Very very helpfull! We are taking a course in Scientific Computing, where we have to implement the Fouriertransform in MRI in MATLAB-code. This 30min video helped us quite a bit.
thank you so so much! amazing presentation! just one question: 8:25 is the phase you are talking about the same phase as when we talk about phaseencoding?
Why acquiring more points in the readout direction (higher kx,max number) requires a stronger readout gradient? If we would have kept gradient strength and sampling window the same, couldn't we just increase the BW to acquire the extra data points within time?
hey, newbie here, so I don't know if this is correct: I think you need a stronger readout so you can go further away from the origin, and that will increase the resolution. To increase the density of points (that will expand your fov) you need to increase sampling (and increasing BW is the answer). So it's two different things, imo.
Wait: Orientation in k-space is orientation of stripes, distance from origin in k-space is frequency of strips BUT where and how is the phase information encoded in k-space? Is it a whole nother k-space that is just phase (if so, again, how is phase encoded?), or is it somehow encoded (how?) on the same k-space that the orientation and frequency are encoded on?
He didn’t describe it very well, but actually, every point in k-space has 2 orthogonal values. From these 2 orthogonal values, you can calculate the magnitude and direction of the vector for that pixel in k-space. Think of it like a colour picture, every pixel in a colour picture actually has 3 values, for the red green and blue amplitudes for that pixel. It’s similar for k-space, every “pixel” in k-space has 2 values, from which you can calculate the magnitude and phase of the vector for that pixel.
@@martinstent5339 Every point in k-space is a complex number of the form a+ib. The modulus defines the intensity of the stripe, the arctan (b/a) the phase angle of stripe relevanto to k-space centre, the direction is perpendicular to stripes to k-space centre and location in k space corresponds to higher k-number (high bin of frequencies). So all of the above describe fully the recipe for stripe patterns that if you draw them in image space , it will form the eventually the image
@@georgiosmenikou8893 Well, it might be a complex number for a mathematician, but really, in an MRI system, it is the X and Y components of the RF signal that was received during the exam which correspond to that point in K-Space. But I am looking at this from the engineering point of view, and you are (probably) looking at it a little more abstractly. I’m thinking we are both right.
First part about k-space is useful but the second part is inaccurate in terms of mri and messing all up, sorry. Longer scan time is not a cause for higher snr. Just making more samples with the same bw and readout gradient will not change the FOV and so on. But thanks anyway.
Well, in concept, it does. Having a longer TR, let's say 4000ms for example, will allow tissue vectors to fully recover back to the longitudinal magnetization. Therefore, during the application of the next excitation pulse (i.e. 90-degrees), will then flip the net magnetization vector 90-degree into the transverse axis. This allows for maximal coherent traverse magnetization. Now that's all assuming we're using a short TE, the appropriate flip angle, and so on. There's a lot of factors that effect the final outcome of the SNR, but *in concept* a larger (longer) TR value will harbor a greater SNR. If you're associating a long TR with a T2-weighted imaging you may be confusing the concept a bit as we also use a long TE which in turn allows for more decay of signals, thus having an inherently low SNR than say a T1-weighted image of Proton Density weighted image. I hope this helps. Take care!
This helped me visualize what K space is and how to think in a different dimension, thank you.
Very well explained. Fits exactly what the professors here are teaching us at Technical Medicine. Thank you.
Best information found on youtube for MRI k-spsace.
Great talk! thanks for making K-Space a little clearer
Very very helpfull! We are taking a course in Scientific Computing, where we have to implement the Fouriertransform in MRI in MATLAB-code. This 30min video helped us quite a bit.
Wonderful presentation! Just enough math to make it clear and explanation of phase
This is a phenomenal video, I wish I had this when I worked as an Engineer for a MRI coil manufacture!
Could you teach us im mri tech i need ur help
@@no-de3lg What is it exactly that you need to know?
