短时傅里叶变换

短时傅里叶变换

Chapter 10 Fourier analysis of signals using discrete Fourier transform10.1 Fourier analysis of signals using the DFT 10.2 DFT analysis of sinusoidal signals 10.3 the time-dependent Fourier transform For finite-length signals, the DFT provides frequencydomain samples of the Discrete-time Fourier transform. In many cases, the signals do not inherently have finite length. The inconsistency between the finite-length requirement of the DFT and the reality of indefinitely long signals can be accommodated exactly or approximately through the concepts of windowing, block processing, and the time-dependent Fourier transform 1 (短时傅立叶变换).

短时傅里叶变换

10.1 Fourier analysis of signals using the DFT

One of the major applications of the DFT is in analyzing the frequency content of continuous-time signals.

Figure 10.1

短时傅里叶变换

真实频谱

抗混迭滤波器频响

由滤波器非理想引入误差

由量化和混迭引入误差

窗序列频谱

时域加窗和频域取样引入误差

Figure 10.2

短时傅里叶变换

采样率与抗混迭滤波器的截止频率的关系:f s= 2 fc

相临谱线间的频率间距与DFT点数的关系:Δω= 2π/ NΔΩ=Δω/ T= 2πf s/ NΔf=ΔΩ/ 2π= f s/ N每条谱线对应的频率:ω k= 2πk/ NΩ k= 2πf s k/ Nfk= fsk/ N

频率分辨率与窗形状和窗长的关系:矩形窗:Δ ml= 4π/ N汉宁/明窗:Δ ml= 8π/ N布莱克曼窗:Δ ml= 12π/ N

短时傅里叶变换

10.2 DFT analysis of sinusoidal signals

10.2.1 the effect of windowing 10.2.2 the effect of spectral sampling

We choose sinusoidal signals as the specific class of examples to discuss, but most of the issues raised apply more generally.

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短时傅里叶变换

10.2.1 the effect of windowingX ( e jω ) V ( e jω )

Before windowing:x[n]= A0 cos(ω0n+θ0 )+ A1 cos(ω1n+θ1)

A0 jθ0 jω0n A0 jθ0 jω0n A1 jθ1 jω1n A1 jθ1 jω1n= e e+ e e+ e e+ e e 2 2 2 2∞< n<∞

A0 jθ0 A0 jθ0 X (e )= 2π[ eδ (ωω0 )+ eδ (ω+ω0 ) 2 2 A1 jθ1 A1 jθ1+ eδ (ωω1 )+ eδ (ω+ω1 )] 2 2jω7

短时傅里叶变换

After windowing:v[n]= x[n]w[n]A0 A0 jθ 0 jω 0 n= w[n]e e+ w[n]e jθ 0 e jω0 n 2 2 A1 A1 jθ1 jω1n+ w[n]e e+ w[n]e jθ1 e jω1n 2 21 V (e )= X (e jω ) *W (e jω ) 2π A0 jθ0 A0 jθ0 j (ωω0 ) n= e W (e )+ e W (e j (ω+ω0 ) n ) 2 2 A A+ 1 e jθ1W (e j (ωω1 ) n )+ 1 e jθ1W (e j (ω+ω1 ) n ) 2 2jω

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短时傅里叶变换

EXAMPLE Example 10.3

| X (e

jω

加窗后)|

加窗前

(1)谱线展宽成窗频谱的主瓣宽 (2)产生旁瓣,衰减等于窗频谱的旁瓣衰减9

Figure 10.3(a)(b)

短时傅里叶变换

Figure 10.3(c)(d)(e)

(3)谱泄露谱线展宽和谱泄露导致:难以确定频率的位置和幅度;降低频率分辨率.产生旁瓣导致:产生假信号; 10淹没小信号.

