数字信号处理lecture17 22页PPT文档

（2）DFT 点数 N L
（ 据频域 采样定 理， N=L 即可从 X(k)通过 内插完 全恢复X ( ) 。
因 此，N=32>L=20 的 离散 频谱已 足够反 映 DTFT 谱 。N=64 有 何 意义？ 为 什么要 提出 computational frequency resolution？ ―― The question of how accurately the DFT represents the peaks in the spectrum. ）
To m ak e the peak at f 0 coin cide with one of th e D F T frequencies, k0 should be an integer in〔 0,N-1〕 :
f0 k0
fs N

k0 N
f0 fs
(9.3.1)
For the peak at f0 (equivalent with f 0 f s ), the
DFT：
linear transformation finite x(n)
(can be implemented by a matrix)
finite X(k)
X ( k )
L 1
x (n )e j kn

L 1
2 j kn
x(n)e N
n0
n0
j 2
记 W N e N L 1
本次课内容：
1. 数值方法求 DTFT 遇到二个问题
DFT
2. 信号的两种频率分辨率的不同含义：
DTFT 谱：f
fw
c
fs L
c 1 TL
，物理频率分辨率
DFT 谱：fbin
fs N
，计算频率分辨率
9.2 DTFT Computation
9.2.1 DTFT at a Single Frequency
It is caused by the lack of adequate physical resolution.
It can only be decreased by increasing the data length L.
● rounding error (p.493 and Fig.9.3.2)
Caused by rounding k The rounding error in the frequencies is less than
fbin f s 2 2N
It decreases with increasing DFT length N.
9.4 Matrix Form of DFT
C alculate the D FT indices in the case of Fig.9.3.1, W hen N＝ 32，
k1 N
f1 ＝ 32x0.2＝ 6.4， N k1 ＝ 25.6 fs
Sim ilarly,
k 2 ＝ 8 （ i n t e g e r ）， N k 2 ＝ 2 4 ，
k3 ＝ 9.6， N k3 ＝ 22.4
Question：
From calculation, DFT indices k ＝ 6 、 10 do not
represent the exact positions of f1 、 f3 .
But in Fig.9.3.1(left, bottom), k ＝ 6,8,10 correctly identifies the three peaks in the spectrum.
f s ＝ 10kH z, f1,2,3＝ 2、 2.5、 3kH z，
f1,2,3 /fs＝ 0.2、 0.25、 0.3）
Physical frequency resolution：


w

c
2 L
， f

fw

c
fs L
（ Hz）
（ 9.1.21）
( the m inim um resolvable frequency separation between two sinusoidal com ponents )
x
(
n
)W
kn N
n0
（ k=0~N -1）
X (0)
X (1)




W W
0 N
0 N
W
0 N
W
1 N


W
0 N
W L 1 N


x(0)

x (1)

X ( N 1) N 1
W
ቤተ መጻሕፍቲ ባይዱ
0 N

X ( ) e jD X ( )
9.3 Physical versus Com putational Resolution
Fig.9.3.1 （ The data is from exam ple 9.1.4：
x (t ) cos( 2 f1t ) cos( 2 f 2t ) cos( 2 f 3t ) ,
k
a
k b a N
a
k bin
bin : bin width
（ k＝ 0~N-1） (9.2.4)
9.2.3 D FT
DFT(discrete Fourier transform ) 离 散 傅 立 叶 变 换
N -poin t D FT of a len gth -L sign al: D TF T at N equ ally spaced
For length－L sequence x(n)，we compute DTFT at any
desired value of by
L 1
X ( ) x(n)e jn （9.2.1） n0
(Nyquist interval
[0,2 ]
, equivalent with
n0
X (k N ) X (k) （p.490）
(一个周期：k=0~N-1)
9.2.4 Zero Padding
padding zeros at the end of x(n):
X ( ) remains the same;
padding zeros to the front of x(n):
[ , ]
)
A more efficient way to evaluate polynomials (Hörner’s rule) :
L1
X (z) x(n)zn n0 x(0) z1(x(1) z1(x(2) z1(x(3) z1)))
and X () X (z) |zej
How to explain it?(p492 中间 for two reasons:)
● biasing error in DTFT spectrum (p.492)
biasing: the maxima of the peaks in the spectrum do not quite correspond to the correct frequencies.
9.2.2 DTFT over Frequency Range
To com pu te th e D TF T over a frequ en cy ran ge [ a , b ) , we can
only com pute the spectrum at N discrete frequencies:
frequencies over the full N yquist interval [0,2 ]
k

2 N
k
（ k＝ 0~N-1）
(9.2.6)
L 1
X ( k ) X ( k ) x ( n ) e j k n n0
(9.2.8)
bin width:
bin
“index” satisfies:

f0

(k0 )
fs N
or

f0

fs

k0
fs N

fs

(N
k0)
fs N
In general, k0 and N k0 are not necessarily integers, and
the DFT will miss the exact peaks.
W
( N
N
1 )(
L
1)

N

L

x
(
L

1
)

L 1
X Ax
(9.4.1)
twiddle factor (旋 转 因 子 ):
j2 N
WN e
(9.4.4)
W
0 N

W
N N
2
1 ，WN N


1
，
W
N N

k

W
k N
,
N=L=2，
A

1 1
1 1
Com putational frequency resolution：
bin

2 N
，
f bin

fs N
（ Hz）
（ 9.2.9）
（ the frequency bin width in the DFT case）
为 了 从 DFT 频 谱 区 分 f1、2、3，
（ 1） 信 号 长 度 L f s 2 f