Transcript 2-dft1
Fourier Analysis of Discrete Time Signals
For a discrete time sequence we define two classes of Fourier Transforms:
• the DTFT (Discrete Time FT) for sequences having infinite duration,
• the DFT (Discrete FT) for sequences having finite duration.
The Discrete Time Fourier Transform (DTFT)
Given a sequence x(n) having infinite duration, we define the DTFT as follows:
X ( ) DTFT x (n)
j n
x
(
n
)
e
n
1
x (n) IDTFT X ( )
2
X ( )e j n d
X ( )
x(n)
…..
…..
N 1
n
continuous frequency
discrete time
Observations:
• The DTFT X ( ) is periodic with period 2
• The frequency
;
is the digital frequency and therefore it is limited to the interval
Recall that the digital frequency is a normalized frequency relative to the sampling frequency,
defined as
2
F
Fs
one period of
Fs
2
Fs / 2
0
0
X ( )
Fs / 2
Fs
2
F
Example:
DTFT
x[n ]
1
0
n
N 1
since
N 1
X ( ) e
n 0
j n
1 e j N
1 e j
e j ( N 1) / 2
sin N / 2
sin / 2
Example:
x[n] A cos( 0 n )
X ( ) A e j ( 0)
A e j ( 0)
Discrete Fourier Transform (DFT)
Definition (Discrete Fourier Transform): Given a finite sequence
x [ x(0), x(1),..., x( N 1)]
its Discrete Fourier Transform (DFT) is a finite sequence
X DFT ( x) [ X (0), X (1),..., X ( N 1)]
where
N 1
X ( k ) x (n) wN kn , wN e j 2 / N
n0
x
DFT
X
Definition (Inverse Discrete Fourier Transform): Given a sequence
X [ X (0), X (1),..., X ( N 1)]
its Inverse Discrete Fourier Transform (IDFT) is a finite sequence
x IDFT ( X ) [ x(0), x(1),..., x( N 1)]
where
1
x ( n)
N
N 1
X (k )w
k 0
X
kn
N
, wN e j 2 / N
IDFT
x
Observations:
• The DFT and the IDFT form a transform pair.
back to the same signal !
x
DFT
x
IDFT
X
X
• The DFT is a numerical algorithm, and it can be computed by a digital computer.
DFT as a Vector Operation
Let
x[0]
x[1]
,
x
x
[
N
1
]
X [0]
X [1]
X DFT {x}
X
[
N
1
]
1
wk
ek N ,
k ( N 1)
wN
Then:
x
X [k ] ek*T x
x
1
X [0]e0 X [1]e1 ... X [ N 1]eN 1
N
ek
1
X [ k ]ek
N
1
1
x[0]
X [0] 1
x[1]
X [1] 1 w
N 1
w
N
N
X DFT {x}
( N 1)( N 1)
N 1
w
w
1
]
1
N
[
x
]
1
N
[
X
N
N
WN
X WN x
1 *T
x WN X
N
WN1
Periodicity: From the IDFT expression, notice that the sequence x(n) can be interpreted as one
period of a periodic sequence x p (n) :
1
x p ( n)
N
N 1
X (k )w
k 0
kn
N
1
N
N 1
X (k )w
k 0
kn
N
wN
kN
1
N
x(n)
k 0
k (n N )
N
x p (n N )
n
x p ( n)
N
X (k )w
original sequence
N 1
2N
N 1
periodic repetition
N
2N
n
This has a consequence when we define a time shift of the sequence.
For example see what we mean with x(n 1) . Start with the periodic extension x p (n)
x p ( n)
A
B
D
N
N
C
n
x p (n 1)
A
D
B
N
N
C
n
If we look at just one period we can define the circular shift
x(n)
x ( n 1) N
A
A
B
D
B
D
C
n
C
A
B
C
D
D
A
B
C
0
1
2
3
0
1
2
3
D
Properties of the DFT:
• one to one
x ( n) X ( k )
DFT x(n m) N wN km X ( k )
x (n m) N is a circular shift
• time shift
where
with no ambiguity;
periodic repetition
x(n)
x( N m) x( N 1)
x(0) x(1) x( N m) x( N m 1) x( N 1)
x (n m) N
x( N m) x( N 1) x(0) x(1) x( N m 1)
X
(
k
)
X
(N k)
• real sequences
| X ( k )| | X ( N k )|
• circular convolution
y (n) x1 (n) x2 (n)
N 1
x1 ( k ) x2 (n k ) N
k 0
circular shift
where both sequences x1 , x2
length N. Then:
must have the same
DFT x1 (n) x2 (n) X 1 ( k ) X 2 ( k ), k 0,..., N 1
Extension to General Intervals of Definition
Take the case of a sequence defined on a different interval:
x[n ]
n0
n0 N 1
How do we compute the DFT, without reinventing a new formula?
First see the periodic extension, which looks like this:
x[n ]
n0 N 1
n0
Then look at the period
n
0 n N 1
x[n ]
n0
n0 N 1
N 1
n
Example: determine the DFT of the finite sequence
x[n] 0.8|n| if 3 n 3
x[n ]
Then take the DFT of the vector
3
3
x x[0], x[1],..., x[3], x[3],..., x[1]
n
x[n ]
3
n