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A.A. 2011/2012
Temporal Data Analysis
3.5 The Sampling Theorem with N = 2 — Page 55
Let us star with the shifted cosine function
{f k } = {0,1}
We expect
The sampling theorem tell us
f (t) = 1 2 + 1 2 cosΩ Nyqt = cos 2 Ω Nyqt
2
f (t) =
+∞∑
k=−∞
f k
sinΩ Nyq (t − k∆t)
Ω Nyq (t − k∆t)
with f k = δ k,even (with periodic continuation)
= sinΩ Nyqt
Ω Nyq t
∞∑ sinΩ Nyq (t − 2l∆t) ∞∑ sinΩ Nyq (t + 2l∆t)
+
+
Ω Nyq (t − 2l∆t) Ω Nyq (t + 2l∆t)
l=1
with the substitution k = 2l
= sinΩ Nyqt
Ω Nyq t
∞∑
+
l=1
with Ω Nyq ∆t = π
= sinΩ Nyqt
Ω Nyq t
= sinΩ Nyqt
Ω Nyq t
= sinΩ Nyqt
Ω Nyq t
[
sin2π(
t
2∆t − l)
2π( t
2∆t − l )
+ 1 ∞∑
2π
l=1
+ sinΩ Nyqt 2t ∞∑
2π 2∆t
l=1
⎛
( )
⎜ ΩNyq t 2 ∞∑
⎝1 + 2
2π
l=1
+ sin2π( t
2∆t + l )
]
2π( t
2∆t + l)
( t
2∆t + l )sinΩ Nyqt + ( t
2∆t − l )sinΩ Nyqt
( t
2∆t + l)( t
2∆t − l )
l=1
( t
2∆t
( ΩNyq t
2π
1
) 2 − l
2
⎞
1 ⎟
) 2
⎠
− l
2
M.Orlandini 129