# 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

### 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

• 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

### 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

• 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

### 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

### 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

### 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

### Bandwidth for QAM

• BW required for QAM is the same as in the case of ASK and PSK
• 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)