Sİr, i just wanted to thank you for your lectures. You have no idea how helpful this channel is. Almost every playlist in this channel is saved to watch later. Much obliged.
Glad it was helpful! Have you seen my video explaining the Fourier Transform? "What is the Fourier Transform?" th-cam.com/video/G74t5az6PLo/w-d-xo.html
The last bit of analysis confused me but I think I got it now: First of all, ‘amplitude’ and ‘phase’ refer to the Fourier coefficients (they are complex: X = amplitude*exp(i*phase)). Secondly, the video is wrong in saying that we discard the amplitude by setting it to zero-we set it to one instead. Otherwise we’d have a zero vector! So we apply inverse DFT to Y = 1*exp(i*phase). When discarding the pase we do set it to zero and thus use Y = amplitude*exp(i*0). Sidenote: We plot here the DFT of the blue part of the image. DFT is applied separately to the red, green and blue component.
Yes, you got it. Sorry for the verbal "typo" (saying the amplitude should be set to zero, when you're right, I meant to say "set to one"). I'll make a note of that in the video's description. Thanks.
Thanks Mr Iain. Great Video. I was very pleased to see a discussion of the phase, I'd still like to see the green and red channels of the image, but I guess that is for us to manipulate with the code you kindly provided. Looking forward to seeing the upcoming videos of the series 👍🏽 Don't know if it was addressed in previous videos, but any futures insights of the math for the phase, as well as more applications of it? Taking the freedom to suggest topics: In seismic recording and ground penetrating radar, when the signal travels through a heterogenous media the velocity is a function of the frequency and this affects the phase as well. Interferometric Aperture Radar Radar is a very interesting application of the phase
I am a little bit confused on the last part when you showed the black image and said that there are actually red, blue and green. From additive synthesis we know that the more red, green and blue you add, the more you have white, so technically in the image at 12:14 there is NO red , NO green and NO blue in order to have black(i.e. [0R; 0G; 0B]). Am I right? It seemed to me that you were implying that in the black picture there was still a quantity of red, green and blue while from what I know it's the complete opposite.
You're right, I made a mistake there. I didn't realise it, but Matlab stores images in a "type" called uint8, and I was thinking that it was Reals. So actually Matlab had done an automatic conversion, that had resulted in all the values being set to zero. So in fact you are right. In that image there is no colour at all. But it's also the wrong image (due to the auto conversion). Anyway, the final image is correct (when I only show the red component). The Matlab code is available here (if you would like to see it for yourself): drive.google.com/file/d/1cncubdpaEvbkCPqEPS0_ybvazYhVpVV7/view and the image is here: drive.google.com/file/d/1zCuwgPnlnVcDFuryRY1NcqEMsyB84E_0/view
Hi Iain, I think the Matlab code would be very useful for me to understand more deeply on your explanation. Would you able to provide the image file: ImageSmall.jpg. Thanks!
Hi sir, I wanna ask a question. In 12:05, you said there was a trick to convert from black image to blue image for getting the phase spectrum. Can you show me the code for the "trick"? It's wonderful if in Python code. Thanks so much.
This TH-cam channel is a hidden gem!!
Sİr, i just wanted to thank you for your lectures. You have no idea how helpful this channel is. Almost every playlist in this channel is saved to watch later.
Much obliged.
Glad you like them!
Katiliyorum
Thank you for this lecture. Fourier Transforms are so unusual and this helped me get a grasp on what is going on.
Glad it was helpful! Have you seen my video explaining the Fourier Transform? "What is the Fourier Transform?" th-cam.com/video/G74t5az6PLo/w-d-xo.html
The last bit of analysis confused me but I think I got it now:
First of all, ‘amplitude’ and ‘phase’ refer to the Fourier coefficients (they are complex: X = amplitude*exp(i*phase)). Secondly, the video is wrong in saying that we discard the amplitude by setting it to zero-we set it to one instead. Otherwise we’d have a zero vector! So we apply inverse DFT to Y = 1*exp(i*phase).
When discarding the pase we do set it to zero and thus use Y = amplitude*exp(i*0).
Sidenote: We plot here the DFT of the blue part of the image. DFT is applied separately to the red, green and blue component.
Yes, you got it. Sorry for the verbal "typo" (saying the amplitude should be set to zero, when you're right, I meant to say "set to one"). I'll make a note of that in the video's description. Thanks.
Thanks Mr Iain. Great Video. I was very pleased to see a discussion of the phase, I'd still like to see the green and red channels of the image, but I guess that is for us to manipulate with the code you kindly provided.
Looking forward to seeing the upcoming videos of the series 👍🏽
Don't know if it was addressed in previous videos, but any futures insights of the math for the phase, as well as more applications of it?
Taking the freedom to suggest topics:
In seismic recording and ground penetrating radar, when the signal travels through a heterogenous media the velocity is a function of the frequency and this affects the phase as well.
Interferometric Aperture Radar Radar is a very interesting application of the phase
Interferometric Synthetic Aperture Radar
Glad you liked the video. Thanks for the radar suggestion, I'll add it to the list.
permission to learn sir.
thank you
You're welcome.
I have one year temperature data. How can I evaluate the amplitude of this temperature by using 2 D FFT? I want the code. thank you
I am a little bit confused on the last part when you showed the black image and said that there are actually red, blue and green. From additive synthesis we know that the more red, green and blue you add, the more you have white, so technically in the image at 12:14
there is NO red , NO green and NO blue in order to have black(i.e. [0R; 0G; 0B]). Am I right?
It seemed to me that you were implying that in the black picture there was still a quantity of red, green and blue while from what I know it's the complete opposite.
You're right, I made a mistake there. I didn't realise it, but Matlab stores images in a "type" called uint8, and I was thinking that it was Reals. So actually Matlab had done an automatic conversion, that had resulted in all the values being set to zero. So in fact you are right. In that image there is no colour at all. But it's also the wrong image (due to the auto conversion). Anyway, the final image is correct (when I only show the red component). The Matlab code is available here (if you would like to see it for yourself): drive.google.com/file/d/1cncubdpaEvbkCPqEPS0_ybvazYhVpVV7/view and the image is here: drive.google.com/file/d/1zCuwgPnlnVcDFuryRY1NcqEMsyB84E_0/view
@@iain_explains thank you for the answer and the Matlab resources sir! I hope you will continue with your great work here on TH-cam! :)
Thanks, I'm glad you like the videos.
Hi Iain, I think the Matlab code would be very useful for me to understand more deeply on your explanation. Would you able to provide the image file: ImageSmall.jpg. Thanks!
I've put a link to the image file on the webpage (under the link for the Matlab code): www.iaincollings.com/image-processing
Thanks!
Hi sir, I wanna ask a question. In 12:05, you said there was a trick to convert from black image to blue image for getting the phase spectrum. Can you show me the code for the "trick"? It's wonderful if in Python code. Thanks so much.
The Matlab code for everything in the video is available at iaincollings.com under the Image Processing tab.
Matt Labs?
Perhaps you haven't seen my webpage? You can find the Matlab code there. iaincollings.com