# Dictionary:Causality

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The property of a sequence such that there is zero energy before some finite starting time. Minimum-phase wavelets are causal but zero-phase wavelets are not.

## Causality and the Fourier Transform

The issue of causality affects data both in the time domain and the frequency domain. This fact is apparent when considering the Fourier transform of causal functions.

A common Fourier Transform convention for forward and inverse temporal transforms is, respectively,

${\displaystyle F(\omega )=\int _{0}^{\infty }f(t)e^{i\omega t}\;dt}$

${\displaystyle f(t)=\int _{-\infty }^{\infty }F(\omega )e^{-i\omega t}\;dt.}$

Here the limits of ${\displaystyle [0,\infty ]}$ imply that ${\displaystyle f(t)}$ is causal and, therefore ${\displaystyle f(t)=0}$ for ${\displaystyle t<0}$.

Thus the inverse Fourier transform must yield a result that agrees with causality.

### The inverse Fourier transform as a contour integral

If we interpret the inverse Fourier transform as a contour integration

${\displaystyle f(t)=\int _{\Gamma }F(\omega )e^{-i\omega t}\;dt,}$

and consider only problems where ${\displaystyle F(\omega )}$ has poles or branch points that exist on or near the real axis, then there is only one particular choice of integration contour ${\displaystyle \Gamma }$ that will yield a causal result for ${\displaystyle f(t).}$

This is the case of the portion of ${\displaystyle \Gamma }$ parallel to the real axis passing over the poles.

To perform a contour integration requires that the integration contour be closed, and that Cauchy's Theorem and, for Fourier integrals, a result known as Jordan's lemma be applied to show that contribution from the contour at infinity vanishes. Closure in the upper half-plane of ${\displaystyle \omega }$ will yield the result of ${\displaystyle f(t)=0}$ (because no pole is included inside the contour) and corresponds to the ${\displaystyle t<0}$ case. Closure of the cuntour in the lower half plane of ${\displaystyle \omega }$ (where one or more poles are included inside the contour) will yield a nonzero value of ${\displaystyle f(t)}$ and will correspond to the values of ${\displaystyle f(t)}$ for ${\displaystyle t>0.}$

To see why this is so, consider ${\displaystyle \omega }$ as a complex variable ${\displaystyle \omega =a+ib}$ and rewrite the exponential in the inverse Fourier transform definition as

${\displaystyle e^{-i\omega t}=e^{-i(a+ib)t}=e^{-iat}e^{bt}.}$

The integral can only converge when the integrand is decaying, so ${\displaystyle b>0}$ (corresponding to closure in the upper half-plane of ${\displaystyle \omega }$) implies that ${\displaystyle t<0}$ and ${\displaystyle b<0}$ (corresponding to closure in the lower half-plane of ${\displaystyle \omega }$) implies that ${\displaystyle t>0.}$

"Figure 1: For the Fourier transform exponent sign conventions here, this is the integration contour that yields a causal result."

For this exponent sign convention, a contour that passes over the singularities of the integrand. Because there are no singularities in the upper half plane of ${\displaystyle \omega }$ causality is identified exactly with analyticity in some half-plane of ${\displaystyle \omega }$ which, in the case of these Fourier transform exponent sign conventions is the upper half-plane.

## Causality and the Hilbert transform

We formally consider a causal function to be ${\displaystyle f(t)H(t)}$ where ${\displaystyle H(t)}$ is the Heaviside step function

${\displaystyle H(t)=\left\{{\begin{array}{ll}0&t<0\\1&t>0.\end{array}}\right.}$

We want to find the Fourier transform of ${\displaystyle f(t)H(t),}$ which may be shown to be a frequency domain convolution

${\displaystyle \int _{-\infty }^{\infty }f(t)H(t)e^{i\omega t}\;dt={\frac {1}{2\pi }}(F\star {\hat {H}})(\omega )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\hat {H}}(\omega -\Omega )F(\Omega )\;d\Omega .}$

