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Worcester Polytechnic Institute

ECE 506: Introduction to LAN’s & WAN’s Solution to Assignment 3 Assigned on: Monday, September 11, 2017 Due on: Monday, September 18, 2017

Problem 7

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Use the Okumura-Hata and COST-231 models to determine the maximum radii of cells at 900 MHz and 1900 MHz respectively having a maximum path loss of 130 dB. Use 𝐚(𝒉𝒎 ) = 𝟑. 𝟐[𝐥𝐨𝐠⁡(𝟏𝟏. 𝟕𝟓𝒉𝒎 )]𝟐 − 𝟒. 𝟗𝟕 for both cases. At 900 MHz we use Okumura-Hata model: 𝐿𝑝 = 69.55 + 26.16 log(𝑓𝑐 ) − 13.82 log(ℎ𝑏 ) − 𝑎(ℎ𝑚 ) + (44.9 − 6.55log⁡(ℎ𝑏 ))log⁡(𝑑) At 1900 MHz we use COST-231 model, for large city,𝐶𝑀 = 0: 𝐿𝑝 = 46.3 + 33.9 log(𝑓𝑐 ) − 13.82 log(ℎ𝑏 ) − 𝑎(ℎ𝑚 ) + (44.9 − 6.55 log(ℎ𝑏 )) log(𝑑) Where 𝑎(ℎ𝑚 ) = 3.2[log(11.75ℎ𝑚 )]2 − 4.97 Plug the following parameters in above models: ℎ𝑏 = 30⁡𝑚, ℎ𝑚 = 2⁡𝑚, 𝐿𝑝 = 130⁡𝑑𝐵, then 𝑎(ℎ𝑚 ) = 1.0454⁡𝑑𝐵 At 900 MHz, 130−69.55−26.16 log(900)+13.82 log(30)+1.0454 44.9−6.55log⁡(2)

= 1.3550⁡𝑘𝑚

130−46.3−33.9 log(1900)+13.82 log(30)+1.0454 44.9−6.55log⁡(2)

= 0.6748⁡𝑘𝑚

𝑑 = 10 At 1900 MHz,

𝑑 = 10

Problem 9

Th

Table P2.1 provides the minimum required RSS for an IEEE 802.11b device to operate at different rates. (a) Calculate the coverage associated with each data rate in the table.

Considering 20 dBm transmit power from an 802.11g access point, we can obtain the path-loss of the signal as: 𝐿𝑝 (𝑑𝐵) = 𝑃𝑡 (𝑑𝐵𝑚) − 𝑅𝑆𝑆(𝑑𝐵𝑚) In Model D, the parameters are 𝛼1 = 2, 𝛼2 = 3.5, and 𝑑𝑏𝑝 = 10⁡𝑚, hence the pathloss model is 10𝛼1 log(𝑑)⁡⁡⁡⁡⁡⁡⁡⁡ ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑑 < 𝑑𝑏𝑝 𝐿𝑝 = 𝐿0 + { 10𝛼1 log(𝑑𝑏𝑝 ) + 10𝛼2 log⁡(𝑑/𝑑𝑏𝑝 )⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑑 ≥ 𝑑𝑏𝑝 When data rate is 11 Mbps and RSS is -82 dBm, 𝐿𝑝 = 20 − (−82) = 102⁡𝑑𝐵,

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𝜆 𝑐 3 ∗ 108 𝐿0 = −20 log ( ) = −20 log ( ) = −20 log ( ) = 40.046 4𝜋 4𝜋𝑓 4 ∗ 3.14 ∗ 2.4 ∗ 109 And coverage can be calculated as 𝐿𝑝 −𝐿0 −10𝛼1 log(𝑑𝑏𝑝 ) 10𝛼2 10

102−40.046−20 log(10)

𝐿𝑝 −𝐿0 −10𝛼1 log(𝑑𝑏𝑝 )

107−40.046−20 log(10)

35 𝑑 = 𝑑𝑏𝑝 ∗ = 10 ∗ 10 = 158⁡𝑚 Similarly, when data rate is 5.5 Mbps and RSS is -87 dBm, 𝐿𝑝 = 20 − (−87) = 107⁡𝑑𝐵, and coverage can be calculated as 10𝛼2 35 𝑑 = 𝑑𝑏𝑝 ∗ 10 = 10 ∗ 10 = 220⁡𝑚 When data rate is 2 Mbps and RSS is -91 dBm, 𝐿𝑝 = 20 − (−91) = 111⁡𝑑𝐵, and coverage can be calculated as 𝐿𝑝 −𝐿0 −10𝛼1 log(𝑑𝑏𝑝 )

