0
$\begingroup$

Given that $\operatorname{G}: \mathbb{R} \rightarrow \mathbb{C}$ is such that $\widehat{\operatorname{G}}(w) = 0$ for $w < 0$, show that $\operatorname{Im}(\operatorname{G})$ is the Hilbert Transform of $\operatorname{Re}(\operatorname{G})$.

So far I have showed that if $\widehat{\operatorname{G}} = \operatorname{h}(w) + i\operatorname{f}(w)$, then:

$$ \operatorname{Re}(\operatorname{G}) = \frac{1}{2\pi} \int_{0}^{\infty}(\operatorname{h}(w)\cos(wt) - \operatorname{f}(w)\sin(wt)) dw $$

$$ \operatorname{Im}(\operatorname{G}) = \frac{1}{2\pi} \int_{0}^{\infty}(\operatorname{f}(w)\cos(wt) + \operatorname{h}(w)\sin(wt)) dw $$

Then I tried calculating the Hilbert Transform of $\operatorname{Re}(\operatorname{G})$, but did not get to $\operatorname{Im}(\operatorname{G})$. Anybody can give me advice on how to proceed with this problem?

Update:

I tried again and got the expected result.

$$\mathcal{F}[\operatorname{Re}(\operatorname{G})] = A + B$$

$$ A = \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) \int_{-\infty}^{\infty} cos(xt)e^{-iwt}dt ~ dx$$

$$ B = - \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{f}(x) \int_{-\infty}^{\infty} sin(xt)e^{-iwt}dt ~ dx$$

Now I will show for A, but the same logic is applied B.

$$ A = \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) \pi[\delta(w+x) + \delta(w-x)] dx$$

Now applying the Hilbert Transform

$$ \mathcal{H}[A] = \mathcal{F}^{-1}[-isign(w)A]$$ $$ \mathcal{H}[A] = \mathcal{F}^{-1} \left[\frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) i\pi[\delta(w+x) - \delta(w-x)] dx \right]$$ $$ \mathcal{H}[A] = \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) sin(xt) dx$$

Doing the same for $B$ we arrive at the desire result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.