倒频谱的范例 倒频谱 （cepstrum ），顾名思义，就是将频谱 （spectrum）的英文前四个字母反过来写。倒频谱是为了某些时候，为了计算方便，将原来信号的频谱先转成类似分贝 的单位，再作逆傅里叶变换 ，把它视为一种新的讯号做处理。倒频谱有复数 倒频谱，及实数 倒频谱。
倒频谱被定义在1963的论文（Bogert等）。定义如下：
字义：倒频谱（信号）是信号频谱取对数的傅立叶变换后的新频谱（信号），有时候会称频谱的倒频谱。
数学上：信号的倒频谱 = IFT ( log ( | FT (信号) | ) + j2πm )（m为实数）
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 复数倒频谱拥有频谱大小跟相位的资讯，实数倒频谱只有频谱大小的资讯，各有各的不同应用。
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倒频谱可以被视为在不同频带上变化速率的资讯，倒频谱一开始被发明在地震 或炸弹 产生的地震回音，现今也被使用在分析雷达 讯号，以及讯号处理 等问题。
自相关倒频谱(autocepstrum)被定义为倒频谱的自相关性，自相关倒频谱有时在分析处理回传讯号时比倒频谱还准确。
倒频谱在处理人声 讯号以及音乐讯号有非常好的效果，例如梅尔频率倒频谱(Mel-Frequency Cepstrum)，用来做声音的辨认，侦测音高等。近年来梅耳倒频谱也被应用在音乐资讯的回复。
倒频谱在声学 中可以将声带 震动的影响去除。
倒频谱用在处理多路径问题时(如声波 的回音 、电磁波 的折、反射等)，如果将其他路径干扰 视为杂讯 ，为了消除杂讯，利用倒频谱，不需测量每条多路径的延迟时间，可以利用传送多次信号，观察其他路径在倒频谱上的效果，并且加以滤除。
语音大致上是由音高、声带 脉冲 、声门波形所组成，我们可以利用倒频谱将这三种元素在倒频域上分开，以利于做语音 讯号的分析。
倒频谱的微分适用于影像处理上的图形辨认(pattern recognition)。
倒频谱与同型声音理论(homomorphic sound theory)有关。
频谱图上的独立变数是频率 ，而倒频谱图上的独立变数为倒频率(quefrency)，倒频率是一种时间的度量单位 。举个例子，声音讯号取样速率等于44100赫兹 ，在倒频谱上有个很大的值在倒频率等于100，代表实际上在44100/100=441赫兹 有很大的值，这值出现在倒频谱上因为频谱上周期 性出现，而频谱上出现的周期与倒频谱很大的值出现的位置有关。
滤波器 (filter)常使用在频谱上，用来保存或删除我们所要或不要的资讯，经过上面的许多讨论，不难猜到，倒滤波器(lifter)就是在倒频谱上所使用的滤波器。低通的倒滤波器跟低通滤波器有点类似，它可以借由在倒频谱上乘以一个window系数，使倒频谱上的高倒频率被压抑，如此依来，当信号转回时域 空间时会变成一个较平滑的信号。
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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转换 所得到的系数, 我们可以利用泰勒展开得到Z的反转换
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无限脉冲响应 )有限脉冲响应 )的情况,
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1. 复数倒频谱至少以
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同态解卷积有非常多应用面，尤其是在声学工程和语音分析方面的实用性
(1) 回声的均衡化
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(2) 声学工程
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(3) 语音分析
透过在complex cepstrum domain中进行滤波，分离这些成分，使得对语音讯号的理解和处理更为精确。
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(4) 地震信号分析
(5) 任意波传播的多路径分析
梅尔频率倒频谱 是倒频谱的一种应用，梅尔频率倒频谱常应用在声音讯号处理，对于声音讯号处理比倒频谱更接近人耳对声音的分析特性，而梅尔频率倒频谱与倒频谱的差别在于:
梅尔频率倒频谱的频带分析是根据人耳听觉特性所设计，人耳对于频率 的分辨能力，是由频率的"比值"决定，也就是说，人耳对200赫兹 和300赫兹之间的差别与2000赫兹和3000赫兹之间的差别是相同的。
梅尔频率倒频谱是针对讯号的能量 取对数 ，而倒频谱是针对讯号原始在频谱上的值取对数 。
梅尔频率倒频谱是使用离散馀弦转换 ，倒频谱是用离散傅立叶变换 。
梅尔频率倒频谱系数足够描述语音 的特征。 梅尔频率倒频谱系数 (MFCCs)的推导步骤：
将信号做傅立叶变换
频谱上的值取绝对值再平方成为能量，在乘上频谱上对应的梅尔频率倒频谱三角重叠窗(window)的系数。
对每个梅尔频率取对数 。
作离散馀弦转换 。
求得梅尔频率倒频谱系数。
梅尔频率倒频谱系数常利用在辨认语音技术上，例如辨认电话中说话的人的身份。
利用每种乐风 、或乐器 在梅尔频域上有不同特性来分析音乐 的种类与类型，并且可以加以分类。
梅尔频率倒频谱系数很容易被外来的杂讯所破坏，因此有些研究结果指出，在求梅尔频率倒频谱系数时，在作离散馀弦转换前，提升适当的能量(大约2或3倍)，以减少杂讯在低能量成份的影响。
相较于原始的倒频谱
倒频谱领域上的一项重要的特性为二信号卷积 之产生，其产生之程序为二倒频谱值(cepstra)之相加：
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{\displaystyle {\widehat {x}}_{d}(n)=Z^{-1}{\frac {X'(Z)}{X(Z)}}}
x
^
d
[
n
]
=
∫
−
1
2
1
2
X
′
(
F
)
X
(
F
)
e
i
2
π
F
d
F
