Homework 4

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Homework 4

ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 4
Solution

Problem 3.6
The mixed signal y(t) is given by
$y(t) = u(t) ⋅ xL(t) = Am (t)cos(2πfct) cos(2 πfct + θ) A = -2 m (t)[cos(2π2fct + θ) + cos(θ)]$
The lowpass filter will cut-off the frequencies above W, where W is the bandwidth of the message signal m(t). Thus, the output of the lowpass filter is $z(t) = A-m (t) cos(θ ) 2$
If the power of m(t) is P_M, then the power of the output signal z(t) is P_out = P_MA² 4 cos ²(θ). The power of the modulated signal u(t) = Am(t) cos(2πf_ct) is P_U = A² 2 P_M. Hence,
$Pout = 1-cos2(θ) PU 2$
A plot of P_out P_U for 0 ≤ θ ≤ π is given in the next figure.

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Problem 3.9
If we let $( Tp) ( Tp) x (t) = - Π ( t +-4-) + Π ( t----4) T2p Tp2$
then using the results of Problem 2.23, we obtain
$∞∑ v(t) = m (t)s(t) = m(t) x(t - nTp) ∞ n= -∞ -1- ∑ n-- j2π nTpt = m (t)Tp n=-∞ X (Tp)e$
where
$⌊ ( ) ( ) ⌋ n t + Tp t - Tp || X ( --) = F ⌈- Π ( --Tp4-) + Π ( --Tp4-) ⌉|| Tp 2- -2 f= nTp- Tp Tp ( Tp Tp) || = --sinc(f---) e- j2πf 4 - ej2πf4 || n 2 2 f= Tp Tp n π = --sinc(--)(- 2j)sin(n -) 2 2 2$
Hence, the Fourier transform of v(t) is $∑∞ V (f) = 1- sinc(n-)(- 2j) sin(n π)M (f - n-) 2 n=-∞ 2 2 Tp$
The bandpass filter will cut-off all the frequencies except the ones centered at 1 _ T_p, that is for n = ±1. Thus, the output spectrum is
$1- 1-- 1- 1-- U (f) = sinc(2 )(- j)M (f - T ) + sinc (2)jM (f + T ) p p = - 2jM (f - 1-) + 2-jM (f + -1-) π Tp π Tp 4 [ 1 1 1 1 ] = -M (f) ⋆ ---δ(f - ---) ----δ(f + ---) π 2j Tp 2j Tp$
Taking the inverse Fourier transform of the previous expression, we obtain $4 1 u(t) = -m (t)sin(2π---t) π Tp$
which has the form of a DSB-SC AM signal, with c(t) = 4 π sin(2π 1 _ T_pt) being the carrier signal.
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Problem 3.11
1) The spectrum of the modulated signal Am(t) cos(2πf_ct) is $A V (f) = -[M (f - fc) + M (f + fc)] 2$
The spectrum of the signal at the output of the highpass filter is
$A- U (f) = 2 [M (f + fc)u- 1(- f - fc) + M (f - fc)u- 1(f - fc)]$
Multiplying the output of the HPF with A cos(2π(f_c + W)t) results in the signal z(t) with spectrum
$A Z (f) = --[M (f + fc)u-1(- f - fc) + M (f - fc)u-1(f - fc)] 2 ⋆A-[δ(f - (fc + W )) + δ(f + fc + W )] 2 A2 = -4-(M (f + fc - fc - W )u-1(- f + fc + W - fc) +M (f + fc - fc + W )u-1(f + fc + W - fc) +M (f - 2fc - W )u- 1(f - 2fc - W ) +M (f + 2f + W )u (- f - 2f - W )) c - 1 c A2- = 4 (M (f - W )u -1(- f + W ) + M (f + W )u -1(f + W ) +M (f - 2fc - W )u- 1(f - 2fc - W ) + M (f + 2fc + W )u-1(- f - 2fc - W ))$
The LPF will cut-off the double frequency components, leaving the spectrum $A2- Y (f) = 4 [M (f - W )u- 1(- f + W ) + M (f + W )u- 1(f + W )]$
The next figure depicts Y (f) for M(f) as shown in Fig. P-5.12.

2) As it is observed from the spectrum Y (f), the system shifts the positive frequency components to the negative frequency axis and the negative frequency components to the positive frequency axis. If we transmit the signal y(t) through the system, then we will get a scaled version of the original spectrum M(f).
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Problem 3.18
The signal x(t) is m(t) + cos(2πf₀t). The spectrum of this signal is X(f) = M(f) + 1 2(δ(f - f₀) + δ(f + f₀)) and its bandwidth equals to W_x = f₀. The signal y₁(t) after the Square Law Device is
$y (t) = x2 (t) = (m (t) + cos(2πf t))2 1 0 = m2 (t) + cos2(2 πf0t) + 2m (t)cos(2πf0t) 2 1 1 = m (t) + --+ --cos(2π2f0t) + 2m (t)cos(2πf0t) 2 2$
The spectrum of this signal is given by $Y1 (f) = M (f)⋆M (f)+ 1δ(f)+ 1-(δ(f - 2f0)+δ(f +2f0 ))+M (f - f0)+M (f +f0) 2 4$
and its bandwidth is W₁ = 2f₀. The bandpass filter will cut-off the low-frequency components M(f) ⋆ M(f) + 1 2δ(f) and the terms with the double frequency components 1 4(δ(f - 2f₀) + δ(f + 2f₀)). Thus the spectrum Y ₂(f) is given by
$Y2(f) = M (f - f0) + M (f + f0)$
and the bandwidth of y₂(t) is W₂ = 2W. The signal y₃(t) is
$y (t) = 2m (t)cos2(2πf t) = m (t) + m (t)cos(2π2f t) 3 0 0$
with spectrum
$Y (t) = M (f ) + 1-(M (f - 2f ) + M (f + 2f )) 3 2 0 0$
and bandwidth W₃ = 2f₀ + W. The lowpass filter will eliminate the spectral components 1 2(M(f - 2f₀) + M(f + 2f₀)), so that y₄(t) = m(t) with spectrum Y ₄ = M(f) and bandwidth W₄ = W. The next figure depicts the spectra of the signals x(t), y₁(t), y₂(t), y₃(t) and y₄(t).

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