Question

A 2000 bps binary information data signal is required to be transmitted in half-duplex mode using BFSK digital modulation technique. If the separation between two carrier frequencies is 4000 Hz, then the minimum bandwidth of the BFSK signal is
1. 4 kHz
2. 6 kHz
3. 8 kHz
4. 12 kHz
5.

Answer 1

Correct Answer - Option 2 : 6 kHz

Concept:

1) Half-duplex mode sender can send the data and also can receive the data but one at a time.it is two-way directional communication but one at a time. So user can use the entire bandwidth.

Bandwidth of BFSK given as:

(B_T)_FSK = (f_c1 - f_c2) + 2(B_T)_{baseband signal}

Where,

(B_T)_FSK→ bandwidth of BFSK, f_c1-f_c2→ separation between two carrier frequency, (B_T)_{baseband signal}→ bandwidth of baseband signal

If pulse shape is not mentioned in the question then consider Nyquist pulse.

For Nyquist pulse (B_T) = R_b/2

 (BT)FSK = (fc1-fc2) + 2(BT)baseband signal

(B_T)_FSK = (fc1-fc2) + 2(Rb/2)

(BT)FSK = (fc1-fc2) + R 

Calculations:

Given (fc1-fc2) = 4000 and R = 2000

(BT)_FSK = 4000 + 2000 = 6 kHz