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Convolution theorem

From Wikipedia, the free encyclopedia

Theorem in mathematics

Inmathematics, theconvolution theorem states that under suitable conditions theFourier transform of aconvolution of two functions (orsignals) is the product of their Fourier transforms. More generally, convolution in one domain (e.g.,time domain) equals point-wise multiplication in the other domain (e.g.,frequency domain). Other versions of the convolution theorem are applicable to variousFourier-related transforms.

Functions of a continuous variable

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Consider two functions $u(x)$ and $v(x)$ withFourier transforms $U {\displaystyle U}$ and $V {\displaystyle V}$ :

{\begin{aligned}U(f)&\triangleq {\mathcal {F}}\{u\}(f)=\int _{-\infty }^{\infty }u(x)e^{-i2\pi fx}\,dx,\quad f\in \mathbb {R} \\V(f)&\triangleq {\mathcal {F}}\{v\}(f)=\int _{-\infty }^{\infty }v(x)e^{-i2\pi fx}\,dx,\quad f\in \mathbb {R} \end{aligned}}

where ${\mathcal {F}}$ denotes theFourier transformoperator. The transform may be normalized in other ways, in which case constant scaling factors (typically $2\pi$ or ${\sqrt {2\pi }}$ ) will appear in the convolution theorem below. The convolution of $u {\displaystyle u}$ and $v {\displaystyle v}$ is defined by:

r(x)=\{u*v\}(x)\triangleq \int _{-\infty }^{\infty }u(\tau )v(x-\tau )\,d\tau =\int _{-\infty }^{\infty }u(x-\tau )v(\tau )\,d\tau .

In this context theasterisk denotes convolution, instead of standard multiplication. Thetensor product symbol $\otimes$ is sometimes used instead.

Theconvolution theorem states that:^[1]^[2]^: eq.8

$R(f)\triangleq {\mathcal {F}}\{r\}(f)=U(f)V(f).\quad f\in \mathbb {R}$

Eq.1a

Applying the inverse Fourier transform ${\mathcal {F}}^{-1},$ produces the corollary:^[2]^{: eqs.7, 10}

Convolution theorem

$r(x)=\{u*v\}(x)={\mathcal {F}}^{-1}\{U\cdot V\}.$

Eq.1b

The theorem also generally applies to multi-dimensional functions.

Multi-dimensional derivation of Eq.1
Consider functions $u, v {\displaystyle u,v}$ inL^p-space $L^{1}(\mathbb {R} ^{n}),$ with Fourier transforms $U, V {\displaystyle U,V}$ : ${\begin{aligned}U(f)&\triangleq {\mathcal {F}}\{u\}(f)=\int _{\mathbb {R} ^{n}}u(x)e^{-i2\pi f\cdot x}\,dx,\quad f\in \mathbb {R} ^{n}\\V(f)&\triangleq {\mathcal {F}}\{v\}(f)=\int _{\mathbb {R} ^{n}}v(x)e^{-i2\pi f\cdot x}\,dx,\end{aligned}}$ where $f\cdot x$ indicates theinner product of $\mathbb {R} ^{n}$ : $f\cdot x=\sum _{j=1}^{n}{f}_{j}x_{j},$ and $dx=\prod _{j=1}^{n}dx_{j}.$ Theconvolution of $u {\displaystyle u}$ and $v {\displaystyle v}$ is defined by: $r(x)\triangleq \int _{\mathbb {R} ^{n}}u(\tau )v(x-\tau )\,d\tau .$ Also: $\iint \|u(\tau )v(x-\tau )\|\,dx\,d\tau =\int \left(\|u(\tau )\|\int \|v(x-\tau )\|\,dx\right)\,d\tau =\int \|u(\tau )\|\,\\|v\\|_{1}\,d\tau =\\|u\\|_{1}\\|v\\|_{1}.$ Hence byFubini's theorem we have that $r\in L^{1}(\mathbb {R} ^{n})$ so its Fourier transform $R {\displaystyle R}$ is defined by the integral formula: ${\begin{aligned}R(f)\triangleq {\mathcal {F}}\{r\}(f)&=\int _{\mathbb {R} ^{n}}r(x)e^{-i2\pi f\cdot x}\,dx\\&=\int _{\mathbb {R} ^{n}}\left(\int _{\mathbb {R} ^{n}}u(\tau )v(x-\tau )\,d\tau \right)\,e^{-i2\pi f\cdot x}\,dx.\end{aligned}}$ Note that $\|u(\tau )v(x-\tau )e^{-i2\pi f\cdot x}\|=\|u(\tau )v(x-\tau )\|,$ Hence by the argument above we may apply Fubini's theorem again (i.e. interchange the order of integration): ${\begin{aligned}R(f)&=\int _{\mathbb {R} ^{n}}u(\tau )\underbrace {\left(\int _{\mathbb {R} ^{n}}v(x-\tau )\ e^{-i2\pi f\cdot x}\,dx\right)} _{V(f)\ e^{-i2\pi f\cdot \tau }}\,d\tau \\&=\underbrace {\left(\int _{\mathbb {R} ^{n}}u(\tau )\ e^{-i2\pi f\cdot \tau }\,d\tau \right)} _{U(f)}\ V(f).\end{aligned}}$

