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

A cleaner way to see the same result without invoking the sign function explicitly: represent a real signal by its analytic signal \(f_a(t) = f(t) + i\hat f(t)\), constructed so that it contains only positive frequencies. Multiplying the analytic signal by \(-i\) rotates it by \(-90^\circ\) in the complex plane at every instant:

\[ -i\,f_a(t) = -i f(t) + \hat f(t) = \hat f(t) - i f(t) \]

But by definition, the analytic signal built from \(\hat f(t)\) is \( \hat f(t) + i\,\mathcal H\{\hat f(t)\} \). Comparing the two expressions for the same quantity \(-i f_a(t)\):

\[ \hat f(t) + i\,\mathcal H\{\hat f(t)\} = \hat f(t) - i f(t) \]

which immediately gives \( \mathcal H\{\hat f(t)\} = -f(t) \), the same result reached via the frequency-domain route, confirming that two successive quadrature (\(90^\circ\)) phase shifts amount to a single \(180^\circ\) (sign-reversing) shift.