How it works
The CTFT pair is X(jω) = ∫x(t)e^(−jωt)dt and x(t) = (1/2π)∫X(jω)e^(jωt)dω. Key transform pairs to memorise: δ(t) ↔ 1 (flat spectrum), u(t) ↔ πδ(ω) + 1/jω, e^(−at)u(t) ↔ 1/(a+jω) for a > 0, and rect(t/τ) ↔ τ·sinc(ωτ/2). The convolution property states that convolution in time maps to multiplication in frequency: x(t)*h(t) ↔ X(jω)H(jω). This is why filter design works — an ideal low-pass filter with cutoff ω_c has a rectangular H(jω), meaning its impulse response h(t) is a sinc function, non-causal and therefore physically unrealisable.
Key points to remember
The CTFT exists for signals satisfying the Dirichlet conditions or that are square-integrable. Time-shifting property: x(t−t₀) ↔ e^(−jωt₀)X(jω) — only phase changes, magnitude is unaffected. Frequency shifting (modulation): x(t)e^(jω₀t) ↔ X(j(ω−ω₀)). Differentiation in time: dx/dt ↔ jω·X(jω), explaining why differentiators amplify high-frequency noise. Parseval's theorem: ∫|x(t)|² dt = (1/2π)∫|X(jω)|² dω links time-domain energy to frequency-domain energy. The duality property — if x(t) ↔ X(jω), then X(jt) ↔ 2πx(−ω) — is a favourite trick question.
Exam tip
The examiner always asks you to find the CTFT of e^(−at)u(t) and then apply the time-shifting or frequency-shifting property to a modified version — show every step of the integration to claim full marks even if the final answer is a standard pair.