倒頻譜 （cepstrum ），顧名思義，就是將頻譜 （spectrum）的英文前四個字母反過來寫。倒頻譜是為了某些時候，為了計算方便，將原來信號的頻譜先轉成類似分貝 的單位，再作逆傅里叶变换 ，把它視為一種新的訊號做處理。倒頻譜有複數 倒頻譜，及實數 倒頻譜。
倒頻譜的範例 倒頻譜被定義在1963的論文（Bogert等）。定義如下：
字義：倒頻譜（信號）是信號頻譜取對數的傅立葉變換後的新頻譜（信號），有時候會稱頻譜的倒頻譜。
數學上：信號的倒頻譜 = IFT ( log ( | FT (信号) | ) + j2πm )（m為實數）
演算法：信号 -> 傅立叶变换 -> 取绝对值 -> 取对数 -> 相位展开 -> 逆傅立叶变换 -> 倒频谱 複數倒頻譜擁有頻譜大小跟相位的資訊，實數倒頻譜只有頻譜大小的資訊，各有各的不同應用。
倒頻譜可以被視為在不同頻帶上變化速率的資訊，倒頻譜一開始被發明在地震 或炸彈 產生的地震回音，現今也被使用在分析雷達 訊號，以及訊號處理 等問題。
自相關倒頻譜(autocepstrum)被定義為倒頻譜的自相關性，自相關倒頻譜有時在分析處理回傳訊號時比倒頻譜還準確。
倒頻譜在處理人聲 訊號以及音樂訊號有非常好的效果，例如梅爾頻率倒頻譜(Mel-Frequency Cepstrum)，用來做聲音的辨認，偵測音高等。近年來梅耳倒頻譜也被應用在音樂資訊的回覆。
倒頻譜在聲學 中可以將聲帶 震動的影響去除。
倒頻譜用在處理多路徑問題時(如聲波 的迴音 、電磁波 的折、反射等)，如果將其他路徑干擾 視為雜訊 ，為了消除雜訊，利用倒頻譜，不需測量每條多路徑的延遲時間，可以利用傳送多次信號，觀察其他路徑在倒頻譜上的效果，並且加以濾除。
語音大致上是由音高、聲帶 脈衝 、聲門波形所組成，我們可以利用倒頻譜將這三種元素在倒頻域上分開，以利於做語音 訊號的分析。
倒頻譜的微分適用於影像處理上的圖形辨認(pattern recognition)。
倒頻譜與同型聲音理論(homomorphic sound theory)有關。
頻譜圖上的獨立變數是頻率 ，而倒頻譜圖上的獨立變數為倒頻率(quefrency)，倒頻率是一種時間的度量單位 。舉個例子，聲音訊號取樣速率等於44100赫茲 ，在倒頻譜上有個很大的值在倒頻率等於100，代表實際上在44100/100=441赫茲 有很大的值，這值出現在倒頻譜上因為頻譜上週期 性出現，而頻譜上出現的週期與倒頻譜很大的值出現的位置有關。
濾波器 (filter)常使用在頻譜上，用來保存或刪除我們所要或不要的資訊，經過上面的許多討論，不難猜到，倒濾波器(lifter)就是在倒頻譜上所使用的濾波器。低通的倒濾波器跟低通濾波器有點類似，它可以藉由在倒頻譜上乘以一個window係數，使倒頻譜上的高倒頻率被壓抑，如此依來，當信號轉回時域 空間時會變成一個較平滑的信號。
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先對信號做Z轉換 , 並整理一下係數, 讓他變成下面的形式
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{\displaystyle X\left(Z\right)={\cfrac {A{Z^{r}}\prod _{k=1}^{m_{i}}(1-{a_{k}}{Z^{-1}})\prod _{k=1}^{m_{0}}(1-{b_{k}}Z)}{\prod _{k=1}^{P_{i}}(1-{c_{k}}{Z^{-1}})\prod _{k=1}^{P_{0}}(1-{d_{k}}Z)}}}
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{\displaystyle {\widehat {X}}\left(Z\right)=logX\left(Z\right)=\log A+r\log Z+\sum _{k=1}^{m_{i}}\log(1-{a_{k}}{Z^{-1}})+\sum _{k=1}^{m_{0}}\log(1-{b_{k}}Z)-\sum _{k=1}^{P_{i}}\log(1-{c_{k}}{Z^{-1}})-\sum _{k=1}^{P_{0}}\log(1-{d_{k}}Z)}
Z轉換 所得到的系數, 我們可以利用泰勒展開得到Z的反轉換
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{\displaystyle {\widehat {x}}\left[n\right]={\begin{cases}\log A&{\mbox{if }}n=0\\-\sum _{k=1}^{m_{i}}{\cfrac {{a_{k}}^{n}}{n}}+\sum _{k=1}^{P_{i}}{\cfrac {{c_{k}}^{n}}{n}}&{\mbox{if }}n>0\\\sum _{k=1}^{m_{0}}{\cfrac {{b_{k}}^{-n}}{n}}-\sum _{k=1}^{P_{0}}{\cfrac {{d_{k}}^{-n}}{n}}&{\mbox{if }}n<0\end{cases}}}
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無限脈衝響應 )有限脈衝響應 )的情況,
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{\displaystyle x[n]=\sum _{k=-\infty }^{\infty }{\frac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]\quad for\ n\neq 0}