111−40.046−20 log(10)

sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m

10𝛼2 35 𝑑 = 𝑑𝑏𝑝 ∗ 10 = 10 ∗ 10 = 286⁡𝑚 When data rate is 1 Mbps and RSS is -94 dBm, 𝐿𝑝 = 20 − (−91) = 114⁡𝑑𝐵, and coverage can be calculated as 𝐿𝑝 −𝐿0 −10𝛼1 log(𝑑𝑏𝑝 )

114−40.046−20 log(10) 35

10𝛼2 𝑑 = 𝑑𝑏𝑝 ∗ 10 = 10 ∗ 10 And we have the new table as follows Data Rate (Mbps) RSS (dBm) 11 -82 5.5 -87 2 -91 1 -94

= 348⁡𝑚

Coverage (m) 158 220 286 348

Table 2.1

(b) Plot the staircase function of the Data Rate vs. RSS

Use 𝑠𝑡𝑎𝑖𝑟𝑠() function in Matlab to plot staircase of Data Rate vs. RSS and the result is shown in Figure 1. (Code is attached in Appendix) (c) Plot the staircase function of the Data Rate vs. Distance

Th

Similarly, we plot staircase of Data Rate vs. Coverage, which is shown in Figure 2

(d) If a mobile terminal moves away from a 802.11 AP and it goes out of the coverage area, calculate the average data rate that terminal observes during the movement in the coverage area of the AP? For up to 158 meters we have 11 Mbps, then for up to 220 meters we have 5.5 Mbps. Next break point is 286 meters which gives 2 Mbps. Finally, we have 1 Mbps for up to 348 meters. Then the probability of having a connection and 11 Mbps data rate 𝜋(158)2

can simply be calculated as 𝑝11 = 𝜋(348)2 = 0.2061, which is the area covered by 11 Mbps over total area covered.

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Similarly, we have 𝑝5.5 = 𝑝1 =

𝜋(348)2 −𝜋(286)2 𝜋(348)2

𝜋(220)2 −𝜋(158)2 𝜋(348)2

= 0.1935, 𝑝2 =

𝜋(286)2 −𝜋(220)2 𝜋(348)2

= 0.2758,

= 0.3246.

sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m

Then we can find the average data rate as 𝑅𝑎𝑣𝑔 = ∑4𝑖=1 𝑝𝑖 𝑅𝑖 and in this case, 𝑅𝑎𝑣𝑔 = 0.2061 ∗ 11 + 0.1935 ∗ 5.5 + 0.2758 ∗ 2 + 0.3246 ∗ 1 = 4.208⁡𝑀𝑏𝑝𝑠

Th

Figure 1 Staircase of Data Rate vs. RSS

Figure 2 Staircase Data Rate vs. Coverage Copyright ©2016 by CWINS. All Rights Reserved. https://www.coursehero.com/file/26639348/Solution-to-HW3-17pdf/

Problem 11 In a mobile communication network, the minimum required signal to noise ratio is 12 dB. The background noise at the frequency of operation is -115 dBm. If the transmit power is 10 W, transmitter antenna gain is 3 dBi, the receiver antenna gain is 3 dBi, the frequency of operation is 800 MHz and the base station and mobile antenna heights are 100 m and 1.4 m respectively, determine the maximum in building penetration loss that is acceptable for a coverage of 5 km if the following path loss models are used. (a) Free space path loss model with 𝜶 = 𝟐.

sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m

In free space model, the path-loss is given by 𝐿𝑝 = 𝐿0 + 20log⁡(𝑑) In this case, we have

3∗108

𝑐

𝐿0 = −𝐺𝑡 − 𝐺𝑟 − 20 log (4𝜋𝑓) = −3 − 3 − 20 log (4∗3.14∗800∗106 ) = 24.5036⁡𝑑𝐵 and we can calculate the path-loss at the distance of 5 km as 𝐿𝑝 = 𝐿0 + 20 log(𝑑) = 24.5036 + 20 log(5000) = 98.4830⁡𝑑𝐵 10𝑊

The transmit power is 10 ∗ log (1𝑚𝑊) = 40⁡𝑑𝐵𝑚. The minimum acceptable received power is 𝑃𝑟𝑚𝑖𝑛 = 𝑆𝑁𝑅 + 𝑁𝑜𝑖𝑠𝑒 = −115⁡𝑑𝐵𝑚 + 12⁡𝑑𝐵 = −103⁡𝑑𝐵𝑚. Then the maximum allowable path-loss is 𝐿𝑝𝑚𝑎𝑥 = 𝑃𝑡 − 𝑃𝑟𝑚𝑖𝑛 = 40 − (−103) = 143⁡𝑑𝐵. If in addition to the free space path-loss, the signal must penetrate a building to reach the receiver, and assuming that this additional loss is a constant referred to a 𝐿𝑏𝑙𝑜𝑠𝑠 , then the total path loss would be 𝐿𝑝 + 𝐿𝑏𝑙𝑜𝑠𝑠 and should not exceed the maximum allowable path loss of 143 dB. Therefore, the maximum allowable building penetrate loss would be 143⁡𝑑𝐵 − 98.4830⁡𝑑𝐵 = 44.5170⁡𝑑𝐵. (b) Two-ray path loss model with 𝜶 = 𝟒.