Multi-dimensional derivation of Eq.1

Consider functions $u, v {\displaystyle u,v}$ inL^p-space $L^{1}(\mathbb {R} ^{n}),$ with Fourier transforms $U, V {\displaystyle U,V}$ :

{\begin{aligned}U(f)&\triangleq {\mathcal {F}}\{u\}(f)=\int _{\mathbb {R} ^{n}}u(x)e^{-i2\pi f\cdot x}\,dx,\quad f\in \mathbb {R} ^{n}\\V(f)&\triangleq {\mathcal {F}}\{v\}(f)=\int _{\mathbb {R} ^{n}}v(x)e^{-i2\pi f\cdot x}\,dx,\end{aligned}}

where $f\cdot x$ indicates theinner product of $\mathbb {R} ^{n}$ : $f\cdot x=\sum _{j=1}^{n}{f}_{j}x_{j},$ and $dx=\prod _{j=1}^{n}dx_{j}.$

Theconvolution of $u {\displaystyle u}$ and $v {\displaystyle v}$ is defined by:

r(x)\triangleq \int _{\mathbb {R} ^{n}}u(\tau )v(x-\tau )\,d\tau .

Also:

\iint |u(\tau )v(x-\tau )|\,dx\,d\tau =\int \left(|u(\tau )|\int |v(x-\tau )|\,dx\right)\,d\tau =\int |u(\tau )|\,\|v\|_{1}\,d\tau =\|u\|_{1}\|v\|_{1}.

Hence byFubini's theorem we have that $r\in L^{1}(\mathbb {R} ^{n})$ so its Fourier transform $R {\displaystyle R}$ is defined by the integral formula:

{\begin{aligned}R(f)\triangleq {\mathcal {F}}\{r\}(f)&=\int _{\mathbb {R} ^{n}}r(x)e^{-i2\pi f\cdot x}\,dx\\&=\int _{\mathbb {R} ^{n}}\left(\int _{\mathbb {R} ^{n}}u(\tau )v(x-\tau )\,d\tau \right)\,e^{-i2\pi f\cdot x}\,dx.\end{aligned}}

Note that $|u(\tau )v(x-\tau )e^{-i2\pi f\cdot x}|=|u(\tau )v(x-\tau )|,$ Hence by the argument above we may apply Fubini's theorem again (i.e. interchange the order of integration):

{\begin{aligned}R(f)&=\int _{\mathbb {R} ^{n}}u(\tau )\underbrace {\left(\int _{\mathbb {R} ^{n}}v(x-\tau )\ e^{-i2\pi f\cdot x}\,dx\right)} _{V(f)\ e^{-i2\pi f\cdot \tau }}\,d\tau \\&=\underbrace {\left(\int _{\mathbb {R} ^{n}}u(\tau )\ e^{-i2\pi f\cdot \tau }\,d\tau \right)} _{U(f)}\ V(f).\end{aligned}}

This theorem also holds for theLaplace transform, thetwo-sided Laplace transform and, when suitably modified, for theMellin transform andHartley transform (seeMellin inversion theorem). It can be extended to the Fourier transform ofabstract harmonic analysis defined overlocally compact abelian groups.

