Questions: Fourier Transform: Definition and Properties

Question 1

If x(t) has Fourier Transform X(f), what is the Fourier Transform of x(t - t₀)?

• A. X(f - t₀)
• B. e^(j2πft₀) X(f)
• C. e^(-j2πft₀) X(f)
• D. X(f) shifted left by t₀

Answer: (C)

The time-shift property states that delaying a signal by t₀ multiplies its spectrum by e^{-j2πft₀}. This is a phase shift — the magnitude |X(f)| is unchanged, but each frequency component acquires a phase proportional to frequency times delay. This is why a pure time delay does not alter a signal's amplitude spectrum.

Question 2

The Fourier Transform can be applied to any periodic signal, just like the Fourier Series.

• T. True
• F. False

Answer: False

The standard Fourier Transform integral requires the signal to be absolutely integrable (∫|x(t)|dt < ∞), which periodic signals generally violate. Periodic signals are handled by the Fourier Series. The Fourier Transform can represent periodic signals only through generalized functions (Dirac deltas in the frequency domain), which is an extension beyond the basic definition.

Question 3

Why is the Fourier Transform more appropriate than the Fourier Series for analyzing aperiodic signals such as a single rectangular pulse?

Model answer: The Fourier Series represents a signal as a sum of harmonics with a discrete set of frequencies, but this requires the signal to repeat periodically forever. An aperiodic signal like a single pulse has a continuous spectrum — energy is spread across all frequencies — and the Fourier Transform captures this as a continuous function X(f), rather than a discrete set of Fourier coefficients.

The Fourier Series can be viewed as the limit of a Fourier Transform when the period T → ∞: the discrete harmonic frequencies 1/T, 2/T, ... become a continuum and the sum becomes an integral. This connection shows that the Fourier Transform is the natural generalization of the Fourier Series to aperiodic signals.