改進型韋格納分佈 (modified Wigner distribution function)，用於時頻分析 的一種方法，屬於信號處理 的範疇。它改進了韋格納分佈原有的相交項(cross term)的問題。尤金·維格納 所提出用於古典力學 ，但是亦可用於時頻分析。韋格納分佈與短時距傅立葉變換 都可用於時頻分析，雖然前者通常擁有較高的解像度且有良好的數學特性，但當有兩個以上的信號成分時，韋格納分佈就會出現相交項問題，這在應用上造成很大的困擾。
W
x
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t
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f
)
=
∫
−
∞
∞
x
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/
2
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x
∗
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t
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τ
f
d
τ
{\displaystyle W_{x}(t,f)=\int _{-\infty }^{\infty }x(t+\tau /2)x^{*}(t-\tau /2)e^{-j2\pi \tau \,f}d\tau }
=
∫
−
∞
∞
X
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f
+
η
/
2
)
⋅
X
∗
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f
−
η
/
2
)
e
j
2
π
t
η
⋅
d
η
{\displaystyle =\int _{-\infty }^{\infty }X(f+\eta /2)\cdot X^{*}(f-\eta /2)e^{j2\pi t\eta }\cdot d\eta }
為了改善韋格納分佈的相交項(cross-term)問題，改進型韋格納分佈在此引入了一個類似遮罩(mask)的函數，將相交項過濾掉。
定義一 ：:
W
x
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t
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f
)
=
∫
−
∞
∞
w
(
τ
/
2
)
w
∗
(
−
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/
2
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x
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2
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x
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/
2
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d
τ
{\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)為遮罩函數. 常為方波，其方波寬度為參數B。可寫成
w
(
t
)
=
{
1
i
f
|
t
|
<
B
0
o
t
h
e
r
w
i
s
e
{\displaystyle w(t)={\begin{cases}1\ \ \ if\ |t|<B\\0\ \ \ otherwise\end{cases}}}
定義二 ：:
W
x
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t
,
f
)
=
∫
−
∞
∞
P
(
θ
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Y
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,
f
+
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/
2
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Y
∗
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t
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/
2
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d
θ
{\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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,
f
)
=
∫
−
∞
∞
w
(
τ
)
x
(
t
+
τ
)
e
−
j
2
π
f
τ
d
τ
{\displaystyle Y(t,f)=\int _{-\infty }^{\infty }w(\tau )x(t+\tau )e^{-j2\pi f\tau }d\tau }
P
(
θ
)
{\displaystyle P(\theta )\,}
P
(
θ
)
≈
1
{\displaystyle P(\theta )\approx 1\ }
P
(
θ
)
≈
0
{\displaystyle P(\theta )\approx 0\ }
適當地選擇
P
(
θ
)
≈
1
{\displaystyle P(\theta )\approx 1\ }
定義三：
W
x
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t
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f
)
=
∫
−
∞
∞
w
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x
L
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2
L
)
⋅
x
L
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t
−
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2
L
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¯
e
−
j
2
π
τ
f
⋅
d
τ
{\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 可以減少相交項(cross-term)的影響(但是不會完全消除)
定義四：
W
x
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∫
−
∞
∞
[
∏
l
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1
q
/
2
x
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t
+
d
l
τ
)
x
∗
(
t
−
d
−
l
τ
)
]
e
−
j
2
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τ
f
d
τ
{\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 }
當 q = 2 和
d
l
=
d
−
l
=
0.5
{\displaystyle d_{l}=d_{-l}=0.5}
當指數函數的次項不超過 q/2 +1時，就可以避免相交項(cross-term)
然而，相交項(cross-term)會介於兩個訊號之間，無法完全被移除。
<說明>
定義四的維格納分佈又稱為多項型維格納分佈 (Polynomial Wigner Distribution Function)
W
x
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t
,
f
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=
∫
−
∞
∞
e
j
2
π
∑
n
=
1
q
2
+
1
n
a
n
t
n
−
1
τ
e
−
j
2
π
τ
f
d
τ
=
∫
−
∞
∞
[
∏
ℓ
=
1
q
/
2
x
(
t
+
d
ℓ
τ
)
x
∗
(
t
−
d
−
ℓ
τ
)
]
e
−
j
2
π
τ
f
d
τ
{\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
(
t
)
=
e
j
2
π
∑
n
=
1
q
2
+
1
n
a
n
t
n
{\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
2
π
∑
n
=
1
q
2
+
1
n
a
n
t
n
−
1
τ
=
∏
ℓ
=
1
q
/
2
x
(
t
+
d
ℓ
τ
)
x
∗
(
t
−
d
−
ℓ
τ
)
{\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 )}
W
x
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t
,
f
)
=
∫
−
∞
∞
e
−
j
2
π
(
f
−
∑
n
=
1
q
2
+
1
)
n
a
n
t
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1
τ
d
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≅
δ
(
f
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=
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+
