# What is proof of Basic Hilbert Transform?

Attached is the question. Need to find the output for signal x(t).

• Can I please ask what have you tried so far? Can you make the question a little bit more specific? – A_A Oct 24 '17 at 9:53
• i believe that the symbol "$\phi(t)$" used for the instantaneous frequency (because it is inside an integral and what comes out of the integral is instantaneous phase offset from a reference phase of cosine) is the wrong symbol. it should be "$\omega(t)$" so as not to confuse. and the result of the integral is $\phi(t)$. (i.e. $\phi(t) = \int\limits_{0}^{t} \omega(u) \, du \qquad$) – robert bristow-johnson Oct 24 '17 at 21:02

Hilbert transform has the definition that its CTFT is:

$$H(\Omega) = \begin{cases} - \frac{\pi}{2} &,& \text{ for } \Omega > 0 \\ + \frac{\pi}{2} &,& \text{ for } \Omega < 0 \\ \end{cases}$$

from which you define the time-domian impulse response $h(t)$ associated Hilbert transformer from the inverse CTFT as:

$$h(t) = \mathcal{IFT} \{ H(\Omega) \} = \frac{1}{2\pi} \int_{-\infty}^{\infty} H(\Omega) e^{j \Omega t} d\Omega$$

You can evaluate the above integral or just use CTFT properties and pairs to conclude that: $$\boxed{ h(t) = \frac{1}{\pi t} }$$

Note that the associated impulse response $h(t)$ of the Hilbert transfomer is noncausal, two-sided and unstable and its Fourier transform can be expressed alternately as $$H(\Omega) = -j \text{sgn}(\Omega)$$ where $\text{sgn}(\cdot)$ is the sign function.

Hilbert transformer is most useful for defining and generating the analytic signal $x_+(t)$ associated with the signal $x(t)$, with has the property that

$$X_+(\Omega) = \begin{cases} 2 X(\Omega) &,& \text{ for } \Omega > 0 \\ 0 &,& \text{ for } \Omega < 0 \\ \end{cases}$$