在数学领域的谐波分析中，连续傅里叶变换 （continuous Fourier transform, CFT）与傅里叶级数 （Fourier series, FS）有非常微妙的关系。而且连续傅里叶变换也与离散时间傅里叶变换 （discrete time Fourier transform, DTFT）和离散傅里叶变换 （discrete Fourier transform, DFT）有很近的关系。傅里叶变换家族 通常就是指这四种变换。
通过利用Dirac delta函数 δ ( t ) {\displaystyle \delta (t)}
在信号理论和数字信号处理 （digital signal processing, DSP）中，DFT扩展用于近似计算连续信号的频谱，其变换的对象只是一个采样点的有限序列，而且可以由快速傅里叶变换 （fast Fourier transform, FFT）实现。
下表中左上、左下、右上和右下分别对应了傅里叶变换家族中CFT、FS、DTFT和DFT四个变换对的定义。
傅里叶变换家族中各种变换的定义 × 连续时间 离散时间 时间非周期 x ( t ) = ∫ − ∞ ∞ X ( f ) e i 2 π f t d f {\displaystyle x(t)=\int _{-\infty }^{\infty }X(f)\ e^{i2\pi ft}\,df} x [ n ] = T s ∫ 1 / T s X ¯ ( f ) e i 2 π f n T s d f {\displaystyle x[n]=T_{s}\int _{1/T_{s}}{\bar {X}}(f)\ e^{i2\pi fnT_{s}}\ df} - X ( f ) = ∫ − ∞ ∞ x ( t ) e − i 2 π f t d t {\displaystyle X(f)=\int _{-\infty }^{\infty }x(t)\ e^{-i2\pi ft}\,dt} X ¯ ( f ) = ∑ n = − ∞ + ∞ x [ n ] e − i 2 π f n T s {\displaystyle {\bar {X}}(f)=\sum _{n=-\infty }^{+\infty }x[n]\ e^{-i2\pi fnT_{s}}} 时间周期 x ¯ ( t ) = ∑ k = − ∞ + ∞ X [ k ] e i 2 π k T 0 t {\displaystyle {\bar {x}}(t)=\sum _{k=-\infty }^{+\infty }\!X[k]\;e^{i{\frac {2\pi k}{T_{0}}}t}} x n = 1 N ∑ k = 0 N − 1 X k e i 2 π N k n , n = 0 , … , N − 1. {\displaystyle x_{n}={\frac {1}{N}}\sum _{k=0}^{N-1}X_{k}\;e^{i{\frac {2\pi }{N}}kn},\quad n=0,\dots ,N-1.} - X [ k ] = 1 T 0 ∫ T 0 x ¯ ( t ) e − i 2 π k T 0 t d t {\displaystyle X[k]={\frac {1}{T_{0}}}\int _{T_{0}}{\bar {x}}(t)\;e^{-i{\frac {2\pi k}{T_{0}}}t}\,dt} X k = ∑ n = 0 N − 1 x n e − i 2 π N k n , k = 0 , … , N − 1. {\displaystyle X_{k}=\sum _{n=0}^{N-1}x_{n}\;e^{-i{\frac {2\pi }{N}}kn},\quad k=0,\dots ,N-1.}
显然，上表是从时域信号的角度来划分的：表的列区分了连续时间和离散时间的信号，而表的行则区分了时间上非周期的信号和时间上周期的信号。其中重要的参量符号解释为：
x [ n ] {\displaystyle x[n]} X [ k ] {\displaystyle X[k]} T s {\displaystyle T_{s}} f 0 = 1 / T 0 {\displaystyle f_{0}=1/T_{0}} x ¯ ( t ) {\displaystyle {\bar {x}}(t)} X ¯ ( f ) {\displaystyle {\bar {X}}(f)} T 0 {\displaystyle T_{0}} f s = 1 / T s {\displaystyle f_{s}=1/T_{s}} x n {\displaystyle x_{n}} X k {\displaystyle X_{k}} N {\displaystyle N} 前面表中的定义都可以通过Dirac delta函数 δ ( t ) {\displaystyle \delta (t)}
1. Dirac comb 函数的傅里叶变换
Dirac comb函数的定义为
Δ T ( t ) = def ∑ n = − ∞ ∞ δ ( t − n T ) {\displaystyle \Delta _{T}(t){\stackrel {\text{def}}{=}}\sum _{n=-\infty }^{\infty }\delta (t-nT)} 在电气工程中通常又称作冲击串（impulse train）或采样函数 （sampling function）。其重要的傅里叶变换为：
