The single tone modulating signal m(t) = m cos(2 pi/mt) is used to generate the

Question

The single tone modulating signal m(t) = m cos(2 pi/mt) is used to generate the VSB signal s(t) = 1/2 aAmAccos[2 pi(fc + fm)t] + 1/2 AmAr(1 - a)cos[2 pi(fc - fm)t] where a is a constant less than unity, representing the attenuation of the upper side frequently. If we represent this VSB signal as a quadrature carrier multiplex s(t) = Acm1(t)cos(2 pi fct) + A1m2(t)sin(2 pi fct) What is m2(t)? The VSB signal, plus the carrier A, cos(2 pi fct), is passed through an envelope detector. Determine the distortion produced by the quadrature component m2(t). What is the value of constant a for which this distortion reaches its worst possible condition? Using the message signal

Explanation / Answer

EXPANDING S(t), we get

s(t)=0.5aAmAccos(2fct)cos(2fmt)

-0.5aAmAcsin(2fct)sin(2fmt)+0.5(1-a)AmAcccos(2fct)cos(2fmt)+0.5(1-a)AmAcsin(2fct)sin(2fmt)

=0.5aAmAccos(2fct)cos(2fmt) +0.5(1-2a)AmAcsin(2fct)sin(2fmt)

after solving

quadrature component

- 0.5(1-2a)AmAcsin(2fmt)

b) after adding the carrier the signal will be

Ac(1+0.5Amcos(2fmt)) cos(2fct) +0.50(1-2a)AmAcsin(2fct)sin(2fmt)

envalop= Ac((1+0.5Amcos(2fmt))2+(0.50(1-2a)AmAcsin(2fmt)2)

=A0[1+0.5Amcos(2fct)]d(t)

where d(t)=(a+[(0.5Am(1-2a)sin(2fmt))/(1+0.5Amcos(2fmt))])

c) d(t) is greatest when a=0

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