 km, 10 km and 20 km. Assuming the ground reflection coefficient of -1. The E-field at 1 km from the transmitter is measured to be 10-3 V/m.
*This problem has been modified from the one in the textbook
Problem 3.6 – Solution Given: ht = 40 m, hr = 3 m, f = 1800 MHz, Gt = Gr = 1, d = 1 km, d = 10 km, d = 20 km, Γ = -1, d 0 = 1 km, E 0 = 10-3 V/m Pr = ? Solution: = λ =
||
120
=
2 2ℎ ℎ
λ
=
λ 2 4
c / f = 3x10 8 / 1800x106 = 0.167 m
Ae = 1 x 0.1672 / 4π = 0.0022 m2 For d = 1 km:
=
||
= 120
(1 × 10− ) 120
0.0022 = 5.8x1012 W = 112.34 dB
Problem 3.6 – Solution – Cont. For d = 10 km:
= =
2 2ℎ ℎ
||
= 120
λ
=
2 × 1 0− × 10 10 ×
(90 × 10− ) 120
10
2 × 40 × 3 0.167 × 10 ×
10
= 90 × 10− /
0.0022 = 4.76 × 10− W = 133.22 dB
For d = 20 km:
= =
2 2ℎ ℎ
||
= 120
λ
=
2 × 1 0− × 10 20 ×
(45 × 10− ) 120
10
2 × 40 × 3 0.167 × 10 ×
10
= 45 × 10− /
0.0022 = 1.19 × 10− W = 139.24 dB
Problem 3.10 If Pt = 10 W, G = 10 dB, Gr = 3 dB, and L = 1 dB at 900 MHz, compute the received power for the knife-edge geometry shown in the figure. Compare this value with the theoretical free space received power if an obstruction did not exist. What is the path loss due to diffraction in this case?
Problem 3.10 – Solution Given: Pt = 10 W , G = 10 dB , Gr = 3 dB , L = 1 dB, f = 900 MHz Pr (knife-edge) = ?
Pr (free space) = ? Gd = ? Solution: λ =
c / f = 3x10 8 / 900x106 = 1/3 m
Subtract all heights by the minimum height hr = 5 m
= tan− = tan−
−
= 0.11285 radians
= 0.195 radians
= = 0.30785 radians =
+
= 0.30785
××
(/) +
= 26.12
Problem 3.10 – Solution – Cont. Using the numerical method to calculate diffraction gain: G d (dB) =
20log
.
= 20log
. .
= -41.3 dB
So, the loss due to diffraction = 41.3 dB For free space: Gt = 10 dB = 10 (linear), Gr = 3 dB = 10 0.3 = 2 (linear), L = 1 dB = 1.26 (linear)
=
(π)
=
×××
(π) ××.
= 4.47 × 10− W = -83.5 dB
The power received with the knife-edge diffraction: Pr (diffraction) = Pr (free-space) + Gd = -83.5 – 41.3 = -124.8 dB = -94.8 dBm
Problem 3.11
Homework
If the geometry and all other system parameters remain exactly the same in problem 3.10, but the frequency is changed, redo Problem 3.10 for the case of (a) f = 50 MHz and (b) f = 1900 MHz.
Problem 3.13* If the received power at a reference distance d 0 = 1 km is equal to 1 micro watt, find the received powers at distances of 2 km, 5 km, 10 km, and 20 km from the same transmitter for the following path loss models: (a) Free space; (b) n = 3; (c) n = 4; (d) 2-ray ground reflection using the approximate expression. Assume f = 1800 MHz, ht = 40 m, hr = 3 m, Gt = Gr = 0 dB.
*This problem has been modified from the one in the textbook
Problem 3.13 – Solution Given: d 0 = 1 km, P0 = 10-6 W Pr = ? for d = 2 km, 5 km, 10 km, 20 km (a) Free space model, Pr = ?
=
For d = 2 km: = For d = 5 km: =
− 10 − 10
For d = 10 km: = 10− For d = 20 km: =
= 0.25 × 10− W = 66 dB = 36 dBm = 0.04 × 10− W = 74 dB = 44 dBm
− 10
= 0.01 × 10− W = 80 dB = 50 dBm = 0.0025 × 10− W = 86 dB = 56 dBm
Problem 3.13 – Solution – Cont. (b) n = 3, Pr = ?
=
For d = 2 km: = For d = 5 km: =
− 10 10−
For d = 10 km: = For d = 20 km: =
= 0.125 × 10− W = 69 dB = 39 dBm = 0.008 × 10− W = 81 dB = 51 dBm
− 10 − 10
= 0.001 × 10− W = 90 dB = 60 dBm = 0.0025 × 10− W = 99 dB = 69 dBm
Problem 3.13 – Solution – Cont. (c) n = 4, Pr = ?
=
For d = 2 km: = For d = 5 km: =
− 10 10−
For d = 10 km: = For d = 20 km: =
= 0.0625 × 10− W = 72 dB = 42 dBm = 0.0016 × 10− W = 88 dB = 58 dBm
− 10 − 10
= 0.0001 × 10− W = 110 dB = 70 dBm = 0.00000625 × 10− W = 112 dB = 82 dBm
Problem 3.13 – Solution – Cont. (d) 2-ray ground reflection model, Pr = ?, f = 1800 MHz, ht = 40 m, hr = 3 m, Gt = Gr = 0 dB
At the reference point d 0 = 1 km, Pr = 10-6 W; Gt = Gr = 0 dB = 1 (linear) λ=
c/ f = 3x108/1800x10 6 = 0.1667 m
= ⇒ =
⇒ 0 =
=
××.
= 5682.6 W the transmitted power
Using the 2-ray ground reflected approximated expression: = For d = 2 km: = 5682.6 For d = 5 km: = 5682.6
× ×
For d = 10 km: = 5682.6 For d = 20 km: = 5682.6
= 5.11 × 10− W = −52.9 dB = −32.9 dBm = 0.13 × 10− W = −68.82 dB = −38.82 dBm
× ×
=× 0.0082 × 10− W = −80.86 dB = −50.86 dBm = 0.000511 × 10− W = −92.9 dB = −62.9 dBm
Problem 6
Homework
Assume a receiver is located 10 km from a 50 W transmitter, the carrier frequency is 1900 MHz, and the antenna gains are 1 and 2 for the transmitter and receiver respectively. Find (a) the power at the receiver assuming free space propagation, (b) the magnitude of the E-field at the receiver antenna (c) the rms voltage applied to the receiver input assuming that the receiver antenna has purely real impedance of 50 Ω and is matched to the receiver, (e) the received power assuming the 2-ray ground reflection model (approximated model).