The Heaviside step is neither even nor odd but may be written in terms of the constant function ${\displaystyle 1/2}$ and the ${\displaystyle \operatorname {sgn}(t)}$

${\displaystyle H(t)={\frac {1}{2}}\operatorname {sgn}(t)+{\frac {1}{2}}.}$

where

${\displaystyle \operatorname {sgn}(t)=\left\{{\begin{array}{ll}-1&t<0\\0&t=0\\1&t>0.\end{array}}\right.}$

We need to compute the Fourier transform of ${\displaystyle H(t).}$

${\displaystyle {\hat {H}}(\Omega )=\int _{-\infty }^{\infty }\left[{\frac {1}{2}}\operatorname {sgn}(t)+{\frac {1}{2}}\right]e^{i\Omega t}\;dt=-{\frac {1}{2}}\int _{-\infty }^{0}e^{i\Omega t}\;dt+{\frac {1}{2}}\int _{0}^{\infty }e^{i\Omega t}\;dt+{\frac {1}{2}}\int _{-\infty }^{\infty }e^{i\Omega t}\;dt.}$

The first two integrals may be performed by elementary means (with the appropriate sign chosen for the imaginary part of ${\displaystyle \Omega }$ to insure exponential decay of the integrand), and may be combined based on the property of analytic continuation, and the third follows from the definition of the Dictionary:Dirac function

${\displaystyle {\hat {H}}(\Omega )=\left[{\frac {-1}{i\Omega }}+\pi \delta (\Omega )\right].}$

Forming the frequency domain convolution

${\displaystyle \int _{-\infty }^{\infty }f(t)H(t)e^{i\omega t}\;dt={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\left[{\frac {-1}{i(\omega -\Omega )}}+\pi \delta (\omega -\Omega )\right]F(\Omega )\;d\Omega .}$

The first integral makes sense only as a Cauchy principal value integral, and the second may be performed by applying the sifting property of the delta function

${\displaystyle \int _{-\infty }^{\infty }f(t)H(t)e^{i\omega t}\;dt={\frac {1}{2\pi i}}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F(\Omega )}{(\Omega -\omega )}}\;d\Omega +{\frac {1}{2}}F(\omega ).}$

Because ${\displaystyle F(\omega )}$ is a complex-valued function we may write this function in terms of its real and imaginary parts ${\displaystyle F=F_{R}+iF_{I}}$

${\displaystyle {\frac {1}{2}}\left[{\frac {1}{\pi }}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F_{I}(\Omega )}{(\Omega -\omega )}}\;d\Omega -{\frac {i}{\pi }}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F_{R}(\Omega )}{(\Omega -\omega )}}\;d\Omega \right]+{\frac {1}{2}}\left[F_{R}(\omega )+iF_{I}(\omega )\right].}$

However, because ${\displaystyle F(\omega )}$ is analytic in the upper half-plane of ${\displaystyle \omega ,}$ the Hilbert transform relations exist between the real and imaginary parts of ${\displaystyle F(\omega )}$

${\displaystyle F_{R}(\omega )={\frac {1}{\pi }}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F_{I}(\Omega )}{(\Omega -\omega )}}\;d\Omega }$

and

${\displaystyle F_{I}(\omega )={\frac {-1}{\pi }}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F_{R}(\Omega )}{(\Omega -\omega )}}\;d\Omega }$

except possibly at a finite collection of values of ${\displaystyle \omega ,}$ where ${\displaystyle F(\omega )}$ fails to be analytic.

### The Kramers-Kronig Relations

Thus, the Fourier transform of a causal function may be written in terms of Hilbert transform pairs

${\displaystyle \int _{-\infty }^{\infty }f(t)e^{i\omega t}\;dt=F(\omega )={\frac {1}{\pi }}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F_{I}(\Omega )}{(\Omega -\omega )}}\;d\Omega +i{\frac {-1}{\pi }}{\mbox{PV}}\int _{-\infty }^{\infty }{\frac {F_{R}(\Omega )}{(\Omega -\omega )}}\;d\Omega .}$

This result is often called the Kramers-Kroenig relation.