{\displaystyle {\widehat {x}}_{d}[n]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}{\frac {X'(F)}{X(F)}}e^{i2\pi F}dF}
(
d
d
Z
X
^
d
(
Z
)
=
d
d
Z
l
o
g
X
(
Z
)
=
X
′
(
Z
)
X
(
Z
)
)
{\displaystyle ({\frac {d}{dZ}}{\widehat {X}}_{d}(Z)={\frac {d}{dZ}}logX(Z)={\frac {X'(Z)}{X(Z)}})}
x
(
n
)
=
x
1
(
n
)
∗
x
2
(
n
)
{\displaystyle x(n)=x_{1}(n)*x_{2}(n)}
X
(
Z
)
=
X
1
(
Z
)
X
2
(
Z
)
{\displaystyle X(Z)=X_{1}(Z)X_{2}(Z)}
X
′
(
Z
)
=
X
1
′
(
Z
)
X
2
(
Z
)
+
X
1
(
Z
)
X
2
′
(
Z
)
{\displaystyle X'(Z)=X_{1}'(Z)X_{2}(Z)+X_{1}(Z)X_{2}'(Z)}
X
′
(
Z
)
X
(
Z
)
=
X
1
′
(
Z
)
X
1
(
Z
)
)
+
X
2
′
(
Z
)
X
2
(
Z
)
)
{\displaystyle {\frac {X'(Z)}{X(Z)}}={\frac {X_{1}'(Z)}{X_{1}(Z)}})+{\frac {X_{2}'(Z)}{X_{2}(Z)}})}
∴
x
^
d
(
n
)
=
x
^
1
d
(
n
)
+
x
^
2
d
(
n
)
{\displaystyle \therefore {\widehat {x}}_{d}(n)={\widehat {x}}_{1d}(n)+{\widehat {x}}_{2d}(n)}
(1)微分倒频谱在shift和scaling时，结果不改变。
y
[
n
]
=
A
X
[
n
−
r
]
{\displaystyle y[n]=AX[n-r]}
⇒
y
^
d
(
n
)
=
{
x
^
d
(
n
)
,
n
≠
1
−
r
+
x
^
d
(
1
)
,
n
=
1
{\displaystyle \Rightarrow {\widehat {y}}_{d}(n)={\begin{cases}{\widehat {x}}_{d}(n),n\neq 1\\-r+{\widehat {x}}_{d}(1),n=1\end{cases}}}
Y
(
z
)
=
A
z
−
r
X
(
z
)
{\displaystyle Y(z)=Az^{-r}X(z)}
Y
(
z
)
=
A
z
−
r
X
′
(
z
)
−
r
A
z
−
r
−
1
X
(
z
)
{\displaystyle Y(z)=Az^{-r}X'(z)-rAz^{-r-1}X(z)}
Y
′
(
z
)
Y
(
z
)
=
X
′
(
z
)
X
(
z
)
−
r
z
−
1
{\displaystyle {\frac {Y'(z)}{Y(z)}}={\frac {X'(z)}{X(z)}}-rz^{-1}}
C
^
[
n
]
{\displaystyle {\widehat {C}}[n]}
x
^
d
[
n
]
{\displaystyle {\widehat {x}}_{d}[n]}
C
^
(
n
)
=
−
x
^
d
(
n
+
1
)
n
,
n
≠
0
{\displaystyle {\widehat {C}}(n)={\frac {-{\widehat {x}}_{d}(n+1)}{n}},n\neq 0}
−
(
n
−
1
)
x
(
n
−
1
)
=
∑
k
=
−
∞
∞
x
^
d
(
n
)
x
(
n
−
k
)
{\displaystyle -(n-1)x(n-1)=\sum _{k=-\infty }^{\infty }{\widehat {x}}_{d}(n)x(n-k)}
⇒
{\displaystyle \Rightarrow }
x
^
d
[
n
]
=
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]}
复数倒频谱的衰减率反比于n
微分倒频谱的衰减率下降
∴
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
![{\displaystyle {\widehat {x}}\left[n\right]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}{\widehat {X}}\left(F\right)e^{j{2\pi }F}dF}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53fd79fd08cb7baa43c180daf38b077280c97a78)
![{\displaystyle {\widehat {X}}\left[F\right]=\log |X(F)|+j\arg[X(F)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b19f21fc1c295029e394a3ae81e0ae6dc5a4d1b)
![{\displaystyle \arg[X[n]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbead016b883d33a91c0f57bac117ce4859ebea2)
![{\displaystyle \arg[X(F)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8be9d5505fb6f5d612ac70f8e630e59db1819743)
![{\displaystyle C\left[n\right]=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}\log |X(F)|e^{j{2\pi }Fn}dF}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34cd5d49bf4f60bb7a029bf52243b16355db9a34)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19942017ae2cdf8b619f798bc865e1f17676dc49)