Periodic convolution (Fourier series coefficients)

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Consider $P {\displaystyle P}$ -periodic functions $u_{_{P}}$ and $v_{_{P}},$ which can be expressed asperiodic summations:

u_{_{P}}(x)\ \triangleq \sum _{m=-\infty }^{\infty }u(x-mP)

and

v_{_{P}}(x)\ \triangleq \sum _{m=-\infty }^{\infty }v(x-mP).

In practice the non-zero portion of components $u {\displaystyle u}$ and $v {\displaystyle v}$ are often limited to duration $P, {\displaystyle P,}$ but nothing in the theorem requires that.

TheFourier series coefficients are:

{\begin{aligned}U[k]&\triangleq {\mathcal {F}}\{u_{_{P}}\}[k]={\frac {1}{P}}\int _{P}u_{_{P}}(x)e^{-i2\pi kx/P}\,dx,\quad k\in \mathbb {Z} ;\quad \quad \scriptstyle {\text{integration over any interval of length }}P\\V[k]&\triangleq {\mathcal {F}}\{v_{_{P}}\}[k]={\frac {1}{P}}\int _{P}v_{_{P}}(x)e^{-i2\pi kx/P}\,dx,\quad k\in \mathbb {Z} \end{aligned}}

where ${\mathcal {F}}$ denotes theFourier series integral.

The product: $u_{_{P}}(x)\cdot v_{_{P}}(x)$ is also $P {\displaystyle P}$ -periodic, and its Fourier series coefficients are given by thediscrete convolution of the $U {\displaystyle U}$ and $V {\displaystyle V}$ sequences:

{\mathcal {F}}\{u_{_{P}}\cdot v_{_{P}}\}[k]=\{U*V\}[k].

The convolution:

{\begin{aligned}\{u_{_{P}}*v\}(x)\ &\triangleq \int _{-\infty }^{\infty }u_{_{P}}(x-\tau )\cdot v(\tau )\ d\tau \\&\equiv \int _{P}u_{_{P}}(x-\tau )\cdot v_{_{P}}(\tau )\ d\tau ;\quad \quad \scriptstyle {\text{integration over any interval of length }}P\end{aligned}}

is also $P {\displaystyle P}$ -periodic, and is called aperiodic convolution.

Derivation of periodic convolution
${\begin{aligned}\int _{-\infty }^{\infty }u_{_{P}}(x-\tau )\cdot v(\tau )\,d\tau &=\sum _{k=-\infty }^{\infty }\left[\int _{x_{o}+kP}^{x_{o}+(k+1)P}u_{_{P}}(x-\tau )\cdot v(\tau )\ d\tau \right]\quad x_{0}{\text{ is an arbitrary parameter}}\\&=\sum _{k=-\infty }^{\infty }\left[\int _{x_{o}}^{x_{o}+P}\underbrace {u_{_{P}}(x-\tau -kP)} _{u_{_{P}}(x-\tau ),{\text{ by periodicity}}}\cdot v(\tau +kP)\ d\tau \right]\quad {\text{substituting }}\tau \rightarrow \tau +kP\\&=\int _{x_{o}}^{x_{o}+P}u_{_{P}}(x-\tau )\cdot \underbrace {\left[\sum _{k=-\infty }^{\infty }v(\tau +kP)\right]} _{\triangleq \ v_{_{P}}(\tau )}\ d\tau \end{aligned}}$

Derivation of periodic convolution

{\begin{aligned}\int _{-\infty }^{\infty }u_{_{P}}(x-\tau )\cdot v(\tau )\,d\tau &=\sum _{k=-\infty }^{\infty }\left[\int _{x_{o}+kP}^{x_{o}+(k+1)P}u_{_{P}}(x-\tau )\cdot v(\tau )\ d\tau \right]\quad x_{0}{\text{ is an arbitrary parameter}}\\&=\sum _{k=-\infty }^{\infty }\left[\int _{x_{o}}^{x_{o}+P}\underbrace {u_{_{P}}(x-\tau -kP)} _{u_{_{P}}(x-\tau ),{\text{ by periodicity}}}\cdot v(\tau +kP)\ d\tau \right]\quad {\text{substituting }}\tau \rightarrow \tau +kP\\&=\int _{x_{o}}^{x_{o}+P}u_{_{P}}(x-\tau )\cdot \underbrace {\left[\sum _{k=-\infty }^{\infty }v(\tau +kP)\right]} _{\triangleq \ v_{_{P}}(\tau )}\ d\tau \end{aligned}}

