I need to calculate cross correlation between 2 images which I read in 2 vectors, both of them uni-dimensional.

As per my understanding I need to do the following operations:

  1. FFT(1D) on vector 1 and vector 2
  2. Multiplication between the outputs of the FFT applied on each vector
  3. IFFT(1D) of the multiplication

If I'd wanted to use 2D FFT in the first step following this method then what would the second step be?


3 Answers 3


Take an image:

Cross-correlating it with impulse should yield itself, and cross-correlating with itself should peak at center. Key points:

  • The operating kernel must be centered about $t=0$. For a discrete sequence $h$ of length $N$, under the FFT, this means $h[0]$, and the second-half of samples are of negative time: $h[n > N/2]$
  • Assuming inputs are normal images, this means time-centering the "template" image: note $x \star h \neq h \star x$.
  • fft2(x) == fft(fft(x, axis=0), axis=1)

cross correlation between 2 images which I read in 2 vectors, both of them uni-dimensional.

That's a problem since the time-centering step isn't straightforward to replicate on a flattened image, I'm unsure how it'd be done. This answer will assume a 2D array, so you can just reshape it into 2D and then back. Finally, you might want to look into padding and boundary effects.

Putting it together, here's cross-correlation of COVID with image-centered unit impulse, and with itself:

additionally, we move it and its flipped copy, and check that the more intense dot spots the unflipped version; note boundary effects:

$$ \texttt{iFFT}_{2d}\bigg( \texttt{FFT}_{2d}\big(x\big) \cdot \overline{\texttt{FFT}_{2d}\big(\texttt{iFFTSHIFT}_{2d}(h)\big)} \bigg) $$

Centering / performance note

Answer assumes we're after circular cross-correlation with no padding, and $x$ and $h$ are "normal" (non-DFT centered) images. With padding/unpadding, the most compute efficient route may be different.

Complete cross-correlation, with padding, is implemented here.

Meta note

This answer was subject of controversy. If readers aren't sure which answer to use, I encourage reading responses to the followup, where I've shown further demos and offered 1000 bounty to invalidate this answer, instead of trusting the votes.

Full code

import numpy as np
import matplotlib.pyplot as plt
from numpy.fft import fft, ifft, fft2, ifft2, ifftshift
from PIL import Image

def cross_correlate_2d(x, h):
    h = ifftshift(ifftshift(h, axes=0), axes=1)
    return ifft2(fft2(x) * np.conj(fft2(h))).real

# load image as greyscale
x = np.array(Image.open("covid.png").convert("L")) / 255.

# make kernels
h0 = np.zeros(x.shape, dtype=x.dtype)
h0[h0.shape[0]//2, h0.shape[1]//2] = 1
h1 = x.copy()

# compute
out0 = cross_correlate_2d(x, h0)
out1 = cross_correlate_2d(x, h1)

# plot
plt.imshow(out0); plt.xticks([]); plt.yticks([]); plt.show()
plt.imshow(out1); plt.xticks([]); plt.yticks([]); plt.show()

second example, same imports:

x = np.array(Image.open("covid_target.png"  ).convert("L")) / 255.
h = np.array(Image.open("covid_template.png").convert("L")) / 255.

# blank regions default to `1`, undo that
x[x==1] = 0
h[h==1] = 0

out = cross_correlate_2d(x, h)
plt.imshow(out, cmap='turbo'); plt.xticks([]); plt.yticks([]); plt.show()
  • $\begingroup$ i do have a question, how do i know if the multiplication (fft2(x) * np.conj(fft2(h)) is done between matrices that can be multipliable (nr of rows first = nr of cols 2nd) ? $\endgroup$ Jan 2 at 17:34
  • $\begingroup$ that is, what if my template image is smaller than the original one? $\endgroup$ Jan 2 at 17:36
  • $\begingroup$ @user3253067 That extends the scope of the question, StackExchange isn't for extended back & forths. If you have a new question, you should post separately. $\endgroup$ Jan 3 at 10:34
  • $\begingroup$ will do, sorry about that $\endgroup$ Jan 3 at 13:13
  • 1
    $\begingroup$ I'm not sure this is correct. For some values of N (even or odd, not sure which yet) conj(fft(ifftshift(h))) may work, but I don't think it does in all cases. I may try simple cases of this with 1D signals when I get a chance. $\endgroup$
    – Gillespie
    Jan 7 at 2:54

Your 2nd step is wrong, it's doing circular convolution. For circular cross-correlation, it should be:

  1. Multiplication between the output of the FFT applied on the first vector and the conjugate of the output of the FFT applied on the second vector.

Aside from that, the steps are the same whether in 1D or 2D:

$$\texttt{iFFT}_{2d}\bigg(\texttt{FFT}_{2d}\big(V_1\big) \cdot \overline{\texttt{FFT}_{2d}\big(V_2\big)} \bigg)$$

You can zero-pad the inputs $V_1$ and $V_2$ to calculate non-periodic cross-correlation using circular cross-correlation.

