倒頻譜的範例 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。
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{\displaystyle {\widehat {x}}\left[n\right]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}{\widehat {X}}\left(F\right)e^{j{2\pi }F}dF}
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奇對稱 ,所以複數倒頻譜的值為實數
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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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{\displaystyle {\widehat {x}}\left[n\right]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}{\widehat {X}}\left(F\right)e^{j{2\pi }F}dF}
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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}
最小相位 (minimum phase) i.e.
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1. 複數倒頻譜至少以
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同態解卷積有非常多應用面，尤其是在聲學工程和語音分析方面的實用性
(1) 回聲的均衡化
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系統函數
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透過對系統函數進行對數轉換，簡化回聲成分的分析和處理
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(2) 聲學工程
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(3) 語音分析
透過在complex cepstrum domain中進行濾波，分離這些成分，使得對語音訊號的理解和處理更為精確。
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(4) 地震信號分析
(5) 任意波傳播的多路徑分析
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。
相較於原始的倒頻譜
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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{\displaystyle x_{1}*x_{2}\rightarrow x'_{1}+x'_{2}}
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{\displaystyle {\widehat {x}}_{d}(n)=Z^{-1}{\frac {X'(Z)}{X(Z)}}}
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{\displaystyle {\widehat {x}}_{d}[n]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}{\frac {X'(F)}{X(F)}}e^{i2\pi F}dF}
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{\displaystyle ({\frac {d}{dZ}}{\widehat {X}}_{d}(Z)={\frac {d}{dZ}}logX(Z)={\frac {X'(Z)}{X(Z)}})}
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{\displaystyle x(n)=x_{1}(n)*x_{2}(n)}
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{\displaystyle X(Z)=X_{1}(Z)X_{2}(Z)}
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{\displaystyle X'(Z)=X_{1}'(Z)X_{2}(Z)+X_{1}(Z)X_{2}'(Z)}
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{\displaystyle {\frac {X'(Z)}{X(Z)}}={\frac {X_{1}'(Z)}{X_{1}(Z)}})+{\frac {X_{2}'(Z)}{X_{2}(Z)}})}
∴
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{\displaystyle \therefore {\widehat {x}}_{d}(n)={\widehat {x}}_{1d}(n)+{\widehat {x}}_{2d}(n)}
(1)微分倒頻譜在shift和scaling時，結果不改變。
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{\displaystyle y[n]=AX[n-r]}
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{\displaystyle \Rightarrow {\widehat {y}}_{d}(n)={\begin{cases}{\widehat {x}}_{d}(n),n\neq 1\\-r+{\widehat {x}}_{d}(1),n=1\end{cases}}}
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{\displaystyle Y(z)=Az^{-r}X(z)}
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{\displaystyle Y(z)=Az^{-r}X'(z)-rAz^{-r-1}X(z)}
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{\displaystyle {\frac {Y'(z)}{Y(z)}}={\frac {X'(z)}{X(z)}}-rz^{-1}}
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{\displaystyle {\widehat {C}}[n]}
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{\displaystyle {\widehat {x}}_{d}[n]}
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{\displaystyle {\widehat {C}}(n)={\frac {-{\widehat {x}}_{d}(n+1)}{n}},n\neq 0}
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{\displaystyle -(n-1)x(n-1)=\sum _{k=-\infty }^{\infty }{\widehat {x}}_{d}(n)x(n-k)}
⇒
{\displaystyle \Rightarrow }
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{\displaystyle {\widehat {x}}_{d}[n]=0}
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{\displaystyle n\leq 0}
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{\displaystyle {\widehat {x}}_{d}[n]=0}
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{\displaystyle n\geq 2}
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{\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
∴
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{\displaystyle \therefore {\widehat {x}}_{d}(n+1)=n{\widehat {c}}(n)\varpropto n{\frac {1}{n}}=1}
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{\displaystyle x[0]=1,x[1]=0.5}
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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]}
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{\displaystyle X(Z)=1+0.5Z^{-1},pole=-0.5}
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{\displaystyle {\widehat {X}}(Z)=\sum _{k=1}^{m_{i}}log(1-(-0.5Z^{-1}))}
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{\displaystyle {\widehat {x}}[n]=\sum _{n=0}^{N}-{\frac {-0.5^{n}}{n}},n>0}
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{\displaystyle {\widehat {x}}[0]=1}
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{\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]}
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{\displaystyle {\widehat {X}}[n]=Z^{-1}}
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{\displaystyle e({\frac {1}{z}})=\sum _{n=0}^{\infty }{\frac {\frac {1}{z^{n}}}{n!}}}
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{\displaystyle x[n]={\begin{cases}{\frac {1}{n!}},n\geq 0\\0,otherwise\\\end{cases}}}
Suppose that an IIR filter is
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{\displaystyle H(Z)={\frac {2z^{3}-4z^{2}-z+2}{2z^{2}-2z+1}}}
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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]}
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{\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})}}}
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{\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})}
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{\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 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)
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![{\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 s[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffe878c033cc70d1aa3ffbd156394791aa8dffde)
![{\displaystyle g[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c5e1d771a2385e9aeb71838a40425bb07c89525)
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![{\displaystyle {\widehat {x}}_{d}[n]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}{\frac {X'(F)}{X(F)}}e^{i2\pi F}dF}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a442cde66a5681fcaec1cf5326223e0d9cedeaa)
![{\displaystyle y[n]=AX[n-r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4224b81109de9905f7a132a4dfeecada6de33f9)
![{\displaystyle {\widehat {C}}[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31005250c7d49f42a1fa823c17a26f2716298630)
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![{\displaystyle x[0]=1,x[1]=0.5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70896529eb43da420d27b4c7fa7fe6c702d39664)
![{\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)
![{\displaystyle {\widehat {x}}[0]=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bdbb01c7d751290c993e714468afa8557faff88)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfd7884d2ca6a4eb0b92e9f16424f89ba72b89f6)
![{\displaystyle {\widehat {X}}[n]=Z^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b5395f3c192a651dbf39001117394100db2c77c)
![{\displaystyle x[n]={\begin{cases}{\frac {1}{n!}},n\geq 0\\0,otherwise\\\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8a7cbd7fa5f686ebb4a438db8ba718475f40a75)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0fa5cdfbbcaef6760dfc4bedb97292af6f1c86d7)
