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CTFT convolution property

Short answer

Convolution in time corresponds to multiplication in frequency: x(t)*h(t) ↔ X(ω)H(ω). The proof substitutes the convolution integral into the Fourier definition, interchanges integration order, uses the time-shift property to obtain e^{-jωτ}H(ω), and integrates over τ to yield X(ω)H(ω).

State and prove the convolution property of the continuous time Fourier transform (show that convolution in time domain results in multiplication in frequency domain).

Statement

If x(t)↔FX(ω)x(t) \xleftrightarrow{F} X(\omega) and h(t)↔FH(ω)h(t) \xleftrightarrow{F} H(\omega), then

y(t)=x(t)∗h(t)=∫−∞∞x(τ)h(t−τ) dτ  ↔F  Y(ω)=X(ω)H(ω)y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau \;\xleftrightarrow{F}\; Y(\omega) = X(\omega)H(\omega)

Convolution in the time domain becomes multiplication in the frequency domain.

Proof

By the definition X(ω)=∫−∞∞x(t)e−jωtdtX(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt:

Y(ω)=∫−∞∞[∫−∞∞x(τ)h(t−τ)dτ]e−jωtdtY(\omega) = \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau\right]e^{-j\omega t}dt

Change the order of integration:

Y(ω)=∫−∞∞x(τ)[∫−∞∞h(t−τ)e−jωtdt]dτY(\omega) = \int_{-\infty}^{\infty} x(\tau)\left[\int_{-\infty}^{\infty} h(t-\tau)e^{-j\omega t}dt\right]d\tau

In the inner integral put σ=t−τ\sigma = t - \tau (dt=dσdt = d\sigma, limits unchanged):

∫−∞∞h(σ)e−jω(σ+τ)dσ=e−jωτH(ω)\int_{-\infty}^{\infty} h(\sigma)e^{-j\omega(\sigma + \tau)}d\sigma = e^{-j\omega\tau}H(\omega)

This is the time-shift property. Substituting back:

Y(ω)=H(ω)∫−∞∞x(τ)e−jωτdτ=H(ω)X(ω)Y(\omega) = H(\omega)\int_{-\infty}^{\infty} x(\tau)e^{-j\omega\tau}d\tau = H(\omega)X(\omega) x(t)∗h(t)↔FX(ω)H(ω)\boxed{x(t)*h(t) \xleftrightarrow{F} X(\omega)H(\omega)}

Significance

  • LTI analysis: the output of an LTI system with impulse response h(t)h(t) is Y(ω)=H(ω)X(ω)Y(\omega) = H(\omega)X(\omega). H(ω)H(\omega) is the frequency response: each frequency component of the input is scaled by ∣H(ω)∣|H(\omega)| and phase-shifted by ∠H(ω)\angle H(\omega).
  • Filtering: an ideal low-pass filter has H(ω)=1H(\omega) = 1 for ∣ω∣<ωc|\omega| < \omega_c and 0 elsewhere. It removes the high-frequency parts of X(ω)X(\omega) simply by multiplication.
  • Cascades: systems in series give H(ω)=H1(ω)H2(ω)H(\omega) = H_1(\omega)H_2(\omega).
  • Computation: a convolution integral can be replaced by: transform, multiply, inverse transform.

Example

x(t)=e−atu(t)x(t) = e^{-at}u(t) is applied to h(t)=e−btu(t)h(t) = e^{-bt}u(t), with a≠ba \ne b and both positive:

Y(ω)=1(a+jω)(b+jω)=1b−a[1a+jω−1b+jω]Y(\omega) = \frac{1}{(a+j\omega)(b+j\omega)} = \frac{1}{b-a}\left[\frac{1}{a+j\omega} - \frac{1}{b+j\omega}\right]

So y(t)=1b−a(e−at−e−bt)u(t)y(t) = \frac{1}{b-a}\left(e^{-at} - e^{-bt}\right)u(t), found without evaluating the convolution integral.

Dual (multiplication/modulation) property: x(t)p(t)↔F12πX(ω)∗P(ω)x(t)p(t) \xleftrightarrow{F} \frac{1}{2\pi}X(\omega) * P(\omega).

✓ Answer checked by Claude Opus 5.5 (high effort)

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