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Modulation Techniques

Question 1. Convert the binary data "011010" into analog waveforms using following modulation techniques:

Answer: Given input pattern:

Analog waveforms.jpg

a. Two level Amplitude Shift Keying output

Analog waveforms1.jpg

b. Two level Frequency Shift Keying output

Analog waveforms2.jpg

c. Two level Phase Shift Keying output

Analog waveforms3.jpg

d. Differential Phase shift keying output

Analog waveforms4.jpg

e. Four level Amplitude Shift Keying output

Analog waveforms5.jpg

f. Four level Phase Shift Keying output

Analog waveforms6.jpg

g. Eight level Amplitude Shift Keying output

Analog waveforms7.jpg

h) Eight level Amplitude Shift Keying output

Analog waveforms8.jpg

Question 2. With fc = 500 kHz, fd = 25 kHz, and M = 16 (L = 4 bits), compute the frequency assignments for each of the sixteen possible 4-bit data combinations.

Answer:

Given fc = 500 kHz and 2fd = 50 kHz

Frequency.jpg

125 175 225 275 325 375 425 475 525 575 625 675 725 775 825 kHz

Bit representation are as follows:
f1= 0000,
f2= 0001,
f3= 0010,
f4= 0011,
f5= 0100,
f6= 0101,
f7= 0110,
f8 = 0111,
f9 = 1000,
f10= 1001,
f11= 1010,
f12= 1011,
f13= 1100,
f14= 1101,
f15= 1110

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Question 3. Draw the approximate Analog Modulation and Frequency Modulation waveforms in complete steps for the following signal:

Answer:

Given input signal

Frequency Modulation waveforms.jpg

Step1:

c(t) = Asin(2Πfct). [fc = carrier frequency, A = amplitude]

Frequency Modulation waveforms1.jpg

Step2:

m(t) = M cos(2Πfmt + Φ) = Am cos(2Πfmt + Φ)  [fm = input sine wave frequency]

Frequency Modulation waveforms2.jpg

Output equation is

y(t) = Asin(2Πfct) + 1/2Am[sin[fc +fm]t + Φ] + sin(2Π[fc - fm]t - Φ)

Frequency Modulation waveforms3.jpg

Question 4. Draw the 16 QAM Constellation Diagram having two different amplitude levels and eight different phase levels.

Answer: 16 QAM Constellation Diagram having two different amplitude levels and eight different phase levels can be shown as

16 QAM Constellation Diagram.jpg

Question 5. Explain and draw the Error Detection Process for Cyclic Redundancy Check.

Answer: The process of finding error in a signal at the receiver end is known as Error Detection Process for Cyclic Redundancy Check.

Let m= number of bit sequence at the transmitter end,
K = bits block

According to this method FCSR is 5 bits for 10bits input pattern and if the required pattern is 6 bits then

Question 6. Compute the frame check sequence for the following information: Message = 10111100, Pattern = 11011

Answer:

 

      1100

11011

10111100

11011

 

   11001

   11011

 

          1000

 

 

Code-word sent = 10110100

 

      1100

11011

10110100

11011

 

   11011

   11011

 

          000

 

 

Received data has no errors

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Question 7. Compute the transmitted signal using Direct Sequence Spread Spectrum for the following information:

Input: 1011, Locally Generated PN bit stream: 101011011010, T = 3Tc

Answer: Given, Input = 1011

And Bit pattern = 101011011010

Then Generated pattern = +1-1+1-1+1+1-1+1+1-1+1-1

Finally signal that can be Transmitted = 101011010000

Question 8. What is the difference between Infrastructure and ad hoc modes in WLAN? Draw their relative diagrams as well.

Answer: Infrastructure modes in WLAN composed of one or more access points whereas ad hoc modes in WLAN composed of only stations.

Infrastructure modes in WLAN used to communicate between stations whereas in other case it is done directly.

Question 9. Compare the differences of TCP and OSI protocols for wired and wireless LANs using diagrams.

Answer:

Application, Presentation, Session, Transport are host layer in TCPprotocols for wired and wireless LANs, whereas application, Session, Application are the layer in OSIprotocols for wired and wireless LANs.

TCP protocols.jpg

Fig: Diagram of TCP protocols for wired and wireless LANs

OSI protocols.jpg

Fig: Diagram of OSIprotocols for wired and wireless LANs

Question 10. Explain why the square and circle shapes cells for cellular communications are not appropriate as compared to hexagonal shape cells.

Answer: Since, the square and circle shapes cells for cellular communications cover an entire area without overlapping i.e. they can cover the entire geographical region without any gaps thus theseare not appropriate as compared to hexagonal shape cells.

Hexagonal shape cells.jpg

Fig: hexagonal shape cells

circle shapes cells.jpg

Fig: circle shapes cells


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