Confusion about convolution theorem in matlab
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Convolution theorem in fourier transform states:
Fourier transform of a convolution of two vectors A and B is pointwise product of Fourier transform of each vector. (Ref: https://en.wikipedia.org/wiki/Convolution_theorem)
Now let A=B=[0.7, 0.2, 0.1], pointwise product of A and B can be obtained through the code:
fft(A).*fft(A)
The result is [1.000000000000000 + 0.000000000000000i, 0.295000000000000 - 0.095262794416288i, 0.295000000000000 + 0.095262794416288i].
Convolution of A and B can be obtained by inverse Fourier transform:
ifft(fft(A).*fft(A))
The result is C=[0.530000000000000 0.290000000000000 0.180000000000000].
However, convolution of A and B can be calculated in the matlab code:
conv(A,A)
The result is C'=[0.490000000000000 0.280000000000000 0.180000000000000 0.040000000000000 0.010000000000000].
The result vectors C is not equal to C', contradictory to the convolution theorem. This example shows my confusion on Fourier tranform in matlab code and convolution theorem. Would anyone help me to understand this?
Thanks very much:)
1 个评论
Arkaprabho Pal
2020-4-21
编辑：Arkaprabho Pal
2020-4-21
The convolution theorem says, roughly, that convolving two sequences is the same as multiplying their Fourier transforms. In order to make this precise, it is necessary to pad the two vectors with zeros and ignore roundoff error. Also make sure to use 'full' keyword in conv function.
t = linspace(0,1,200)
x = sin(2*pi*t)
y = exp(-2*t)
X = fft([x zeros(1,length(y)-1)])
Y = fft([y zeros(1,length(x)-1)])
then conv(x,y,'full') = ifft(X.*Y)
采纳的回答
There are differences between the continuous-domain convolution theorem and the discrete one. In particular, the discrete domain theorem says that ifft(fft(A).*fft(B)) gives the circulant convolution of A with B. You can get obtain a linear convolution result from a circulant convolution if you do sufficient zero-padding:
>> ifft(fft(A,5).*fft(A,5),5)
ans =
0.4900 0.2800 0.1800 0.0400 0.0100
5 个评论
Felix Wong
2017-9-17
Matt J
2017-9-17
n needs to be large enough to hold the result of the convolution.
Felix Wong
2017-9-18
Image Analyst
2017-9-18
Every time you convolve A with something, it adds a length of A to the something. So A**A is twice as long A**(A**A) is 3 times as long, etc. Don't do the padding yourself or else it will be even longer. Just specify the 'full' option in conv() or conv2() and it will enlarge the array by the correct amount for you.
Felix Wong
2017-9-19
编辑：Felix Wong
2017-9-19
更多回答（1 个）
Image Analyst
2017-9-16
Try this:
A = [0.7, 0.2, 0.1]
B = [0.7, 0.2, 0.1]
Cconv = conv(A, B)
% Now if C = A ** B
% Then fftC = fftA * fftB
fftA = fft(A)
fftB = fft(B)
fftC = fftA .* fftB
% If we inverse transform this to get C back:
% ifft(fftC) = C = ifft(fftA .* fftB)
C2 = ifft(fftC)
C3 = ifft(fftA .* fftB)
I think the difference you're seeing in Cconv and C2 or C3 is that CConv is the full convolution - 5 elements - while doing it the fft way, you are clipping to the same size as your input arrays - 3 elements - and so inherently you have a rect function in there multiplying your A and B. And the convolution of a rect is a triangle function, which, because your array is so small, wreaks havoc on your values. With much larger arrays, it won't be as noticeable.
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