Page 453 - Numerical Methods for Chemical Engineering

442 9 Fourier analysis

Applying quadrature, we write (9.37) as
N
N
1 ( t) −im( ω)(k−1)( t)
F(ω m ) ≈ √ f (t k )e −iω m t k ( t) = √ f k e (9.38)
2π 2π
k=1 k=1
We next show that the F(ω m ) values have ω-periodicity,
F(ω m + l2ω max ) = F(ω m ) l = 0, ±1, ±2,... (9.39)

Using (9.31) and (9.32),

N
ω m + l2ω max = m( ω) + l2 ( ω) = [m + lN]( ω) (9.40)
2
a b
We next use (9.38) and e a+b = e e to obtain
N
( t) −im( ω)(k−1)( t) −ilN( ω)(k−1)( t)
F(ω m + l2ω max ) = √ f k e e (9.41)
2π
k=1
Now, as
π 2P 2π
$ %
( ω)( t) = =
P N N
we have
−ilN( ω)(k−1)( t) −ilN(k−1)(2π/N) −i(l2π)(k−1)
e = e = e (9.42)
a b
Using e ab = (e ) , this becomes
. / (k−1)
−i(l2π)(k−1) −i(l2π) (k−1)
e = e = [cos(−l2π) + i sin (−l2π)] = 1 (9.43)
Therefore, we have ω-periodicity,
N
( t) −im( ω)(k−1)( t)
F(ω m + l2ω max ) = √ f k e = F(ω m ) (9.44)
2π
k=1
Rather than evaluating F(ω) at the frequencies ω m = m( ω)in (−ω ,ω max ), we thus
max
could instead evaluate F(ω)at ω n = (n − 1)( ω)in [0, 2ω max ), and then obtain the values
at “negative” frequencies using F(ω n − 2ω max ) = F(ω n ). We then have
N
( t) −i(n−1)( ω)(k−1)( t) (n − 1)π
F(ω n ) ≈ √ f k e ω n = (9.45)
2π k=1 P
a b
Again using e ab = (e ) ,
N
( t) −i( ω)( t) (n−1)(k−1)

F(ω n ) ≈ √ f k e (9.46)
2π
k=1
As ( ω)( t) = 2π/N,wehave e −i( ω)( t) = e −i2π/N ≡ W and thus
N
( t) (n−1)(k−1)
F(ω n ) ≈ √ f k W (9.47)
2π
k=1