The corresponding convolution theorem is:

${\mathcal {F}}\{u_{_{P}}*v\}[k]=\ P\cdot U[k]\ V[k].$

Eq.2

Derivation of Eq.2
${\begin{aligned}{\mathcal {F}}\{u_{_{P}}*v\}[k]&\triangleq {\frac {1}{P}}\int _{P}\left(\int _{P}u_{_{P}}(\tau )\cdot v_{_{P}}(x-\tau )\ d\tau \right)e^{-i2\pi kx/P}\,dx\\&=\int _{P}u_{_{P}}(\tau )\left({\frac {1}{P}}\int _{P}v_{_{P}}(x-\tau )\ e^{-i2\pi kx/P}dx\right)\,d\tau \\&=\int _{P}u_{_{P}}(\tau )\ e^{-i2\pi k\tau /P}\underbrace {\left({\frac {1}{P}}\int _{P}v_{_{P}}(x-\tau )\ e^{-i2\pi k(x-\tau )/P}dx\right)} _{V[k],\quad {\text{due to periodicity}}}\,d\tau \\&=\underbrace {\left(\int _{P}\ u_{_{P}}(\tau )\ e^{-i2\pi k\tau /P}d\tau \right)} _{P\cdot U[k]}\ V[k].\end{aligned}}$

Derivation of Eq.2

{\begin{aligned}{\mathcal {F}}\{u_{_{P}}*v\}[k]&\triangleq {\frac {1}{P}}\int _{P}\left(\int _{P}u_{_{P}}(\tau )\cdot v_{_{P}}(x-\tau )\ d\tau \right)e^{-i2\pi kx/P}\,dx\\&=\int _{P}u_{_{P}}(\tau )\left({\frac {1}{P}}\int _{P}v_{_{P}}(x-\tau )\ e^{-i2\pi kx/P}dx\right)\,d\tau \\&=\int _{P}u_{_{P}}(\tau )\ e^{-i2\pi k\tau /P}\underbrace {\left({\frac {1}{P}}\int _{P}v_{_{P}}(x-\tau )\ e^{-i2\pi k(x-\tau )/P}dx\right)} _{V[k],\quad {\text{due to periodicity}}}\,d\tau \\&=\underbrace {\left(\int _{P}\ u_{_{P}}(\tau )\ e^{-i2\pi k\tau /P}d\tau \right)} _{P\cdot U[k]}\ V[k].\end{aligned}}

Functions of a discrete variable (sequences)

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By a derivation similar to Eq.1, there is an analogous theorem for sequences, such as samples of two continuous functions, where now ${\mathcal {F}}$ denotes thediscrete-time Fourier transform (DTFT) operator. Consider two sequences $u[n]$ and $v[n]$ with transforms $U {\displaystyle U}$ and $V {\displaystyle V}$ :

{\begin{aligned}U(f)&\triangleq {\mathcal {F}}\{u\}(f)=\sum _{n=-\infty }^{\infty }u[n]\cdot e^{-i2\pi fn}\;,\quad f\in \mathbb {R} ,\\V(f)&\triangleq {\mathcal {F}}\{v\}(f)=\sum _{n=-\infty }^{\infty }v[n]\cdot e^{-i2\pi fn}\;,\quad f\in \mathbb {R} .\end{aligned}}

The§ Discrete convolution of $u {\displaystyle u}$ and $v {\displaystyle v}$ is defined by:

r[n]\triangleq (u*v)[n]=\sum _{m=-\infty }^{\infty }u[m]\cdot v[n-m]=\sum _{m=-\infty }^{\infty }u[n-m]\cdot v[m].

Theconvolution theorem for discrete sequences is:^[3]^[4]^{: p.60 (2.169)}

$R(f)={\mathcal {F}}\{u*v\}(f)=\ U(f)V(f).$

Eq.3

Periodic convolution

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$U(f)$ and $V(f),$ as defined above, are periodic, with a period of 1. Consider $N {\displaystyle N}$ -periodic sequences $u_{_{N}}$ and $v_{_{N}}$ :

u_{_{N}}[n]\ \triangleq \sum _{m=-\infty }^{\infty }u[n-mN]

and

v_{_{N}}[n]\ \triangleq \sum _{m=-\infty }^{\infty }v[n-mN],\quad n\in \mathbb {Z} .

