Hardware: mE5 marathon VCL.
When we use the operator FFT to process an image in the frequency domain,Input of FFT operator two branches of an image or something else? What is the meaning of the information in the output graph of operator FFT Forward and Inverse ?
(PS:Original and bandpass images greater than 1M cannot be uploaded)
Hello zhuomuke
due to holidays we won't be able to give you an answer within this week. We will have a look at it next week.
BR
Johannes
Dear Zhuomuke,
thank you for your question. The meaning "Foward" as value of Parameter "Transformation Mode" of the operator "FFT" means transformation to frequency space. With value "Inverse" the FFT operator transforms the frequency space to "normal"image space. The functionality of FFT operator in this mode corresponds to Inverse Fast Fourier Transform. To transform an image with no imaginary part to frequency space set the lower (imaginary ) input node of operator "FFT" to constant value "Zero". In your VisualApplets installation directory under Examples/processing/Advanced/FastFourierTransform you can find an example implementation.
Dear Zhuomuke,
a convulution of an image with a filter in the frequency domain can be performed via multiplication.
In the VisualApplets design "FFtandbandpass.va" this is done correctly. After the image is transformed via FFT to frequency space a multiplication with a filter image (buffered in operator CoefficientBuffer) is performed. The filter image ( e.g. bandpass filter) buffered in the operator Coefficient Buffer can be generated using external software.
Dear CarmenZ:
Thank you for your reply.
As shown in the attached drawing.There are two defects in the OriginImage, and the coordinate is (779, 237) (884, 123).The processing in Halcon and the result is shown in dstimage.If we want use the operator FFT to process an image in the frequency domain.How do we implement it?