短时傅里叶变换

We can find that windowing smears or broadens the impulse in theoretical Fourier representation, and thus reduces the ability to resolve sinusoidal signals that are closely spaced in frequency . The amplitude of one spectrum is affected by the amplitude of another

and vice versa when two components are closely spaced in frequency. This interaction is called leakage(泄露). The component at one frequency leaks into the vicinity of another component due to the spectral smearing introduced by the window. So reduced resolution and leakage are the two primary effects on the spectrum as a result of applying a window to the signal. The resolution is influenced primarily by the width of the main lobe of W (e jω ),while the degree of leakage depends on the relative amplitude of the main lobe and the side lobes of W (e jω ) . We define the frequency resolution(频率分辨率) is equal to the width of the main lobe of W (e jω ) . 11注意:频率分辨率=主瓣宽>DFT谱线间距

短时傅里叶变换

EXAMPLE

V (e jω )

L= 32

example 10 .8: 2π x[ n]= (cos( n) 14 4π n ))+ 0 .75 cos( 15 w[ n]= kaiser ( L= 32~ 64,β= 5 .48 )

L= 42

L= 54

L= 64

Conclusion: increase L can increase resolution Figure 10.10

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短时傅里叶变换

EXAMPLE

2π v[ n]= (3 .5 * cos( n) 14 2π n ))+ 3 .5 * 0 .75 cos( 25 w R[ n]: red w hanning[ n]: blue L= 32

13 Conclusion: shape of window has effect on frequency resolution

短时傅里叶变换

Determine window's shape and length(1) for Kaiser windows:0.12438( Asl+ 6.3) β= 0.76609( Asl 13.26) 0.4+ 0.09834( Asl 13.26) 0L= 24π ( Asl+ 12)+1 155Δ mlΔ ml: main lobe width Asl: relative side lobe level

60< Asl< 120 13.26≤ Asl≤ 60 Asl< 13.26

(2)for Blackman window: look up the table14

短时傅里叶变换

10.2.2 the effect of spectral samplingV ( e jω ) V[ k]

The DFT of the windowed sequence provides samples of V (e jω ) . Spectral sampling can sometimes produce misleading results.

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短时傅里叶变换

EXAMPLE

Before sampling

2π example 10 .4: x[ n]= (cos( n) 14 4π+ 0 .75 cos( n )) R 64[ n] 15 w[ n]' s length L= 64

峰值未取到

After samplingN=64

N=128

Figure 10.5(a)(b)(f)18

增加N峰值取到

短时傅里叶变换

EXAMPLEexample 10 .5: x[ n]= (cos(+ 0.75 cos( 2π n )) R64[ n] 8 w[ n]' s length L= 64 2π n 16

V[k]= X (e jω )≠ V (e jω )

N=64增加N其他值也取到 N=128

只取到峰值和零值 Figure 10.6 Conclusion: increase N can fine the sampling of the spectrum. Figure 10.719

短时傅里叶变换

N=32

EXAMPLE

example 10 .7:

N=64

N=128

2π x[ n]= (cos( n) 14 4π+ 0 .75 cos( n )) 15 w[ n]= kaiser ( L= 32,β= 5 .48 ) N= 32~ 1024

N=1024

Conclusion: increase N can't increase frequency resolution. Figure 10.920

短时傅里叶变换

EXAMPLE

MATLAB分析窗长对 DFT的影响

f ( t )= cos( 2π f 1t )+ cos( 2π f 2 t ), f 1= 2 Hz, f 2= 2 . 5 Hz, f s= 64 Hz, 0≤ n≤ 63 f ( t )|t= nT f1[ n]=,T= 1/ fs 64≤ n< 128 0 f 2[ n]= f ( t )|t= nT 0≤ n< 128分别作 128点 DFT,比较二者的不同.

F1[n]:窗长64,矩形窗,DFT点数128 F2[n]:窗长128,矩形窗,DFT点数128

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短时傅里叶变换

L1=64; L2=128; N=128; T=1/64 n1=0:L1-1; x1=cos(2*pi*2*n1*T)+ cos(2*pi*2.5*n1*T) n2=0:L2-1; x2=cos(2*pi*2*n2*T)+ cos(2*pi*2.5*n2*T) k=0:N-1; X1=fft(x1,N); X2=fft(x2,N) stem(k,abs(X1)); hold on; stem(k,

abs(X2),'r.');

思考:如何用加窗DFT求周期为N的周期序列的DFS?即使峰值正好被取样到,且谱线间隔=主瓣宽的一半.条件:采用矩形窗;DFT点数=窗长.22

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