∑ n = − ∞ ∞ δ ( t − n T ) = 1 T ∑ k = − ∞ ∞ e i 2 π k T t ⟷ F 1 T ∑ k = − ∞ ∞ δ ( f − k T ) = ∑ n = − ∞ ∞ e − i 2 π n T f {\displaystyle \sum _{n=-\infty }^{\infty }\delta (t-nT)={\frac {1}{T}}\sum _{k=-\infty }^{\infty }e^{i{\frac {2\pi k}{T}}t}\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad {\frac {1}{T}}\sum _{k=-\infty }^{\infty }\delta \left(f-{\frac {k}{T}}\right)=\sum _{n=-\infty }^{\infty }e^{-i2\pi nTf}} 这个变换在傅里叶变换家族中各个变换之间转换上起关键作用。
2. 傅里叶变换的卷积定理 （convolution theorem）
这包括了傅里叶变换的时域卷积和频域卷积：
x 1 ( t ) ∗ x 2 ( t ) ⟷ F X 1 ( f ) ⋅ X 2 ( f ) x 1 ( t ) ⋅ x 2 ( t ) ⟷ F X 1 ( f ) ∗ X 2 ( f ) {\displaystyle {\begin{aligned}x_{1}(t)\ast x_{2}(t)&\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad X_{1}(f)\cdot X_{2}(f)\\x_{1}(t)\cdot x_{2}(t)&\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad X_{1}(f)\ast X_{2}(f)\end{aligned}}} 3. 泊松求和公式 （Poisson summation formula）
由Dirac comb函数的傅里叶变换和卷积定理，容易证明泊松求和公式：
1. ∑ n = − ∞ ∞ x ( t − n T 0 ) = 1 T 0 ∑ k = − ∞ ∞ X ( k T 0 ) e i 2 π k T 0 t 2. ∑ n = − ∞ ∞ x ( n T ) e − i 2 π n T f = 1 T ∑ k = − ∞ ∞ X ( f − k T ) {\displaystyle {\begin{aligned}1.\qquad &\sum _{n=-\infty }^{\infty }x(t-nT_{0})={\frac {1}{T_{0}}}\sum _{k=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}\right)e^{i{\frac {2\pi k}{T_{0}}}t}\\2.\qquad &\sum _{n=-\infty }^{\infty }x(nT)e^{-i2\pi nTf}={\frac {1}{T}}\sum _{k=-\infty }^{\infty }X\left(f-{\frac {k}{T}}\right)\end{aligned}}} 若第1和第2公式中分别取 t = 0 {\displaystyle t=0} f = 0 {\displaystyle f=0}
∑ n = − ∞ ∞ x ( n T ) = 1 T ∑ k = − ∞ ∞ X ( k T ) {\displaystyle \sum _{n=-\infty }^{\infty }x(nT)={\frac {1}{T}}\sum _{k=-\infty }^{\infty }X\left({\frac {k}{T}}\right)} 这表明，傅里叶变换时时域函数 x ( t ) {\displaystyle x(t)} X ( f ) {\displaystyle X(f)} T {\displaystyle T} 1 / T {\displaystyle 1/T} 1 / T {\displaystyle 1/T}
图 1. 此“立方体”图形表示了连续傅里叶变换 、离散时间傅里叶变换 、傅里叶级数 和离散傅里叶变换 之间的关系。
X ~ k {\displaystyle {\tilde {X}}_{k}} X k {\displaystyle X_{k}} X k = T 0 X ~ k {\displaystyle X_{k}=T_{0}{\tilde {X}}_{k}} 图中粗的双箭头（ ↔ {\displaystyle \leftrightarrow } 总的说来，各种变换之间的转换是一个周期扩展或采样的过程：
如果时域进行周期扩展，则频域为采样；如果时域进行采样，则频域为周期扩展； 一个转换中，周期扩展的周期与采样的间隔有倒数关系； 频域的周期扩展或者采样，都有一个周期或采样间隔作系数； 这里的周期扩展就是与Dirac comb函数相卷积，而采样则是与Dirac comb函数相乘。
从CFT分别到FS和DTFT的转换都容易推导，下面具体说明FS和DTFT到DFT/DFT'转换的推导，最后说明连续FT与DFT/DFT'的关系。
设DTFT，及对应的CFT为：
x [ n ] ⟷ D T F T X ¯ ( f ) ∑ n = − ∞ ∞ x [ n ] ⋅ δ ( t − n T ) ⟷ F X ¯ ( f ) {\displaystyle {\begin{aligned}x[n]&\quad {\stackrel {\mathcal {DTFT}}{\longleftrightarrow }}\quad {\bar {X}}(f)\\\sum _{n=-\infty }^{\infty }x[n]\cdot \delta (t-nT)&\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad {\bar {X}}(f)\end{aligned}}} 在时域作周期为 T 0 {\displaystyle T_{0}}
