Hilbert transform of sinusoid -- apparent contradiction

The arrow "$\xrightarrow{\mathcal H}$" denotes a Hilbert transform:

$$\cos(\omega t)\xrightarrow{\mathcal H} \sin(\omega t)$$

happiness.

But

$$\cos(\omega t) = \cos(-\omega t) \xrightarrow{\mathcal H} \sin(-\omega t) = -\sin(\omega t)$$

This is a contradiction. What have I done wrong?

• hm, hard to say what you've done wrong. How did you arrive at $\cos(-\omega t) \rightarrow \sin(-\omega t)$? Commented Dec 10, 2017 at 14:20
• @MarcusMüller I just applied the same rule as in the first line. I could write $\Omega = -\omega$ then have $\cos(\Omega t) \rightarrow \sin(\Omega t)$ then substitute back in for $\omega$. Commented Dec 10, 2017 at 14:27
• no, you did not just apply the same rule! You also assumed something that doesn't apply to the Hilbert Transform: universal linearity. Commented Dec 10, 2017 at 14:29

The error lies in the assumption that if $g(t)$ is the Hilbert transform of $f(t)$, then the Hilbert transform of $f(-t)$ must be $g(-t)$. This is not the case.

Let $f^-(t)=f(-t)$. Then we have

$$g(t)=\mathcal{H}\{f\}(t)=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(\tau)}{t-\tau}d\tau\tag{1}$$

and

\begin{align}\mathcal{H}\{f^-\}(t)&=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(-\tau)}{t-\tau}d\tau\\&=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(\tau)}{t+\tau}d\tau\\&=-\mathcal{H}\{f\}(-t)\\&=-g(-t)\end{align}\tag{2}

So if $f(t)=\cos(\omega t)$ and $g(t)=\mathcal{H}\{f\}(t)=\sin(\omega t)$, then the Hilbert transform of $f^-(t)=\cos(-\omega t)$ equals $-g(-t)=-\sin(-\omega t)=\sin(\omega t)$. Happiness again.

EDIT:

In response to your comment about substituting for $\omega$ instead of reversing time, note that

$$\mathcal{H}\{\cos(\omega_0 t)\}=\sin(\omega_0 t)$$

is only valid for $\omega_0>0$. This can most easily be seen in the frequency domain, where the Hilbert transform corresponds to a multiplication with $-j\;\text{sign}(\omega)$. The Fourier transform of $f(t)=\cos(\omega_0t)$ is given by

$$F(\omega)=\pi\left[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\right]$$

If $\omega_0>0$, multiplication by $-j\;\text{sign}(\omega)$ gives the Fourier transform of $\sin(\omega_0t)$. However, if $\omega_0<0$, we obtain the Fourier transform of $-\sin(\omega_0t)$, because then $\delta(\omega+\omega_0)$ appears at positive frequencies instead of $\delta(\omega-\omega_0)$.

• ah yay for happiness! Did this on paper in the time you TeXed it, as it's been quite a while since I've been doing Hilbert transforms. Nice! Commented Dec 10, 2017 at 14:29
• I think I accepted this answer too quickly. I agree with everything you say, but my intent is to change the sign on $\omega$ rather than reverse time. Please see my comment to my original question. I still can't understand why it goes wrong when I substitute $\Omega$ with $-\omega$. Thanks. Commented Dec 10, 2017 at 16:18
• @user3856370: If you change the sign of $\omega$ you do reverse time, simply because $(-\omega)t=\omega(-t)$. Looking at formula (1) in my answer, you will understand that you can't just substitute $at$ for $-at$. The Hilbert transforms transforms $t$ to a "new" $t$. Commented Dec 10, 2017 at 16:29
• @user3856370: Maybe an easier formulation would be: $\mathcal{H}\{\cos(\omega_0t)\}=\sin(\omega_0t)$ only for $\omega_0>0$! Commented Dec 10, 2017 at 16:43
• @user3856370: See my updated answer. Commented Dec 10, 2017 at 16:57

There is not contradiction. If you are more familiar with the Fourier transform, you may remember the time reversal property: if $$\mathcal{F} [x(t)] = X(\omega)$$ then:

$$\mathcal{F} [x(-t)] = X(-\omega)$$ but you cannot say in general that

$$\mathrm{(FALSE)\quad}\mathcal{F} [x(-t)] = -X(\omega)\mathrm{\quad (FALSE)}$$

One cannot pull the minus sign out of formula so easily. Fourier and Hilbert are linear transforms. This tells you about what an outer sum and multiplication can do: $$\mathcal{T}(a.x(t)+b.y(t)) = a.\mathcal{T}(x(t))+b. \mathcal{T}(y(t))$$ but nothing about a sign change on the inner variable $t$.