Question

If \(\hat{f}(t)\) denotes the Hilbert transform of \(f(t)\), then the Hilbert transform of \(\hat{f}(t)\) is equal to

• A. \( -\hat{f}(t) \)
• B. \( f(t) \)
• C. \( -f(t) \)
• D. \( -\dfrac{d}{dt}f(t) \)

Hint:

In the frequency domain the Hilbert transform multiplies by -i·sgn(ω); squaring this factor gives -1, so applying it twice just negates the signal.

Answer 1

The Correct Option is C

The Hilbert transform is most easily analysed in the frequency domain. If \(F(\omega)\) is the Fourier transform of \(f(t)\), the Hilbert transform corresponds to multiplying the spectrum by \(-i\,\text{sgn}(\omega)\):

\[ \hat{f}(t) \;\longleftrightarrow\; -i\,\text{sgn}(\omega)\,F(\omega) \]

This operator leaves the amplitude spectrum unchanged and only shifts every frequency component's phase by \(-90^\circ\) for \(\omega>0\) and \(+90^\circ\) for \(\omega<0\) (a quadrature filter).

Applying the Hilbert transform a second time multiplies the spectrum by the same factor again:

\[ \mathcal{H}\{\hat f(t)\} \;\longleftrightarrow\; \big(-i\,\text{sgn}(\omega)\big)^2 F(\omega) = -i^2\,\text{sgn}^2(\omega)\,F(\omega) = -(-1)(1)\,F(\omega) \]

Wait — carefully: \(i^2=-1\) and \(\text{sgn}^2(\omega)=1\) for \(\omega\neq0\), so \(\big(-i\,\text{sgn}(\omega)\big)^2 = i^2\,\text{sgn}^2(\omega) = -1\). Hence

\[ \mathcal{H}\{\hat f(t)\} \;\longleftrightarrow\; -F(\omega) \]

Taking the inverse Fourier transform of both sides gives \( \mathcal{H}\{\hat{f}(t)\} = -f(t) \), i.e. applying the \(90^\circ\) phase-shift operator twice produces a net \(180^\circ\) phase shift, which is simply a sign flip in the time domain.

\(\boxed{\mathcal{H}\{\hat f(t)\} = -f(t)}\)