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Digital Fourier Transform(DFT)

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Digital Fourier
Transform(DFT)
Fourier Series
 Signal
used in daily life are
continuous or discrete in nature.
Discrete signal are used mainly in
digital communication. Signals may
be periodic or aperiodic also in
nature. For these type of signals,
we can use Fourier analysis to
extract the information. Periodic
signal, which may be represented
by a function f(X) called as Fourier
series. Where we
for extract
Fourier Series

But
in the case of aperiodic
signals Fourier transform and
inverse Fourier transform are
used to analysis the signals.
Fourier transform f(w) is in
generally in frequency domain,
whereas as inverse F(t) is in the
real time domain.
Fourier Series
F(ω)
F(ω)
Periodic signals
ω
2ω
ω
Aperiodic signals
F(t)
F(ω)
Fourier Transform
Inverse Fourier Transform
t
t
ω
ω
F(x)
Fourier Series
Period a
Complex continuous
periodic function
by fourier series.
x
F(𝝎)
Each terms contains
frequency
as
a
integral multiple of
original
complex
wave frequency.
𝜶𝟑
𝜶𝟏
𝜶𝟐 𝜶𝟒
𝝎
Fourier Series
∞
𝒇 𝒙 = 𝒂𝟎 +
𝜶𝒏 𝒄𝒐𝒔 𝒏𝝎𝒕 + 𝜷𝒏 𝒔𝒊𝒏 𝒏𝝎𝒕
𝒏=𝟏
∞
𝒇 𝒙 = 𝒂𝟎 +
𝒏=𝟏
𝟐
𝜷𝒏 =
𝑻
𝟐
𝜶𝒏 =
𝑻
𝒂/𝟐
n=1,2,3,4………………
𝟐𝒏𝝅𝒙
𝟐𝒏𝝅𝒙
𝜶𝒏 𝐜𝐨𝐬
+ 𝜷𝒏 𝐬𝐢𝐧
𝒂
𝒂
n=1,2,3,4………………
𝟐𝒏𝝅𝒙
𝒇 𝒙 𝐬𝐢𝐧(
)𝒅𝒙
𝒂
−𝒂/𝟐
𝒂/𝟐
𝟐𝒏𝝅𝒙
𝒇 𝒙 𝐬𝐢𝐧
𝒅𝒙
𝒂
−𝒂/𝟐
Fourier Series
Using the Euler's relation,
e=cosθ+isinθ
∞
we can modify the Fourier series as
𝒊𝟐𝝅𝒏𝒙/𝒂
𝒇 𝒙 =
𝒆
𝒇𝒏
𝒏=−∞
The above summation of exponential
summation series extends from negative
infinity and the 𝑓𝑛
is the Fourier
coefficient .If we can found out 𝑓𝑛 then the
series is determined. we use the
kronecker delta function to solve.
𝒂/𝟐
𝒅𝒙
𝒏=−𝒂/𝟐
a𝜹𝒏𝒎 =
Due to this property the
exponential
function
with
integer value n is orthogonal
−𝒊𝟐𝝅𝒏𝒙/𝒂 𝒊𝟐𝝅𝒏𝒙/𝒂
𝒆
𝒆
𝜹𝒏𝒎
𝟏, 𝒊𝒇 𝒎 = 𝒏
=
𝟎, 𝒊𝒇 𝒎 ≠ 𝒏
𝒂/𝟐
. We will use
this to determine
𝒇𝒏 .
