Attached is the question. Need to find the output for signal x(t).
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3$\begingroup$ Can I please ask what have you tried so far? Can you make the question a little bit more specific? $\endgroup$ – A_A Oct 24 '17 at 9:53
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$\begingroup$ 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 $) $\endgroup$ – robert bristow-johnson Oct 24 '17 at 21:02
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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
$$ X_+(\Omega) = \begin{cases} 2 X(\Omega) &,& \text{ for } \Omega > 0 \\ 0 &,& \text{ for } \Omega < 0 \\ \end{cases} $$
