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I'm having trouble calculating the Fourier Transform of the sin function. Specifically, the function

$ G(\omega)=\int _{-\infty}^{\infty} g(t)\ e^{-i \omega t} dt $

For the fourier transform of

$ g(t)=\sin (2 \pi A t)$

the answer is:

$ \dfrac{\pi}{i}[ \delta(\omega - 2 \pi A)-\delta(\omega + 2 \pi A)] $.

My only problem is that I don't understand why the factor in the front is $\pi/i $ , and not just the $ 1/(2i) $ that falls out of the expansion of the sine function.

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    $\begingroup$ ${\cal F} (t \mapsto e^{iat})(\omega) = 2 \pi \delta(\omega-a)$. $\endgroup$ – copper.hat Dec 3 '14 at 5:32
  • $\begingroup$ Hey copperhat, where does the 2*pi come from? When i do the FT, i just get the dirac delta.. $\endgroup$ – user63602 Dec 3 '14 at 5:38
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    $\begingroup$ ${\cal F}(\delta)(\omega) = 1$, then use the inverse transform (${\cal F} \hat{f} = 2 \pi f$) which gives ${\cal F} (t \mapsto 1) = 2 \pi \delta$. $\endgroup$ – copper.hat Dec 3 '14 at 6:41
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The short answer is that the $2 \pi$ comes from the inversion formula.

Here is an informal perspective which gives as hint as to how this can be made more rigorous:

A distribution is defined by its action on a space of nicely behaved test functions.

For a function $f$ from a restricted class of ordinary functions, we can define a distribution $T_f$ by $T_f(\phi) = \int f \phi$. For the function $t \mapsto 1$, we get $T_1(\phi) = \int \phi$.

The '$\delta$ function' is defined by the distribution $T_\delta(\phi) = \phi(0)$. That is, it takes a test function $\phi$ and returns its value at $0$.

The Fourier transform of a distribution is defined by $\hat{T_f}(\phi) = T_f(\hat{\phi})$, where $\hat{\phi}$ is the ordinary Fourier transform of $\phi$.

We see that ${\hat{T}_1}(\phi) = T_1(\hat{\phi}) = \int \hat{\phi}$.

The standard inversion formula shows that $\int \hat{\phi} = 2 \pi \phi(0)$, which gives ${\hat{T}_1}(\phi) = 2 \pi T_\delta(\phi)$, or more succinctly, ${\hat{T}_1} = 2 \pi T_\delta$.

The same sort of analysis shows that ${\hat{T}_{t \mapsto e^{iat}}}(\phi) = 2 \pi \phi(a) = 2 \pi T_\delta(\omega \mapsto \phi(\omega+a))$. The last expression may be written informally as $T_{ \omega \mapsto 2 \pi \delta(\omega-a) } (\phi)$, which is the desired result.

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Remember

$ \sin( 2 \pi A t) = (e^ { i\ 2 \pi A t} - e^ {- i\ 2 \pi A t} )/ ( 2 i) $

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Use the concept of duality property

$$x(t) \rightleftharpoons X(\omega$$

$$X(t) \rightleftharpoons 2\pi x(-\omega)$$

$$\delta(t) \rightleftharpoons 1$$

$$ 1 \rightleftharpoons 2\pi \delta(-\omega)$$

Since $\delta(\omega)$ is a even function

$$ 1 \rightleftharpoons 2\pi \delta(\omega)$$

$$e^{j\omega_0t} \rightleftharpoons 2\pi \delta(\omega-\omega_0)$$

$$e^{-j\omega_0t} \rightleftharpoons 2\pi \delta(\omega+\omega_0)$$

$$sin(\omega_0t) \rightleftharpoons \frac{\pi}{j} [ \delta(\omega-\omega_0) - \delta(\omega+\omega_0)]$$

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