2

Given are $N$ complex numbers $x_0,x_1,\cdots,n_{N-1}$. The discrete Fourier transformation is defined as:

$$k_m=\sum_{n=0}^{N-1} x_n e^{-i\frac{2\pi}{N}mn}~~~(1)$$

The inverse transformation is defined as:

$$x_n=\sum_{m=0}^{N-1}k_m e^{i\frac{2\pi}{N} n}~~~(2)$$

My goal is to perform the continuum limit. Then I should obtain the normal Fourier transformation formulas $f(x)=\int_{-\infty}^{\infty} dk e^{-ikx} f(k)$ and $f(k)=\int_{-\infty}^{\infty} dx f(x) e^{-ikx}$.

To do the proof I draw the following cartoons. Left: each $n$ is related to some $x$ value that will in the end be a continuous variable. The $x_n$'s will become the $f(x)$. Before performing the continuum limit the $x$ interval is finite. It reaches from $-L/2$ to $L/2$. The distance between two $x$ values before the continuum limit is $\Delta x=\frac{L}{N-1}$. The $x$ which will become the continuous variable after the continuum limit can be parameterized as $x=-\frac{L}{2}+n\Delta x$.

Analogous relations are also true in Fourier space. $m=0$ is related to a $k$ value of $-4\pi/L$ and $m=N-1$ is related to $4\pi/L$. the variable that will become continuous after taking the continuum limit is parameterized as $k=-\frac{4\pi}{L}+m\Delta k$ with $\Delta k=\frac{8\pi}{L(N-1)}$.

enter image description here

Now the continuum limit is $N\to \infty$ and $L\to \infty$ where $\Delta x=\frac{L}{N-1}\to 0$, i.e. $N$ diverges faster than $L$. One thing that is unclear right now is that in this case the $k$ range in Fourier space shrinks to zero since it goes from $-4\pi/L$ to $4\pi/L$?

Anyways, ignoring this for a moment my problem is more how I can rewrite $(1)$ as $f(k)=\int_{-\infty}^{\infty} dx f(x) e^{-ikx}$ after performing the continuum limit. It is trivial that $k_m$ becomes $f(k)$ and $x_n$ becomes $f(x)$. But what happens with the $m n$ factor in the exponential? Using the expressions for $n$ and $m$ I get

$mn=\frac{k+\frac{4\pi}{L}}{\Delta k}\frac{x+\frac{L}{2}}{\Delta x}=\left(x k + x\frac{4\pi}{L}+k\frac{L}{2}+2\pi\right)\frac{(N-1)^2}{8\pi}$

But this does not really make sense to me since first the result diverges in the continuum limit ($L\to \infty$) and second there are additional terms on top of the desired $xk$ term?

Also to convert the sum $\sum_{n=0}^{N-1}$ to an integral we need something like $\sum_{n=0}^{N-1}\Delta k\to \int dk$, but I am missing the $\Delta k$?

jojo123456
  • 123

1 Answers1

1

$\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ As OP showed, when take the continuum limit we let $L\rightarrow\infty,N\rightarrow\infty$ and keep $\delta L \approx L/N\rightarrow 0$. Then for the discrete Fourier transformation,

\begin{aligned} & k_{m}=\sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1}x_{n}e^{-i\frac{2\pi}{N}mn}\\ \Rightarrow & k_{m}=\sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1}x_{n}e^{-i\frac{2\pi}{N\delta L}m\left(n\delta L\right)}\\ \xRightarrow[n\delta L=x,dx/\delta L=dn]{N\delta L=L\rightarrow\infty,f(x)=x_{n}} & k_{m}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\dfrac{dx}{\delta L}f(x)e^{-i\frac{2\pi}{N\delta L}mx}\\ \xRightarrow[k=\frac{m}{N\delta L}]{\hat{f}(k)=\delta Lk_{m}} & \hat{f}(k)=\int_{-\infty}^{\infty}dxf(x)e^{-i2\pi kx} \end{aligned}

for the inverse discrete Fourier transformation

\begin{aligned} & x_{n}=\dfrac{1}{N}\sum_{m=-\frac{N}{2}}^{\frac{N}{2}-1}k_{m}e^{i\frac{2\pi}{N}mn}\\ \Rightarrow & x_{n}=\dfrac{1}{N}\sum_{m=-\frac{N}{2}}^{\frac{N}{2}-1}k_{m}e^{i\frac{2\pi}{N\delta L}m\left(n\delta L\right)}\\ \xRightarrow[f(x)=x_{n}]{n\delta L=x} & f(x)=\dfrac{1}{N}\sum_{m=-\frac{N}{2}}^{\frac{N}{2}-1}k_{m}e^{i\frac{2\pi}{N\delta L}mx}\\ \xRightarrow[k=\frac{m}{N\delta L},dkN\delta L=dm]{\hat{f}(k)=\delta Lk_{m}} & f(x)=\dfrac{1}{N}\int_{-\infty}^{\infty}dkN\delta L\dfrac{\hat{f}(k)}{\delta L}e^{i2\pi kx}\\ \Longrightarrow & f(x)=\int_{-\infty}^{\infty}dk\hat{f}(k)e^{i2\pi kx} \end{aligned} The results can be checked by comparing to defination of Fourier transform. Some remarks:

  1. Notice I have shifted "n" summation range to make it symmetric around zero, it can be rewrite to original summation range as follows: $$k_{m}=\sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1}x_{n}e^{-i\frac{2\pi}{N}mn}=\sum_{n=0}^{\frac{N}{2}-1}x_{n}e^{-i\frac{2\pi}{N}mn}+\sum_{n=-\frac{N}{2}}^{-1}x_{n}e^{-i\frac{2\pi}{N}m(n+N)}=\sum_{n=0}^{\frac{N}{2}-1}x_{n}e^{-i\frac{2\pi}{N}mn}+\sum_{n=\frac{N}{2}}^{N}x_{n-N}e^{-i\frac{2\pi}{N}mn}$$, this shift may be helpful when we approximate CFT(Discrete Fourier Transformation) using DFT(Discrete Fourier Transformation)/FFT, since softwares like Matlab usually use range $0\to N-1$ to calculate DFT, while our data $x_n$ may be symmetric around zero in time domain(i.e. for $x_n, n=-N/2 \cdots N/2-1$). Under this circumstance to directly compare results of DFT and CFT we need to shift the n range("fftshift" command in Matlab) .
  2. There are some errors use DFT to approximate CFT, see the related post.Finally, a kind reminder regarding the scaling factor between the DFT and the CFT: $\hat{f}(k) = \delta L k_m.$
Zane Chen
  • 26