@@louvoodoo What is k space in summary
@@RatoniCats I wish I could easily summarize it myself, but my brain isn't that big. I would suggest just watching this video.
Thank You for your research and bringing more facts in.😘
At the start of the lecture you hinted at a precursor to this lecture, is there a link to it?
This is very helpful. Thank you!
Great explanation. This helps a lot!
This video is fantastic, but I can't find anymore of his lectures on MRI. By chance does anyone have a link to any?
Only been binging MRIQUIZ, need to see this wonder with my own two eyes, MRI registry being taken on Saturday 🙃 wish me luck.
thank you so so much! amazing presentation!
just one question:
8:25 is the phase you are talking about the same phase as when we talk about phaseencoding?
where is the previous lecture.please share the link
So interesting
Dont understand all but stunning video
Why acquiring more points in the readout direction (higher kx,max number) requires a stronger readout gradient? If we would have kept gradient strength and sampling window the same, couldn't we just increase the BW to acquire the extra data points within time?
hey, newbie here, so I don't know if this is correct: I think you need a stronger readout so you can go further away from the origin, and that will increase the resolution. To increase the density of points (that will expand your fov) you need to increase sampling (and increasing BW is the answer). So it's two different things, imo.
Wait: Orientation in k-space is orientation of stripes, distance from origin in k-space is frequency of strips BUT where and how is the phase information encoded in k-space? Is it a whole nother k-space that is just phase (if so, again, how is phase encoded?), or is it somehow encoded (how?) on the same k-space that the orientation and frequency are encoded on?
He didn’t describe it very well, but actually, every point in k-space has 2 orthogonal values. From these 2 orthogonal values, you can calculate the magnitude and direction of the vector for that pixel in k-space. Think of it like a colour picture, every pixel in a colour picture actually has 3 values, for the red green and blue amplitudes for that pixel. It’s similar for k-space, every “pixel” in k-space has 2 values, from which you can calculate the magnitude and phase of the vector for that pixel.
@@martinstent5339 Every point in k-space is a complex number of the form a+ib. The modulus defines the intensity of the stripe, the arctan (b/a) the phase angle of stripe relevanto to k-space centre, the direction is perpendicular to stripes to k-space centre and location in k space corresponds to higher k-number (high bin of frequencies). So all of the above describe fully the recipe for stripe patterns that if you draw them in image space , it will form the eventually the image
@@georgiosmenikou8893 Well, it might be a complex number for a mathematician, but really, in an MRI system, it is the X and Y components of the RF signal that was received during the exam which correspond to that point in K-Space. But I am looking at this from the engineering point of view, and you are (probably) looking at it a little more abstractly. I’m thinking we are both right.
very helpful!
Never seen a good looking physicist before.
First part about k-space is useful but the second part is inaccurate in terms of mri and messing all up, sorry. Longer scan time is not a cause for higher snr. Just making more samples with the same bw and readout gradient will not change the FOV and so on. But thanks anyway.
I didn't watch it all, but what I think is meant is that the longer the TR, the higher the SNR value, therefore, the longer scan time.
James McAnally that's it, longer TR will not always affect SNR in better way
Well, in concept, it does. Having a longer TR, let's say 4000ms for example, will allow tissue vectors to fully recover back to the longitudinal magnetization. Therefore, during the application of the next excitation pulse (i.e. 90-degrees), will then flip the net magnetization vector 90-degree into the transverse axis. This allows for maximal coherent traverse magnetization. Now that's all assuming we're using a short TE, the appropriate flip angle, and so on. There's a lot of factors that effect the final outcome of the SNR, but *in concept* a larger (longer) TR value will harbor a greater SNR.
If you're associating a long TR with a T2-weighted imaging you may be confusing the concept a bit as we also use a long TE which in turn allows for more decay of signals, thus having an inherently low SNR than say a T1-weighted image of Proton Density weighted image.
I hope this helps. Take care!