  • $\begingroup$ This is true for continuous signals, but for digital signals, conjugating after taking the FFT does not flip the signal in time. It does a modulo N flip. So I think you'd need to shift by one sample (via a linear phase in the frequency domain) first. $\endgroup$
    – Gillespie
    Dec 29, 2022 at 2:21
  • 1
    $\begingroup$ Nice \text{} alt, I steal $\endgroup$ Dec 29, 2022 at 12:22
  • $\begingroup$ This should be disclaimed for incompleteness, the centering is often more important than conjugation. That or added discussion of padding & unpadding. $\endgroup$ May 4 at 16:45
  • $\begingroup$ @OverLordGoldDragon feel free to edit :) $\endgroup$
    – Jdip
    May 4 at 16:51

As Jdip correctly pointed out, your implementation is doing circular convolution, not cross-correlation. But I want to add a nuance of discrete circular cross-correlation that makes his answer slightly incorrect.

Frequency Conjugation for Discrete vs. Continuous Signals

There are two key differences between cross-correlation and convolution:

  • In cross-correlation, one of the vectors is conjugated (in the time domain)
  • In convolution, one of the vectors is reversed/flipped

Thus, to perform cross-correlation via FFT-implemented circular convolution, we must pre-flip and conjugate one of the vectors: cross_correlation(x,y) = convolution(flip(conjugate(x)), y).

The conjugation property of the continuous time Fourier transform says that conjugating in the frequency domain conjugates and flips in the time domain: $\mathcal{F}^{-1} (\overline{X(\omega)}) = \overline{x(-t)}$ . So you should be able to perform cross-correlation via FFT by just conjugating one of the FFTed vectors, right? Wrong!

The reason is that for discrete signals such as images, conjugation in the Fourier domain does not equate to time reversal. Rather, it performs modulo N time reversal: $ [x_1, x_2, x_3,... x_N] => [x_1, x_N, x_{N-1}, x_{N-2},... x_2] $.


Because of this, I recommend doing the conjugation and flipping in the time domain.1 Here are the steps:

  1. Flip and conjugate $V_2$: $W_2 = \overline{\texttt{flip}(V_2)}$
  2. Take the FFT of $V_1$ and $W_2$
  3. Multiply FFTed vectors
  4. Take the IFFT of the product

$$ \texttt{IFFT}_{2D}(\texttt{FFT}_{2D}(V_1) \cdot \texttt{FFT}_{2D}(W_2)) $$


$$ \texttt{IFFT}_{2D}(\texttt{FFT}_{2D}(V_1) \cdot \texttt{FFT}_{2D}(\overline{\texttt{flip}(V_2)})) $$


OverLordGoldDragon has argued in the comments that we shouldn't reverse/flip $V_2$ to perform cross-correlation via convolution. I offer a simple proof in MATLAB that we should:

%Create vectors and cross-correlate them
N = 8; M = 3; 
v1 = randi(9, N, 1);        %Gives a vector of random integers less than 10
v2 = v1(3:3+M-1);           %Take a small subset of v1 as our second vector
P = N + M - 1;              %Length of full convolution 

y = xcorr(v1, v2);          %Cross-correlation 
y = y(N-M+1:end);           %Get rid of extra xcorr lags (zeros)

%Now, do this with convolution 
%NOTE: No time domain conjugation because v2 is real: v2 == conj(v2)
y2 = conv(flip(v2), v1);    %y2 and y are identical 

%Try conjugating in the frequency domain instead of flipping
%y3 and y are not the same!
y3 = conv(ifft(conj(fft(v2))), v1); 


1More efficient alternative: Use a linear phase multiply in the frequency domain to circularly shift the time domain vector by 1 sample before (or after, depending on the direction of the shift) conjugating in the frequency domain.

  • 1
    $\begingroup$ subtle nuance! thanks! $\endgroup$
    – Jdip
    Dec 29, 2022 at 9:24
  • 2
    $\begingroup$ While this answer is onto the right idea, I don't know that it's correct. Firstly, "modulo N reversal" is time reversal, $x_1 = x[n=0]$ so $x[n] = x[-n]$ holds. More importantly, to cross-correlate or convolve, the operating kernel (so one of the images) must be DFT-centered. flip as in x[::-1] does not achieve this, in fact it makes more of a mess. The correct steps involve ifftshift. $\endgroup$ Dec 29, 2022 at 10:53
  • 1
    $\begingroup$ flip(x) isn't time-reversal for DFT, which is the point you're making anyway, but important to distinguish "reversal we want" vs math, as it suggests $x[n] \rightarrow x[-n]$ isn't achieved. I made your point a while back, got warm reception. $\endgroup$ Dec 29, 2022 at 13:59
  • 1
    $\begingroup$ For circular cross-correlation (which the beginning of the answer says you write about) it suffices to do what you call modulo N time reversal. Your approach of using flip effectively just shifts the output of circular cross-correlation to the left by one step. $\endgroup$ May 9 at 8:55
  • 2
    $\begingroup$ @Dan Boschen & Olli, I'm busy at work lately but I'll come back to this eventually and clarify. $\endgroup$
    – Gillespie
    May 10 at 3:09

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