Page 452 - Numerical Methods for Chemical Engineering

Fourier series and transforms in one dimension 441

discrete frequencies
mπ
ω m = m( ω) ω = m = 0, ±1, ±2,... (9.30)
P
While there are an infinite number of such frequencies, we have available only N data {f 1 ,
f 2 ,..., f N }. Let us choose the {ω n } to be the lowest ones satisfying (9.30),
π N N
ω m = m( ω) ω = m = 0, ±1, ±2,..., ± − 1 , + (9.31)
P 2 2
The frequency resolution ω is determined by the length of the sampling period 2P. The
highest frequency that we resolve from the sampled data is

N Nπ π π
ω max = ω = = = (9.32)
2 2P (2P/N) t
Toresolvehigherfrequencies,weneedtosample f (t)moreoften(decrease t).If F(ω) = 0
forall |ω| >ω max , f (t)issaidtobebandwidth-limited,andthesampleddataaresufficientto
characterize F(ω). Otherwise, if F(ω) = 0 for some |ω| >ω max , there exist high-frequency
components of f (t) that are sampled incorrectly and that can corrupt the values of f (ω m ).
This is known as aliasing, a subject that we discuss later in more detail.
To obtain the normalization constant A in (9.28), we match
+∞ +∞
' ∞ '

F(ω)G(ω)dω = A F(ω m ) δ(ω − ω m )G(ω)dω
m=−∞
−∞ −∞
∞

= A F(ω m )G(ω m ) (9.33)
m=−∞
to the quadrature formula
+∞
∞
'

F(ω)G(ω)dω ≈ F(ω m )G(ω m )( ω) (9.34)
m=−∞
−∞
to yield
π
A = ω = (9.35)
P
From (9.17) and (9.29), we have
2P
( ω) 1 ' −iω m t
√ F(ω m ) = c m = f (t)e dt (9.36)
2π 2P
0
Using
√ √
( 2π) 1 ( 2π) P 1
= = √
2P ( ω) 2P π 2π
the Fourier transform values are
2P
1 ' −iω m t
F(ω m ) = √ f (t)e dt (9.37)
2π
0