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The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with

$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$

can be calculated in the following way. First, note that

$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$

From this table we know that

$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$

We also know that

$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$

Combining $(3)$ and $(4)$ we get

$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$

So, using $(2)$,

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$

Now we also know that the Hilbert transform operator and the differentiation operator commute:

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$

which yields

$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$

Integrating $(8)$ finally gives

$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$

Note that this result can also be obtained using Mathematica (which I don't have available). According to this thread, the command

Integrate[-1/2*Log[1 + (\[Tau]*\[Nu])^2]/(\[Nu] - \[Omega]), {\[Nu], -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> \[Tau] >0 && \[Omega] >0, GenerateConditions -> False]/Pi 

gives

-ArcTan[\[Tau] \[Omega]]

The negative sign comes from the different definition of the Hilbert transform, as can be seen in the denominator of the integral in the Mathematica command.

I would like to add that the causality of the inverse Fourier transform of $\log H(j\omega)$, i.e., the causality of the complex cepstrum for a minimum-phase system $H(j\omega)$ can also be understood intuitively. Note that any zero of $H(s)$ in the right half-plane causes a singularity in $\log H(s)$ in the right half-plane, and consequently, the corresponding inverse Fourier transform must be two-sided because the region of convergence is a strip including the imaginary axis. Only if there are no zeros in the right half-plane (i.e., the system is minimum-phase) will $\log H(s)$ have all its singularities in the left half-plane, and the inverse transform yields a right-sided causal function.

From $(4)$ we can see another nice property of the Hilbert transform, namely that the inverse transform is just given by the (forward) transform with a negative sign:

$$\mathcal{H}^{-1}\{f\}=-\mathcal{H}\{f\}\tag{10}$$

That means that for every Hilbert transform pair we find, we get another one for free:

$$\mathcal{H}\{f\}=g\Longrightarrow\mathcal{H}\{g\}=-f\tag{11}$$

Applying $(11)$ to $(9)$ we find

$$\mathcal{H}\{\arctan(\omega)\}=-f(\omega)=\frac12\log(1+\omega^2)\tag{12}$$

The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with

$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$

can be calculated in the following way. First, note that

$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$

From this table we know that

$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$

We also know that

$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$

Combining $(3)$ and $(4)$ we get

$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$

So, using $(2)$,

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$

Now we also know that the Hilbert transform operator and the differentiation operator commute:

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$

which yields

$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$

Integrating $(8)$ finally gives

$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$

Note that this result can also be obtained using Mathematica (which I don't have available). According to this thread, the command

Integrate[-1/2*Log[1 + (\[Tau]*\[Nu])^2]/(\[Nu] - \[Omega]), {\[Nu], -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> \[Tau] >0 && \[Omega] >0, GenerateConditions -> False]/Pi 

gives

-ArcTan[\[Tau] \[Omega]]

The negative sign comes from the different definition of the Hilbert transform, as can be seen in the denominator of the integral in the Mathematica command.

I would like to add that the causality of the inverse Fourier transform of $\log H(j\omega)$, i.e., the causality of the complex cepstrum for a minimum-phase system $H(j\omega)$ can also be understood intuitively. Note that any zero of $H(s)$ in the right half-plane causes a singularity in $\log H(s)$ in the right half-plane, and consequently, the corresponding inverse Fourier transform must be two-sided because the region of convergence is a strip including the imaginary axis. Only if there are no zeros in the right half-plane (i.e., the system is minimum-phase) will $\log H(s)$ have all its singularities in the left half-plane, and the inverse transform yields a right-sided causal function.

From $(4)$ we can see another nice property of the Hilbert transform, namely that the inverse transform is simply given by the (forward) transform with a negative sign:

$$\mathcal{H}^{-1}\{f\}=-\mathcal{H}\{f\}\tag{10}$$

That means that for every Hilbert transform pair that we find, we get another one for free:

$$\mathcal{H}\{f\}=g\Longrightarrow\mathcal{H}\{g\}=-f\tag{11}$$

Applying $(11)$ to $(9)$ we find

$$\mathcal{H}\{\arctan(\omega)\}=-f(\omega)=\frac12\log(1+\omega^2)\tag{12}$$

The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with

$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$

can be calculated in the following way. First, note that

$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$

From this table we know that

$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$

We also know that

$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$

Combining $(3)$ and $(4)$ we get

$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$

So, using $(2)$,

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$

Now we also know that the Hilbert transform operator and the differentiation operator commute:

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$

which yields

$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$

Integrating $(8)$ finally gives

$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$

Note that this result can also be obtained using Mathematica (which I don't have available). According to this thread, the command

Integrate[-1/2*Log[1 + (\[Tau]*\[Nu])^2]/(\[Nu] - \[Omega]), {\[Nu], -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> \[Tau] >0 && \[Omega] >0, GenerateConditions -> False]/Pi 

gives

-ArcTan[\[Tau] \[Omega]]

The negative sign comes from the different definition of the Hilbert transform, as can be seen in the denominator of the integral in the Mathematica command.

I would like to add that the causality of the inverse Fourier transform of $\log H(j\omega)$, i.e., the causality of the complex cepstrum for a minimum-phase system $H(j\omega)$ can also be understood intuitively. Note that any zero of $H(s)$ in the right half-plane causes a singularity in $\log H(s)$ in the right half-plane, and consequently, the corresponding inverse Fourier transform must be two-sided because the region of convergence is a strip including the imaginary axis. Only if there are no zeros in the right half-plane (i.e., the system is minimum-phase) will $\log H(s)$ have all its singularities in the left half-plane, and the inverse transform yields a right-sided causal function.