These functions occur as the result of sampling $U {\displaystyle U}$ and $V {\displaystyle V}$ at intervals of $1/N$ and performing an inversediscrete Fourier transform (DFT) on $N {\displaystyle N}$ samples (see§ Sampling the DTFT). The discrete convolution:

\{u_{_{N}}*v\}[n]\ \triangleq \sum _{m=-\infty }^{\infty }u_{_{N}}[m]\cdot v[n-m]\equiv \sum _{m=0}^{N-1}u_{_{N}}[m]\cdot v_{_{N}}[n-m]

is also $N {\displaystyle N}$ -periodic, and is called aperiodic convolution. Redefining the ${\mathcal {F}}$ operator as the $N {\displaystyle N}$ -length DFT, the corresponding theorem is:^[5]^[4]^{: p. 548}

${\mathcal {F}}\{u_{_{N}}*v\}[k]=\ \underbrace {{\mathcal {F}}\{u_{_{N}}\}[k]} _{U(k/N)}\cdot \underbrace {{\mathcal {F}}\{v_{_{N}}\}[k]} _{V(k/N)},\quad k\in \mathbb {Z} .$

Eq.4a

And therefore:

$\{u_{_{N}}*v\}[n]=\ {\mathcal {F}}^{-1}\{{\mathcal {F}}\{u_{_{N}}\}\cdot {\mathcal {F}}\{v_{_{N}}\}\}.$

Eq.4b

Under the right conditions, it is possible for this $N {\displaystyle N}$ -length sequence to contain a distortion-free segment of a $u*v$ convolution. But when the non-zero portion of the $u(n)$ or $v(n)$ sequence is equal or longer than $N, {\displaystyle N,}$ some distortion is inevitable. Such is the case when the $V(k/N)$ sequence is obtained by directly sampling the DTFT of the infinitely long§ Discrete Hilbert transform impulse response.^[A]

For $u {\displaystyle u}$ and $v {\displaystyle v}$ sequences whose non-zero duration is less than or equal to $N, {\displaystyle N,}$ a final simplification is:

Circular convolution

$\{u_{_{N}}*v\}[n]=\ {\mathcal {F}}^{-1}\{{\mathcal {F}}\{u\}\cdot {\mathcal {F}}\{v\}\}.$

Eq.4c

This form is often used to efficiently implement numerical convolution bycomputer. (see§ Fast convolution algorithms and§ Example)

As a partial reciprocal, it has been shown^[6]that any linear transform that turns convolution into a product is the DFT (up to a permutation of coefficients).

Derivations of Eq.4
A time-domain derivation proceeds as follows: ${\begin{aligned}\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}v\}[k]&\triangleq \sum _{n=0}^{N-1}\left(\sum _{m=0}^{N-1}u_{_{N}}[m]\cdot v_{_{N}}[n-m]\right)e^{-i2\pi kn/N}\\&=\sum _{m=0}^{N-1}u_{_{N}}[m]\left(\sum _{n=0}^{N-1}v_{_{N}}[n-m]\cdot e^{-i2\pi kn/N}\right)\\&=\sum _{m=0}^{N-1}u_{_{N}}[m]\cdot e^{-i2\pi km/N}\underbrace {\left(\sum _{n=0}^{N-1}v_{_{N}}[n-m]\cdot e^{-i2\pi k(n-m)/N}\right)} _{\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\quad \scriptstyle {\text{due to periodicity}}}\\&=\underbrace {\left(\sum _{m=0}^{N-1}u_{_{N}}[m]\cdot e^{-i2\pi km/N}\right)} _{\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]}\left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right).\end{aligned}}$ A frequency-domain derivation follows from§ Periodic data, which indicates that the DTFTs can be written as: ${\mathcal {F}}\{u_{_{N}}v\}(f)={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}v\}[k]\right)\cdot \delta \left(f-k/N\right).\quad \scriptstyle {\mathsf {(Eq.5a)}}$ ${\mathcal {F}}\{u_{_{N}}\}(f)={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot \delta \left(f-k/N\right).$ The product with $V(f)$ is thereby reduced to a discrete-frequency function: ${\begin{aligned}{\mathcal {F}}\{u_{_{N}}v\}(f)&=G_{_{N}}(f)V(f)\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot V(f)\cdot \delta \left(f-k/N\right)\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot V(k/N)\cdot \delta \left(f-k/N\right)\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot \left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right)\cdot \delta \left(f-k/N\right),\quad \scriptstyle {\mathsf {(Eq.5b)}}\end{aligned}}$ where the equivalence of $V(k/N)$ and $\left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right)$ follows from§ Sampling the DTFT. Therefore, the equivalence of (5a) and (5b) requires: $\scriptstyle {\rm {DFT}}\displaystyle {\{u_{_{N}}v\}[k]}=\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot \left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right).$ We can also verify the inverse DTFT of (5b): ${\begin{aligned}(u_{_{N}}v)[n]&=\int _{0}^{1}\left({\frac {1}{N}}\sum _{k=-\infty }^{\infty }\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\cdot \delta \left(f-k/N\right)\right)\cdot e^{i2\pi fn}df\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\cdot \underbrace {\left(\int _{0}^{1}\delta \left(f-k/N\right)\cdot e^{i2\pi fn}df\right)} _{{\text{0, for}}\ k\ \notin \ [0,\ N)}\\&={\frac {1}{N}}\sum _{k=0}^{N-1}{\bigg (}\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]{\bigg )}\cdot e^{i2\pi {\frac {n}{N}}k}\\&=\ \scriptstyle {\rm {DFT}}^{-1}\displaystyle {\bigg (}\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}{\bigg )}.\end{aligned}}$