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{\displaystyle x[n]=\sum _{k=-\infty }^{\infty }{\frac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]\quad for\ n\neq 0}
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{\displaystyle {\widehat {x}}\left[n\right]={\begin{cases}0&{\mbox{if }}n<0\\{\cfrac {x[n]}{x[0]}}-\sum _{k=0}^{n-1}{\cfrac {k}{n}}{\widehat {x}}\left[k\right]{\cfrac {x[n-k]}{x[0]}}&{\mbox{if }}n>0\\\log A&{\mbox{if }}n=0\end{cases}}}
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1. 複數倒頻譜至少以
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编辑
梅爾頻率倒頻譜 是倒頻譜的一種應用，梅爾頻率倒頻譜常應用在聲音訊號處理，對於聲音訊號處理比倒頻譜更接近人耳對聲音的分析特性，而梅爾頻率倒頻譜與倒頻譜的差別在於:
梅爾頻率倒頻譜的頻帶分析是根據人耳聽覺特性所設計，人耳對於頻率 的分辨能力，是由頻率的"比值"決定，也就是說，人耳對200赫茲 和300赫茲之間的差別與2000赫茲和3000赫茲之間的差別是相同的。
梅爾頻率倒頻譜是針對訊號的能量 取對數 ，而倒頻譜是針對訊號原始在頻譜上的值取對數 。
梅爾頻率倒頻譜是使用離散餘弦轉換 ，倒頻譜是用離散傅立葉變換 。
梅爾頻率倒頻譜係數足夠描述語音 的特徵。 梅爾頻率倒頻譜係數 (MFCCs)的推導步驟：
將信號做傅立葉變換
頻譜上的值取絕對值再平方成為能量，在乘上頻譜上對應的梅爾頻率倒頻譜三角重疊窗(window)的係數。
對每個梅爾頻率取對數 。
作離散餘弦轉換 。
求得梅爾頻率倒頻譜係數。
梅爾頻率倒頻譜係數常利用在辨認語音技術上，例如辨認電話中說話的人的身份。
利用每種樂風 、或樂器 在梅爾頻域上有不同特性來分析音樂 的種類與類型，並且可以加以分類。
梅爾頻率倒頻譜係數很容易被外來的雜訊所破壞，因此有些研究結果指出，在求梅爾頻率倒頻譜係數時，在作離散餘弦轉換前，提升適當的能量(大約2或3倍)，以減少雜訊在低能量成份的影響。
倒頻譜領域上的一項重要的特性為二信號卷積 之產生，其產生之程序為二倒頻譜值(cepstra)之相加：
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(1)微分倒頻譜在shift和scaling時，結果不改變。
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{\displaystyle \Rightarrow }
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]
=
0
{\displaystyle {\widehat {x}}_{d}[n]=0}
n
≤
0
{\displaystyle n\leq 0}
x
^
d
[
n
]
=
0
{\displaystyle {\widehat {x}}_{d}[n]=0}
n
≥
2
{\displaystyle n\geq 2}
x
^
d
[
n
]
{\displaystyle {\widehat {x}}_{d}[n]}
複數倒頻譜的衰減率反比於n
微分倒頻譜的衰減率下降
∴
x
^
d
(
n
+
1
)
=
n
c
^
(
n
)
∝
n
1
n
=
1
{\displaystyle \therefore {\widehat {x}}_{d}(n+1)=n{\widehat {c}}(n)\varpropto n{\frac {1}{n}}=1}
x
[
0
]
=
1
,
x
[
1
]
=
0.5
{\displaystyle x[0]=1,x[1]=0.5}
x
[
n
]
⟶
Z
t
r
a
n
s
f
o
r
m
X
(
Z
)
⟶
l
o
g
X
^
(
Z
)
⟶
Z
−
1
x
^
[
n
]
{\displaystyle x[n]\quad {\stackrel {Ztransform}{\longrightarrow }}\quad X(Z)\quad {\stackrel {log}{\longrightarrow }}\quad {\widehat {X}}(Z)\quad {\stackrel {Z^{-1}}{\longrightarrow }}\quad {\widehat {x}}[n]}
X
(
Z
)
=
1
+
0.5
Z
−
1
,
p
o
l
e
=
−
0.5
{\displaystyle X(Z)=1+0.5Z^{-1},pole=-0.5}
X
^
(
Z
)
=
∑
k
=
1
m
i
l
o
g
(
1
−
(
−
0.5
Z
−
1
)
)
{\displaystyle {\widehat {X}}(Z)=\sum _{k=1}^{m_{i}}log(1-(-0.5Z^{-1}))}
x
^
[
n
]
=
∑
n
=
0
N
−
−
0.5
