In a QAM receiver using synchronous detection, the local oscillator produces an

Question

In a QAM receiver using synchronous detection, the local oscillator produces an accurate cosine wave 2cos(omega_c t), but the local oscillator phase shifter has an error of pi/6 radians. What is the component of the quadrature channel message appearing in the in-phase output, and what is the component of the in-phase message appearing in the quadrature output? In a QAM receiver using synchronous detection, the local oscillator produces an accurate cosine wave 2cos(omega_c t), but the local oscillator phase shifter has an error of pi/6 radians. What is the component of the quadrature channel message appearing in the in-phase output, and what is the component of the in-phase message appearing in the quadrature output?

Explanation / Answer

The QAM signal is the sum of the two DSB-modulated signals, i.e.

QAM (t) = m1 (t) cos Ct + m­2 (t) sin Ct, where m1 (t) and m2 (t) are the baseband message signals to be transmitted.

The in-phase channel message component is,

xIN-PHASE (t) = QAM (t) (2 cos Ct) = 2 ( m1 (t) cos Ct + m­2 (t) sin Ct ) cos Ct

                                                           = 2 m1 (t) [ cos2 Ct ] + 2 m­2 (t) sin Ct cos Ct

                                                           = 2 m1 (t) [(1+ cos 2Ct)/2] + m­2 (t) sin 2Ct

                                                           = m1 (t) + m1 (t) cos 2Ct + m­2 (t) sin 2Ct   

The above signal is fed through the Low pass filter, which will filter the 2Ct components.

Hence the in-phase message in quadrature output will be m1 (t).

The quadrature channel message component is,

xQUADRATURE (t) = QAM (t) (2 sin (Ct + /6) ) = 2 ( m1 (t) cos Ct + m­2 (t) sin Ct ) sin (Ct + /6)

                                                 = 2 m1 (t) [cos Ct x sin (Ct + /6)] + 2 m­2 (t) [sin Ct x sin (Ct + /6)]

2 cos A sin B = sin (A+B) – sin (A-B)

i.e. 2 cos Ct x sin (Ct + /6) = sin (2Ct + /6) – sin (-/6) = sin (2Ct + /6) + 0.5

2 sin A sin B = sin (A+B) + sin (A-B)

i.e. sin Ct x sin (Ct + /6) = sin (2Ct + /6) + sin (-/6) = sin (2Ct + /6) - 0.5

xQUADRATURE (t) = m1 (t) [sin (2Ct + /6) + 0.5] + m­2 (t) [sin (2Ct + /6) - 0.5]

                        = 0.5 [ m1 (t) + m2 (t) ] + sin 2Ct [ m1 (t) + m2 (t) ]

The above signal is fed through the Low pass filter, which will filter the 2Ct components.

Hence the quadrature message in quadrature output will be 0.5 [ m1 (t) + m2 (t) ]

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