1
$\begingroup$

Is there a way to 'bypass' the condition $\Delta k=\frac{1}{x_{max}-x_{min}}$, which requires the range of $x$ to be very large when approximating the CFT encountered in Michealson Interometery as a DFT?


To contextualise my question:

Suppose the the continuous Fourier Transform (CFT) is defined as $$\hat{f}(k)=\frac{1}{2}\int_{-\infty}^{\infty} f(x) e^{-i2\pi kx}dx$$

Suppose that $f \to 0$ for sufficiently large and small values of $x$, then $$\hat{f}(k)=\frac{1}{2}\int_{-\infty}^{\infty} f(x) e^{-i2\pi kx}dx$$ $$\hat{f}(k_m)\approx \frac{1}{2}\sum_{n=0}^{N-1}f(x_n)e^{-i2\pi k_mx_n}\Delta x$$ $$=\frac{1}{2}\sum_{n=0}^{N-1}f(x_0+n\Delta x)e^{-2i(m\Delta k)(x_0+n \Delta x)}\Delta x$$ $$=\frac{1}{2}e^{-2i m\Delta k x_0 }\sum _{n=0}^{N-1}f(x_0+n\Delta x)e^{-i2mn \Delta x \Delta k }$$ where $x_n=x_0+n\Delta x$, $k_m=m \Delta k$ (so $k_0=0$)

The Discrete Fourier Transform is defined as follows $$F_m=\sum^{N-1}_{n=0}f_ne^{-i2mn/N}$$ We conclude that for the last sum appearing in the CFT to approximate to the DFT, $\Delta x \Delta k=1/N $ must hold. $$\hat{f}(k_m)\approx \frac{1}2{e^{-2im\Delta kx_0}F_m} \tag{1}$$

since $\Delta x=\frac{x_{max}-x_{min}}{N-1}$, $\Delta k=\frac{1}{N \Delta x} \approx \frac{1}{x_{max}-x_{min}}$ for N sufficiently large.

This means that if I use the approximation (1) in the nano-metre scale, it will fail since $\Delta k$ (separation between $k_{i+1}$ and $k_{i}$)will be very large. So (1) must be used in the range, say $-100<x<100$, a range like $-10^-{9}<x<10^{-9}$ (visible wavelength scale) will cause the approximation to break down.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.