( ∑ n = − ∞ ∞ x [ n ] ⋅ δ ( t − n T ) ) ∗ ( ∑ n = − ∞ ∞ δ ( t − n T 0 ) ) = ∑ n = − ∞ ∞ ∑ i = − ∞ ∞ x ( n T ) ⋅ δ ( t − n T − i N T ) = ∑ n = − ∞ ∞ ( ∑ i = − ∞ ∞ x ( n T − i N T ) ) δ ( t − n T ) {\displaystyle {\begin{aligned}&\left(\sum _{n=-\infty }^{\infty }x[n]\cdot \delta (t-nT)\right)\ast \left(\sum _{n=-\infty }^{\infty }\delta (t-nT_{0})\right)\\&=\sum _{n=-\infty }^{\infty }\sum _{i=-\infty }^{\infty }x(nT)\cdot \delta (t-nT-iNT)\\&=\sum _{n=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }x(nT-iNT)\right)\delta (t-nT)\end{aligned}}} 其中代入了 T 0 = N T {\displaystyle T_{0}=NT} n {\displaystyle n} i {\displaystyle i} − ∞ {\displaystyle -\infty } ∞ {\displaystyle \infty } n − i N {\displaystyle n-iN} n {\displaystyle n}
x n = ∑ i = − ∞ ∞ x ( n T − i N T ) = ∑ i = − ∞ ∞ x [ n − i N ] = ∑ i = − ∞ ∞ x ( n T − i T 0 ) {\displaystyle {\begin{aligned}x_{n}&=\sum _{i=-\infty }^{\infty }x(nT-iNT)=\sum _{i=-\infty }^{\infty }x[n-iN]\\&=\sum _{i=-\infty }^{\infty }x(nT-iT_{0})\end{aligned}}} 在频域作带系数 1 / T 0 {\displaystyle 1/T_{0}} 1 / T 0 {\displaystyle 1/T_{0}}
X ¯ ( f ) ⋅ 1 T 0 ∑ k = − ∞ ∞ δ ( f − k T 0 ) = 1 T 0 T ( ∑ k = − ∞ ∞ X ( f − k T ) ) ⋅ ( ∑ k = − ∞ ∞ δ ( f − k T 0 ) ) = 1 T 0 T ∑ k = − ∞ ∞ ( ∑ i = − ∞ ∞ X ( f − i T ) ⋅ δ ( f − k T 0 ) ) = 1 T 0 T ∑ k = − ∞ ∞ ( ∑ i = − ∞ ∞ X ( k T 0 − i T ) ) ⋅ δ ( f − k T 0 ) {\displaystyle {\begin{aligned}&{\bar {X}}(f)\cdot {\frac {1}{T_{0}}}\sum _{k=-\infty }^{\infty }\delta \left(f-{\frac {k}{T_{0}}}\right)\\&={\frac {1}{T_{0}T}}\left(\sum _{k=-\infty }^{\infty }X\left(f-{\frac {k}{T}}\right)\right)\cdot \left(\sum _{k=-\infty }^{\infty }\delta \left(f-{\frac {k}{T_{0}}}\right)\right)\\&={\frac {1}{T_{0}T}}\sum _{k=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }X\left(f-{\frac {i}{T}}\right)\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)\right)\\&={\frac {1}{T_{0}T}}\sum _{k=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}-{\frac {i}{T}}\right)\right)\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)\end{aligned}}} 取：
X ~ k = 1 T 0 T ∑ i = − ∞ ∞ X ( k T 0 − i T ) = 1 T 0 X ¯ ( k T 0 ) {\displaystyle {\tilde {X}}_{k}={\frac {1}{T_{0}T}}\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}-{\frac {i}{T}}\right)={\frac {1}{T_{0}}}{\bar {X}}\left({\frac {k}{T_{0}}}\right)}
设FS，及对应的CFT为：
