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The equation of inverse Fourier transform is the following: $$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t} d\omega $$ $$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \big[\cos(\omega t) + i\sin(\omega t)\big] d\omega $$ Now, if I have only real value in the time domain $f(t)$, then the imaginary part of $\int e^{i\omega t}$ must go to zero. In our signal processing course, I always had real valued time domain function. Then why am I adding this imaginary part of $\int e^{i\omega t}$ in inverse Fourier transform since it always goes to zero?

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then the imaginary part must go to zero

No. That's a fallacy. There's really not much to explain here other than this intuition is wrong.

The imaginary part of a Fourier transform is only zero if the transformed signal has conjugate symmetry (which is the same as symmetry for real signals) around t=0.

Think about it. You decomposed your $e^{-ix}=\cos x + i\sin x$, and then used an integral to project your signal onto these two things. (since the integral of a product between a signal and another (conjugated) signal is the inner product of the signal vector space, and hence the projection, if you're after the mathematical formalism.)

The part of your signal that is symmetrical to t=0 projects onto the cosine, and the part that is antisymmetrical to t=0 to the sine. The sine has an imaginary unit in front.

This is really among the first things that are typically illustrated in textbooks when they introduce the Fourier transform, so this might be a great point to go back to that!

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  • $\begingroup$ I think the OP is clear that he "always had real valued time domain function" ?! $\endgroup$ – AlexTP Aug 16 at 10:06
  • $\begingroup$ yes, but the point is that Swakshar assumes that there's no imaginary part in the frequency-domain function, so he'd be "adding" an imaginary part, which isn't happening. $\endgroup$ – Marcus Müller Aug 16 at 10:07
  • $\begingroup$ @MarcusMüller if $F(t)$ is a real-valued function in the time domain, then how can imaginary part of $\int e^{iwt}$ not goes to zero? $\endgroup$ – Swakshar Deb Aug 16 at 10:32
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    $\begingroup$ because it simply doesn't. If it has any components that aren't symmetrical to t=0, then the part of the signal that doesn't project on the cosine, but on the sine, isn't zero. $\endgroup$ – Marcus Müller Aug 16 at 11:08
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    $\begingroup$ @SwaksharDeb because $i \int F(w) sin(wt) dw$ is also real. Marcus is right. $\endgroup$ – AlexTP Aug 16 at 11:14
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Let $x(t)$ be a real-valued function with Fourier integral

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\tag{1}$$

Clearly, if $x(t)$ is real-valued, the integral on the the right-hand side of $(1)$ must be real-valued too, even though $e^{j\omega t}$ is complex-valued and $X(\omega)$ is generally complex-valued too.

Since $x(t)=x^*(t)$ (where $^*$ denotes complex conjugation), we have

$$\begin{align}\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega&=\frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega\\&=\frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(-\omega)e^{j\omega t}d\omega\end{align}\tag{2}$$

Consequently, for real-valued $x(t)$, the Fourier transform $X(\omega)$ satisfies

$$X(\omega)=X^*(-\omega)\tag{3}$$

With $X(\omega)=X_R(\omega)+jX_I(\omega)$, the imaginary part of the right-hand side of $(1)$ can be written as

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[X_R(\omega)\sin(\omega t)+X_I(\omega)\cos(\omega t)\right]d\omega\tag{4}$$

From $(3)$ it follows that $X_R(\omega)$ is an even function and $X_I(\omega)$ is an odd function. Consequently, both expressions in the integrand of $(4)$ are odd, and hence the integral $(4)$ equals zero, as must be the case for real-valued $x(t)$.

In sum, for real-valued $x(t)$, even though $(1)$ is of course correct, the Fourier integral can also be written as a purely real-valued integral:

$$\begin{align}x(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[X_R(\omega)\cos(\omega t)-X_I(\omega)\sin(\omega t)\right]d\omega\\&=\frac{1}{\pi}\int_{0}^{\infty}\left[X_R(\omega)\cos(\omega t)-X_I(\omega)\sin(\omega t)\right]d\omega\end{align}\tag{5}$$

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You are not adding an imaginary part (given strictly real signals). F(w) can be imaginary (for any non-even-symmetric signals) . And the product of an imaginary part (from F) and and imaginary part (from e) is a real (possibly negative) value.

Thus you are adding real values (which are the results from the complex or imaginary multiplications).

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