For a binary FSK signal with a mask frequency of 49 kHz, a space-

For a binary FSK signal with a mask frequency of 49 kHz, a space-
| 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

A. 0.5 kHz

B. 1.0 kHz

C. 2.0 kHz

D. 4.0 kHz

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

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.

  • 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

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