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Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE) PDF Download

INTRODUCTION

The slides cover the following topics:

1. Digital to Analog Conversion

2. Types of Digital to Analog Conversion

3. Bit Rate and Baud Rate

4. Amplitude Shift Keying (ASK)

5. Frequency Shift Keying (FSK)

6. Phase Shift Keying (PSK)

7. Constellation Diagrams

 

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Digital-to-Analog Conversion

Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.

 

Topics discussed in this section:

1. Aspects of Digital-to-Analog Conversion
2. Amplitude Shift Keying
3. Frequency Shift Keying
4. Phase Shift Keying
5. Quadrature Amplitude Modulation
 
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Digital to Analog Conversion
 
1. Digital data needs to be carried on an analog signal.
2. A carrier signal (frequency fc) performs the function of transporting the digital data in an analog waveform.
3. The analog carrier signal is manipulated to uniquely identify the digital data being carried.
 
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Figure 5.1  Digital-to-analog conversion

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Figure 5.2  Types of digital-to-analog conversion

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Note:

Bit rate, N, is the number of bits per second (bps).

Baud rate is the number of signal elements per second (bauds).

In the analog transmission of digital data, the signal or baud rate is less than or equal to the bit rate.

S=Nx1/r bauds

Where r is the number of data bits per signal element.

 

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Example 5.1

An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.

 

Solution

In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

  Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Example 5.2

An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?

 

Solution

In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Amplitude Shift Keying (ASK)

1. ASK is implemented by changing the amplitude of a carrier signal to reflect amplitude levels in the digital signal.
 
2. For example: a digital “1” could not affect the signal, whereas a digital “0” would, by making it zero.
 
3. The line encoding will determine the values of the analog waveform to reflect the digital data being carried.
 
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Bandwidth of ASK
 
1. The bandwidth B of ASK is proportional to the signal rate S.

B = (1+d)S

 
2. “d” is due to modulation and filtering, lies between 0 and 1.
 
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Figure 5.3  Binary amplitude shift keying

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Figure 5.4  Implementation of binary ASK

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Example 5.3

We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?

 

Solution

The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at                 fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Example 5.4

In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.

 

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Figure 5.5  Bandwidth of full-duplex ASK used in Example 5.4

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Frequency Shift Keying

 
1. The digital data stream changes the frequency of the carrier signal, fc .
 
2. For example, a “1” could be represented by f1=fc +Δf, and a “0” could be represented by f2=fc - Δf.
 
 
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Figure 5.6  Binary frequency shift keying

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Bandwidth of FSK

 

If the difference between the two frequencies (f1 and f2) is 2Δf, then the required BW B will be:

B = (1+d)xS + 2Δf

 

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Example 5.5

We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?

 

Solution

This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

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Coherent and Non Coherent

1. In a non-coherent FSK scheme, when we change from one frequency to the other, we do not adhere to the current phase of the signal.
 
2. In coherent FSK, the switch from one frequency signal to the other only occurs at the same phase in the signal.
 

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Multi level FSK

1. Similarly to ASK, FSK can use multiple bits per signal element.
 
2. That means we need to provision for multiple frequencies, each one to represent a group of data bits.
 
3. The bandwidth for FSK can be higher

B = (1+d)xS + (L-1)/2Δf = LxS

 

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Figure 5.7  Bandwidth of MFSK used in Example 5.6

 

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

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Example 5.6

We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.

 

Solution

We can have L = 23 = 8. The baud rate is S = 3 Mbps/3 = 1 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1M = 8M. Figure 5.8 shows the allocation of frequencies and bandwidth.

 

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Figure 5.8  Bandwidth of MFSK used in Example 5.6

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Phase Shift Keyeing

1. We vary the phase shift of the carrier signal to represent digital data.
 
2. The bandwidth requirement, B is:

B = (1+d)xS

3. PSK is much more robust than ASK as it is not that vulnerable to noise, which changes amplitude of the signal.

 

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Figure 5.9  Binary phase shift keying

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

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Figure 5.10  Implementation of BASK

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Quadrature PSK

1. To increase the bit rate, we can code 2 or more bits onto one signal element.
 
2. In QPSK, we parallelize the bit stream so that every two incoming bits are split up and PSK a carrier frequency. One carrier frequency is phase shifted 90o from the other - in quadrature.
 
3. The two PSKed signals are then added to produce one of 4 signal elements. L = 4 here.
 
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Figure 5.11  QPSK and its implementation

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Example 5.7

Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.

 

Solution

For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.

 

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Constellation Diagrams

 
1. A constellation diagram helps us to define the amplitude and phase of a signal when we are using two carriers, one in quadrature of the other.
 
2. The X-axis represents the in-phase carrier and the Y-axis represents quadrature carrier.
 

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Figure 5.12  Concept of a constellation diagram

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

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Example 5.8

Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.

 

Solution

Figure 5.13 shows the three constellation diagrams.

 

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Figure 5.13  Three constellation diagrams

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

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Note

Quadrature amplitude modulation is a combination of ASK and PSK.

 

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Figure 5.14  Constellation diagrams for some QAMs

Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

 

 

 

 

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FAQs on Chapter 5 - Analog Transmission PPT, Data Communications and Networking, ECE - Computer Science Engineering (CSE)

1. What is analog transmission?
Ans. Analog transmission is a method of transmitting data signals in the form of continuous waves. In this type of transmission, the data is represented by varying the amplitude, frequency, or phase of the carrier signal.
2. What are the advantages of analog transmission?
Ans. Some advantages of analog transmission include: - It can carry a large amount of information in a given bandwidth. - Analog signals are less susceptible to noise and interference compared to digital signals. - It is compatible with existing analog systems and devices. - Analog transmission allows for smooth transmission of continuous signals, making it suitable for applications like audio and video transmission.
3. What are the limitations of analog transmission?
Ans. Analog transmission has some limitations, including: - It is more prone to signal degradation over long distances compared to digital transmission. - Analog signals are more susceptible to noise and interference, which can result in signal distortion. - It is difficult to separate multiple analog signals transmitted simultaneously without using complex techniques. - Analog transmission is less secure compared to digital transmission, as it is easier to intercept and modify analog signals.
4. How does analog transmission differ from digital transmission?
Ans. Analog transmission differs from digital transmission in the way data is represented and transmitted. In analog transmission, data is represented by continuously varying the properties of the carrier signal, such as amplitude, frequency, or phase. On the other hand, digital transmission represents data as discrete binary values (0s and 1s) and transmits them using digital signals. Digital transmission offers better noise immunity, easier error detection and correction, and higher transmission accuracy compared to analog transmission.
5. What are some examples of analog transmission?
Ans. Some examples of analog transmission include: - AM and FM radio broadcasting: The audio signals are modulated onto carrier signals using amplitude modulation (AM) or frequency modulation (FM) techniques and transmitted over the airwaves. - Analog telephone systems: Voice signals are transmitted using analog signals over copper wire networks. - Analog CCTV systems: Video signals are transmitted using analog signals over coaxial cables to monitor and record surveillance footage.
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