082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

796 Chapter 24 The Fourier transform
The inverse d.c.t. of a sequenceF[k],k = 0,1,2,...,N −1, is another sequence
f[n],alsohavingN terms,defined by
f[n] = √ 1 ∑N−1
[ ( π
{F[0] +2 F[k]cos
N N k n + 1 )] }
2
k=1
forn=0,1,2, ... ,N−1
Again,aderivationofthisinversewillnotbepresentedhere.However,wewilldemonstrate
the inverse property for a specific example.
Example24.25 (a) Findthe d.c.t.,F[k], ofthe sequence f[n] = 2,4,6.
(b) Apply the inverse d.c.t. to F[k] and show that the original sequence, f[n], is
obtained.
Solution (a) We use the formula
F[k] = √ 1 ∑N−1
N
n=0
[ π
f[n] cos
N
Herethe number of terms,N, isthree.
Whenk=0
F[0] = 1 √
3
2∑
Whenk=1
n=0
(
n + 1 ) ]
k
2
[ ( π
f[n]cos n + 1 ) ]
×0
3 2
= 1 √
3
[2cos0 +4cos0+6cos0]= 12 √
3
F[1] = 1 √
3
2∑
= √ 1 [
3
Whenk=2
n=0
[ ( π
f[n]cos n + 1 ) ]
×1
3 2
]
fork=0,1,2, ... ,N−1
2cos π 6 +4cosπ 2 +6cos5π 6
√ )]
3
= 1 [√ √ ]
√ 3−3 3 = −2
2 3
= √ 1
√ (
3
[2 ×
3 2 +0+6× −
F[2] = 1 √
3
2∑
= √ 1 [
3
n=0
[ ( π
f[n]cos n + 1 ) ]
×2
3 2
]
2cos π 3 +4cosπ+6cos5π 3
= √ 1 [
2 × 1
3 2 +4× (−1) +6× 1 ]
=1−4+3=0
2
So the d.c.t. of the sequence 2,4,6is 12 √
3
, −2, 0.