Problem 3.1 The demodulated output, in general, is y D (t) = Lpfx c (t) 2 cos[! c t + (t)]g where... more Problem 3.1 The demodulated output, in general, is y D (t) = Lpfx c (t) 2 cos[! c t + (t)]g where Lp ffg denotes the lowpass portion of the argument. With x c (t) = A c m (t) cos [! c t + 0 ] the demodulated output becomes y D (t) = Lp f2A c m (t) cos [! c t + 0 ] cos [! c t + (t)]g Performing the indicated multiplication and taking the lowpass portion yields y D (t) = A c m (t) cos [ (t) 0 ] If (t) = 0 (a constant), the demodulated output becomes y D (t) = A c m (t) cos [ 0 0 ] Letting A c = 1 gives the error " (t) = m (t) [1 cos (0 0)]
Problem 3.1 The demodulated output, in general, is y D (t) = Lpfx c (t) 2 cos[! c t + (t)]g where... more Problem 3.1 The demodulated output, in general, is y D (t) = Lpfx c (t) 2 cos[! c t + (t)]g where Lp ffg denotes the lowpass portion of the argument. With x c (t) = A c m (t) cos [! c t + 0 ] the demodulated output becomes y D (t) = Lp f2A c m (t) cos [! c t + 0 ] cos [! c t + (t)]g Performing the indicated multiplication and taking the lowpass portion yields y D (t) = A c m (t) cos [ (t) 0 ] If (t) = 0 (a constant), the demodulated output becomes y D (t) = A c m (t) cos [ 0 0 ] Letting A c = 1 gives the error " (t) = m (t) [1 cos (0 0)]
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