Homework 11

[prev] [prev-tail] [tail] [up]

Homework 11

ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 11
Solution

Problem 7.3

a) To show that the waveforms ψ_n(t), n = 1,…, 3 are orthogonal we have to prove that
$∫ ∞ -∞ ψm (t)ψn (t)dt = 0, m ⁄= n$
Clearly,
$∫ ∞ ∫ 4 c12 = ψ1 (t)ψ2 (t)dt = ψ1 (t)ψ2 (t)dt ∫-∞ ∫ 0 2 4 = 0 ψ1 (t)ψ2(t)dt + 2 ψ1(t)ψ2(t)dt 1 ∫ 2 1 ∫ 4 1 1 = -- dt - -- dt = --× 2 - -× (4 - 2 ) 4 0 4 2 4 4 = 0$
Similarly,
$∫ ∫ ∞ 4 c13 = - ∞ ψ1(t)ψ3(t)dt = 0 ψ1(t)ψ3(t)dt 1∫ 1 1 ∫ 2 1 ∫ 3 1 ∫ 4 = -- dt - -- dt - -- dt + -- dt 4 0 4 1 4 2 4 3 = 0$
and
$∫ ∞ ∫ 4 c23 = ψ2(t)ψ3(t)dt = ψ2(t)ψ3(t)dt -∫∞1 ∫ 2 0∫ 3 ∫ 4 = 1- dt - 1- dt + 1- dt - 1- dt 4 0 4 1 4 2 4 3 = 0$
Thus, the signals ψ_n(t) are orthogonal.
b) We first determine the weighting coefficients
$∫ x = ∞ x (t)ψ (t)dt, n = 1,2,3 n -∞ n$

$∫ 4 1∫ 1 1 ∫ 2 1 ∫ 3 1 ∫ 4 x1 = x(t)ψ1(t)dt = - -- dt +-- dt - -- dt + -- dt = 0 ∫0 2∫ 0 2 1 2 2 2 3 4 1- 4 x2 = 0 x(t)ψ2(t)dt = 2 0 x(t)dt = 0 ∫ 4 1∫ 1 1 ∫ 2 1 ∫ 3 1 ∫ 4 x3 = x(t)ψ3(t)dt = - -- dt --- dt + -- dt + -- dt = 0 0 2 0 2 1 2 2 2 3$
As it is observed, x(t) is orthogonal to the signal waveforms ψ_n(t), n = 1, 2, 3 and thus it can not represented as a linear combination of these functions.
____________________________________________________________________________________________________
Problem 7.5

a) As an orthonormal set of basis functions we consider the set
${ { 1 0 ≤ t < 1 1 1 ≤ t < 2 ψ1(t) = 0 o.w ψ2(t) = 0 o.w { { ψ (t) = 1 2 ≤ t < 3 ψ (t) = 1 3 ≤ t < 4 3 0 o.w 4 0 o.w$
In matrix notation, the four waveforms can be represented as $( ) ( ) ( ) s1(t) 2 - 1 - 1 - 1 ψ1(t) || s2(t) || || - 2 1 1 0 || || ψ2(t) || |( s3(t) |) = |( 1 - 1 1 - 1 |) |( ψ3(t) |) s4(t) 1 - 2 - 2 2 ψ4(t)$
Note that the rank of the transformation matrix is 4 and therefore, the dimensionality of the waveforms is 4
b) The representation vectors are
$[ ] s1 = 2 - 1 - 1 - 1 [ ] s2 = - 2 1 1 0 [ ] s3 = 1 - 1 1 - 1 [ ] s4 = 1 - 2 - 2 2$

c) The distance between the first and the second vector is
$∘ --------- ∘ ||[-----------------]||2 √ --- d1,2 = |s1 - s2|2 = | 4 - 2 - 2 - 1 | = 25$
Similarly we find that
$∘ --------- ∘ |[--------------]|2- √ -- d1,3 = |s1 - s3|2 = || 1 0 - 2 0 || = 5 ∘ --------- ∘ |[--------------]|- --- d = |s - s |2 = || 1 1 1 - 3 ||2 = √ 12 1,4 1 4 ∘ ------------------ ∘ --------2 ||[ ]||2 √ --- d2,3 = |s2 - s3| = | - 3 2 0 1 | = 14 ∘ --------- ∘ |[----------------]|2- √ --- d2,4 = |s2 - s4|2 = || - 3 3 3 - 2 || = 31 ∘ --------- ∘ ------------------ 2 ||[ 0 1 3 - 3 ]||2 √ --- d3,4 = |s3 - s4| = | | = 19$
Thus, the minimum distance between any pair of vectors is d_min = .
____________________________________________________________________________________________________
Problem 7.8

Assuming that E[n²(t)] = σ_n², we obtain
$[( ) ( ) ] ∫ T ∫ T E [n1n2 ] = E s1(t)n(t)dt s2(v)n(v)dv ∫ ∫ 0 0 T T = 0 0 s1(t)s2(v)E[n(t)n(v)]dtdv ∫ T = σ2n s1(t)s2(t)dt 0 = 0$
where the last equality follows from the orthogonality of the signal waveforms s₁(t) and s₂(t).
____________________________________________________________________________________________________
Problem 7.14

a) The impulse response of the filter matched to s(t) is
$h(t) = s (T - t) = s (3 - t) = s(t)$
where we have used the fact that s(t) is even with respect to the t = T 2 = 3 2 axis.
b) The output of the matched filter is
$∫ t y(t) = s(t) ⋆ s(t) = s(τ)s(t - τ )dτ ( 0 || 0 t < 0 |||| A2t 0 ≤ t < 1 |||| 2 |||{ A (22 - t) 1 ≤ t < 2 = 2A (t - 2 ) 2 ≤ t < 3 ||| 2A2 (4 - t) 3 ≤ t < 4 |||| A2(t - 4) 4 ≤ t < 5 |||| A2(6 - t) 5 ≤ t < 6 ||( 0 6 ≤ t$
A scetch of y(t) is depicted in the next figure

c) At the output of the matched filter and for t = T = 3 the noise is
$∫ T nT = 0 n (τ )h (T - τ)dτ ∫ T ∫ T = n (τ )s (T - (T - τ))dτ = n(τ)s(τ)d τ 0 0$
The variance of the noise is
$[∫ T∫ T ] σ2nT = E n(τ)n (v )s (τ )s(v )d τdv 0 0 ∫ T ∫ T = s(τ)s(v)E[n(τ)n (v)]dτ dv 0 ∫0T ∫ T = N0- s(τ)s(v)δ(τ - v)dτ dv 2 ∫0 0 N0- T 2 2 = 2 0 s (τ )dτ = N0A$

d) For antipodal equiprobable signals the probability of error is
$⌊ ⌋ ∘ (---)-- P (e) = Q ⌈ S-- ⌉ N o$
where _o is the output SNR from the matched filter. Since
$(S ) y2(T ) 4A4 --- = ------ = ------ N o E [n2T] N0A2$
we obtain
$⌊ ----⌋ ∘ 4A2 P (e) = Q ⌈ ---⌉ N0$

[prev] [prev-tail] [front] [up]