n
n
,
n
>
0
{\displaystyle {\widehat {x}}[n]=\sum _{n=0}^{N}-{\frac {-0.5^{n}}{n}},n>0}
x
^
[
0
]
=
1
{\displaystyle {\widehat {x}}[0]=1}
x
^
[
n
]
⟶
Z
t
r
a
n
s
f
o
r
m
X
^
(
Z
)
⟶
e
x
p
X
(
Z
)
⟶
Z
−
1
x
[
n
]
{\displaystyle {\widehat {x}}[n]\quad {\stackrel {Ztransform}{\longrightarrow }}\quad {\widehat {X}}(Z)\quad {\stackrel {exp}{\longrightarrow }}\quad {X}(Z)\quad {\stackrel {Z^{-1}}{\longrightarrow }}\quad {x}[n]}
X
^
[
n
]
=
Z
−
1
{\displaystyle {\widehat {X}}[n]=Z^{-1}}
e
(
1
z
)
=
∑
n
=
0
∞
1
z
n
n
!
{\displaystyle e({\frac {1}{z}})=\sum _{n=0}^{\infty }{\frac {\frac {1}{z^{n}}}{n!}}}
x
[
n
]
=
{
1
n
!
,
n
≥
0
0
,
o
t
h
e
r
w
i
s
e
{\displaystyle x[n]={\begin{cases}{\frac {1}{n!}},n\geq 0\\0,otherwise\\\end{cases}}}
Suppose that an IIR filter is
H
(
Z
)
=
2
z
3
−
4
z
2
−
z
+
2
2
z
2
−
2
z
+
1
{\displaystyle H(Z)={\frac {2z^{3}-4z^{2}-z+2}{2z^{2}-2z+1}}}
x
[
n
]
⟶
Z
t
r
a
n
s
f
o
r
m
X
(
Z
)
⟶
l
o
g
X
^
(
Z
)
⟶
Z
−
1
x
^
[
n
]
{\displaystyle x[n]\quad {\stackrel {Ztransform}{\longrightarrow }}\quad X(Z)\quad {\stackrel {log}{\longrightarrow }}\quad {\widehat {X}}(Z)\quad {\stackrel {Z^{-1}}{\longrightarrow }}\quad {\widehat {x}}[n]}
H
(
Z
)
=
(
−
2
)
(
z
)
(
z
−
2
2
z
−
1
)
(
z
+
2
2
z
−
1
)
(
1
−
1
2
z
)
(
1
−
1
+
j
2
z
−
1
)
(
1
−
1
−
j
2
z
−
1
)
{\displaystyle H(Z)={\frac {(-2)(z)(z-{\frac {\sqrt {2}}{2}}z^{-1})(z+{\frac {\sqrt {2}}{2}}z^{-1})(1-{\frac {1}{2}}z)}{(1-{\frac {1+j}{2}}z^{-1})(1-{\frac {1-j}{2}}z^{-1})}}}
H
^
(
Z
)
=
l
o
g
(
−
2
)
+
3
l
o
g
(
z
)
+
l
o
g
(
1
±
2
2
z
−
1
)
+
l
o
g
(
1
−
1
2
z
)
−
l
o
g
(
1
−
1
±
j
2
z
−
1
)
{\displaystyle {\widehat {H}}(Z)=log(-2)+3log(z)+log(1\pm {\frac {\sqrt {2}}{2}}z^{-1})+log(1-{\frac {1}{2}}z)-log(1-{\frac {1\pm j}{2}}z^{-1})}
h
^
[
n
]
=
{
l
o
g
(
−
2
)
,
n
=
0
−
(
2
2
)
n
+
(
−
2
2
)
n
n
+
(
1
+
j
2
)
n
+
(
1
−
j
2
)
n
n
,
n
>
0
(
1
2
)
−
n
n
,
n
<
0
{\displaystyle {\widehat {h}}[n]={\begin{cases}log(-2),n=0\\\displaystyle {-{\frac {{({\frac {\sqrt {2}}{2}})}^{n}+{({\frac {-{\sqrt {2}}}{2}})}^{n}}{n}}+{\frac {{({\frac {1+j}{2}})}^{n}+{({\frac {1-j}{2}})}^{n}}{n}},n>0}\\\displaystyle {\frac {{({\frac {1}{2}})}^{-n}}{n}},n<0\\\end{cases}}}
{\displaystyle }
B. P. Bogert, M. J. R. Healy, and J. W. Tukey : "The quefrency analysis of time series for echoes: cepstrum, pseudo-autocovariance , cross-cepstrum, and saphe cracking". Proceedings of the Symposium on Time Series Analysis (M. Rosenblatt, Ed) Chapter 15, 209-243. New York: Wiley, 1963.
