written 2.8 years ago by | modified 2.8 years ago by |
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
written 2.8 years ago by | modified 2.8 years ago by |
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
written 2.8 years ago by | • modified 2.8 years ago |
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 -
Modulation Scheme | Bit Rate ($f_b$) | Baud Rate (N) | Minimum Bandwidth (B) | Bandwidth Efficiency $B_η$ |
---|---|---|---|---|
BPSK | 2400 bps | 2400 Bd | 2400 Hz | 1 bit per second per cycle of bandwidth |
QPSK | 2400 bps | 1200 Bd | 1200 Hz | 2 bits per second per cycle of bandwidth |
8 - PSK | 2400 bps | 800 Bd | 800 Hz | 3 bits per second per cycle of bandwidth |
16 - PSK | 2400 bps | 600 Bd | 600 Hz | 4 bits per second per cycle of bandwidth |