FT of Impulse Function

The Fourier transform of the Dirac delta function is a constant equal to 1 for all frequencies, showing that an ideal impulse contains equal energy at every frequency. This property makes the impulse the standard test signal for measuring system frequency response.

Core Concept

The Dirac delta delta(t) is not a conventional function but a distribution defined by two properties: it is zero for all t not equal to 0, and its integral over all time equals 1. Its most useful property is the sifting property: the integral of f(t)*delta(t - t0) over all time equals f(t0).

To find its Fourier transform, apply the sifting property directly to the transform integral. The FT of delta(t) is the integral of delta(t)*e^(-j2*pi*f*t) dt. Sifting at t = 0 gives e^(-j2*pi*f*0) = e^0 = 1. So the transform is simply 1 for all f.

This flat spectrum makes the impulse ideal for system identification. Feeding an impulse into a linear system and observing the output (the impulse response) in the frequency domain immediately gives the full frequency response. No other test signal reveals all frequencies simultaneously.

Key Formula

FT pair: delta(t) <-> 1. The shifted impulse pair is delta(t - t0) <-> e^(-j2*pi*f*t0). The time shift introduces a linear phase term but does not change the magnitude spectrum, which remains 1 for all f. By duality, 1 in time transforms to delta(f), and e^(j2*pi*f0*t) transforms to delta(f - f0).

Problem: Find the FT of x(t) = delta(t - 3)

Step 1: Apply the FT definition
  X(f) = integral from -inf to inf of delta(t-3) * e^(-j2*pi*f*t) dt

Step 2: Use the sifting property
  Sifting picks out the integrand at t = 3
  X(f) = e^(-j2*pi*f*3)
        = e^(-j6*pi*f)

Step 3: Interpret the result
  |X(f)| = |e^(-j6*pi*f)| = 1 for all f  (flat magnitude)
  angle(X(f)) = -6*pi*f  (linear phase)

Final Answer: X(f) = e^(-j6*pi*f)
Magnitude spectrum: 1 (flat)
Phase spectrum:    -6*pi*f (linear in f)

Exam Tip: Two key duality results appear often. The FT of delta(t) is 1, and by duality, the FT of the constant signal x(t) = 1 is delta(f). Similarly, FT{delta(t-t0)} = e^(-j2*pi*f*t0) and by duality FT{e^(j2*pi*f0*t)} = delta(f-f0). Memorise these four pairs and you can quickly handle sinusoids and constants in the frequency domain.

Properties Summary

• Basic pair: delta(t) <-> 1, flat spectrum with unit magnitude at all frequencies.
• Time shift: delta(t - t0) <-> e^(-j2*pi*f*t0), magnitude stays 1 but phase becomes linear.
• Duality: 1 (constant signal) <-> delta(f), a constant in time is a single spike in frequency.
• Frequency shift: e^(j2*pi*f0*t) <-> delta(f - f0), a complex sinusoid is a single spectral spike.
• Sifting property: integral x(t)*delta(t - t0) dt = x(t0), valid for any continuous x(t).
• Scaling: delta(at) = (1/|a|)*delta(t), so FT{delta(at)} = 1/|a|.

Quick Revision

• delta(t) has infinite height, zero width, and unit area.
• FT{delta(t)} = 1 follows directly from the sifting property.
• FT{delta(t - t0)} = e^(-j2*pi*f*t0), only phase changes.
• FT{1} = delta(f) by duality.
• FT{e^(j2*pi*f0*t)} = delta(f - f0) by the frequency-shift duality.
• FT{cos(2*pi*f0*t)} = (1/2)[delta(f-f0) + delta(f+f0)], two symmetric impulses.
• Exam trap: writing FT{delta(t-t0)} as 1 instead of e^(-j2*pi*f*t0). The time shift adds a phase factor; without it the answer is the FT of an unshifted impulse.