Wireless Data Link

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Question 1: Calculate the FSPL in dB for a distance of 5Km and 10 Km for two different frequencies 5.8GHz, 2.4GHz and construct a table.

Answer: Given Distance d= 5 km

Given Frequency f=2.4 GHZ

when the distance (d is in km) and the frequency (f is in MHz) , the following formula is used to calculate the FSPL

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45 (1)

The given frequency is in GHZ, so we have to convert it in MHZ to use the above formula,

Frequency=2400 MHZ

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45

Substitute value of frequency and distance in equation (1),we get

FPSL(dB) = 20log10(5) + 20log10(2400) + 32.45

FPSL= 114.0336 dB.

Given Distance d= 5 km

Frequency f=5.8 GHZ

when the distance (d is in km) and the frequency (f is in MHz) , the following formula is used to calculate the FSPL

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45 (1)

The given frequency is in GHZ, so we have to convert it in MHZ to use the above formula,

Frequency=5800 MHZ

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45

Substitute value of frequency and distance in equation (1),we get

FP(dB) = 20log10(d) + 20log10(f) + 32.45

FP(dB) = 20log10(5) + 20log10(5800) + 32.45

FPSL= 121.698 dB.

Given Distance d= 10 km

Frequency f=2.4 GHZ

when the distance (d is in km) and the frequency (f is in MHz) , the following formula is used to calculate the FSPL

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45 (1)

The given frequency is in GHZ, so we have to convert it in MHZ to use the above formula,

Frequency=2400 MHZ

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45

Substitute value of frequency and distance in equation (1),we get

FP(dB) = 20log10(d) + 20log10(f) + 32.45

FP(dB) = 20log10(10) + 20log10(2400) + 32.45

FPSL= 120.05 dB

Given Distance d= 10 km

Frequency f=5.8 GHZ

when the distance (d is in km) and the frequency (f is in MHz) , the following formula is used to calculate the FSPL

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45 (1)

The given frequency is in GHZ, so we have to convert it in MHZ to use the above formula,

Frequency=5800 MHZ

FPSL(dB) = 20log10(d) + 20log10(f) + 32.45

Substitute value of frequency and distance in equation (1),we get

FP(dB) = 20log10(d) + 20log10(f) + 32.45

FP(dB) = 20log10(10) + 20log10(5800) + 32.45

FPSL= 127.72 dB.

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Question 2: For the information given below, calculate the link budget analysis for the wireless system, which will involve calculating the following:

• Received Power

• Link Margin

Distance: 5 km

Frequency: 5.8GHz

Link Type: Point-to-Point, Line-of-Sight

Tx power: +23dBm

Antenna gain is 24dBi

Assume negligible loss for cabling and connectors

Receiver Sensitivity: -72dBm

Answer: Given Distance=5 km

Given Frequency = 5.8GHz

Link Type is given in Point-to-Point and Line-of-Sight

Given, the value of Transmission power is equal to +23dBm

Given, Antenna gain is equal to 24dBi

We have to assume negligible loss for both cabling and connectors

The Receiver Sensitivity is equal to -72dBm

The gain is calculated by using the following formula

Gains(dB)=Transmit Antenna gain+ Receiver Antenna Gain (2)

Given Same antenna is used at the transmitter and the receiver side, also given the link budget calculations will be the same as the link budget in both directions is expected to be symmetrical.

So that the gain will be,

Gains(dB)=24+ 24=48 dBi

The Received Power is calculated by using formula given in equation (2), which as follows

Received Power (dBm) = Transmitted Power (dBm) + Gains (dB) - Losses (dB) (2)

From the above table we can see that loss is equal to 121.698 dB when the distance is equal to 5km and the frequency is equal to 5.8 GHZ.

Substitute the value of Transmission power, Gain and loss in equation (2), we get

Received Power = 23 + 48 - 121.698

Received Power = -50.70 dB.

The Link Margin is calculated by using formula given in equation (3), which as follows, the link margin is obtained by subtracting receiver sensitivity from the received power,

Link Margin=Received Power-Receiver sensitivity (3)

Substitute the value of Received Power and Receiver sensitivity in equation (3), we get

Link Margin=-50.70 -(-72)

Link Margin=21.3 dB

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Question 3: For the calculated Link Margin at 5Km what can you say about the availability/reliability of the link based on the following Table which shows the relationship between the available link margin and link availability as a percentage of time.

Answer: The fade margin or the link margin is defined by how much margin is in dB between the ratio of received signal strength level and the receiver sensitivity, that is what percentage of amount, the received signal level reduced without affecting the performance of the system..Link margin or fade margin is directly related with link availability. The table given in question c) consists time availability and their related fade margin. The relationship between the available link margin and link availability is given in, percentage of time. From those values we can see that whenever the time increases, the fade margin also increases. if the fade margin is high then the link has system outage where as if the link has less or no fade margin, then it has a periodic outages because of path fading. Therefore a fade margin is a important factor in the link, to achieve the good link availability in dry or rocky area. A fading channel is a communication channel which experiences fading. Fading in a channel occurs because of multipath propagation which is also referred as multipath induced fading, rain or wave propagation affected by obstacles because of shadowing. In wireless systems, multipath propagation results in multiple copies of a signal to arrive at different signal phases at the receiver. If these signals add up destructively, the resulting signal power can be lower by a factor of 100 or 1000 (20 or 30 dB). The signal level relative to the noise declines making signal detection at the receiver more difficult. It is therefore highly recommended to keep a link margin of 30 dB when designing a wireless system.

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