x ¯ ( t ) ⟷ F S X [ k ] x ¯ ( t ) ⟷ F ∑ k = − ∞ ∞ X [ k ] ⋅ δ ( f − k T 0 ) . {\displaystyle {\begin{aligned}{\bar {x}}(t)&\quad {\stackrel {\mathcal {FS}}{\longleftrightarrow }}\quad X[k]\\{\bar {x}}(t)&\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad \sum _{k=-\infty }^{\infty }X[k]\cdot \delta \left(f-{\frac {k}{T_{0}}}\right).\end{aligned}}} 在时域作间隔为 T {\displaystyle T}
x ¯ ( t ) ⋅ ∑ n = − ∞ ∞ δ ( t − n T ) = ( ∑ n = − ∞ ∞ x ( t − n T 0 ) ) ⋅ ( ∑ n = − ∞ ∞ δ ( t − n T ) ) = ∑ n = − ∞ ∞ ( ∑ i = − ∞ ∞ x ( t − i T 0 ) ⋅ δ ( t − n T ) ) = ∑ n = − ∞ ∞ ( ∑ i = − ∞ ∞ x ( n T − i T 0 ) ) ⋅ δ ( t − n T ) {\displaystyle {\begin{aligned}&{\bar {x}}(t)\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)\\&=\left(\sum _{n=-\infty }^{\infty }x(t-nT_{0})\right)\cdot \left(\sum _{n=-\infty }^{\infty }\delta (t-nT)\right)\\&=\sum _{n=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }x(t-iT_{0})\cdot \delta (t-nT)\right)\\&=\sum _{n=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }x(nT-iT_{0})\right)\cdot \delta (t-nT)\end{aligned}}} 取：
x n = ∑ i = − ∞ ∞ x ( n T − i T 0 ) = x ¯ ( n T ) {\displaystyle x_{n}=\sum _{i=-\infty }^{\infty }x(nT-iT_{0})={\bar {x}}(nT)} 在频域作带系数 1 / T {\displaystyle 1/T} 1 / T {\displaystyle 1/T}
( ∑ k = − ∞ ∞ X [ k ] ⋅ δ ( f − k T 0 ) ) ∗ ( 1 T ∑ k = − ∞ ∞ δ ( f − k T ) ) = 1 T 0 T ∑ k = − ∞ ∞ ∑ i = − ∞ ∞ X ( k T 0 ) ⋅ δ ( f − k T 0 − i N T 0 ) = 1 T 0 T ∑ k = − ∞ ∞ ( ∑ i = − ∞ ∞ X ( k − i N T 0 ) ) ⋅ δ ( f − k T 0 ) {\displaystyle {\begin{aligned}&\left(\sum _{k=-\infty }^{\infty }X[k]\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)\right)\ast \left({\frac {1}{T}}\sum _{k=-\infty }^{\infty }\delta \left(f-{\frac {k}{T}}\right)\right)\\&={\frac {1}{T_{0}T}}\sum _{k=-\infty }^{\infty }\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}\right)\cdot \delta \left(f-{\frac {k}{T_{0}}}-{\frac {iN}{T_{0}}}\right)\\&={\frac {1}{T_{0}T}}\sum _{k=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }X\left({\frac {k-iN}{T_{0}}}\right)\right)\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)\end{aligned}}} 其中也代入了 T 0 = N T {\displaystyle T_{0}=NT} k {\displaystyle k} i {\displaystyle i} − ∞ {\displaystyle -\infty } ∞ {\displaystyle \infty } k − i N {\displaystyle k-iN} k {\displaystyle k}
