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Parseval's Theorem says that:

$$\int_{-\infty}^{\infty}g(t)f(t)^\ast dt = \frac{1}{2\pi} \int_{-\infty}^{\infty}G(\omega)F(\omega)^\ast d\omega$$

Although I know how to prove it, it's difficult to imagine how the two integrals can be equal. If we define $g(t)$ and $f(t)$ as time-domain real-number signals (so $f(t)$'s conjugate is also a real-number signal), then the left side is also real numbered. But the $G(\omega)$ and $F(\omega)^\ast$ can be complex numbered, so the integral on the right side is also a complex number.

Then how could a real number equal to a complex number? There must be something wrong with my above understanding. Could anyone help me to point it out?

Glorfindel
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Weigh
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  • What do you know about the properties of Fourier transforms of real functions? – Brian Borchers May 04 '17 at 13:47
  • I don't think the $\frac{1}{2\pi}$ should be on the right hand side. This should be equality between the $L^2$ inner products of two (presumably Schwartz) functions and their transforms. To prove it, begin with the right hand side, substitute in for one of the transforms the explicit integral representation involving the original function, and make a further simplification involving the inversion formula to obtain your result. – Sawyer May 04 '17 at 13:52
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    @Sawyer there are many different variations of the Fourier transform. Not all are unitary but they can be made unitary. – Cameron L. Williams May 04 '17 at 13:55
  • The integral of the imaginary part of $FG^*$ equals $0$ if $f$ and $g$ are real. – Mark Viola May 04 '17 at 14:03
  • @BrianBorchers I know something... I normally use the power spectral density function to estimate systems' FRF. But I think I am poor on theoretical things. There must be something wrong with my understanding of this equation, perhaps the integral of complex number or something else. – Weigh May 04 '17 at 14:05
  • @Dr.MV why the integral of the imaginary part equals 0?... Could you please help me to prove it, or give me a simple example? Thank you very much! – Weigh May 04 '17 at 14:09
  • @SawyerR. I use the definition of inverse Fourier transform as: $g(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)e^{j\omega t}d\omega$, the unit of $\omega$ is rad/s. So there is a $\frac{1}{2\pi}$ on the right side. – Weigh May 04 '17 at 14:14
  • @SawyerR. The factor of $1/(2\pi)$ is irrelevant to the issue herein. – Mark Viola May 04 '17 at 14:35

2 Answers2

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Assume that $f(t)$ and $g(t)$ are real valued functions. Then, we have

$$\begin{align} F(\omega)&=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\ &=\underbrace{\int_{-\infty}^\infty f(t) \cos(\omega t)\,dt}_{\text{Even in}\,\omega}+i\underbrace{\int_{-\infty}^\infty f(t) \sin(\omega t)\,dt}_{\text{Odd in }\,\omega} \end{align}$$

$$\begin{align} G(\omega)&=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\ &=\underbrace{\int_{-\infty}^\infty g(t) \cos(\omega t)\,dt}_{\text{Even in}\,\omega}+i\underbrace{\int_{-\infty}^\infty g(t) \sin(\omega t)\,dt}_{\text{Odd in }\,\omega} \end{align}$$

Now, note that the product of $F$ and the complex conjugate of $G$ is given by

$$\begin{align} F(\omega)G^*(\omega)&=\left(\text{Re}(F(\omega))\text{Re}(G(\omega))+\text{Im}(F(\omega))\text{Im}(G(\omega))\right)\\\\ &-i\left(\text{Re}(F(\omega))\text{Im}(G(\omega))-\text{Im}(F(\omega))\text{Re}(G(\omega))\right) \end{align}$$

Therefore, the imaginary part of $F(\omega)G^*(\omega)$ is an odd function. Since convergent integrals are equal to their Cauchy Principal Values, the integral of any odd convergent function over symmetric limits vanishes. And we are done!

Mark Viola
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Here's a way to understand it using essentially only one fact: the integral of the product of an even function ($f(x)=f(-x)$) and an odd function ($f(x)=-f(-x)$) is zero:

If $f$ is even and $g$ is odd, then $$ \int_{-\infty}^{\infty} f(x)g(x) \, dx = 0. $$

Any function $f$ can be written as the sum of an even function and an odd function: $$ f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}. $$ Call the former fraction $f_e$, the latter $f_o$.

How does the Fourier transform work on the even and odd part? We can split $e^{ikx}=\cos{kx}+i\sin{kx}$, and the cosine is even, the sine odd. Therefore, if we write $F(k) = \int_{-\infty}^{\infty} e^{ikx} f(x) \, dx$, $$ F(k) = \int_{-\infty}^{\infty} (\cos{kx}+i\sin{kx})(f_e(x)+f_o(x)) \, dx \\ = \int_{-\infty}^{\infty} (\cos{kx}) f_e(x) \, dx +i\int_{-\infty}^{\infty} (\sin{kx}) f_o(x) \, dx, $$ since the other integrals vanish. But both these integrands are even, so they are equal to twice the integral over $(0,\infty)$: $$ F(k) = 2\int_{0}^{\infty} (\cos{kx}) f_e(x) \, dx +2i\int_{0}^{\infty} (\sin{kx}) f_o(x) \, dx = F_c(k) + iF_o(k), $$ say. But the evenness of cosine means that $F_c$ is even, and in the same way $F_o$ is odd. Moreover, both are obviously real. But then, considering the right-hand side of Parseval, $$ \int_{-\infty}^{\infty} F(k) G(k)^* \, dk = \int_{-\infty}^{\infty} (F_c(k)+F_s(k)) (G_c(k)-G_s(k)) \, dk = \int_{-\infty}^{\infty} F_c(k) G_c(k) \, dk + \int_{-\infty}^{\infty} F_s(k) G_s(k) \, dk; $$ the other two terms vanish because $F_c,G_c$ are even and $F_s,F_s$ are odd.

Chappers
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