改进型韦格纳分布 (modified Wigner distribution function)，用于时频分析 的一种方法，属于信号处理 的范畴。它改进了韦格纳分布原有的相交项(cross term)的问题。尤金·维格纳 所提出用于古典力学 ，但是亦可用于时频分析。韦格纳分布与短时距傅立叶变换 都可用于时频分析，虽然前者通常拥有较高的解析度且有良好的数学特性，但当有两个以上的信号成分时，韦格纳分布就会出现相交项问题，这在应用上造成很大的困扰。
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }x(t+\tau /2)x^{*}(t-\tau /2)e^{-j2\pi \tau \,f}d\tau }
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{\displaystyle =\int _{-\infty }^{\infty }X(f+\eta /2)\cdot X^{*}(f-\eta /2)e^{j2\pi t\eta }\cdot d\eta }
为了改善韦格纳分布的相交项(cross-term)问题，改进型韦格纳分布在此引入了一个类似遮罩(mask)的函数，将相交项过滤掉。
定义一 ：:
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }w(\tau /2)w^{*}(-\tau /2)x(t+\tau /2)x^{*}(t-\tau /2)e^{-j2\pi \tau \,f}d\tau }
，其中w(t)为遮罩函数. 常为方波，其方波宽度为参数B。可写成
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e
{\displaystyle w(t)={\begin{cases}1\ \ \ if\ |t|<B\\0\ \ \ otherwise\end{cases}}}
定义二 ：:
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }P(\theta )Y(t,f+\theta /2)Y^{*}(t,f-\theta /2)d\theta }
Y
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{\displaystyle Y(t,f)=\int _{-\infty }^{\infty }w(\tau )x(t+\tau )e^{-j2\pi f\tau }d\tau }
P
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{\displaystyle P(\theta )\,}
P
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≈
1
{\displaystyle P(\theta )\approx 1\ }
P
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0
{\displaystyle P(\theta )\approx 0\ }
适当地选择
P
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≈
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{\displaystyle P(\theta )\approx 1\ }
定义三：
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }w(\tau )x^{L}(t+{\tfrac {\tau }{2L}})\cdot {\overline {x^{L}(t-{\tfrac {\tau }{2L}})}}e^{-j2\pi \tau f}\cdot d\tau }
增加 L 可以减少相交项(cross-term)的影响(但是不会完全消除)
定义四：
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }[\prod _{l=1}^{q/2}x(t+d_{l}\tau )x^{*}(t-d_{-l}\tau )]e^{-j2\pi \tau f}d\tau }
当 q = 2 和
d
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−
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0.5
{\displaystyle d_{l}=d_{-l}=0.5}
当指数函数的次项不超过 q/2 +1时，就可以避免相交项(cross-term)
然而，相交项(cross-term)会介于两个讯号之间，无法完全被移除。
<说明>
定义四的维格纳分布又称为多项型维格纳分布 (Polynomial Wigner Distribution Function)
W
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]
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n-1}\tau }\ e^{-j2\pi \tau f}d\tau =\int _{-\infty }^{\infty }[\prod _{\ell =1}^{q/2}x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )]e^{-j2\pi \tau f}d\tau }
If
x
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j
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∑
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a
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{\displaystyle x(t)=e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n}}}
所以
d
ℓ
{\displaystyle d_{\ell }}
e
j
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{\displaystyle e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n-1}\tau }=\prod _{\ell =1}^{q/2}x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )}
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }e^{-j2\pi (f-\sum _{n=1}^{{\tfrac {q}{2}}+1})na_{n}t^{n-1}\tau }d\tau \cong \delta (f-\sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n-1}\tau )}
其中
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{\displaystyle \sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n-1}}
x
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{\displaystyle x(t)}
接下来，我们来看
d
ℓ
{\displaystyle d_{\ell }}
(1) 当
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{\displaystyle q=2}
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{\displaystyle x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )=e^{j2\pi \sum _{n=1}^{2}na_{n}t^{n-1}\tau }}
如果我们把
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{\displaystyle x(t)=e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}a_{n}t^{n}}}
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{\displaystyle a_{2}(t+d_{1}\tau )^{2}+a_{1}(t+d_{1}\tau )-a_{2}(t-d_{-1}\tau )-a_{1}(t-d_{-1}\tau )=2a_{2}t\tau +a_{1}\tau }
{
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{\displaystyle {\begin{cases}d_{1}+d_{-1}=1\\d_{1}-d_{-1}=0\end{cases}}}
⟹
{
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{\displaystyle \Longrightarrow {\begin{cases}d_{1}={\tfrac {1}{2}}\\d_{-1}={\tfrac {1}{2}}\end{cases}}}
由此可以知道，当
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{\displaystyle q=2}
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{\displaystyle d_{-1}=d_{1}={\tfrac {1}{2}}}
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{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }[\prod _{l=1}^{q/2}x(t+d_{l}\tau )x^{*}(t-d_{-l}\tau )]e^{-j2\pi \tau f}d\tau =\int _{-\infty }^{\infty }x(t+\tau /2)x^{*}(t-\tau /2)e^{-j2\pi \tau \,f}d\tau ,\quad if\quad q=2,\ d_{-1}=d_{1}={\tfrac {1}{2}}}
(2) 当
q
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4
{\displaystyle q=4}