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a
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τ
)
{\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
∑
n
=
1
q
2
+
1
n
a
n
t
n
−
1
{\displaystyle \sum _{n=1}^{{\tfrac {q}{2}}+1}na_{n}t^{n-1}}
x
(
t
)
{\displaystyle x(t)}
接下來，我們來看
d
ℓ
{\displaystyle d_{\ell }}
(1) 當
q
=
2
{\displaystyle q=2}
x
(
t
+
d
ℓ
τ
)
x
∗
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t
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=
e
j
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∑
n
=
1
2
n
a
n
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−
1
τ
{\displaystyle x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )=e^{j2\pi \sum _{n=1}^{2}na_{n}t^{n-1}\tau }}
如果我們把
x
(
t
)
=
e
j
2
π
∑
n
=
1
q
2
+
1
a
n
t
n
{\displaystyle x(t)=e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}a_{n}t^{n}}}
a
2
(
t
+
d
1
τ
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2
+
a
1
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−
a
2
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τ
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−
a
1
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t
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−
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)
=
2
a
2
t
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+
a
1
τ
{\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 }
{
d
1
+
d
−
1
=
1
d
1
−
d
−
1
=
0
{\displaystyle {\begin{cases}d_{1}+d_{-1}=1\\d_{1}-d_{-1}=0\end{cases}}}
⟹
{
d
1
=
1
2
d
−
1
=
1
2
{\displaystyle \Longrightarrow {\begin{cases}d_{1}={\tfrac {1}{2}}\\d_{-1}={\tfrac {1}{2}}\end{cases}}}
由此可以知道，當
q
=
2
{\displaystyle q=2}
d
−
1
=
d
1
=
1
2
{\displaystyle d_{-1}=d_{1}={\tfrac {1}{2}}}
W
x
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t
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f
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=
∫
−
∞
∞
[
∏
l
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1
q
/
2
x
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t
+
d
l
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x
∗
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l
τ
)
]
e
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2
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τ
f
d
τ
=
∫
−
∞
∞
x
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t
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/
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x
∗
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t
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/
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e
−
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2
π
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f
d
τ
,
i
f
q
=
2
,
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1
=
d
1
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1
2
{\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) 當
q
=
4
{\displaystyle q=4}
x
(
t
+
d
ℓ
τ
)
x
∗
(
t
−
d
−
ℓ
τ
)
=
e
j
2
π
∑
n
=
1
3
n
a
n
t
n
−
1
τ
{\displaystyle x(t+d_{\ell }\tau )x^{*}(t-d_{-\ell }\tau )=e^{j2\pi \sum _{n=1}^{3}na_{n}t^{n-1}\tau }}
如果我們把
x
(
t
)
=
e
j
2
π
∑
n
=
1
q
2
+
1
a
n
t
n
{\displaystyle x(t)=e^{j2\pi \sum _{n=1}^{{\tfrac {q}{2}}+1}a_{n}t^{n}}}
a
3
(
t
+
d
1
τ
)
3
+
a
2
(
t
+
d
1
τ
)
2
+
a
1
(
t
+
d
1
τ
)
a
3
(
t
+
d
2
τ
)
3
+
a
2
(
t
+
d
2
τ
)
2
+
a
1
(
t
+
d
2
τ
)
−
a
3
(
t
+
d
−
1
τ
)
3
−
a
2
(
t
+
d
−
1
τ
)
2
−
a
1
(
t
+
d
−
1
τ
)
−
a
3
(
t
+
d
−
2
τ
)
3
−
a
2
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t
+
d
−
2
τ
)
2
−
a
1
(
t
+
d
−
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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
3
t
2
τ
+
2
a
2
t
τ
+
a
1
τ
{\displaystyle =3a_{3}t^{2}\tau +2a_{2}t\tau +a_{1}\tau }
{
a
3
(
t
+
d
1
τ
)
3
+
a
3
(
t
+
d
2
τ
)
3
−
a
3
(
t
+
d
−
1
τ
)
3
−
a
3
(
t
+
d
−
2
τ
)
3
a
2
(
t
+
d
1
τ
)
2
+
a
2
(
t
+
d
2
τ
)
2
−
a
2
(
t
+
d
−
1
τ
)
2
−
a
2
(
t
+
d
−
2
τ
)
2
a
1
(
t
+
d
1
τ
)
+
a
1
(
t
+
d
2
τ
)
−
a
1
(
t
+
d
−
1
τ
)
−
a
1
(
t
+
d
−
2
τ
)
{\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
3
t
2
τ
2
a
2
t
τ
a
1
τ
{\displaystyle ={\begin{cases}3a_{3}t^{2}\tau \\2a_{2}t\tau \\a_{1}\tau \end{cases}}}
所以我們可以得到
{
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 }}
在此有兩個例子來說明改進型韋格納分佈 確實能消除相交項。
x
(
t
)
=
{
cos
(
3
π
t
)
t
≤
−
4
cos
(
6
π
t
)
−
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}}}
左圖是韋格納分佈；右圖是改進型韋格納分佈。可以很明顯地看出，改進型韋格納分佈大大地改進相交項的問題，相對地增加清晰度。
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)}
左圖是韋格納分佈；右圖是改進型韋格納分佈。明顯地看出，改進型韋格納分佈確實可改進相交項的問題，同時增加清晰度。
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)
