# Meaning of frequency in DFT

I am trying to get a better understanding of DFT of an image. Assume that we have an image of size $$N\times N$$, denoted as $$f(m, n), m, n = 0, 1, \ldots, N$$, and its Fourier transform $$F(k, l), k, l = 0,1, \ldots, N-1$$. It is often said that $$F(N-1, N-1)$$ represents the highest frequency. But what exactly does it mean?

If I was transforming a $$1D$$ signal, the "base" functions (I quote "base" because from my understanding of linear algebra DFT is not really a change of basis) are $$e^{-i2\pi \frac{km}{N}}$$ where $$m$$ is index of signal in space domain and $$k$$ is index in Fourier domain. For $$k=N-1$$ I get $$F(N-1) = f(0)e^{0}+f(1)e^{-i2\pi \frac{N-1}{N}} + f(2)e^{-i4 \pi \frac{N-1}{N}} + \ldots f(N-1)e^{-i(N-1) \pi \frac{N-1}{N}}$$. Now, some of the "base" functions in $$F(N-1)$$ are contained in other $$F(k), k < N-1$$, but definitely not the last one $$e^{-i(N-1) \pi \frac{N-1}{N}}$$. So does the "highest frequency" mean that the range of the arguments of the exponentials in $$F(N-1)$$ is larger than in any other $$F(k), k?

It might be helpful to think of the DFT as correlating the signal $$f(n)$$ with a set of “basis” functions, these being complex exponentials of the form: $$W_k(n) = e^{-i2\pi \frac{k}{N}n}, \quad 0\leq n where $$\frac{k}{N}$$ is the discrete frequency used to build a particular function of this set.
For example, the 3rd bin of the DFT of $$f(n)$$ contains the result of correlating $$f(n)$$ with a complex exponential of frequency $$\frac{3}{N}$$.

The higher $$k$$ is, the higher the base exponential’s frequency is. When the DFT gets to the case where $$k = N-1$$, it simply means it’s correlating $$f(n)$$ with the complex exponential of highest frequency allowed by the DFT size $$N$$.

• So can I understand W_k(n) as a discretized signal starting at time n=0 and evolving to n=N-1?
– Avec
Aug 23 at 12:58
• I feel like I can imagine it for signals evolving in time, because there is some start and an end, but what does it mean for images that have spatial coordinates instead?
– Avec
Aug 23 at 14:30
• Mathematically, a 2-dimensional DFT can be obtained by applying a bunch of 1-dimensional DFTs. This is outside of the scope of this question, but you can consider each row of your matrix (your image) to be a 1D signal. See this
– Jdip
Aug 23 at 14:53