Understanding the Complex Conjugate Property in Fourier Transform

In summary, the conversation discusses the properties of complex conjugates and Fourier transforms, using the definitions of ##f^*## and ##\widetilde{f}(k)##. The conversation also addresses a sign error in the calculations and clarifies the difference between using eikx and e-ikx. One mistake is also pointed out between equations (7) and (8), which is resolved by using the property ##\widetilde{u}(-k) = \widetilde{u}(k)^*##.
  • #1
arpon
235
16
[##f^*## represents complex conjugate of ##f##. ]

[##\widetilde{f}(k)## represents Fourier transform of the function ##f(x)##.]

$$\begin{align}
\int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}f^*(x)\left(e^{-ikx}\right)^*\,dx\\
&=\int_{-\infty}^{\infty}\left(f(x)e^{-ikx}\right)^*\,dx\\
&=\left(\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right)^*\\
&=\left[\widetilde{f}(k)\right]^*\\
\end{align}
$$
Now, let
$$f(x)=u(x)+iv(x)$$
where ##u(x)## and ##v(x)## are the real and imaginary parts of ##f(x)##.
Again, we have,
$$\begin{align}
\int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}\left(u(x)-iv(x)\right)e^{ikx}\,dx\\
&=\int_{-\infty}^{\infty}u(x)e^{-i(-k)x}\,dx-i\int_{-\infty}^{\infty}v(x)e^{-i(-k)x}\,dx\\
&=\widetilde{u}(-k)-i\widetilde{v}(-k)\\
&=\left[\widetilde{u}(-k)+i\widetilde{v}(-k)\right]^*\\
&=\left[\widetilde{f}(-k)\right]^* \text{ [using linearity property of Fourier transform]}

\end{align}
$$
So, I am getting different results. What is wrong with this calculation.
 
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  • #2
There is a sign error between (3) and (4).
 
  • #3
mfb said:
There is a sign error between (3) and (4).
I used the defination,
$$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)$$
 
  • #4
arpon said:
I used the defination,
$$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)$$
Well, not quite. The definition is [itex] \widetilde f(k)= \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/itex]
 
  • #5
Svein said:
Well, not quite. The definition is [itex] \widetilde f(k)= \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/itex]

There are actually many non-equivalent definitions. Most just differ in the constant in front.
 
  • #6
Well, it doesn't matter which definition you use, but keep it consistent. Either eikx or e-ikx. Using the former, (3) is the Fourier transform for -k instead of k.
 
  • #7
mfb said:
Well, it doesn't matter which definition you use, but keep it consistent. Either eikx or e-ikx. Using the former, (3) is the Fourier transform for -k instead of k.
I did not understand why (3) is the Fourier transform of -k instead of k. Look, I used the defination,
$$
\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)\\
$$
Taking complex conjugate on both sides,
$$
\left(\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right)^*=\left[\widetilde{f}(k)\right]^*\\
$$
So (3) and (4) are justified.
 
  • #8
Ah, I misread the signs, sorry.

There is a mistake between (7) and (8). u and v are real, but their Fourier transformations in general won't be real. You can use ##\widetilde{u}(-k) = \widetilde{u}(k)^*##:

$$\begin{align}
\widetilde{u}(-k)-i\widetilde{v}(-k)
&= \widetilde{u}(k)^* -i \widetilde{v}(k)^*\\
\end{align}$$

Not sure how to simplify that.
 
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What is the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It allows us to represent a complex signal as a sum of simpler sinusoidal functions.

Why is the Fourier transform important?

The Fourier transform is important because it is a fundamental tool in signal processing and data analysis. It is used to analyze and manipulate signals in many different fields, including engineering, physics, and mathematics.

What is the difference between the Fourier transform and the Fourier series?

The Fourier transform is a continuous representation of a signal, while the Fourier series is a discrete representation. The Fourier transform is used for signals that are defined over a continuous range, while the Fourier series is used for signals that are defined over a finite set of values.

How is the Fourier transform calculated?

The Fourier transform is calculated using an integral equation, which involves multiplying the signal by a series of sinusoidal functions and integrating over the entire range of the signal. This process is repeated for different frequencies to obtain the complete representation of the signal.

What are some practical applications of the Fourier transform?

The Fourier transform has many practical applications, including filtering and smoothing of signals, image processing, data compression, and spectral analysis. It is also used in various fields such as telecommunications, audio and video processing, and medical imaging.

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