X ~ k = 1 T 0 T ∑ i = − ∞ ∞ X ( k − i N T 0 ) = 1 T ∑ i = − ∞ ∞ X [ k − i N ] = 1 T 0 T ∑ i = − ∞ ∞ X ( k T 0 − i T ) {\displaystyle {\begin{aligned}{\tilde {X}}_{k}&={\frac {1}{T_{0}T}}\sum _{i=-\infty }^{\infty }X\left({\frac {k-iN}{T_{0}}}\right)={\frac {1}{T}}\sum _{i=-\infty }^{\infty }X[k-iN]\\&={\frac {1}{T_{0}T}}\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}-{\frac {i}{T}}\right)\end{aligned}}}
前面FS到DFT和DTFT到DFT的推导都得到相同的 x n {\displaystyle x_{n}} X ~ k {\displaystyle {\tilde {X}}_{k}} x n {\displaystyle x_{n}} X ~ k {\displaystyle {\tilde {X}}_{k}}
X ~ k = 1 T 0 ∑ n = 0 N − 1 x n e − 2 π i N k n x n = T ∑ k = 0 N − 1 X ~ k e 2 π i N k n {\displaystyle {\begin{aligned}{\tilde {X}}_{k}&={\frac {1}{T_{0}}}\sum \limits _{n=0}^{N-1}\ x_{n}e^{-{\frac {2\pi i}{N}}kn}\\x_{n}&=T\sum \limits _{k=0}^{N-1}\ {\tilde {X}}_{k}e^{{\frac {2\pi i}{N}}kn}\end{aligned}}} 记为：
x n ⟷ D F T ′ X ~ k {\displaystyle x_{n}\quad {\stackrel {\mathcal {DFT'}}{\longleftrightarrow }}\quad {\tilde {X}}_{k}} 对比传统DFT变换对的 x n {\displaystyle x_{n}} X k {\displaystyle X_{k}}
X k = T 0 X ~ k . {\displaystyle X_{k}=T_{0}{\tilde {X}}_{k}.} 这一对变换的等式右边系数的乘积为 T / T 0 = 1 / N {\displaystyle T/T_{0}=1/N} DFT 中的说明，因而完全可以将这里的DFT'看作传统DFT的另一种变换形式 。
而由前面转换的推导过程可得到：
∑ n = − ∞ ∞ x n ⋅ δ ( t − n T ) ⟷ F ∑ k = − ∞ ∞ X ~ k ⋅ δ ( f − k T 0 ) = 1 T 0 ∑ k = − ∞ ∞ X k ⋅ δ ( f − k T 0 ) {\displaystyle \sum _{n=-\infty }^{\infty }x_{n}\cdot \delta (t-nT)\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad {\begin{matrix}\displaystyle {\sum _{k=-\infty }^{\infty }{\tilde {X}}_{k}\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)}\\\displaystyle {={\frac {1}{T_{0}}}\sum _{k=-\infty }^{\infty }X_{k}\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)}\end{matrix}}} 为一对CFT，其中要求 T 0 = N T {\displaystyle T_{0}=NT} x ( t ) ⟷ F X ( f ) {\displaystyle x(t){\stackrel {\mathcal {F}}{\longleftrightarrow }}X(f)}
x n = ∑ i = − ∞ ∞ x ( n T − i T 0 ) ⟷ D F T ′ X ~ k = 1 T 0 T ∑ i = − ∞ ∞ X ( k T 0 − i T ) ⟷ D F T X k = 1 T ∑ i = − ∞ ∞ X ( k T 0 − i T ) {\displaystyle x_{n}=\sum _{i=-\infty }^{\infty }x(nT-iT_{0}){\begin{matrix}\displaystyle {\quad {\stackrel {\mathcal {DFT'}}{\longleftrightarrow }}\quad {\tilde {X}}_{k}={\frac {1}{T_{0}T}}\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}-{\frac {i}{T}}\right)}\\\displaystyle {\quad {\stackrel {\mathcal {DFT}}{\longleftrightarrow }}\quad X_{k}={\frac {1}{T}}\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}-{\frac {i}{T}}\right)}\end{matrix}}} 其中可以任选 T 0 {\displaystyle T_{0}} T {\displaystyle T} n , k ∈ Z {\displaystyle n,k\in \mathbb {Z} }