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{\displaystyle x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )=e^{j2\pi \sum _{n=1}^{3}na_{n}t^{n-1}\tau }}
如果我们把
x
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t
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j
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∑
n
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a
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t
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{\displaystyle x(t)=e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}a_{n}t^{n}}}
a
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1
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{\displaystyle a_{3}(t+d_{1}\tau )^{3}+a_{2}(t+d_{1}\tau )^{2}+a_{1}(t+d_{1}\tau )a_{3}(t+d_{2}\tau )^{3}+a_{2}(t+d_{2}\tau )^{2}+a_{1}(t+d_{2}\tau )-a_{3}(t+d_{-1}\tau )^{3}-a_{2}(t+d_{-1}\tau )^{2}-a_{1}(t+d_{-1}\tau )-a_{3}(t+d_{-2}\tau )^{3}-a_{2}(t+d_{-2}\tau )^{2}-a_{1}(t+d_{-2}\tau )}
=
3
a
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{\displaystyle =3a_{3}t^{2}\tau +2a_{2}t\tau +a_{1}\tau }
{
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2
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+
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2
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a
2
(
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+
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2
a
1
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t
+
d
1
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+
a
1
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t
+
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2
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−
a
1
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−
a
1
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{\displaystyle {\begin{cases}a_{3}(t+d_{1}\tau )^{3}+a_{3}(t+d_{2}\tau )^{3}-a_{3}(t+d_{-1}\tau )^{3}-a_{3}(t+d_{-2}\tau )^{3}\\a_{2}(t+d_{1}\tau )^{2}+a_{2}(t+d_{2}\tau )^{2}-a_{2}(t+d_{-1}\tau )^{2}-a_{2}(t+d_{-2}\tau )^{2}\\a_{1}(t+d_{1}\tau )+a_{1}(t+d_{2}\tau )-a_{1}(t+d_{-1}\tau )-a_{1}(t+d_{-2}\tau )\end{cases}}}
=
{
3
a
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t
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a
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a
1
τ
{\displaystyle ={\begin{cases}3a_{3}t^{2}\tau \\2a_{2}t\tau \\a_{1}\tau \end{cases}}}
所以我们可以得到
{
d
1
+
d
2
+
d
−
1
+
d
−
2
=
1
d
1
2
+
d
2
2
−
d
−
1
2
−
d
−
2
2
=
0
d
1
3
+
d
2
3
+
d
−
1
3
+
d
−
2
3
=
0
{\displaystyle {\begin{cases}d_{1}+d_{2}+d_{-1}+d_{-2}=1\\{d_{1}}^{2}+{d_{2}}^{2}-{d_{-1}}^{2}-{d_{-2}}^{2}=0\\{d_{1}}^{3}+{d_{2}}^{3}+{d_{-1}}^{3}+{d_{-2}}^{3}=0\end{cases}}}
可以看到如果
q
{\displaystyle q}
d
ℓ
{\displaystyle d_{\ell }}
在此有两个例子来说明改进型韦格纳分布 确实能消除相交项。
x
(
t
)
=
{
cos
(
3
π
t
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t
≤
−
4
cos
(
6
π
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)
−
4
<
t
≤
4
cos
(
4
π
t
)
t
>
4
{\displaystyle x(t)={\begin{cases}\cos(3\pi t)\ \ \ t\leq -4\\\cos(6\pi t)\ \ \ -4<t\leq 4\ \ \ \\\cos(4\pi t)\ \ \ t>4\end{cases}}}
左图是韦格纳分布；右图是改进型韦格纳分布。可以很明显地看出，改进型韦格纳分布大大地改进相交项的问题，相对地增加清晰度。
x
(
t
)
=
exp
(
j
⋅
(
t
−
5
)
3
−
j
⋅
6
π
⋅
t
)
{\displaystyle x(t)=\exp(j\cdot (t-5)^{3}-j\cdot 6\pi \cdot t)}
左图是韦格纳分布；右图是改进型韦格纳分布。明显地看出，改进型韦格纳分布确实可改进相交项的问题，同时增加清晰度。
Jian-Jiun Ding, class lecture of Time Frequency Analysis and Wavelet transform, Graduate Institute of Communication Engineering, National Taiwan University, Taipei, Taiwan, 2007.
L. J. Stankovic, S. Stankovic, and E. Fakultet, “An analysis of instantaneous frequency representation using time frequency distributions-generalized Wigner distribution,” IEEE Trans. on Signal Processing, pp. 549-552, vol. 43, no. 2, Feb. 1995
Jian-Jiun Ding, Time frequency analysis and wavelet transform class notes, Graduate Institute of Communication Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2017.
Jian-Jiun Ding, Time frequency analysis and wavelet transform class notes, Graduate Institute of Communication Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2018. 
![{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }[\prod _{l=1}^{q/2}x(t+d_{l}\tau )x^{*}(t-d_{-l}\tau )]e^{-j2\pi \tau f}d\tau }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cf68cacbcd3a707a04e7c68d64eed52a9256b44)
![{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n-1}\tau }\ e^{-j2\pi \tau f}d\tau =\int _{-\infty }^{\infty }[\prod _{\ell =1}^{q/2}x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )]e^{-j2\pi \tau f}d\tau }](https://wikimedia.org/api/rest_v1/media/math/render/svg/319e854a87bf973c40e5b22c6eb33abeba7ad171)
![{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }[\prod _{l=1}^{q/2}x(t+d_{l}\tau )x^{*}(t-d_{-l}\tau )]e^{-j2\pi \tau f}d\tau =\int _{-\infty }^{\infty }x(t+\tau /2)x^{*}(t-\tau /2)e^{-j2\pi \tau \,f}d\tau ,\quad if\quad q=2,\ d_{-1}=d_{1}={\tfrac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8de2918f29ab8da6fff362afd8333b95b36846db)
