In a ASK system, the amplitude of the carrier signal is 10V and t

In a ASK system, the amplitude of the carrier signal is 10V and t
| In a ASK system, the amplitude of the carrier signal is 10V and the binary to be transmitted is {1, 0, 1, 1, 1, 1, 0, 0, …}. The amplitude of the output will be:

A. {± 5V, 0, ± 5V, ± 5 V, ± 5V, ± 5V, 0, 0,…}

B. {0, ± 10V, 0, 0, 0, 0, ± 10 V, ± 10 V, …}

C. {± 10 V, 0, ± 10 V, ± 10 V, ± 10 V, ± 10 V, 0, 0, …}

D. {± 1 V, 0, ± 1 V, ± 1 V, ± 1V, ± 1V, 0, 0,…}

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

ASK System:

1. For ASK Transmitter on-off keying is used.

2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.

1 : s1(t) = Ac cos 2πfct

0 : s2 (t) = 0

 

Hence,

ASK = [± 10 V, 0, ± 10 V, ± 10 V, ± 10 V, ± 10 V, 0, 0]

FSK (Frequency Shift Keying):

In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S1 (A) = Acos 2π fHt

For binary ‘0’ → S2 (t) = A cos 2π fLt . The constellation diagram is as shown:

ASK(Amplitude Shift Keying):

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (t) = 0

The Constellation Diagram Representation is as shown:

 

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

PSK(Phase Shift Keying):

In PSK (phase shift keying) binary 1 is represented with a carrier signal and binary O is represented with 180° phase shift of a carrier

For binary ‘1’ → S1 (A) = Acos 2π fct

For binary ‘0’ → S2 (t) = A cos (2πfct + 180°) = - A cos 2π fct

The Constellation Diagram Representation is as shown:

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