Derivations of Eq.4

A time-domain derivation proceeds as follows:

{\begin{aligned}\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}*v\}[k]&\triangleq \sum _{n=0}^{N-1}\left(\sum _{m=0}^{N-1}u_{_{N}}[m]\cdot v_{_{N}}[n-m]\right)e^{-i2\pi kn/N}\\&=\sum _{m=0}^{N-1}u_{_{N}}[m]\left(\sum _{n=0}^{N-1}v_{_{N}}[n-m]\cdot e^{-i2\pi kn/N}\right)\\&=\sum _{m=0}^{N-1}u_{_{N}}[m]\cdot e^{-i2\pi km/N}\underbrace {\left(\sum _{n=0}^{N-1}v_{_{N}}[n-m]\cdot e^{-i2\pi k(n-m)/N}\right)} _{\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\quad \scriptstyle {\text{due to periodicity}}}\\&=\underbrace {\left(\sum _{m=0}^{N-1}u_{_{N}}[m]\cdot e^{-i2\pi km/N}\right)} _{\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]}\left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right).\end{aligned}}

A frequency-domain derivation follows from§ Periodic data, which indicates that the DTFTs can be written as:

{\mathcal {F}}\{u_{_{N}}*v\}(f)={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}*v\}[k]\right)\cdot \delta \left(f-k/N\right).\quad \scriptstyle {\mathsf {(Eq.5a)}}

{\mathcal {F}}\{u_{_{N}}\}(f)={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot \delta \left(f-k/N\right).

The product with $V(f)$ is thereby reduced to a discrete-frequency function:

{\begin{aligned}{\mathcal {F}}\{u_{_{N}}*v\}(f)&=G_{_{N}}(f)V(f)\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot V(f)\cdot \delta \left(f-k/N\right)\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot V(k/N)\cdot \delta \left(f-k/N\right)\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot \left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right)\cdot \delta \left(f-k/N\right),\quad \scriptstyle {\mathsf {(Eq.5b)}}\end{aligned}}

where the equivalence of $V(k/N)$ and $\left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right)$ follows from§ Sampling the DTFT. Therefore, the equivalence of (5a) and (5b) requires:

\scriptstyle {\rm {DFT}}\displaystyle {\{u_{_{N}}*v\}[k]}=\left(\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\right)\cdot \left(\scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\right).

