4 a] Waveform of the Input and Modulated QPSK

Input Stream = 101101

4 b] 8 - PSK

The given data -

Time of Signalling Element = $ t_s$ = 0.1 ms

To find -

i] Bit Rate = $f_b$ = ?

ii] Baud Rate = N = ?

iii] Minimum Bandwidth = B = ?

iv] Truth Time Table and its Constellation Diagram

Formulae -

Baud Rate -

$$ N = \frac {1}{t_s}$$

Where $t_s$ is in Seconds.

Bit Rate -

$$ f_b = 3 \times N$$

Minimum Bandwidth -

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

That is nothing but Baud Rate (N) = Minimum Bandwidth (B).

Because in PSK Modulation Schemes Baud Rate (N) is the same as Minimum Bandwidth (B).

Solution -

i] Baud Rate -

$$ N = \frac {1}{t_s} = \frac {1}{0.1 \times 10^{-3}} = 10 \ Kbps$$

ii] Bit Rate -

$$ f_b = 3 \times N = 3 \times 10 = 30 \ Kbps$$

iii] Minimum Bandwidth -

$$ B = \frac{f_b}{3} = \frac{30}{3} = 10 \ kHz $$

iv] Truth Table and Constellation Diagram for 8 - PSK

• In 8 - PSK have 8 = $2^3$ different phases, each phase can represent by using 3 bits called Tribit.
• It can be extended, by varying the signal by shifts of 45 deg (instead of 90 deg in 4-PSK).

Truth Table for 8 - PSK -

Constellation Diagram for 8 - PSK -