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Modulation Techniques

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

a. Two level Amplitude Shift Keying output

b. Two level Frequency Shift Keying output

c. Two level Phase Shift Keying output

d. Differential Phase shift keying output

e. Four level Amplitude Shift Keying output

f. Four level Phase Shift Keying output

g. Eight level Amplitude Shift Keying output

h) Eight level Amplitude Shift Keying output

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.

Given fc = 500 kHz and 2fd = 50 kHz

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:

Given input signal

Step1:

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

Step2:

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

Output equation is

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

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

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

 1100 11011 10111100 11011 11001    11011 1000

Code-word sent = 10110100

 1100 11011 10110100 11011 11011    11011 000

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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

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.

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.

Fig: Diagram of TCP protocols for wired and wireless LANs

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.

Fig: hexagonal shape cells

Fig: circle shapes cells

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