ENC3250 Solutions pg 26

ENC3250 Solutions pg 26 - 6.13 slope overload distortions...

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Unformatted text preview: 6.13 slope overload distortions DM output 1 0 6.14 s(t) = A c cos[2πf ct + φ(t)] = 10 cos [(108 )πt + 5 sin 2 π(10 3 )t] Therefore, φ(t) = 5 sin 2π (103 )t, and the maximum phase deviation is 5 radians. For frequency deviation, recognize that the change in frequency is determined by the derivative of the phase: φ'(t) = 5 (2π) (10 3 ) cos 2π(10 3 )t which yields a frequency deviation of Δf = (1/2π )[ 5 (2π) (10 3 )] = 5 kHz 6.15 a. s(t) = A c cos[2πf ct + npm(t)] = 10 cos [2π(106 )t + 0.1 sin (103 )πt] Ac = 10; fc = 106 10 m(t) = 0.1 sin (103 )πt, so m(t) = 0.01 sin (10 3 )πt b. s(t) = A c cos[2πf ct + φ(t)] = 10 cos [2π(106 )t + 0.1 sin (103 )πt] Ac = 10; fc = 106 φ(t) = 0.1 sin (103 )πt, so φ'(t) = 100π cos (10 3 )πt = nfm(t) = 10 m(t) Therefore m(t) = 10 π cos (10 3 )πt 6.16 a. For AM, s(t) = [1 + m(t)] cos(2π fct) s1 (t) = [1 + m1 (t)] cos(2π fct); s2 (t) = [1 + m2 (t)] cos(2π fct) For the combined signal m c(t) = m1 (t) + m2 (t), sc(t) = [1 + m1 (t) + m2 (t)] cos(2π fct) = s1 (t) + s2 (t) – 1, which is a linear combination of s1 (t) and s2 (t). -26- ...
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.

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