−𝒂/𝟐
𝒂/𝟐
in nature
𝒅𝒙
𝒊𝟐𝝅𝒏𝒙/𝒂
𝒆
𝒏=−∞
𝒅𝒙 𝒆
𝒏=−∞ −𝒂/𝟐
∞
𝒊𝟐𝝅𝒎𝒙
−
𝒆 𝒂 𝒇(𝒙)
𝒂/𝟐
∞−𝒂/𝟐
=
𝒅𝒙
𝒊𝟐𝝅𝒎𝒙
−
𝒆 𝒂 𝒇(𝒙)=
𝒊𝟐𝝅𝒎𝒙/𝒂 𝒊𝟐𝝅𝒏𝒙/𝒂
𝒆
𝒇𝒏
∞
𝒇𝒏 =
𝒂𝜹𝒏𝒎 𝒇𝒏
𝒏=−∞
= 𝒂𝒇𝒏
∞
𝒇𝒏 =
𝒆
𝒊𝒌𝒏 𝒙
𝒇𝒏
𝒏=−∞
𝟏
𝒂
𝒂/𝟐
𝒅𝒙 𝒆
−𝒊𝒌𝒏 𝒙
𝒇(𝒙) = 𝒇𝒏
−𝒂/𝟐
Where 𝒌𝒏 = 𝟐𝝅𝒏/𝒂 is the wave number
Fourier
𝒇 𝒙 = 𝐥𝐢𝐦
Transform
𝒂→∞
∞
𝒆𝒊𝟐𝝅𝒏𝒙/𝒂 𝒇𝒏
𝒏=−∞
∞
𝒆𝒊𝒌𝒏𝒙 𝒇𝒏
𝒇 𝒙 = 𝐥𝐢𝐦
𝒂→∞
𝒏=−∞
∞
𝒇 𝒙 = 𝐥𝐢𝐦
𝒂→∞
𝒏=−∞
∞
Δk 𝒊𝒌 𝒙
𝒆 𝒏 𝑭(𝒌𝒏 ) 𝑭(𝒌𝒏 )=
𝟐𝝅
𝒇 𝒙 = 𝐥𝐢𝐦
𝒂→∞
𝒌𝒏 =nΔk,
Δk=
𝟐𝝅/𝒂
𝒏=−∞
Δk 𝒊𝒌 𝒙 𝟐𝝅𝒇𝒏
𝒆 𝒏
𝟐𝝅
Δk
dk 𝒊𝒌 𝒙
𝒇 𝒙 =
𝒆 𝒏 𝑭(𝒌)
−∞ 𝟐𝝅
∞
𝟐𝝅𝒇𝒏
Δk
𝟐𝝅𝒇𝒏
𝑭(𝒌𝒏 ) = 𝐥𝐢𝐦
𝒂→∞ Δk
𝟐𝝅 𝟏
𝑭(𝒌𝒏 ) = 𝐥𝐢𝐦
𝒂→∞ 𝟐𝝅/𝒂 𝒂
𝒂/𝟐
𝒅𝒙 𝒆−𝒊𝒌𝒏 𝒙 𝒇(𝒙)
−𝒂/𝟐
∞
𝑭(𝒌𝒏 ) =
𝒅𝒙 𝒆
−𝒊𝒌𝒙
𝒇(𝒙)
−∞
∞
dk 𝒊𝒌 𝒙
𝒇 𝒙 =
𝒆 𝒏 𝑭(𝒌)
−∞ 𝟐𝝅
∞
𝑭(𝒌) =
𝒅𝒙 𝒆
−∞
−𝒊𝒌𝒙
𝒇(𝒙)
Now let us consider a digital sequence
that is periodic in nature. The sequence
contain n datas and have a period N. Then
we can express the amplitude of each
data point as x[1], x[2]… x[n].
Since the time period for a periodic
signal is finite, we can represent this
as a summation of exponential series
as did for Fourier series equation 1
Frequency Analysis of Discrete time signals
. Exponential series contains frequency
components as an integral multiple of
its fundamental frequency 1/N. Then
the difference between successive
harmonics is 1/N Hz or 2pi/N radians.
X[n]
𝑿[𝒏] =
𝒙[𝒌]𝒆
𝒋𝟐𝝅𝒌𝒏/𝑵
,k=0,1,23………..(1)
N(period)
n data points
n
2
1
t
𝒋𝟐𝝅𝒌𝒏/𝑵
𝒙[𝒌]𝒆
,k=0,1,2…(1)
X[n]
𝐗𝐧 =
figure x[n] is periodic in nature, so
X[n]=x[n+rN], where r is an
integer
N(period)
n data points
n
2
1
t
Fourier series of a continuous periodic
signals contains frequency up to a
infinite range in the summation series but
for Fourier series of discrete periodic
signals only frequency in the range
(0,1,…N) is needed. The proof is that the
exponential function is identical for the
kth and (k+N)th frequency. That is,
𝒆
𝒋𝟐𝝅𝒌𝒏/𝑵
=𝒆
𝒋𝟐𝝅(𝒌+𝑵)𝒏/𝑵
𝑵−𝟏
𝒙[𝒌]𝒆𝒋𝟐𝝅𝒌𝒏/𝑵 ………………..2
X[n]=
𝟎
To find out x[k], we multiply
𝒋𝟐𝝅𝒓𝒏/𝑵
equation 2 with 𝒆
on both
sides and summing n=0 to N-1.
