Homework 2

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Homework 2

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ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 2
Solution

Problem 2.17
(Convolution theorem:) $F [x(t) ⋆ y(t)] = F [x(t)]F [y(t)] = X (f)Y (f)$
Thus
$-1 sinc(t) ⋆ sinc (t) = F [F [sinc (t) ⋆ sinc(t)]] = F -1[F [sinc (t)] ⋅ F[sinc(t)]] -1 -1 = F [Π (f)Π (f)] = F [Π (f)] = sinc (t)$

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Problem 2.18
$∫ ∞ F [x(t)y(t)] = x(t)y(t)e -j2πftdt ∫- ∞∞ (∫ ∞ ) = X (θ)ej2πθtdθ y(t)e -j2πftdt -∞ - ∞( ) ∫ ∞ ∫ ∞ -j2π(f-θ)t = -∞ X (θ) -∞ y(t)e dt dθ ∫ ∞ = X (θ)Y (f - θ)dθ = X (f) ⋆ Y (f) -∞$

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Problem 2.27
Let the response of the LTI system be h(t) with Fourier transform H(f). Then, from the convolution theorem we obtain $Y (f) = H (f)X (f ) =⇒ Λ (f) = Π (f)H (f)$
However, this relation cannot hold since Π(f) = 0 for 1 2 < |f| whereas Λ(f)0 for 1 < |f|≤ 1∕2.
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Problem 2.56
$1- 1- x(t) = sinct cos2πf0t =⇒ X (f) = 2Π (f + f0)) + 2Π (f - f0)) 2 1 1 h(t) = sinc tsin 2πf0t =⇒ H (f) = - --Λ (f + f0)) +---Λ(f - f0)) 2j 2j$
The lowpass equivalents are
$Xl (f) = 2u (f + f0)X (f + f0) = Π (f) 1- Hl (f) = 2u (f + f0)H (f + f0) = jΛ (f ) ( 1- 1 1 |{ 2j(f + 1) - 2 < f ≤ 0 Yl(f) = -Xl(f )Hl (f) = 12j(- f + 1) 0 ≤ f < 12 2 |( 0 otherwise$
Taking the inverse Fourier transform of Y _l(f) we can find the lowpass equivalent response of the system. Thus,
$- 1 yl(t) = F [Yl(f)] 1-∫ 0 j2πft -1-∫ 12 j2πft = 2j - 1(f + 1)e df + 2j 0 (- f + 1)e df [ 2 ] | | 1-- -1--- j2πft --1---j2πft ||0 -1--1---j2πft||0 = 2j j2πtf e + 4π2t2e |- 1 + 2j j2πte |- 1 [ ] 2|1 |12 -1- -1--- j2πft --1---j2πft ||2 1---1-- j2πft||2 - 2j j2πtfe + 4π2t2e |0 + 2jj2πt e |0 [ 1 1 ] = j - ----sin πt + -----(cos πt - 1) 4πt 4π2t2$
The output of the system y(t) can now be found from y(t) = Re[y_l(t)e^j2πf₀t]. Thus
$[ ] y (t) = Re (j[- -1--sinπt + --1---(cosπt - 1)])(cos 2πf0t + j sin2πf0t ) 4πt 4 π2t2 --1--- -1-- = [4π2t2(1 - cosπt) + 4πt sin πt]sin2πf0t$

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