Determine the minimum bandwidth, baud, and bandwidth efficiency for the following bit rates and modulation schemes-BPSK, QPSK, 8-PSK, and 16-PSK.where fb = 2400 bps

The given data -

$f_b$ = Bit Rate = 2400 bps

To find -

B = Minimum Bandwidth = ?

N = Baud Rate = ?

$B_η$ = Bandwidth Efficiency = ?

for Modulation Schemes like BPSK, QPSK, 8 - PSK & 16 - PSK

Formulae -

Important Point to Remember - in PSK Modulation Schemes Baud Rate (N) is the same as Minimum Bandwidth (B).

BPSK Modulation Scheme -

$$B = Minimum\ Bandwidth = f_b$$

$$N = Baud \ Rate = f_b$$

QPSK Modulation Scheme -

$$B = Minimum \ Bandwidth = \frac{f_b}{2}$$

$$N = Baud \ Rate = \frac{f_b}{2}$$

8 - PSK Modulation Scheme -

$$B = Minimum \ Bandwidth = \frac{f_b}{3}$$

$$N = Baud \ Rate = \frac{f_b}{3}$$

16 - PSK Modulation Scheme -

$$B = Minimum \ Bandwidth = \frac{f_b}{4}$$

$$N = Baud\ Rate = \frac{f_b}{4}$$

$$B_η = Bandwidth \ Efficiency = \frac{Transmission \ Bit \ Rate (bps)}{Minimum \ Bandwidth (Hz)} = \frac{f_b}{B}$$

Solution -

1] BPSK Modulation Scheme -

$$B = Minimum \ Bandwidth = f_b = 2400 Hz$$

$$N = Baud \ Rate = f_b = 2400 Bd$$

$$B_η = Bandwidth\ Efficiency = \frac{2400 bps}{2400 Hz} = 1$$

$B_η$ = Bandwidth Efficiency for BPSK = 1 bit per second per cycle of bandwidth

2] QPSK Modulation Scheme -

$$B = Minimum \ Bandwidth = \frac{f_b}{2} = \frac{2400}{2}= 1200 Hz$$

$$N = Baud \ Rate = \frac{f_b}{2} = \frac{2400}{2} = 1200 Bd$$

$$B_η = Bandwidth \ Efficiency = \frac{2400 bps}{1200 Hz} = 2$$

$B_η$ = Bandwidth Efficiency for QPSK = 2 bits per second per cycle of bandwidth

3] 8 - PSK Modulation Scheme -

$$B = Minimum\ Bandwidth = \frac{f_b}{3} = \frac{2400}{3} = 800 Hz$$

$$N = Baud \ Rate = \frac{f_b}{3} = \frac{2400}{3} = 800 Bd$$

$$B_η = Bandwidth \ Efficiency = \frac{2400 bps}{800 Hz} = 3$$

$B_η$ = Bandwidth Efficiency for 8 - PSK = 3 bits per second per cycle of bandwidth

4] 16 - PSK Modulation Scheme -

$$B = Minimum\ Bandwidth = \frac{f_b}{4} = \frac{2400}{4} = 600 Hz$$

$$N = Baud \ Rate = \frac{f_b}{4} = \frac{2400}{4}= 600 Bd$$

$$B_η = Bandwidth\ Efficiency = \frac{2400 bps}{600 Hz} = 4$$

$B_η$ = Bandwidth Efficiency for 16 - PSK = 4 bits per second per cycle of bandwidth

Final Summarization of the above Findings -