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CS601 - Data Communication - Lecture Handout 18

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Frequency Shift Keying (FSK)

  • Frequency of signal is varied to represent binary 1 or 0
  • The frequency of the signal during each bit duration is constant and depends on the bit (0 or 1)
  • Both peak amplitude and phase remains constan

Frequency Shift Keying (FSK)

Effect of Noise on FSK

  • Avoids most of the Noise problems of ASK
  • Because Rx 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 carrier

BW of FSK

BW of FSK

  • Although FSK shifts between two carrier frequencies, it is easier to analyze as two co-existing frequencies
  • BW required for FSK is equal to the Baud rate of the signal plus the frequency shift
  • Frequency Shift=Difference b/w two carrier frequencies
  • BW= (fc1 – fc0) +Nbaud

Example 5.11

Find the minimum BW for an FSK signal transmitted at 2000 bps. TX is in half duplex mode and carrier must be separated by 3000 Hz

Solution:

For FSK, if fc1 and fc2 are the carrier frequencies, then:
BW=Baud Rate + (fc1 – fc0)
Baud rate is the same as bit rate
BW=2000 + (fc1 – fc0) = 2000 + 3000 = 5000 Hz

Phase Shift Keying (PSK)

  • In PSK, phase of carrier is varied to represent binary 1 or 0
  • Both peak amplitude and frequency remains constant as the phase changes
  • For Example: if we start with a phase of 0 degrees to represent binary 0 , then we can change the phase to 180 degrees to send binary 1
  • The phase of signal during duration is constant and its value depends upon the bit (0 or

2PSK

  • The above method is often called 2 PSK, or Binary PSK, because two different phases ( 0 and 180 degrees) are used
  • Figure makes this point clear by showing the relationship of phase to bit value
  • A second diagram called constellation diagram or phase state diagram shows same relationship by illustrating only the phases

2PSK

Effect of Noise on PSK

  • PSK is not susceptible to the noise degradation that affects ASK, nor to the bandwidth limitations of FSK
  • Smaller variations in signal can be detected reliably by the receiver

4PSK

4PSK

  • Based on the above two facts, instead of utilizing only two variations of a signal, each representing one bit, we can use four variations and let each phase shift represent two bits

4PSK

4PSK1

BW for PSK

  • Minimum bandwidth required for PSK transmission is the same as ASK
  • As we have seen max bit rate in PSK is much greater than that of ASK
  • So while max baud rate of ASK and PSK are the same for a given BW, PSK bit rate using the same BW can be two or more times greater

8 PSK

8 PSK

QAM

  • 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
    • So far we have been changing only of the characteristics of the sine wave, But what if we alter two

 

  • What should these two characteristics be?
  • BW limitations make combination of FSK with other changes practically useless
  • Why not combine ASK and PSK?
  • ‘x’ variation in phase and ‘y variations in amplitude result into a total of x * y variations and corresponding no. of bits per variation

Quadrature Amplitude Modulation (QAM)

Quadrature Amplitude Modulation (QAM)

Variation of QAM

  • Variations of QAM are numerous
  • Any measurable amount changes in amplitude can be combined with any measurable no. of changes in Phase

4 QAM & 8 QAM (Figure)

  • In both case no. of amplitude shifts is more than the no. of phase shifts
  • Because amplitude changes are susceptible to Noise , number of phase shifts used by QAM is always larger than the amplitude shifts

Time domain plot of 8 QAM

Time domain plot of 8 QAM

Three possible variations of 16 QAM

Three possible variations of 16 QAM

Bandwidth for QAM

  • BW required for QAM is the same as in the case of ASK and PSK
  • QAM has the same advantages as PSK over ASK
  • Bit Baud Comparison

Bit Baud Comparison

Consult book section 5.3

Example 5.11

A constellation diagram consists of eight equally spaced points on a circle. If bit rate is 4800 bps, what is the Baud Rate?

Solution:

Constellation indicates 8 PSK with the points 45 degree apart
Baud Rate= 4800 / 3 = 1600 baud

Summary

  • Digital to Analog Conversion
  • Frequency Shift Keying (FSK)
  • Phase Shift Keying (PSK)
  • Quadrature Amplitude Modulation (QAM)

Reading Sections

  • Section 5.3, “Data Communications and Networking” 4th Edition by Behrouz A. Forouzan