D. G. Childers, D. P. Skinner, R. C. Kemerait, "The Cepstrum: A Guide to Processing （页面存档备份 ，存于互联网档案馆 ）," Proceedings of the IEEE , Vol. 65, No. 10, October 1977, pp. 1428-1443.
Jian-Jiun Ding, Advanced Digital Signal Processing class note,the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2008
Jian-Jiun Ding, Advanced Digital Signal Processing class note,the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2024
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![{\displaystyle x[n]={\widehat {x}}\left[n\right]=0,n>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81495c66d4414a7c87ca5a67565810ec87979661)
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![{\displaystyle {\widehat {x}}\left[n\right]={\begin{cases}0&{\mbox{if }}n>0\\{\cfrac {x[n]}{x[0]}}-\sum _{k=n+1}^{0}{\cfrac {k}{n}}{\widehat {x}}\left[k\right]{\cfrac {x[n-k]}{x[0]}}&{\mbox{if }}n<0\\\log A&{\mbox{if }}n=0\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f740b6047d4c0205a92e5114685a9807c2b3527c)
![{\displaystyle |{\widehat {x}}\left[n\right]|=c|{\frac {{\alpha }^{n}}{n}}|\quad -\infty <n<\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e336e1c3ee0bbb56cd37840d6eb070f51b230316)
![{\displaystyle {\widehat {x}}\left[n\right]=0\quad for\ all\ n<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1be4cfda70976ead8352817f095e5687a6b19bb1)
![{\displaystyle {\widehat {x}}\left[n\right]=0\quad for\ all\ n>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92a992c190980f58cb79553e4bf42fb5f6e38f20)
![{\displaystyle y[n]=x[n]+{\alpha }x[n-N_{p}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65d53e471788f953f19bd12bc11a10b4189a3cbc)
![{\displaystyle y[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/305428e6d1fb59cd0163a7a96ace52292a262afa)
![{\displaystyle x[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/864cbbefbdcb55af4d9390911de1bf70167c4a3d)
![{\displaystyle p[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a8ded9fe52b8f86424447c30b547440ba6bbb67)
![{\displaystyle p[n]={\delta }[n]+{\alpha }{\delta [n-N_{p}]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7d0445c72d307f6eee5a1773bb8368ad7b5eb32)
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![{\displaystyle {\hat {p}}[n]={\sum _{k=1}^{\infty }}(-1)^{k+1}{\frac {\alpha ^{k}}{k}}{\delta }(n-k{\cdot }N_{P})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/025c1205c8b39af66a9955b72affadad4418d008)
![{\displaystyle y[n]=x[n]*h[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9151e7a5bd07493b61827b52aefd550b3acc5414)
![{\displaystyle h[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89981bbbb05ffd469eeadb828c18359965985e46)
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![{\displaystyle {\widehat {C}}[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31005250c7d49f42a1fa823c17a26f2716298630)
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![{\displaystyle x[n]\quad {\stackrel {Ztransform}{\longrightarrow }}\quad X(Z)\quad {\stackrel {log}{\longrightarrow }}\quad {\widehat {X}}(Z)\quad {\stackrel {Z^{-1}}{\longrightarrow }}\quad {\widehat {x}}[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c59b1842420ee6a92a756488936bf3ffb0bb6a1f)
![{\displaystyle {\widehat {x}}[n]=\sum _{n=0}^{N}-{\frac {-0.5^{n}}{n}},n>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28673a650671e36f64c1a34dda570a95e20a1781)
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