We can also verify the inverse DTFT of (5b):

{\begin{aligned}(u_{_{N}}*v)[n]&=\int _{0}^{1}\left({\frac {1}{N}}\sum _{k=-\infty }^{\infty }\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\cdot \delta \left(f-k/N\right)\right)\cdot e^{i2\pi fn}df\\&={\frac {1}{N}}\sum _{k=-\infty }^{\infty }\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]\cdot \underbrace {\left(\int _{0}^{1}\delta \left(f-k/N\right)\cdot e^{i2\pi fn}df\right)} _{{\text{0, for}}\ k\ \notin \ [0,\ N)}\\&={\frac {1}{N}}\sum _{k=0}^{N-1}{\bigg (}\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}[k]\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}[k]{\bigg )}\cdot e^{i2\pi {\frac {n}{N}}k}\\&=\ \scriptstyle {\rm {DFT}}^{-1}\displaystyle {\bigg (}\scriptstyle {\rm {DFT}}\displaystyle \{u_{_{N}}\}\cdot \scriptstyle {\rm {DFT}}\displaystyle \{v_{_{N}}\}{\bigg )}.\end{aligned}}

Convolution theorem for inverse Fourier transform

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There is also a convolution theorem for the inverse Fourier transform:

Here, " $\cdot$ " represents theHadamard product, and " $*$ " represents a convolution between the two matrices.

{\begin{aligned}&{\mathcal {F}}\{u*v\}={\mathcal {F}}\{u\}\cdot {\mathcal {F}}\{v\}\\&{\mathcal {F}}\{u\cdot v\}={\mathcal {F}}\{u\}*{\mathcal {F}}\{v\}\end{aligned}}

so that

{\begin{aligned}&u*v={\mathcal {F}}^{-1}\left\{{\mathcal {F}}\{u\}\cdot {\mathcal {F}}\{v\}\right\}\\&u\cdot v={\mathcal {F}}^{-1}\left\{{\mathcal {F}}\{u\}*{\mathcal {F}}\{v\}\right\}\end{aligned}}

Convolution theorem for tempered distributions

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The convolution theorem extends totempered distributions. Here, $v {\displaystyle v}$ is an arbitrary tempered distribution:

{\begin{aligned}&{\mathcal {F}}\{u*v\}={\mathcal {F}}\{u\}\cdot {\mathcal {F}}\{v\}\\&{\mathcal {F}}\{\alpha \cdot v\}={\mathcal {F}}\{\alpha \}*{\mathcal {F}}\{v\}.\end{aligned}}

But $u=F\{\alpha \}$ must be "rapidly decreasing" towards $-\infty$ and $+\infty$ in order to guarantee the existence of both, convolution and multiplication product. Equivalently, if $\alpha =F^{-1}\{u\}$ is a smooth "slowly growing" ordinary function, it guarantees the existence of both, multiplication and convolution product.^[7]^[8]^[9]

In particular, every compactly supported tempered distribution, such as theDirac delta, is "rapidly decreasing". Equivalently,bandlimited functions, such as the function that is constantly $1 {\displaystyle 1}$ are smooth "slowly growing" ordinary functions. If, for example, $v\equiv \operatorname {\text{Ш}}$ is theDirac comb both equations yield thePoisson summation formula and if, furthermore, $u\equiv \delta$ is the Dirac delta then $\alpha \equiv 1$ is constantly one and these equations yield theDirac comb identity.

Notes

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^An example is theMATLAB function,hilbert(u,N).

References

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^McGillem, Clare D.; Cooper, George R. (1984).Continuous and Discrete Signal and System Analysis (2 ed.). Holt, Rinehart and Winston. p. 118 (3–102).ISBN 0-03-061703-0.
^^a ^bWeisstein, Eric W."Convolution Theorem".From MathWorld--A Wolfram Web Resource. Retrieved8 February 2021.
^Proakis, John G.; Manolakis, Dimitri G. (1996),Digital Signal Processing: Principles, Algorithms and Applications (3 ed.), New Jersey: Prentice-Hall International, p. 297,Bibcode:1996dspp.book.....P,ISBN 9780133942897, sAcfAQAAIAAJ
^^a ^bOppenheim, Alan V.;Schafer, Ronald W.; Buck, John R. (1999).Discrete-time signal processing (2nd ed.). Upper Saddle River, N.J.: Prentice Hall.ISBN 0-13-754920-2.
^Rabiner, Lawrence R.; Gold, Bernard (1975).Theory and application of digital signal processing. Englewood Cliffs, NJ: Prentice-Hall, Inc. p. 59 (2.163).ISBN 978-0139141010.
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