MITS5003 Wireless Networks & Communication - 2019S2 Assignment No 2 Submission Due Date: 04/10/2019 before 5 pm Submission Guidelines: All submissions are to be submitted through turn-it-in....

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MITS5003 Wireless Networks & Communication - 2019S2 Assignment No 2 Submission Due Date: 04/10/2019 before 5 pm Submission Guidelines: All submissions are to be submitted through turn-it-in. Drop-boxes linked to turn-it-in will be set up in the Unit of Study Moodle account. Assignments not submitted through these drop- boxes will not be considered. Submissions must be made by the due date and time. The turn-it-in similarity score will be used in determining the level if any of plagiarism. Turn-it-in will check conference web-sites, Journal articles, the Web and your own class member submissions for plagiarism. You can see your turn-it-in similarity score when you submit your assignment to the appropriate drop-box. If this is a concern you will have a chance to change your assignment and re-submit. However, re-submission is only allowed prior to the submission due date and time. After the due date and time have elapsed you cannot make re-submissions and you will have to live with the similarity score as there will be no chance for changing. Thus, plan early and submit early to take advantage of this feature. You can make multiple submissions, but please remember we only see the last submission, and the date and time you submitted will be taken from that submission. Your document should be a single word or pdf document containing your report. 1. Convert the binary data “110101” into analog waveforms using following modulation techniques: a. Two level Amplitude Shift Keying b. Two level Frequency Shift Keying c. Two level Phase Shift Keying d. Differential Phase shift keying e. Four level Amplitude Shift Keying f. Four level Phase Shift Keying g. Eight level Amplitude Shift Keying 2. With fc = 1000 kHz, fd = 50 kHz, and M = 16 (L = 4 bits), using Multiple Frequency-Shift Keying (MFSK), compute the frequency assignments for each of the sixteen possible 4- bit data combinations. 3. Draw the approximate Analog Modulation and Frequency Modulation waveforms in complete steps for the following signal: 4. Draw the 16 QAM Constellation Diagram having four different amplitude levels and four different phase levels. 5. Explain and draw the Error Detection Process for Cyclic Redundancy Check (CRC). 6. Compute the frame check sequence using Cyclic Redundancy Check (CRC) for the following information: Message = 111010110, Pattern = 101110 7. Compute the transmitted signal using Direct Sequence Spread Spectrum for the following information: Input: 101, Locally Generated PN bit stream: 011011010110, T = 4Tc 8. Explain why the square and circle shapes cells for cellular communications are not appropriate as compared to hexagonal shape cells.
Answered Same DaySep 25, 2021MITS5003

Answer To: MITS5003 Wireless Networks & Communication - 2019S2 Assignment No 2 Submission Due Date: 04/10/2019...

Sudipta answered on Oct 04 2021
139 Votes
Subject – Wireless Network and communications
1) Analog waveform for the binary data 110101 is represented below:
Two level Amplitude Shift Keying

Two level Frequency Shift Keying

Two level Phase Shift Keying
Differential Phase shift keying
Four level Amplitude Shift Keying
Four level Phase Shift Keying
Eight level Amplitude Shift Keying


2.) Given: fc = 1000 kHz, fd = 50 kHz, and M = 16, L = 4 bits
So for the frequency assignment of each of the 16 possible 4 bit data combinations we use the given formula:
Fi=fc+(2i-L-M)*fd where fc is carrier frequency and fd is the deviation frequency
Where i range from 1 to 16, So the solutions we get are:F9=1000+(18-16-4)*50
=900kHz
F10=1000+(20-16-4)*50
=1000kHz
F11=1000+(22-16-4)*50
=1100kHz
F12=1000+(24-16-4)*50
=1200kHz
F13=1000+(26-16-4)*50
=1300kHz
F14=1000+(28-16-4)*50
=1400kHz
F15=1000+(30-16-4)*50
=1500kHz
F16=1000+(32-16-4)*50
=1600kHz
F1= 1000+(2-16-4)*50
=100kHz
F2=1000+(4-16-4)*50
=200kHz
F3=1000+(6-16-4)*50
=300kHz
F4=1000+(8-16-4)*50
=400kHz
F5=1000+(10-16-4)*50
=500kHz
F6=1000+(12-16-4)*50
=600kHz
F7=1000+(14-16-4)*50
=700kHz
F8=1000+(16-16-4)*50
=800kHz
3.) In Frequency modulation the amplitude remains the same for the whole curve only the frequency variation changes. It is less for negative half and more for positive half. The below images shows the amplitude modulation and frequency modulation of the given waveform.

FREQUENCY...
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