In two-ray path-loss model with 𝛼 = 4, we have 𝑃𝑟 = 𝑃𝑡 𝐺𝑡 𝐺𝑟

ℎ𝑏 2 ℎ𝑚 2 𝑑4

and

Th

𝑃𝑡 ℎ𝑏 2 ℎ𝑚 2 𝐿𝑝 = ( ) = −10 log (𝐺𝑡 𝐺𝑟 ) = −10 log(𝐺𝑡 𝐺𝑟 ℎ𝑏 2 ℎ𝑚 2 ) + 40log⁡(𝑑) 𝑃𝑟 𝑑𝐵 𝑑4 Hence, we have 𝐿𝑝 = −3 − 3 − 20 log(100) − 20 log(1.4) + 40 log(5000) = 99.0362⁡𝑑𝐵 And the maximum allowable building penetration loss would be 143⁡𝑑𝐵 − 99.0362⁡𝑑𝐵 = 43.9638⁡𝑑𝐵

(c) Okumura-Hata model for a small city.

The Okumora-Hata macrocell path loss model for a small city is given by: 𝐿𝑝 = 69.55 + 26.16 log(𝑓𝑐 ) − 13.82 log(ℎ𝑏 ) − 𝑎(ℎ𝑚 ) + (44.9 − 6.55log⁡(ℎ𝑏 ))log⁡𝑑

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With 𝑓𝑐 = 800⁡𝑀𝐻𝑧, ℎ𝑏 = 100⁡𝑚, ℎ𝑚 = 1.4⁡𝑚, 𝑑 = 5000⁡𝑚 and for a small city, 𝑎(ℎ𝑚 ) = 1.1(log(𝑓𝑐 ) − 0.7)ℎ𝑚 − (1.56 log(𝑓𝑐 ) − 0.8) = 1.1(log(800) − 0.7) ∗ 1.4 − (1.56 log(800) − 0.8) = −0.3361⁡𝑑𝐵 Hence, 𝐿𝑝 = 69.55 + 26.16 log(800) − 13.82 log(100) − (−0.3361) + (44.9 − 6.55 log(100)) ∗ log(5000) = 235.8181⁡𝑑𝐵 We can see that the path loss without the additional building penetration loss is already higher than the maximum allowable path loss of 143 dB. This is because the Okumura-Hata model was based on measurements that considered scenarios of having to go through buildings to receiver and explains the much higher path-loss values coming from this model. Problem 13

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A model system is to provide 95% successful communication at the fringe of coverage with a location variability having a zero mean Gaussian distribution with standard deviation of 8 dB. What fade margin is required? The solution to the fading margin is obtained by equating the tail probability to 0.05 ( 95% of the cell edge is adequately covered). 𝐹𝜎 By equation 1 − 𝛾 = 0.5𝑒𝑟𝑓𝑐(𝜎√2 ), we can calculate the fading margin as below: 𝐹𝜎 𝐹𝜎 1 − 0.95 = 0.5𝑒𝑟𝑓𝑐 ( ) => ⁡𝑒𝑟𝑓𝑐 ( ) = 0.1 8√2 8√2 By using inverse complementary error function in Matlab, we can find that 𝐹𝜎 = 1.1631 => 𝐹𝜎 = 13.1588⁡𝑑𝐵 8√2 Problem 16 The IEEE 802.11 WLANs operate at a maximum transmission power of 100 mW (20 dBm) using multiple channels with different carrier frequencies. The IEEE 802.11g uses 2.402-2.480 GHz bands and the IEEE 802.11a uses 5.150-5.825 GHz bands. Both standards use OFDM modulation with a bandwidth of 20 MHz.

