Introduction / Context:
The Fourier transform of the unit impulse δ(t) is fundamental in signals and systems. This property explains why δ(t) is used as an excitation for LTI systems: it excites all frequencies equally.
Given Data / Assumptions:
- Signal under consideration is δ(t).
- Fourier transform pair: δ(t) ↔ 1 (constant for all frequencies).
- Phase considerations: the transform is real and constant, hence zero phase across all frequencies.
Concept / Approach:
The Fourier transform of δ(t) is ∫ δ(t) e^{−jωt} dt = 1 for all ω. This means δ(t) contains all frequency components with equal weight, phase constant at zero. Thus δ(t) has infinite bandwidth and excites every frequency equally in the system frequency response.
Step-by-Step Solution:
Start with definition: F(ω) = ∫ δ(t) e^{−jωt} dt.Using sifting property: evaluates to e^{−jω*0} = 1.Thus F(ω) = 1 for all ω.This implies flat spectrum (all-pass) with zero phase.
Verification / Alternative check:
Impulse response testing in practice: every system's output to δ(t) reveals its frequency response because δ(t) excites all frequencies.
Why Other Options Are Wrong:
Pure DC: would correspond to constant function in time domain, not impulse.Pure AC: not defined meaningfully; impulse is broadband, not single-tone.Infinite bandwidth with linear phase variations: bandwidth is infinite, but phase is constant, not linear.Narrowband sinusoid: opposite of δ(t) properties.
Common Pitfalls:
Confusing δ(t) (impulse in time) with δ(ω) (impulse in frequency); they are duals.
Final Answer:
Entire frequency range with constant phase
Discussion & Comments