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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.

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2 Answers 2

up vote 6 down vote accepted

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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is there a shortcut way instead of integrating. i cant find a fourier table that deals with f instead of w (omega) –  IvanMatala Feb 20 '12 at 12:40
    
integrating the Dirac $\delta$ function is the short-cut way. Typically looking things up in a Fourier table takes longer... –  Willie Wong Feb 20 '12 at 12:59

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 $2\pi \frac{1}{2\pi}e^{i2\cdot 2\pi t} = e^{i4\pi t}$.

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