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Mumbai University > Electronics and Telecommunication Engineering > Sem 6 > Data Communication

Marks: 10 Marks

Year: Dec 2016

 modified 2.3 years ago  • written 2.3 years ago by Sayali Bagwe • 2.1k
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It is combination of PSK and ASK (Hybrid Scheme)

To pure PSk though need hybrid 1, distance between two signal. To increase 'd' we have to heavily energy of every symbol. This is possible by changing amplitude of signal alongwith phase constellation diagram.

Energy of signal $S_1(t) = \sqrt{ES_1} = \sqrt{a^2 + a^2}$

Energy of signal $s_2(t) = \sqrt{ES_2} = \sqrt{qa^2} + a^2$

Energy of signal $S_3(t) = \sqrt{ES_3} = \sqrt{qa^2 + Qa^2}$

Energy of signal $S_4(t) = \sqrt{ES_4} = \sqrt{a^2 + qa^2}$

Distance between two signal points d = 2a

Average energy in quadrant $ES = \frac{ES_1 + ES_2 + ES_3 + ES_4}{4}$

$ES = 10a^2, \hspace{2cm} a = \sqrt{0.1ES}$

For 16 QASK, M = 16, N = 4, ES = 4 Eb , $a = \sqrt{0.4E_b}$

$d_{16QASK} = 2 \sqrt{0.4E_b} \hspace{1cm} [\because d = 2a]\\ d_{16QASK} = 1.264 \sqrt{E_b}$

$d_{16PSK} = 2 \sqrt{4E_b} sin \bigg( \frac{\pi}{16}\bigg) = 0.7 \sqrt{E_b}$

$S_1(t) = a \phi_1 (t) - a \phi_2 (t)\\ S_2(l) = a \phi_1(t) - 3a \phi_2(t)\\ S_3(l) = 3a \phi_1(t) - 3a \phi_2 (t)\\ S_4(t) = 3a \phi_1(t) - a \phi_2(t)$

$s(t) = k_1 a \phi_1 (t) + K_2 a \phi_2 (t)$

Where $K_1$ & $K_2 = \pm 1 or \pm 3$

But $\phi_1(t) = \sqrt{\frac{2}{T_3}} cos(2 \pi fct), \phi_2(t) = \frac{2}{t_3} sin (2 \pi fct)$

$s(t) = K_1 \sqrt{0.1ES}\sqrt{\frac{2}{T_3}} cos (2 \pi fct) + K_2 \sqrt{0.1 Es} \sqrt{\frac{2}{T_3}} sin (2 \pi fct)\\ s(t) = K_1 \sqrt{0.2 F_s} cos (2 \pi fct) + K_2 \sqrt{0.2 F_s} sin (2 \pi Fct)$

Output Transmitter:

$s(t) = A_o(t) \sqrt{P_s} cos (2 \pi fct) + A_e(t) \sqrt{P_s} sin (2 \pi Fct)$

Comparing with required output.

$A_o(t) = K_1 \sqrt{0.2} = \pm \sqrt{0.2} / \pm 3 \sqrt{0.2}\\ A_e(t) = K_2 \sqrt{0.2} = \pm \sqrt{0.2} / \pm 3 \sqrt{0.2}$