Th

(a) Calculate received signal strength in dBm at one meter distance of an IEEE 802.11g access point for the smallest and the largest possible carrier frequencies in the band. Assume that transmitter and receiver antenna gains are one and in one meter distance signal propagation follows the free space propagation rule. 4𝜋

In free space, the path loss can be calculated by 𝐿0 = 20 log ( 𝜆 ) = 20log⁡( the transmit power is 𝑃𝑡 = 100⁡𝑚𝑊 = 20⁡𝑑𝐵𝑚, 𝑃0 = 𝑃𝑡 − 𝐿0 4𝜋∗2.402∗109

At 2.402 GHz, we have 𝐿0 = 20 log ( 40.0532 = −20.0532⁡𝑑𝐵𝑚.

3∗108 4𝜋∗2.480∗109

At 2.480 GHz, we have 𝐿0 = 20 log ( 40.3308 = −20.3308⁡𝑑𝐵𝑚.

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3∗108

4𝜋𝑓 𝑐

) and

) = 40.0532⁡𝑑𝐵 , and 𝑃0 = 20 − ) = 40.3308⁡𝑑𝐵 , and 𝑃0 = 20 −

(b) Repeat (a) for the IEEE 802.11a WLANs. 4𝜋∗5.150∗109

Similarly, at 5.150 GHz, we have 𝐿0 = 20 log ( 𝑃0 = 20 − 46.6779 = −26.6779⁡𝑑𝐵𝑚. 4𝜋∗5.825∗109

At 5.150 GHz, we have 𝐿0 = 20 log ( 47.7477 = −27.7477⁡𝑑𝐵𝑚.

3∗108

3∗108

) = 46.6779⁡𝑑𝐵 , and

) = 47.7477⁡𝑑𝐵 , and 𝑃0 = 20 −

(c) Compare the received signal strengths at one meter distance of the IEEE 802.11g and IEEE 802.11a devices. Use the middle of the allocated band for each standard as the carrier frequency in your calculation.

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Considering the middle of the frequency band for IEEE 802.11g, we have 𝑓1 + 𝑓2 2.402 + 2.48 𝑓= = = 2.441⁡𝐺𝐻𝑧 2 2 4𝜋∗2.441∗109 And𝐿0 = 20 log ( 3∗108 ) = 40.1931⁡𝑑𝐵, and 𝑃0 = 20 − 40.1931 = −20.1931⁡𝑑𝐵𝑚. Similarly, for IEEE 802.11a, we have 𝑓1 + 𝑓2 5.15 + 5.825 𝑓= = = 5.4875⁡𝐺𝐻𝑧 2 9 2 4𝜋∗5.4875∗10 And 𝐿0 = 20 log ( ) = 47.2293⁡𝑑𝐵 , and 𝑃0 = 20 − 47.2293 = 3∗108 −27.2293⁡𝑑𝐵𝑚. It can be observed that increasing the carrier frequency effectively drops the received power. In the case of higher frequencies, the receivers have to be equipped with more sensitive receive antennas in order to receive the weaker signal. (d) Compare the rate of the received signal fluctuations (Doppler shift), due to the change in frequency of operation, for the IEEE 802.11g and the IEEE 802.11a. Use the middle of the allocated for each standard as the carrier frequency in your calculations.

Th

Assume the moving speed is 𝑉𝑚 , we have 𝑉 𝑉 ∗𝑓 𝑉𝑚 For IEEE 802.11g, 𝑓𝑚1 = 𝜆𝑚 = 𝑚𝑐 1 = 0.1229 For IEEE 802.11a, 𝑓𝑚2 =

1

𝑉𝑚 𝜆2

=

𝑉𝑚 ∗𝑓2 𝑐

𝑉

𝑚 = 0.0547

𝑓

0.0547

The rate of Doppler shift will be 𝑟𝑎𝑡𝑒 = ⁡ 𝑓𝑚1 = 0.1229 = 0.4451 𝑚2

Problem 17

A multiple channel has three paths at 0, 50, and 100 nsec with the relative strengths of 0, -10, and -15 dBm, respectively, (a) What is the multipath spread of the channel?

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The multipath spread is defined as the difference between the minimum and maximum path delays: 𝛕 = 𝛕𝐦𝐚𝐱 − 𝛕𝐦𝐢𝐧 = 𝟏𝟎𝟎 − 𝟎 = 𝟏𝟎𝟎⁡𝐧𝐬𝐞𝐜 (b) Calculate the RMS multipath spread of the channel. Power in Watts of three paths can be calculated as: 𝟎

