Homework 6

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Homework 6

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ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 6
Solution

Problem 3.35
1) If the signal m(t) = m₁(t) + m₂(t) DSB modulates the carrier A_c cos(2πf_ct) the result is the signal
$u (t) = Acm (t)cos(2πfct) = Ac(m1 (t) + m2 (t))cos(2πfct) = Acm1 (t)cos(2πfct) + Acm2 (t)cos(2πfct) = u1(t) + u2(t)$
where u₁(t) and u₂(t) are the DSB modulated signals corresponding to the message signals m₁(t) and m₂(t). Hence, AM modulation satisfies the superposition principle.
2) If m(t) frequency modulates a carrier A_c cos(2πf_ct) the result is
$∫ ∞ u (t) = A cos(2 πf t + 2πk (m (τ) + m (τ))dτ) c c f - ∞ 1 2 ∫ ∞ ⁄= Ac cos(2 πfct + 2πkf - ∞ m1 (τ)dτ) ∫ ∞ +Ac cos(2πfct + 2πkf m2(τ )d τ) -∞ = u1(t) + u2(t)$
where the inequality follows from the nonlinearity of the cosine function. Hence, angle modulation is not a linear modulation method.
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Problem 3.36
The transfer function of the FM discriminator is $R- H (s) = -----R-------= -----L-s------ R + Ls + C1s- s2 + RLs + L1C-$
Thus,
$( )2 4π2 R- f2 |H (f )|2 = (------------L)2------------- 1LC-- 4π2f 2 + 4π2 (RL)2f2$
As it is observed |H(f)|² ≤ 1 with equality if
$f = --√1---- 2π LC$
Since this filter is to be used as a slope detector, we require that the frequency content of the signal, which is [80 - 6, 80 + 6] MHz, to fall inside the region over which |H(f)| is almost linear. Such a region can be considered the interval [f₁₀,f₉₀], where f₁₀ is the frequency such that |H(f₁₀)| = 10% max[|H(f)|] and f₉₀ is the frequency such that |H(f₁₀)| = 90% max[|H(f)|].
With max[|H(f)| = 1, f₁₀ = 74 × 10⁶ and f₉₀ = 86 × 10⁶, we obtain the system of equations
$50 × 103 1 1 4π2f120 + --------2 πf10[1 - 0.12]2 - ----= 0 L LC 2 2 50-×-103- 2 12 -1-- 4π f90 + L 2 πf90[1 - 0.9 ] - LC = 0$
Solving this system, we obtain $L = 14.98 mH C = 0.018013 pF$
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Problem 3.38
The frequency deviation is given by $fd(t) = fi(t) - fc = kfm (t)$
whereas the phase deviation is obtained from
$∫ t ϕd (t) = 2πkf m (τ )dτ -∞$
In the next figure we plot the frequency and the phase deviation when m(t) is as in Fig. P-3.38 with k_f = 25.

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Problem 3.39
Using Carson’s rule we obtain $( kf max [|m (t)|] |{ 20020 kf = 10 Bc = 2(β + 1)W = 2 (--------------+ 1 )W = | 20200 kf = 100 W ( 22000 kf = 1000$

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