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Analog Transmission-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

Birla Institute of Technology and ScienceData Communication Systems and Computer Networks

This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: Analog, Transmission, Solutions, Digital, Aplitude, Frequency, Phase, Peak, BPSK, Calculate, Origin, QPSK

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

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CHAPTER 5
Analog Transmission
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. Normally, analog transmission refers to the transmission of analog signals using a
band-pass channel. Baseband digital or analog signals are converted to a complex
analog signal with a range of frequencies suitable for the channel.
3. The process of changing one of the characteristics of an analog signal based on the
information in digital data is called digital-to-analog conversion. It is also called
modulation of a digital signal. The baseband digital signal representing the digital
data modulates the carrier to create a broadband analog signal.
5. We can say that the most susceptible technique is ASK because the amplitude is
more affected by noise than the phase or frequency.
7. The two components of a signal are called I and Q. The I component, called in-
phase, is shown on the horizontal axis; the Q component, called quadrature, is
shown on the vertical axis.
9.
a. AM changes the amplitude of the carrier
b. FM changes the frequency of the carrier
c. PM changes the phase of the carrier
Exercises
11. We use the formula S = (1/r) × N, but first we need to calculate the value of r for
each case.
a. r = log22 = 1 → S = (1/1) × (2000 bps) = 2000 baud
b. r = log22 = 1 → S = (1/1) × (4000 bps) = 4000 baud
c. r = log24 = 2 → S = (1/2) × (6000 bps) = 3000 baud
d. r = log264 = 6 → S = (1/6) × (36,000 bps) = 6000 baud
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1

CHAPTER 5

Analog Transmission

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. Normally, analog transmission refers to the transmission of analog signals using a band-pass channel. Baseband digital or analog signals are converted to a complex analog signal with a range of frequencies suitable for the channel.
  2. The process of changing one of the characteristics of an analog signal based on the information in digital data is called digital-to-analog conversion. It is also called modulation of a digital signal. The baseband digital signal representing the digital data modulates the carrier to create a broadband analog signal.
  3. We can say that the most susceptible technique is ASK because the amplitude is more affected by noise than the phase or frequency.
  4. The two components of a signal are called I and Q. The I component, called in- phase, is shown on the horizontal axis; the Q component, called quadrature, is shown on the vertical axis.

a. AM changes the amplitude of the carrier b. FM changes the frequency of the carrier c. PM changes the phase of the carrier

Exercises

  1. We use the formula S = (1/r) × N , but first we need to calculate the value of r for each case.

a. r = log 22 = 1 → S = (1/1) × (2000 bps) = 2000 baud b. r = log 22 = 1 → S = (1/1) × (4000 bps) = 4000 baud c. r = log 24 = 2 → S = (1/2) × (6000 bps) = 3000 baud d. r = log 264 = 6 → S = (1/6) × (36,000 bps) = 6000 baud

  1. We use the formula r = log 2 L to calculate the value of r for each case.
  2. See Figure 5.

a. This is ASK. There are two peak amplitudes both with the same phase ( degrees). The values of the peak amplitudes are A 1 = 2 (the distance between the first dot and the origin) and A 2 = 3 (the distance between the second dot and the origin). b. This is BPSK, There is only one peak amplitude (3). The distance between each dot and the origin is 3. However, we have two phases, 0 and 180 degrees. c. This can be either QPSK (one amplitude, four phases) or 4-QAM (one ampli- tude and four phases). The amplitude is the distance between a point and the origin, which is (2 2 + 2^2 ) 1/2^ = 2.83. d. This is also BPSK. The peak amplitude is 2, but this time the phases are 90 and 270 degrees.

  1. We use the formula B = (1 + d) × (1/r) × N , but first we need to calculate the value of r for each case.

a. log 24 = 2 b. log 28 = 3 c. log 24 = 2 d. log 2128 = 7

Figure 5.1 Solution to Exercise 15

a. r = 1 → B= (1 + 1) × (1/1) × (4000 bps) = 8000 Hz b. r = 1 → B = (1 + 1) × (1/1) × (4000 bps) + 4 KHz = 8000 Hz c. r = 2 → B = (1 + 1) × (1/2) × (4000 bps) = 2000 Hz d. r = 4 → B = (1 + 1) × (1/4) × (4000 bps) = 1000 Hz

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