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# Lecture 11

## on Mar 16, 2014

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## Lecture 11Presentation Transcript

• DATA COMMUNICATIONS & NETWORKING LECTURE-11 Course Instructor : Sehrish Rafiq Department Of Computer Science University Of Peshawar
• LECTURE OVERVIEW  Digital-To –Analog Modulation  ASK  FSK  PSK  QAM
• DIGITAL-TO-ANALOG MODULATION  Modulation of binary data or digital to analog modulation is the process of changing one of the characteristics of an analog signal based on the information a digital signal(0’s and 1’s).  When you transmit data from one computer to another across a public phone line, the original data is digital but because telephone lines carry analog signals, the data must be converted.  The digital data must be modulated on an analog signal that has been manipulated to look like two distinct values corresponding to binary 1 and binary 0.
• DIGITAL-TO-ANALOG MODULATION
• BIT RATE BAUD RATE & CARRIER FREQUENCY  Bit rate is the number of bits per second.  Baud rate is the number of signal units per second.  Baud rate < Bit rate.  If an analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate???  The bit rate of a signal is 3000.If each signal unit carries 6 bits, what is the baud rate??? Example 1Example 1
• SOLUTIONS  Baud rate=number of signal units per second  =>Baud rate =1000 bauds per second(bauds/s).  Bit Rate = baud rate x number of bits per signal unit  =>Bit rate=1000x 4=4000bps.  Baud rate=bit rate/number of bits per signal unit  =>Baud rate =3000/6=500 baud/s. Solution 1Solution 1 Solution 2Solution 2
• CARRIER FREQUENCY  In analog transmission, the sending device produces a high –frequency signal that acts a basis for the information signal.  This base signal is called the carrier signal or carrier frequency.  The receiving device is tuned to the frequency of the carrier signal that it expects from the sender.
• CARRIER FREQUENCY  Digital information then modulates the carrier signal by modifying one or more of its characteristics( amplitude, frequency or phase).  This kind of modification is called modulation or shift keying.  The information signal is called the modulating signal.
• TYPES OF DIGITAL-TO-ANALOG MODULATION
• ASK(AMPLITUDE SHIFT KEYING)  In amplitude shift keying,the strength of the carrier signal is varied to represent binary 1 or 0.  Both frequency and phase remain constant while the amplitude changes.  Which voltage represents 1 and which represents 0 are left to the system designers.  A bit duration is the period of time that defines 1 bit.  The peak amplitude of the signal during each bit duration is constant and its value depends on the bit (0 or 1).
• PROBLEMS WITH ASK  ASK transmission is highly susceptible to noise interference.  The term noise refers to unintentional voltages introduced on to a line by various phenomena such as heat or electromagnetic induction created by other sources.  These unintentional voltages combine with the signal to change the amplitude.  A 0 can be changed to 1 and a 1 can be changed to 0.
• PROBLEMS WITH ASK CONTINUED…  High amount of energy is required.  A popular ASK technique is called on/off keying(OOK).  In OOK one of the values is represented by no voltage.  The advantage is a reduction in the amount of energy required to transmit information.
• ON-OFF KEYING OR BINARY AMPLITUDE SHIFT KEYING
• BAND WIDTH FOR ASK  Although there is only one carrier frequency, the process of modulation produces a complex signal that is a combination of many simple signals, each with a different frequency.  When we decompose an ASK-modulated signal ,we get a spectrum of many simple frequencies.  However the most significant ones are those between fc - N baud/2 and fc + N baud/2 with fc as the carrier frequency and N baud as the baud rate.
• BAND WIDTH FOR ASK  The Bandwidth requirements of ASK are calculated using the formula:  BW=(1+d) N baud  d is a constant factor whose value lies between 0 and 1.  The value of d depends on the modulation and filtration process and the medium.
• Example 3Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half- duplex. SolutionSolution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.
• Example 4Example 4 Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? SolutionSolution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.
• FREQUENCY SHIFT KEYING  In FSK, the frequency of the carrier signal is varied to represent binary 1 or 0.  The frequency of the signal during each bit duration is constant and its value depends on the bit (0 or 1).  Both peak amplitude and phase remain constant.
• HOW FSK HANDLES NOISE PROBLEM???  FSK avoids most of the problems from noise.  Because the receiving device is looking for specific frequency changes over a given number of periods, it can ignore voltage spikes.  The limiting factors of FSK are the physical capabilities of the medium.
• BAND WIDTH REQUIREMENT FOR FSK  Although there are only two carrier frequencies, the process of modulation produces a composite signal that is a combination of many simple signals each with a different frequency.  We can say that the FSK spectrum is a combination of two ASK spectra centered on fc0 and fc1 .  BW= fc1 - fc0 + N baud .
• PSK  In Phase shift keying, the phase of the carrier is varied to represent binary 1 or 0.  Both peak amplitude and frequency remain constant as the phase changes.  For example we start with a phase of 0 degree to represent binary 0 then we can change the phase to 180 degree to represent binary 1.  The phase of the signal during each bit duration is constant and the value depends on the bit (0 or 1).  This method is often called 2-PSK or binary PSK , because two different phases(0 degree and 180 degree are used).
• 2-PSK OR BINARY PSK METHOD
• 2-PSK OR BINARY PSK METHOD
• 4-PSK METHOD
• 4- PSK METHOD
• 8-PSK
• BAND WIDTH FOR PSK  The minimum bandwidth required for PSK transmission is the same as that required for ASK Transmission.
• BIT RATE IN PSK  The maximum baud rates of ASK and PSK are same for a given bandwidth.  PSK bit rates using the same bandwidth as ASK can be 2 or more times greater.
• LIMITATIONS OF PSK  PSK is limited by the ability of the equipment to distinguish small differences in phase.  This factor limits its potential bit rate.
• Example 5Example 5 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. SolutionSolution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
• Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Example 6Example 6 SolutionSolution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
• QAM(QUADRATURE AMPLITUDE MODULATION)  Band width limitations makes combinations of FSK practically useless.  Why not combine ASK and PSK?  If we have x variations in phase and y variations in amplitude, giving us x times y possible variations and the corresponding number of bits per variation.  Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit(bit, dibit, tribit
• QAM CONTINUED…  Theoretically, any measurable number of changes in amplitude can be combined with any measurable number of changes in phase.
• 4-QAM & 8-QAM CONSTELLATIONS
• 8-QAM SIGNAL
• BIT AND BAUD RATE COMPARISON
• BIT AND BAUD RATE COMPARISON ModulationModulation UnitsUnits Bits/BauBits/Bau dd Baud rateBaud rate Bit Rate ASK, FSK, 2-PSKASK, FSK, 2-PSK Bit 1 N N 4-PSK, 4-QAM4-PSK, 4-QAM Dibit 2 N 2N 8-PSK, 8-QAM8-PSK, 8-QAM Tribit 3 N 3N 16-QAM16-QAM Quadbit 4 N 4N 32-QAM32-QAM Pentabit 5 N 5N 64-QAM64-QAM Hexabit 6 N 6N 128-QAM128-QAM Septabit 7 N 7N 256-QAM256-QAM Octabit 8 N 8N
• Example 7Example 7 A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? SolutionSolution The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud
• Example 8Example 8 Compute the baud rate for a 72,000-bps 64-QAM signal. SolutionSolution A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000 / 6 = 12,000 baud
• THANKS !!!