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The single-tone modulating signal \( m(t) = A_m \cos(2\pi f_m t) \) is used to generate the VSB signal:
\[
s(t) = \frac{1}{2} a A_m A_c \cos[2\pi(f_c + f_m)t] + \frac{1}{2} A_m A_c(1 - a) \cos[2\pi(f_c - f_m)t]
\]
where \( a \) is a constant, less than unity, representing the attenuation of the upper side frequency.
(a) Find the quadrature component of the VSB signal \( s(t) \).
(b) The VSB signal, plus the carrier \( A_c \cos(2\pi f_c t) \), is passed through an envelope detector. Determine the distortion produced by the quadrature component.
(c) What is the value of constant \( a \) for which this distortion reaches its worst possible condition?
Most Upvoted Answer
The single-tone modulating signal \( m(t) = A_m \cos(2\pi f_m t) \) is...
Quadrature Component of the VSB Signal
The quadrature component of the VSB signal is given by the second term in the expression for \( s(t) \):
\[ \frac{1}{2} A_m A_c (1 - a) \cos[2\pi(f_c - f_m)t] \]

Distortion Produced by the Quadrature Component
When the VSB signal is passed through an envelope detector along with the carrier, the quadrature component leads to distortion in the demodulated signal. This distortion occurs because the envelope detector responds to the sum of the carrier and the VSB signal, resulting in interference from the quadrature component.

Worst Possible Condition of Distortion
The distortion produced by the quadrature component is maximum when the phase difference between the carrier and the quadrature component is 90 degrees. This happens when the value of \( a \) is such that the amplitude of the quadrature component is equal to that of the main VSB signal. In this case, the interference caused by the quadrature component is at its peak, leading to the worst possible distortion in the demodulated signal.
Therefore, the value of constant \( a \) for which the distortion reaches its worst possible condition is when \( a = 0.5 \). At this value, the amplitude of the quadrature component is half that of the main VSB signal, resulting in maximum interference and distortion in the demodulated signal.
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The single-tone modulating signal \( m(t) = A_m \cos(2\pi f_m t) \) is used to generate the VSB signal:\[s(t) = \frac{1}{2} a A_m A_c \cos[2\pi(f_c + f_m)t] + \frac{1}{2} A_m A_c(1 - a) \cos[2\pi(f_c - f_m)t]\]where \( a \) is a constant, less than unity, representing the attenuation of the upper side frequency.(a) Find the quadrature component of the VSB signal \( s(t) \).(b) The VSB signal, plus the carrier \( A_c \cos(2\pi f_c t) \), is passed through an envelope detector. Determine the distortion produced by the quadrature component.(c) What is the value of constant \( a \) for which this distortion reaches its worst possible condition?
Question Description
The single-tone modulating signal \( m(t) = A_m \cos(2\pi f_m t) \) is used to generate the VSB signal:\[s(t) = \frac{1}{2} a A_m A_c \cos[2\pi(f_c + f_m)t] + \frac{1}{2} A_m A_c(1 - a) \cos[2\pi(f_c - f_m)t]\]where \( a \) is a constant, less than unity, representing the attenuation of the upper side frequency.(a) Find the quadrature component of the VSB signal \( s(t) \).(b) The VSB signal, plus the carrier \( A_c \cos(2\pi f_c t) \), is passed through an envelope detector. Determine the distortion produced by the quadrature component.(c) What is the value of constant \( a \) for which this distortion reaches its worst possible condition? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The single-tone modulating signal \( m(t) = A_m \cos(2\pi f_m t) \) is used to generate the VSB signal:\[s(t) = \frac{1}{2} a A_m A_c \cos[2\pi(f_c + f_m)t] + \frac{1}{2} A_m A_c(1 - a) \cos[2\pi(f_c - f_m)t]\]where \( a \) is a constant, less than unity, representing the attenuation of the upper side frequency.(a) Find the quadrature component of the VSB signal \( s(t) \).(b) The VSB signal, plus the carrier \( A_c \cos(2\pi f_c t) \), is passed through an envelope detector. Determine the distortion produced by the quadrature component.(c) What is the value of constant \( a \) for which this distortion reaches its worst possible condition? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The single-tone modulating signal \( m(t) = A_m \cos(2\pi f_m t) \) is used to generate the VSB signal:\[s(t) = \frac{1}{2} a A_m A_c \cos[2\pi(f_c + f_m)t] + \frac{1}{2} A_m A_c(1 - a) \cos[2\pi(f_c - f_m)t]\]where \( a \) is a constant, less than unity, representing the attenuation of the upper side frequency.(a) Find the quadrature component of the VSB signal \( s(t) \).(b) The VSB signal, plus the carrier \( A_c \cos(2\pi f_c t) \), is passed through an envelope detector. Determine the distortion produced by the quadrature component.(c) What is the value of constant \( a \) for which this distortion reaches its worst possible condition?.
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