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

784 Chapter 24 The Fourier transform
24.9.1 Definitionofthed.f.t.
We consider a sequence ofN terms, f[0], f[1], f[2],..., f[N −1].
Thed.f.t.ofasequence f[n],n = 0,1,2,...,N −1,isanothersequenceF[k],also
havingN terms,defined by
F[k] =
∑N−1
n=0
f[n]e −2jnkπ/N
fork=0,1,2,...,N−1
We writeF[k] = D{f[n]}todenote the d.f.t. of the sequence f[n].
There are a number of variations of this definition and you need to be aware that
different authors may use slightly different formulae. This can cause confusion when
thed.f.t.isfirstmet.Inparticular,someauthorsincludeafactor 1 N inthedefinition,and
others use a positive exponential term instead of the negative one given above. What is
crucial isthatyou know which formula isbeing used and you apply itconsistently.
We now give an example toshow the application of the formula.
Example24.19 Findthe d.f.t. ofthe sequence f[n] = 1,2,−5,3.
Solution We use the formula
D{f[n]} =F[k] =
∑N−1
n=0
f[n]e −2jnkπ/N
fork=0,1,2,...,N−1
Herethe number ofterms,N, isfour:
3∑
F[k] =
n=0
f[n]e −2jnkπ/4
fork=0,1,2,3
So,whenk = 0,
3∑
F[0] = f[n]e 0
n=0
= 1
Whenk = 1,
3∑
F[1] =
=1+2+(−5)+3
n=0
f[n]e −2jnπ/4
= 1 +2e −2jπ/4 + (−5)e −2j2π/4 +3e −2j3π/4
=1+2e −πj/2 −5e −πj +3e −3jπ/2
= 1 +2(−j) −5(−1) +3j
=6+j