𝑵−𝟏 𝑵−𝟏
𝑵−𝟏
−𝒋𝟐𝝅𝒓𝒏/𝑵
𝒙[𝒏] 𝒆
=
𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵
𝒙[𝒌]𝒆
𝒏=𝟎 𝒌=𝟎
𝒏=𝟎
Recall the exponential summation of
complex function as given in below
𝑵−𝟏
𝒋𝟐𝝅𝒌𝒏/𝑵
𝒆
𝒏=𝟎
𝑵 𝒊𝒇 𝒌 = 𝟎, ±𝑵, ±𝟐𝑵
=
𝟎
𝒊𝒇 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
𝑵−𝟏
𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵
𝒆
𝑵 𝒊𝒇 𝒌 − 𝒓 = 𝟎, ±𝑵, ±𝟐𝑵
=
𝟎
𝒊𝒇 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
𝒏=𝟎
𝑵−𝟏
𝒙[𝒏] 𝒆
−𝒋𝟐𝝅𝒓𝒏/𝑵 =
𝒏=𝟎
𝑵−𝟏 𝑵−𝟏
𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵
𝒙[𝒌]𝒆
𝒏=𝟎 𝒌=𝟎
If k-r=0,then k=r and the series is modified
𝟏
=
𝑿[𝒌]
𝑵
𝑵−𝟏
𝒙[𝒏] 𝒆
𝒏=𝟎
−𝒋𝟐𝝅𝒌𝒏/𝑵
k=0,1,2…N1…….(2)
𝑵−𝟏
𝒙[𝒏]
𝒏=𝟎
−𝒋𝟐𝝅𝒓𝒏/𝑵
𝒆
𝑵−𝟏 𝑵−𝟏
=
𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵
𝒙[𝒌]𝒆
𝒌=𝟎 𝒏=𝟎
𝑵−𝟏
𝑵−𝟏
𝒋𝟐𝝅(𝟏−𝒓)𝒏/𝑵
𝒙[𝟏]𝒆
𝒏=𝟎
𝑵−𝟏
+
𝒙[𝟐]𝒆
𝒏=𝟎
𝒙[𝑵
−
𝟏]𝒆
+……..
𝒌=𝟎
𝒋𝟐𝝅(𝟐−𝒓)𝒏/𝑵
𝒋𝟐𝝅(𝑵−𝟏−𝒓)𝒏/𝑵
If k-r=0, only
this term is ≠0
𝑵−𝟏
𝑵−𝟏
𝒙[𝒏]
𝒏=𝟎
𝑵−𝟏
𝒏=𝟎
𝑵−𝟏
𝒙[𝟐]𝒆
𝒋𝟐𝝅(𝟏−𝒓)𝒏/𝑵
+
𝒙[𝟏]𝒆
−𝒋𝟐𝝅𝒓𝒏/𝑵
=
𝒆
𝒋𝟐𝝅(𝟐−𝒓)𝒏/𝑵
+…. 𝒙[𝒌]𝒆𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵
+….
𝒌=𝟎
𝒏=𝟎
If k-r=0,then k=r and then
𝟏
𝑿[𝒌] =
𝑵
𝑵−𝟏
−𝒋𝟐𝝅𝒌𝒏/𝑵
𝒙[𝒏] 𝒆
𝒏=𝟎
k=0,1,2…N-
X[k]N
𝟏
𝑿[𝒌] =
𝑵
𝑵−𝟏
−𝒋𝟐𝝅𝒌𝒏/𝑵
𝒙[𝒏] 𝒆
𝒏=𝟎
k=0,1,2…..N-1…………….(2)
DFTS can be represented by
equation 2 and
corresponding
Fourier coefficient X[k] gives its
frequency domain spectra. X[k]
represents the phase and amplitude
of corresponding frequency.
k > N?
𝑵−𝟏
𝟏
−𝒋𝟐𝝅(𝒌+𝑵)𝒏/𝑵
𝑿[𝒌 + 𝑵]=
𝒙[𝒏] 𝒆
𝑵
𝑵−𝟏
𝒏=𝟎
𝟏
−𝒋𝟐𝝅𝒌𝒏/𝑵 −𝒋𝟐𝝅𝑵𝒏/𝑵 =x[k]
𝒙[𝒏] 𝒆
𝒆
𝑵
𝒏=𝟎
−𝒋𝟐𝝅𝑵𝒏/𝑵
𝒆
=
cos 𝟐𝝅𝒏 + 𝒋𝒔𝒊𝒏𝟐𝝅𝒏 = 𝟏
𝒇𝒐𝒓 𝒏 = 𝟎, 𝟏 … 𝑵 − 𝟏
Thus ,x[k] is discrete variable
but with a period of N
Discrete
Time Fourier
Series(DTFS)
X[n] =
𝑥[𝑘]𝑒
𝑗2𝜋𝑘𝑛/𝑁
𝑋[𝑘]
1
=
𝑁
𝑁−1
𝑥[𝑛]
𝑛=0
−𝑗2𝜋𝑘𝑛/𝑁
𝑒
Frequency Analysis of Discrete aperiodic signals
Since the time period for a aperiodic signal is infinite
(N=∞), the time domain spectra contain the number
of data points (-∞ to ∞) extends to infinite. we can
add this to summation of exponential series to
determine x[k].
∞
𝒙[𝒏] 𝒆−𝒋𝟐𝝅𝒌𝒏/𝑵
𝑿[𝒌] =
𝒏=−∞
As in the figure x[n] is
periodic in nature, we can
write
X[n]=x[n+rN], where r
is an integer
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