The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with

$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$

can be calculated in the following way. First, note that

$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$

From this table we know that

$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$

We also know that

$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$

Combining $(3)$ and $(4)$ we get

$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$

So, using $(2)$,

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$

Now we also know that the Hilbert transform operator and the differentiation operator commute:

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$

which yields

$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$

Integrating $(8)$ finally gives

$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$

Note that this result can also be obtained using Mathematica (which I don't have available). According to this thread, the command

Integrate[-1/2*Log[1 + (\[Tau]*\[Nu])^2]/(\[Nu] - \[Omega]), {\[Nu], -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> \[Tau] >0 && \[Omega] >0, GenerateConditions -> False]/Pi 

gives

-ArcTan[\[Tau] \[Omega]]

The negative sign comes from the different definition of the Hilbert transform, as can be seen in the denominator of the integral in the Mathematica command.

I would like to add that the causality of the inverse Fourier transform of $\log H(j\omega)$, i.e., the causality of the complex cepstrum for a minimum-phase system $H(j\omega)$ can also be understood intuitively. Note that any zero of $H(s)$ in the right half-plane causes a singularity in $\log H(s)$ in the right half-plane, and consequently, the corresponding inverse Fourier transform must be two-sided because the region of convergence is a strip including the imaginary axis. Only if there are no zeros in the right half-plane (i.e., the system is minimum-phase) will $\log H(s)$ have all its singularities in the left half-plane, and the inverse transform yields a right-sided causal function.

From $(4)$ we can see another nice property of the Hilbert transform, namely that the inverse transform is just given by the (forward) transform with a negative sign:

$$\mathcal{H}^{-1}\{f\}=-\mathcal{H}\{f\}\tag{10}$$

That means that for every Hilbert transform pair we find, we get another one for free:

$$\mathcal{H}\{f\}=g\Longrightarrow\mathcal{H}\{g\}=-f\tag{11}$$

Applying $(11)$ to $(9)$ we find

$$\mathcal{H}\{\arctan(\omega)\}=-f(\omega)=\frac12\log(1+\omega^2)\tag{12}$$

Nice question! The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with

$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$

can be calculated in the following way. First, note that

$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$

From this table we know that

$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$

We also know that

$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$

Combining $(3)$ and $(4)$ we get

$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$

So, using $(2)$,

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$

Now we also know that the Hilbert transform operator and the differentiation operator commute:

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$

which yields

$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$

Integrating $(8)$ finally gives

$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$

Note that this result can also be obtained using Mathematica (which I don't have available). According to this thread, the command

Integrate[-1/2*Log[1 + (\[Tau]*\[Nu])^2]/(\[Nu] - \[Omega]), {\[Nu], -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> \[Tau] >0 && \[Omega] >0, GenerateConditions -> False]/Pi 

gives

-ArcTan[\[Tau] \[Omega]]

The negative sign comes from the different definition of the Hilbert transform, as can be seen in the denominator of the integral in the Mathematica command.

I would like to add that the causality of the inverse Fourier transform of $\log H(j\omega)$, i.e., the causality of the complex cepstrum for a minimum-phase system $H(j\omega)$ can also be understood intuitively. Note that any zero of $H(s)$ in the right half-plane causes a singularity in $\log H(s)$ in the right half-plane, and consequently, the corresponding inverse Fourier transform must be two-sided because the region of convergence is a strip including the imaginary axis. Only if there are no zeros in the right half-plane (i.e., the system is minimum-phase) will $\log H(s)$ have all its singularities in the left half-plane, and the inverse transform yields a right-sided causal function.

The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with

$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$

can be calculated in the following way. First, note that

$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$

From this table we know that

$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$

We also know that

$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$

Combining $(3)$ and $(4)$ we get

$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$

So, using $(2)$,

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$

Now we also know that the Hilbert transform operator and the differentiation operator commute:

$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$

which yields

$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$

Integrating $(8)$ finally gives

$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$

Note that this result can also be obtained using Mathematica (which I don't have available). According to this thread, the command

Integrate[-1/2*Log[1 + (\[Tau]*\[Nu])^2]/(\[Nu] - \[Omega]), {\[Nu], -Infinity, Infinity}, PrincipalValue -> True, Assumptions -> \[Tau] >0 && \[Omega] >0, GenerateConditions -> False]/Pi 

gives

-ArcTan[\[Tau] \[Omega]]

The negative sign comes from the different definition of the Hilbert transform, as can be seen in the denominator of the integral in the Mathematica command.

I would like to add that the causality of the inverse Fourier transform of $\log H(j\omega)$, i.e., the causality of the complex cepstrum for a minimum-phase system $H(j\omega)$ can also be understood intuitively. Note that any zero of $H(s)$ in the right half-plane causes a singularity in $\log H(s)$ in the right half-plane, and consequently, the corresponding inverse Fourier transform must be two-sided because the region of convergence is a strip including the imaginary axis. Only if there are no zeros in the right half-plane (i.e., the system is minimum-phase) will $\log H(s)$ have all its singularities in the left half-plane, and the inverse transform yields a right-sided causal function.