Homework 3

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Homework 3

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ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 3
Solution

Problem 2.12
a) We can write x(t) as x(t) = 2Π( t 4) - 2Λ( t 2). Then $t t F [x(t)] = F[2Π (-)] - F[2Λ (-)] = 8sinc (4f ) - 4sinc2(2f ) 4 2$

b)
$x (t) = 2Π (t-) - Λ (t) =⇒ F [x(t)] = 8sinc(4f) - sinc2(f) 4$

c)
$∫ ∞ ∫ 0 ∫ 1 X (f) = x (t)e-j2πftdt = (t + 1)e-j2πftdt + (t - 1)e-j2πftdt (-∞ ) -1 0 j 1 -j2πft||0 j - j2πft||0 = 2πf-t + 4-π2f2 e || + 2-πfe || ( ) - 1| -|1 -j-- --1--- - j2πft||1 --j- - j2πft||1 + 2πf t + 4π2f2 e |0 - 2πf e |0 = -j-(1 - sin (πf )) πf$

d) We can write x(t) as x(t) = Λ(t + 1) - Λ(t - 1). Thus
$2 j2πf 2 - j2πf 2 X (f ) = sinc (f)e - sinc (f )e = 2jsinc (f) sin(2πf )$

e) We can write x(t) as x(t) = Λ(t + 1) + Λ(t) + Λ(t - 1). Hence,
$X (f ) = sinc2(f)(1 + ej2πf + e- j2πf) = sinc2(f)(1 + 2cos(2πf )$

f) We can write x(t) as
$[ ( ) ( )] 1 1 x (t) = Π 2f0(t - 4f-) - Π 2f0(t - 4f-) sin (2πf0t) 0 0$
Then
$[ ( ) ( ) ] 1 f -j2π-1f 1 f j2π 1-f X (f ) = 2f--sinc 2f-- e 4f0 - 2f--sinc 2f-) e 4f0 0 0 0 0 ⋆-j(δ(f + f ) - δ (f + f )) 2 ( 0 ) ( 0 ) ( ) ( ) 1 f + f0 f + f0 1 f - f0 f - f0 = ----sinc ------- sin π ------- - ---sinc ------- sin π ------- 2f0 2f0 2f0 2f0 2f0 2f0$

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Problem 3.2
$u(t) = m(t)c(t) = A(sinc(t) + sinc2(t)) cos(2 πfct)$
Taking the Fourier transform of both sides, we obtain
$U(f ) = A- [Π (f) + Λ(f)] ⋆ (δ(f - fc) + δ(f + fc)) 2 A- = 2 [Π (f - fc) + Λ (f - fc) + Π (f + fc) + Λ(f + fc)]$
Π(f -f_c)0 for |f -f_c| < 1 2, whereas Λ(f -f_c)0 for |f -f_c| < 1. Hence, the bandwidth of the bandpass filter is 2.
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Problem 3.3
The following figure shows the modulated signals for A = 1 and f₀ = 10. As it is observed both signals have the same envelope but there is a phase reversal at t = 1 for the second signal Am₂(t) cos(2πf₀t) (right plot). This discontinuity is shown clearly in the next figure where we plotted Am₂(t) cos(2πf₀t) with f₀ = 3.

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Problem 3.4
$1 y(t) = x(t) + -x2(t) 2 ( ) = m (t) + cos(2πf t) + 1 m2 (t) + cos2(2πf t) + 2m (t) cos(2πf t) c 2 c c 1 2 1 1 = m (t) + cos(2πfct) + 2m (t) + 4-+ 4-cos(2π2fct) + m (t) cos(2 πfct)$
Taking the Fourier transform of the previous, we obtain
$Y(f ) = M (f ) + 1M (f ) ⋆ M (f) + 1-(M (f - f ) + M (f + f )) 2 2 c c 1- 1- 1- + 4δ(f ) + 2 (δ(f - fc) + δ(f + fc)) + 8 (δ(f - 2fc) + δ(f + 2fc))$
The next figure depicts the spectrum Y (f)

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