digital communications 5th - instructor solution manual

If you are a student using this Manual, you are using it without permission... heφnt, eφmti = 0 for all n 6= mUsing the fact that the energy in lowpass equivalent signal is twice the ene

for Digital Communications, 5th Edition

(Chapter 2) 1

Prepared byKostas StamatiouJanuary 11, 2008

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b In exactly the same way as in part (a) we prove :

ˆx(t) = ˆx(−t)

Exploiting the phase-shifting property (2-1-4) of the Hilbert transform :

ˆ

2[−jδ(f − f0) + jδ(f + f0)] = 1

2j[δ(f − f0)− δ(f + f0)] = F−1{sin 2πf0t}Hence, ˆx(t) = sin ω0t

d In a similar way to part (c) :

x(t) = sin ω0t⇒ X(f) = 2j1 [δ(f − f0)− δ(f + f0)]⇒ ˆX(f ) = 1

2[−δ(f − f0)− δ(f + f0)]

2[δ(f− f0) + δ(f + f0)] =−F−1{cos 2πω0t} ⇒ ˆx(t) = − cos ω0t

we have : ˆx(t) =ˆ −x(t)

have that :

ˆX(f )

= |H(f )| |X(f )| = |X(f )| Hence :

−∞

ˆX(f ) ... class="text_page_counter">Trang 4

Problem 2.3

A well-known result in estimation theory based on the minimum mean-squared-error criterion states

series expansion... class="text_page_counter">Trang 43

for Digital Communications, 5th Edition

(Chapter 3) 1

Prepared... class="page_container" data-page="8">

Problem 2.9

We know from Fourier transform properties that if a signal x(t) is real-valued then its Fourier

is real-valued is : Sl(−f)