For a binary FSK signal with a mask frequency of 49 kHz, a space-frequency of 51 kHz and on the input bit rate of 2 kbps, the peak frequency deviation will be

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  1. 0.5 kHz
  2. 1.0 kHz
  3. 2.0 kHz
  4. 4.0 kHz

Answer (Detailed Solution Below)

Option 2 : 1.0 kHz
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Detailed Solution

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Concept:

In FSK (Frequency Shift Keying), binary 1 is represented with a high-frequency carrier signal, and binary 0 is represented with a low-frequency carrier, i.e. in FSK, the carrier frequency is switched between 2 extremes.

F2 S.B Madhu 21.10.19 D 6

  • Frequency measurements of the FSK signal are usually stated in terms of “shift” and center frequency. The shift is the frequency difference between the mark and space frequencies.
  • The nominal center frequency is halfway between the mark and space frequencies.
  • Frequency deviation is equal to the absolute value of the difference between the center frequency and the mark or space frequencies.​

 

The deviation is also equal, numerically, to one-half of the shift, i.e.

|fs - fm|= 2 Δf

Δf = frequency deviation

fs = space-frequency

fm = mask frequency

Calculation:

With fs = 51 kHz and fm = 49 kHz

\({\rm{\Delta }}f = \frac{{\left| {{f_s} - {f_m}\;} \right|}}{2}\) 

\(= \frac{{\left| {51 - 49} \right|}}{2}\;kHz\) 

Δf = 1 kHz

26 June 1

The bandwidth of FSK is given by:

\(\left( {{f_s} + \frac{1}{{{T_b}}}} \right) - \left( {{f_m} - \frac{1}{{{T_b}}}} \right)\) 

\(= \left( {{f_s} - {f_m}} \right) + \frac{2}{{{T_b}}}\) 

|fs - fm|= 2 Δf

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