Problem -- Fourier transform

  • #1
236
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.
 

Answers and Replies

  • #2
35,127
11,363
There is a sign error between (3) and (4).
 
  • #3
236
16
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
Svein
Science Advisor
Insights Author
2,125
679
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
22,089
3,292
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
35,127
11,363
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
236
16
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
35,127
11,363
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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