The Hilbert transform is applied to x(t) = cos(2πft). What is the result?
Acos(2πft + π/4) — the signal is shifted 45° in phase
Bsin(2πft) — the signal is shifted 90° in phase
C−cos(2πft) — the signal is phase-inverted (180° shift)
Dcos(2πft + π) — the signal is shifted 180° in phase
The Hilbert transform applies a 90° phase lag to all positive frequencies (multiplication by −j in the frequency domain). Since cos(2πft) has a positive-frequency component at +f and a negative-frequency component at −f, applying −j to the positive and +j to the negative gives exactly sin(2πft). This is the defining property: H[cos(2πft)] = sin(2πft) and H[sin(2πft)] = −cos(2πft). The Hilbert transform is a wideband 90° phase shifter.
Question 2 Multiple Choice
An engineer wants to extract the envelope m(t) from an AM radar pulse x(t) = m(t)·cos(2πfct). Which operation correctly recovers m(t)?
ATake the absolute value of x(t) directly
BForm the analytic signal z(t) = x(t) + j·H[x(t)] and compute |z(t)| = √(x² + H[x]²)
CApply a bandpass filter centered at fc and take the real part
DDifferentiate x(t) with respect to time
For x(t) = m(t)·cos(2πfct), the analytic signal is z(t) ≈ m(t)·e^{j2πfct} (when m(t) is bandlimited below fc). Taking the magnitude: |z(t)| = m(t). This is an ideal envelope detector. Option A gives |m(t)·cos(2πfct)|, which retains fast oscillations at 2fc — it does not cleanly separate envelope from carrier. Option C preserves the carrier. Option D gives instantaneous frequency, not envelope.
Question 3 True / False
The analytic signal z(t) = x(t) + j·H[x(t)] contains twice as much information as the original real signal x(t) because it has both real and imaginary parts.
TTrue
FFalse
Answer: False
The analytic signal contains the same information as x(t). For a real signal, negative-frequency components are always the complex conjugate of positive-frequency components — they carry no additional information. The analytic signal suppresses these redundant negative frequencies. The imaginary part H[x(t)] is fully determined by x(t): knowing x(t) is sufficient to compute x̂(t) via convolution with 1/(πt). No new information is added; the representation is mathematically reorganized to make instantaneous amplitude and phase accessible.
Question 4 True / False
For an FM signal, the instantaneous frequency can be recovered from the analytic signal by differentiating the instantaneous phase.
TTrue
FFalse
Answer: True
Writing z(t) = A(t)·e^{jφ(t)}, the instantaneous phase is φ(t) = arctan[H[x(t)]/x(t)], and the instantaneous frequency is fi(t) = (1/2π)·dφ/dt. For an FM signal, the message is encoded in frequency variations, so dφ/dt directly recovers the modulating signal. Phase unwrapping — adding or subtracting 2π at ±π discontinuities — is needed to make φ(t) a smooth, continuously increasing function before differentiation.
Question 5 Short Answer
Why do negative frequencies in the spectrum of a real-valued signal carry no independent information, and how does the analytic signal exploit this property?
Think about your answer, then reveal below.
Model answer: For any real-valued signal x(t), the Fourier transform satisfies conjugate symmetry: X(−f) = X*(f). The negative-frequency component is always the complex conjugate of the positive-frequency component, so it is completely determined by positive frequencies and adds no new information. The analytic signal exploits this by zeroing out all negative-frequency components and doubling the positive-frequency ones, producing a one-sided spectrum Z(f) = 2X(f) for f > 0. This creates a unique polar representation z(t) = A(t)e^{jφ(t)} with unambiguous instantaneous amplitude A(t) = |z(t)| and phase φ(t) = arg(z(t)), which would be ill-defined if negative frequencies were retained.
If negative frequencies were kept, the representation A(t)e^{jφ(t)} would not be unique — different combinations of A and φ could produce the same signal. Removing the redundant half creates the mathematical structure needed for clean instantaneous parameter extraction.