−𝟏𝟎

−𝟏𝟓

sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m

𝑷𝟏 = 𝟏𝟎𝟏𝟎 = 𝟏⁡𝒎𝑾, 𝑷𝟐 = 𝟏𝟎 𝟏𝟎 = 𝟎. 𝟏⁡𝒎𝑾, 𝑷𝟑 = 𝟏𝟎 𝟏𝟎 = 𝟎. 𝟎𝟑𝟏𝟔⁡𝒎𝑾 In this case ∑𝟑𝒊=𝟏 𝝉𝒊 𝟐 𝑷𝒊 𝟎 ∗ 𝟏 + 𝟓𝟎𝟐 ∗ 𝟎. 𝟏 + 𝟏𝟎𝟎𝟐 ∗ 𝟎. 𝟎𝟑𝟏𝟔 ̅̅̅ 𝟐 𝝉 = = = 𝟓𝟎𝟎. 𝟏𝟕𝟔𝟕 𝟏 + 𝟎. 𝟏 + 𝟎. 𝟎𝟑𝟏𝟔 ∑𝟑𝒊=𝟏 𝑷𝒊 And ∑𝟑𝒊=𝟏 𝝉𝒊 𝑷𝒊 𝟎 ∗ 𝟏 + 𝟓𝟎 ∗ 𝟎. 𝟏 + 𝟏𝟎𝟎 ∗ 𝟎. 𝟎𝟑𝟏𝟔 𝝉̅ = 𝟑 = = 𝟕. 𝟐𝟏𝟏𝟎 𝟏 + 𝟎. 𝟏 + 𝟎. 𝟎𝟑𝟏𝟔 ∑𝒊=𝟏 𝑷𝒊 Then we can calculate RMS multipath delay spread as 𝝉𝒓𝒎𝒔 = √̅̅̅ 𝝉𝟐 − (𝝉̅)𝟐 = √𝟓𝟎𝟎. 𝟏𝟕𝟔𝟕 − (𝟕. 𝟐𝟏𝟏𝟎)𝟐 = 𝟐𝟏. 𝟏𝟕𝟎𝟐⁡𝒏𝒔𝒆𝒄

(c) What would be the difference between multipath spreads and RMS multipath spreads of this three-path channel and a two-path channel formed by the first and the third path of this profile?

Th

The procedure for finding the multipath spread and RMS multipath spread of the new two path model is the same as that for the three path model except that there are now only two paths. The new multipath spread is calculated to be 𝝉 = 𝝉𝒎𝒂𝒙 − 𝝉𝒎𝒊𝒏 = 𝟏𝟎𝟎 − 𝟎 = 𝟏𝟎𝟎⁡𝒏𝒔𝒆𝒄 In the new model 𝟎 ∗ 𝟏 + 𝟏𝟎𝟎𝟐 ∗ 𝟎. 𝟎𝟑𝟏𝟔 ̅̅̅ 𝟐 𝝉 = = 𝟑𝟎𝟔. 𝟑𝟐𝟎𝟑 𝟏 + 𝟎. 𝟎𝟑𝟏𝟔 And 𝟎 ∗ 𝟏 + 𝟏𝟎𝟎 ∗ 𝟎. 𝟎𝟑𝟏𝟔 𝝉̅ = = 𝟑. 𝟎𝟔𝟑𝟐 𝟏 + 𝟎. 𝟎𝟑𝟏𝟔 The new RMS multipath spread delay is calculated to be 𝝉𝒓𝒎𝒔 = √̅̅̅ 𝝉𝟐 − (𝝉̅)𝟐 = √𝟑𝟎𝟔. 𝟑𝟐𝟎𝟑 − (𝟑. 𝟎𝟔𝟑𝟐)𝟐 = 𝟏𝟕. 𝟐𝟑𝟏𝟗⁡𝒏𝒔𝒆𝒄

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Appendix: Matlab Code

Th

sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m

clear all; close all; %% Problem 2.9 % Part (b) DataRate=[11, 5.5, 2, 1]; RSS=[-82, -87, -91, -94]; Coverage=[158, 220, 286, 348]; figure(1) stairs(DataRate, RSS, 'Linewidth', 2); axis([0 12 -92 -80]); grid on xlabel('Data Rate (Mbps)'); ylabel('RSS (dBm)'); % Part (c) figure(2) stairs(DataRate, Coverage, 'Linewidth', 2); axis([0 12 150 350]); grid on xlabel('Data Rate (Mbps)'); ylabel('Coverage (m)'); % Part (d) p11=Coverage(1)^2/Coverage(4)^2; p5_5=(Coverage(2)^2-Coverage(1)^2)/Coverage(4)^2; p2=(Coverage(3)^2-Coverage(2)^2)/Coverage(4)^2; p1=(Coverage(4)^2-Coverage(3)^2)/Coverage(4)^2; disp([p11 p5_5 p2 p1]); DataRate_avg=p11*DataRate(1)+p5_5*DataRate(2)+p2*DataRate(3)+p1*DataRat e(4);

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