Part I: Encoding and error control
Sam is a computer science student and working on a new micro robot design. The robot continuously sends every two seconds a status string comprising the accelerometer reading (4 bits), ultrasound obstacle detection (6 bits), motor functionality (4 bits) and battery power level (2 bits).
Bits sent per second= data rate
Bits sent on two seconds = 2+4+6+4
=16 bits
16 bits is in the Interval of 2 seconds so in one second it will be 8bits.
Non return to zero
in this technique one level of voltage zero corresponds to another which corresponds to One
Pseudoternary
Involves encoding the ones with flat lines and zeros with negative and positive voltages alternating
Non return to zero invert
Encoding depends on the bit that is before. It should have some starting point
Status string= readings in accelerometer + motor status functionality + battery level status + obstacle detection in ultrasound reading
Binary convertion
Ultrasound= 48cm
Binary = 110000
Acceleration = 5m/s2
Binary = 0101
Motor status = 1111
Level of the battery 75/100 which is 0.75
Binary = 0.11
Combined status string will be a sequence of the individual strings status
= 0101+110000+1111+0.11
= 0101110000111111
FSK
FSK is the switching of the sinusoidal career wave frequency, depending on the digital input signal
Bit string = 010111000011111
FSK is achieved through changing the frequency of the sinosidal carrier according to the digital input signal
ASK
Bit string = 01011100001111
Carrier wave is transmitted when the input is one but it is not transmitted when the input is 0
PSK
A phase shift of 180 degrees takes place whenever there is a change in the input signal e.g from 1 to 0 or from 0 to 1
First step
Input bit sequence = Data word
010111000011111 =
No of divisor digits -1= additional digits
8-1 = 8
7
Second step
The seven digits are added to the data word to form augmented dataword
Augmented data word= data word + additional digit
= 0101110000111111+ 0000000
= 01011100001111110000000
Third step
Augmented data is changed to polynomial
01011100001111110000000
X21+X20+X19+X18+X16+X15+X14+X9+X8+X7+X6+X5+X4
Forth step
The divisor is changed in to polynomial
11001101
X7+X6+X3+X2+1
Fifth step
Sliding window flow control
The sender and receiver need to have a window of frames. Every particular frame is numbered according to the sliding window for a window of size p , the number 0 to p-1 is given to the frame. Many frames can be sent by the sender as long as it fits in a window. A slide is when a receiver sends an acknowledgement after receiving the frames sent up to a certain point. For example p=8 the size of the window is (p-1) 7 .
Stop and wait Flow control
For every frame that has been sent the sender has to wait for an acknowledgement because the other frame can only be sent after the frame has been received.This process goes on until the sender sends an (EOT)End of Transmission frame .
Vertical Redundancy Check (VRC)
Parity bit which is aredundant bit is applied in every data unit in order for the total number of 1’s to be even.
Checksum
Two algorithms are in place the checksum generator at the receiver and checksum generator at the sender. The following steps are followed by the sender the data units are divided in to sections then units . The sections are then added using 1’s compliment so as to get the sum. The checksum is achieved when the sum is completed after that the data is sent with the checksum. The unit received is divided into sections each of specific bits therefore an addition is done of the section using 1’s complement to get the sum the end of the calculation the data is only accepted if it is zero. (Thomas, Sharma, Banerjee, Lee, & Dalal, 2013)
Part II: Multiplexing and multiple access
Multiplexing plays an important role in communication as it allows the combination of multiple streams together. Different multiplexing techniques are used for various applications.
TDM(Time Division Multiplexing)
This is a technique which allows the whole frequency bandwidth to be accessed but for a short period of time. Co-channel interference occurs when two different transmissions overlap
FDM( Frequency Division Multiplexing)
FDM allows the channel to occupy a fraction of the frequency. Channels are defined through the center of the bandwidth and the frequency. Example of this technique is radio and television signals (Holma & Toskala, 2009)
CDMA(Code Division Multiple Access)
Channels sharing frequency simultaneously occurs in this technique. An example of CDMA is GPS(Global Positioning System). CDMA uses analog to digital conversion together with spread spectrum technology (Palanki, Khandekar, & Sutivong, 2014) .
OFDM has better performance and tolerance furthermore it allows more subscribers in, multiple cellular application. The signal used in OFDM is subdivided into small chunks called sub-carriers which are supposed to be orthogonal during transmission this gives room for consistency and reach of the waves in real time (Hwang, Yang, Wu, Li, & Li, 2009).
fb= allowable number of sub carriers
Channel bandwidth = 40MHz
Allowable subcarriers = 128 sub carriers
fb=40MHz/128subcarriers
= 0.3125MHz
= 312.5 KHz
T= 1/f(b)
fb= 0.3125
T=1/0.3125
=0.0032 micro seconds
iii. Explain how OFDM overcomes the issue of inter symbol interference (ISI)
It modulates data in the frequency domain over short subcarriers and this results to long OFDM symbols which are longer than the delay spread. A cyclic prefix which has an appropriate length is chosen to avoid inter-OFDM-symbol interference (Thomas et al., 2013)
Part III: Wi-Fi
Room 1
Area= L*W
=10*10
=100m2
BSS size=100m2
Room 2
Area= L*W
=10*10
=100m2
BSS size=100m2
Room 3
Area= L*W
=10*10
=100m2
BSS size=100m2
Room 4
Area= L*W
=10*10
=100m2
BSS size=100m2
Room 5
Area= L*W
=10*10
=100m2
BSS size=100m2
Lounge and reception
Area= L*W
=20*10
=100m2
BSS size=100m2
Extended Service Set(ESS)
ESS= Total coverage area of all the access points connected to the Distribution switch
=Area(Lounge+room5+room4+room3+room2+room1)
= (200+100+100+100+100+100)
=700m2
The actual data passing through the network per second
For every room throughput(100*8)mbps
=800mbps
The rooms are five so(800*5)
=4000mbps
Lounge thoughput (100*25)
=2500mbps
Total throughput = 4000+2500
= 6500
IEEE wifi is good for the above setup that is IEE 802.11ac because it allows video streaming , the buffering is less and surfing will be faster.
Use only wpa2 encryption, WPS to be turned off, make a point of changing Access point password.
References
Holma, H., & Toskala, A. (2009). LTE for UMTS: OFDMA and SC-FDMA based radio access. John Wiley & Sons.
Hwang, T., Yang, C., Wu, G., Li, S., & Li, G. Y. (2009). OFDM and its wireless applications: A survey. IEEE Transactions on Vehicular Technology, 58(4), 1673–1694.
Palanki, R., Khandekar, A., & Sutivong, A. (2014). Code division multiplexing in a single-carrier frequency division multiple access system. Google Patents.
Thomas, D. A., Sharma, P., Banerjee, S., Lee, S.-J., & Dalal, A. C. (2013). Method and system for content downloads via an insecure communications channel to devices. Google Patents.
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