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I have the function

$ \delta(f-2) $

How can we inverse Fourier transform it? It's easy if $f$ is replaced with $w$. But based on my knowledge, $w = 2\pi f$.

The correct answer is

$ e^{4\pi i t} $

Can somebody explain to me what happened? Thanks.

2 Answers 2

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The inverse Fourier transform of $\delta(f-2)$ is $\mathcal F^{-1}[\delta](t) = \int \delta(f-2) e^{i2\pi ft} \, df = e^{i2\pi2t} = e^{i4\pi t}$ The 2nd equality holds by definition of the delta function.

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    integrating the Dirac $\delta$ function **is** the short-cut way. Typically looking things up in a Fourier table takes longer...2012-02-20
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Two commonly used definitions of the Fourier transform of $x(t)$ are $\begin{align*} X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t} \mathrm dt, &x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega\\ \hat{X}(f) &= \int_{-\infty}^{\infty} x(t) e^{-i 2\pi ft} \mathrm dt, &x(t) = \int_{-\infty}^{\infty} X(f) e^{i 2\pi f t}\mathrm df. \end{align*}$ The two functions are related as $\hat{X}(f) = X(2\pi f)$ and $X(\omega) = \hat{X}(f/2\pi)$.

I think your question essentially is: if you have a table that tells you the inverse Fourier transform of $X(\omega) = \delta(\omega-\omega_0)$ is $\frac{1}{2\pi}e^{i\omega_0t}$ from which it is easy to deduce that the inverse Fourier transform of $\delta(\omega-2)$ is $\frac{1}{2\pi}e^{i2t}$, how do you deduce from this that the inverse Fourier transform of $\hat{X}(f-f_0)=\delta(f-f_0)$ is $e^{i 2\pi f_0 t}$ in general, and that the inverse Fourier transform of $\delta(f-2)$ is $e^{i 4\pi t}$? As williamdemeo showed you, and Willie Wong emphasized to you, just computing the Fourier integral $x(t) = \int_{-\infty}^{\infty} \hat{X}(f) e^{i2\pi f t} \mathrm df = \int_{-\infty}^{\infty} \delta(f-2) e^{i2\pi f t} \mathrm df = e^{i 2\pi 2 t} = e^{i4\pi t}$ is far easier than fussing with tables. But if you are dead set on using tables only, then note that if $x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega$ is known to you, then since it does not matter what we call the variable of integration $ x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(f) e^{i f t}\mathrm df = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(f) e^{i 2 \pi f (t/2\pi)}\mathrm df $ Let $y(t)$ denote the rightmost integral. Then we have that $y(t)$ is the inverse Fourier transform of $X(f)$ evaluated at $t/2\pi$, and it happens to equal $2\pi x(t)$. So,

given that $x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t}\mathrm d\omega$ is the inverse Fourier transform of $X(\omega)$, the inverse Fourier transform of $X(f)$ is $\int_{-\infty}^{\infty} X(f) e^{i 2 \pi f t}\mathrm df = 2\pi \cdot x(2\pi t).$

In particular, given that the inverse the inverse Fourier transform of $\delta(\omega-2)$ is $\frac{1}{2\pi}e^{i2t}$, the inverse Fourier transform of $\delta(f-2)$ is $2\pi \frac{1}{2\pi}e^{i2\cdot 2\pi t} = e^{i4\pi t}$.