Is the $2\pi$ factor correct in the Inverse Fourier transform formula?

Question

These two formulae were given in my circuit analysis textbook:

Fourier Transform: $$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$

Inverse Fourier Transform: $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega$$

I have a bit of a doubt as to whether the second formula is true. I don't understand where the $2\pi$ factor is coming from in the second formula?

If the above hold formulae true, is is the following provable?

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\right)e^{i\omega t}d\omega$$

Answer 1

The factor comes from the fact that

$$ \delta (x-\alpha )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{ip(x-\alpha )}\ dp \tag{1} $$

Consider your last integral

\begin{eqnarray} \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t')e^{-i\omega t'}dt'\right) e^{i\omega t} d\omega &=& \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t')e^{-i\omega t'}e^{i\omega t} dt'\right) d\omega \\ &=& \int_{-\infty}^{\infty}f(t')\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega (t - t')}d\omega \right) dt' \\ &\stackrel{(1)}{=}& \int_{-\infty}^{\infty}f(t')\delta(t-t')dt' \\ &=& f(t) \tag{2} \end{eqnarray}

Answer 2

I don't understand where the $2\pi$ factor is coming from in the second formula?

Hint. A way to see why from $$ F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt \tag1 $$ one has to divide by $2\pi$ to define $$ f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega. \tag2 $$

Observe that pluging $f(t)=e^{-t^2}$ yields, with the standard gaussian evaluation, $$ F(\omega)=e^{\large-\frac{\omega^2}4}\cdot \sqrt{\pi} $$ and $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega=\frac{1}{\color{red}{2\pi}}\cdot \color{red}{2\pi}\cdot e^{-t^2}=f(t). $$

Answer 3

Take the fourier transform with oscillatory factor $-1$ of the rect window of height $A$ of width $2T$ centred about the origin:

$$\int_{-T}^{T} Ae^{-i\omega x} dx = \frac{2A\sin(-T \omega)}{-\omega} = \frac{2A\sin(T \omega)}{\omega} $$ $$\int_{-\infty}^{\infty} \frac{2A\sin(T \omega)}{\omega} d\omega = 2\pi A \operatorname{sgn} (T). $$

If you perform this integral you will get the $2\pi$ factor as mathematical fact.

$$\int_{-\infty}^{\infty} \frac{2A\sin(T \omega)}{\omega} e^{i\omega x'} d\omega$$

$$= \int_{-\infty}^{\infty} \frac{2A\sin(T \omega)}{\omega}i\sin(\omega x')d\omega + \int_{-\infty}^{\infty} \frac{2A\sin(T \omega)}{\omega}\cos(\omega x')d\omega $$

$$= 0 + A\pi (\operatorname{sgn} (T - x') + \operatorname{sgn} (T + x')). $$ Which is the rect window function of $x'$ at a height of $2\pi A$ and width of $2T$.

Clearly this now extrapolates to any signal within the unit rect window $A=1$, any sinusoid in the original rect window of height $1$ will now be $2\pi$ greater in magnitude. This holds for any $T$ which shows that:

$$\lim_{T\to \infty} \int_{-T}^{T} Ae^{-i\omega x} dx = 2\pi A\delta(-\omega) = 2\pi A\delta(\omega). $$

There is a similar case for uncentred rect windows which yield imaginary components due to multiple sin frequencies no longer being an odd function within the window i.e. multiple sin frequencies now no longer have an area of 0 between the bounds of the uncentred rect window – the frequency response is complex.

The positive frequency domain of the fourier transform of $\cos (x)$ with an oscillatory factor of $-1$ and preferred variable to represent angular frequency being $\omega$ is $\pi \delta(\omega-1)$.

The positive frequency domain of the fourier transform of $\cos (2\pi x)$ with an oscillatory factor of $-1$ is $\pi \delta(\omega-2\pi)$.

The positive frequency domain of the fourier transform of $\cos (x)$ with an oscillatory factor of $-2\pi$ and preferred variable for frequency is $f$ is $\pi \delta(2\pi f-1)$ i.e. $\frac{1}{2}\delta(f-\frac{1} {2\pi})$.

The positive frequency domain of the fourier transform of $\cos (2\pi x)$ with an oscillatory factor of $-2\pi$ is $\pi \delta(2\pi f -2\pi)$ I.e. $\frac{1}{2}\delta (f - 1)$.

The latter 2 show how using an oscillatory factor of $2\pi$ makes the $2\pi$ factor disappear. The use of $\omega$ and $f$ does not affect the integral other than the semantic difference – if you're using an oscillatory factor of $-1$ it makes sense to use $\omega$ as it would be confusing to represent the angular frequency domain as a function of $f$. You can get the $f$ frequency domain from the angular frequency by simply substituting $\omega$ with $2\pi f$ in the result, which can be done without u-substitution because it is not the variable of integration, and therefore $\pi \delta(\omega-1) = \pi \delta(2\pi f-1) = \frac{1}{2} \delta(f-\frac{1}{2\pi})$.