Continuous-Time Fourier Transform (CTFT)

This is getting into the weeds of signal processing.

Convolution is very expensive. This is convolution

(f∗g)(t)≜∫−∞∞f(τ) g(t−τ) dτ(f * g)(t) \triangleq \int_{-\infty}^{\infty} f(\tau) \, g(t - \tau) \, d\tau

Let g(t)=eiωtg(t)=e^{i\omega t}

(f∗g)(t)=∫f(τ) eiω(t−τ) dτ=eiωt∫f(τ) e−iωτ dτ(f * g)(t) = \int f(\tau) \, e^{i\omega(t-\tau)} \, d\tau = e^{i\omega t} \int f(\tau) \, e^{-i\omega \tau} \, d\tau

We now call

f~(ω)≜∫−∞∞f(τ) e−iωτ dτ\boxed{\tilde{f}(\omega) \triangleq \int_{-\infty}^{\infty} f(\tau) \, e^{-i\omega \tau} \, d\tau}

So

(f∗eiωt)(t)=eiωtf~(ω)(f*e^{i\omega t})(t)=e^{i\omega t}\tilde{f}(\omega)

The inverse of f~\tilde{f} is

f(t)≜12π∫−∞∞f^(ω) eiωt dω\boxed{f(t) \triangleq \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\omega) \, e^{i\omega t} \, d\omega}

We can write

f(t)↔f^(ω)f(t) \leftrightarrow \hat{f}(\omega)

Cheat sheet

δ(x)↔1\delta(x) \leftrightarrow 1 1↔2π δ(k)1 \leftrightarrow 2\pi\,\delta(k) eik0x↔2π δ(k−k0)e^{ik_0 x} \leftrightarrow 2\pi\,\delta(k - k_0) cos⁡(k0x)↔π[δ(k−k0)+δ(k+k0)]\cos(k_0 x) \leftrightarrow \pi\big[\delta(k - k_0) + \delta(k + k_0)\big] sin⁡(k0x)↔−iπ[δ(k−k0)−δ(k+k0)]\sin(k_0 x) \leftrightarrow -i\pi\big[\delta(k - k_0) - \delta(k + k_0)\big] e−ax2↔πa e−k2/(4a)(Gaussian↔Gaussian)e^{-a x^2} \leftrightarrow \sqrt{\tfrac{\pi}{a}}\, e^{-k^2/(4a)} \qquad (\text{Gaussian} \leftrightarrow \text{Gaussian}) e−a∣x∣↔2aa2+k2(Lorentzian)e^{-a|x|} \leftrightarrow \frac{2a}{a^2 + k^2} \qquad (\text{Lorentzian}) f(x−a)↔e−ika f~(k)(shift↔phase)f(x - a) \leftrightarrow e^{-ika}\,\tilde{f}(k) \qquad (\text{shift} \leftrightarrow \text{phase}) eik0xf(x)↔f~(k−k0)(phase↔shift)e^{ik_0 x} f(x) \leftrightarrow \tilde{f}(k - k_0) \qquad (\text{phase} \leftrightarrow \text{shift}) f(ax)↔1∣a∣f~(k/a)(narrow↔wide)f(ax) \leftrightarrow \tfrac{1}{|a|}\tilde{f}(k/a) \qquad (\text{narrow} \leftrightarrow \text{wide}) f′(x)↔ik f~(k)xf(x)↔i f~′(k)f'(x) \leftrightarrow ik\,\tilde{f}(k) \qquad x f(x) \leftrightarrow i\,\tilde{f}'(k)

Convolution

(f∗g)(x)↔f~(k) g~(k)f(x)g(x)↔12π(f~∗g~)(k)(f * g)(x) \leftrightarrow \tilde{f}(k)\,\tilde{g}(k) \qquad f(x)g(x) \leftrightarrow \tfrac{1}{2\pi}(\tilde{f}*\tilde{g})(k)