![{\displaystyle {\widehat {x}}\left[n\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca03af6e680e2941e245d067160a55907703f309)
![{\displaystyle nx[n]=\sum _{k=-\infty }^{\infty }k{\widehat {x}}\left[k\right]x[n-k]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6053e45be601a109313ea254e6b277b57462b82a)
![{\displaystyle x[n]=\sum _{k=-\infty }^{\infty }{\frac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]\quad for\ n\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/386f0809c5dcf9db9063c484700063ab375aa886)
![{\displaystyle x[n]={\widehat {x}}\left[n\right]=0,n<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1514630ccb026c54daad133a2d74bb842e37ff1a)
![{\displaystyle x[n]=\sum _{k=0}^{\infty }{\frac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]\quad for\ n>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b4951bd7acfad55c12edc885966d193f70f57d6)
![{\displaystyle x[n]={\widehat {x}}\left[n\right]x[0]+\sum _{k=0}^{n-1}{\frac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7edc8cf408a7d7ad9140920a64c1844b161d3ed9)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed67360fdb129413613e9166fecd8b6501f4e738)
![{\displaystyle x[n]={\widehat {x}}\left[n\right]=0,n>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81495c66d4414a7c87ca5a67565810ec87979661)
![{\displaystyle {\begin{aligned}x[n]&=\sum _{k=n}^{0}{\cfrac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]\quad for\ n<0\\&={\widehat {x}}\left[n\right]x[0]+\sum _{k=n+1}^{0}{\cfrac {k}{n}}{\widehat {x}}\left[k\right]x[n-k]\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c90ffd6365339c19df9be6406e90b8af6ae5e890)
![{\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)
![{\displaystyle y[n]=x[n]+{\alpha }x[n-N_{p}]=x[n]*p[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/887e12b559c77334cb4ebfc42f8f750c5ac3a1b1)
![{\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)
![{\displaystyle s[n]=g[n]*v[n]*p[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7658554b89ba6d45a618ac59e6b76c18cf87d5ab)
![{\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)
![{\displaystyle v[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cbc20699310c20ae48d376fc4f6365c67966405)
![{\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)
![{\displaystyle {\widehat {x}}_{d}[n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f7bd246bfa08d8e619a68a4e0f03675881ab74d)
![{\displaystyle {\widehat {x}}_{d}[n]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e3061ee1f32e533cbfe1517948a38ea63e5e4b8)
![{\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)