Oppenheim, Alan V.; Schafer, R. W.; and Buck, J. R., (1999). Discrete-time signal processing , Upper Saddle River, N.J. : Prentice Hall. ISBN 0137549202 Sklar, B., (2001). Digital Communications: Foundamentals and Applicatons, 2nd Edition , Prentice Hall PTR. ISBN 0130847887 
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![{\displaystyle x[n]=T_{s}\int _{1/T_{s}}{\bar {X}}(f)\ e^{i2\pi fnT_{s}}\ df}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16e1629a685dcb9f18e2614945520bb903229bc5)
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![{\displaystyle {\begin{aligned}x[n]&\quad {\stackrel {\mathcal {DTFT}}{\longleftrightarrow }}\quad {\bar {X}}(f)\\\sum _{n=-\infty }^{\infty }x[n]\cdot \delta (t-nT)&\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad {\bar {X}}(f)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e64cb6f15277601c425fc89b5ed8dc9f787552c)
![{\displaystyle {\begin{aligned}&\left(\sum _{n=-\infty }^{\infty }x[n]\cdot \delta (t-nT)\right)\ast \left(\sum _{n=-\infty }^{\infty }\delta (t-nT_{0})\right)\\&=\sum _{n=-\infty }^{\infty }\sum _{i=-\infty }^{\infty }x(nT)\cdot \delta (t-nT-iNT)\\&=\sum _{n=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }x(nT-iNT)\right)\delta (t-nT)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f89f084800bfe66de3fb03abfc1bdcebf39b684e)
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![{\displaystyle {\begin{aligned}{\bar {x}}(t)&\quad {\stackrel {\mathcal {FS}}{\longleftrightarrow }}\quad X[k]\\{\bar {x}}(t)&\quad {\stackrel {\mathcal {F}}{\longleftrightarrow }}\quad \sum _{k=-\infty }^{\infty }X[k]\cdot \delta \left(f-{\frac {k}{T_{0}}}\right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c1dead2aaac05dec0ae6a850950d51a1e02265d)
![{\displaystyle {\begin{aligned}&\left(\sum _{k=-\infty }^{\infty }X[k]\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)\right)\ast \left({\frac {1}{T}}\sum _{k=-\infty }^{\infty }\delta \left(f-{\frac {k}{T}}\right)\right)\\&={\frac {1}{T_{0}T}}\sum _{k=-\infty }^{\infty }\sum _{i=-\infty }^{\infty }X\left({\frac {k}{T_{0}}}\right)\cdot \delta \left(f-{\frac {k}{T_{0}}}-{\frac {iN}{T_{0}}}\right)\\&={\frac {1}{T_{0}T}}\sum _{k=-\infty }^{\infty }\left(\sum _{i=-\infty }^{\infty }X\left({\frac {k-iN}{T_{0}}}\right)\right)\cdot \delta \left(f-{\frac {k}{T_{0}}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c642ba2021c20ecaafafcd76ac6931d08274eaf)
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