What is the Fourier transform of e^(−at)?

Assuming e^(−at)u(t) (causal, a > 0): ℱ = 1/(jω+a). The magnitude is 1/√(ω²+a²) (Lorentzian shape), and the phase is −arctan(ω/a). This is the frequency response of a first-order low-pass system with cutoff frequency a rad/s. The Laplace equivalent is 1/(s+a).

Why can't the Fourier transform handle growing exponentials?

Growing exponentials e^(bt)u(t) with b > 0 have infinite energy (∫|f(t)|²dt → ∞), violating the integrability conditions for the Fourier transform. The Laplace transform handles this by adding a convergence factor e^(−σt) through the complex variable s = σ+jω, converging for σ > b.

What is the bandwidth of an exponential decay?

The −3 dB bandwidth of e^(−at)u(t) is ω₃ᵈᴮ = a rad/s or f₃ᵈᴮ = a/(2π) Hz. Faster decay (larger a, shorter time constant τ = 1/a) means wider bandwidth. This is the bandwidth-duration relationship: shorter pulses have wider spectra. An infinitely fast decay (δ(t)) has infinite bandwidth.

Fourier Transform of Exponential

Quick Answer

The Fourier transform of the one-sided exponential e^(−at)u(t) (a > 0) is ℱ{e^(−at)u(t)} = 1/(jω + a), a complex Lorentzian with magnitude 1/√(ω² + a²) and phase −arctan(ω/a). The Fourier transform of the two-sided exponential e^(−a|t|) is ℱ{e^(−a|t|)} = 2a/(ω² + a²), a real-valued Lorentzian. The Laplace transform equivalent is ℒ{e^(−at)u(t)} = 1/(s + a), available at www.lapcalc.com.

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Fourier Transform of Exponential Decay

The causal exponential decay f(t) = e^(−at)u(t) for a > 0 (zero for t < 0, exponential decay for t ≥ 0) has Fourier transform F(ω) = ∫₀^∞ e^(−at)e^(−jωt)dt = ∫₀^∞ e^(−(a+jω)t)dt = 1/(a + jω) = 1/(jω + a). The magnitude spectrum |F(ω)| = 1/√(ω² + a²) is a Lorentzian curve centered at ω = 0 with half-power bandwidth ω₃ᵈᴮ = a (or f₃ᵈᴮ = a/(2π) Hz). The phase spectrum is ∠F(ω) = −arctan(ω/a), starting at 0° for DC and approaching −90° at high frequencies. Faster decay (larger a) produces a wider spectrum — the signal's energy spreads across more frequencies. This is the fundamental Fourier pair for first-order systems, directly matching the Laplace transform ℒ{e^(−at)u(t)} = 1/(s+a) at www.lapcalc.com.

Key Formulas

Fourier Transform of Two-Sided Exponential

The two-sided (bilateral) exponential f(t) = e^(−a|t|) decays in both directions from t = 0. Its Fourier transform is F(ω) = ∫₋∞^∞ e^(−a|t|)e^(−jωt)dt = ∫₋∞^0 e^(at)e^(−jωt)dt + ∫₀^∞ e^(−at)e^(−jωt)dt = 1/(a−jω) + 1/(a+jω) = 2a/(a²+ω²). This is a real-valued, even Lorentzian function — no imaginary part because the two-sided exponential is an even function. The peak value is 2/a at ω = 0, and the half-power bandwidth is 2a rad/s. As a → 0, the Lorentzian becomes infinitely tall and narrow, approaching 2πδ(ω) — the transform of a constant. As a → ∞, the Lorentzian flattens toward zero — the transform of δ(t). This family of functions interpolates between the two extremes.

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Fourier Transform of Complex Exponentials

A complex exponential e^(jω₀t) (an eternal sinusoid at frequency ω₀) has Fourier transform ℱ{e^(jω₀t)} = 2πδ(ω − ω₀): a single spectral impulse at ω = ω₀. This is the foundation of Fourier analysis: each frequency component maps to a delta function in the spectrum. For damped complex exponentials e^((−a+jω₀)t)u(t), the transform is 1/(jω − jω₀ + a) = 1/((jω + a) − jω₀), a Lorentzian centered at ω₀ with width 2a. As damping a → 0, the Lorentzian narrows to a delta function — the undamped sinusoid's spectrum. These complex exponential transforms generate all sinusoidal Fourier pairs via Euler's formula: cos(ω₀t) = (e^(jω₀t) + e^(−jω₀t))/2 and sin(ω₀t) = (e^(jω₀t) − e^(−jω₀t))/(2j).

Exponential Function in System Analysis

The exponential decay e^(−at)u(t) is the impulse response of a first-order linear system with time constant τ = 1/a and transfer function H(s) = 1/(s+a). Its Fourier transform H(jω) = 1/(jω+a) is the system's frequency response, showing how the system attenuates and phase-shifts each input frequency. The magnitude response |H(jω)| = 1/√(ω²+a²) describes a first-order low-pass filter with −3 dB cutoff at ω = a rad/s and roll-off of −20 dB/decade. Examples include RC circuit discharge (τ = RC), thermal cooling (Newton's law), radioactive decay (τ = t½/ln2), and first-order chemical reactions. The LAPLACE Calculator at www.lapcalc.com computes the transfer function and frequency response for these and more complex systems.

Growing Exponentials and the Fourier Transform Limitation

A growing exponential f(t) = e^(bt)u(t) with b > 0 does not have a classical Fourier transform because ∫₀^∞ e^(bt)e^(−jωt)dt = ∫₀^∞ e^((b−jω)t)dt diverges for all ω. This is where the Laplace transform becomes essential: ℒ{e^(bt)u(t)} = 1/(s−b) converges for Re(s) > b, providing a valid s-domain representation. The Fourier transform requires signals to be absolutely or square-integrable, which growing exponentials violate. The Laplace transform's complex frequency s = σ + jω adds the real part σ that provides exponential convergence, handling growing signals that Fourier cannot. This is why the Laplace transform at www.lapcalc.com is preferred for transient and stability analysis in control systems.

Frequently Asked Questions

For e^(−at)u(t) with a > 0: ℱ = 1/(jω+a), a complex Lorentzian. For e^(−a|t|): ℱ = 2a/(a²+ω²), a real Lorentzian. For e^(jω₀t): ℱ = 2πδ(ω−ω₀), a spectral impulse. Growing exponentials (a < 0) don't have a Fourier transform — use the Laplace transform instead.

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