**Related Content:** CS601 - VU Lectures, Handouts, PPT Slides, Assignments, Quizzes, Papers & Books of Data Communication

- 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

- 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

- 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

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

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

- 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

- 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

- 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

- 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

- 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

**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

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

- 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

- 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

Consult book section 5.3

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

Constellation indicates 8 PSK with the points 45 degree apart

Baud Rate= 4800 / 3 = 1600 baud

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

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