Could you teach us im mri tech i need ur help

@@no-de3lg What is it exactly that you need to know?

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.

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.

where is the previous lecture.please share the link

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!