Acquiring a k-space line

[Inaki4]

I've been wondering for a while about how making an image dealt with the fact that acquiring several samples of the k-space implies that all of them are acquired at different times, so the T2* relaxation would cause that there is a distortion because different samples are affected by different decay of the transverse magnetization measured.

I read in the book Fundamentals of Medical Imaging by Paul Suetens (Page 74) that this indeed is true, but the acquisition is based on the assumption that the measuring time is much shorter than the relaxation time, so it can be neglected.

Still, I wonder:

1) If we are considering an spin-echo sequence we are trying to get a T2-weighted image by means of cancelling the effect of T2* at t=TE. But since we measure before and after t=TE, we are affected in those measures by T2*, so we would require that the measuring time TM is much shorter than T2*, that is, TM<<T2*. If we are considering a gradient-echo sequence, we also need TM<<T2*.

2) What is more, if we are dealing with an EPI sequence (gradient echo or spin-echo), this implies that the measurement time is increased by a factor K (the number of lines acquired within the same pulse). So we need K·TM<<T2*.

I don't know the typical values of the sampling time during the echo, but is it reasonable to make all these assumptions?

Iñaki Rabanillo Viloria

[AndrewBworth]

The intensity of the different parts of k-space vary a great deal. So the parts of k-space that have the greatest intensity, make the largest contribution to the overall contrast. In general, the center of k-space contains most of the information about the INTENSITY of the information in the picture, while the whole of the k-space determines it's distribution. Fine details can't be discerned until they separate in k-space - so the outer edges of k-space contain the information about finer details.
The general intensity of a point in k-space is given by a function called a sinc function, sin(x)/(x), with the center of k-space given the value kx, ky = 0, where the value is greatest. (For 2D, the envelope of k-space would be given by a sinc function on the x dimension multiplied by a sinc function on the y, for 3D, three sinc functions). You'll see this function drops off quite rapidly from 0.
Sometimes a method known as keyhole imaging is done, where an entire image is acquired, but then only the center of the k-space is replaced - this is useful for dynamic imaging.

[Aaliyah]

Keyhole imaging is used for dynamic imaging with contrast medium. The advantage is that the keyhole technique increases temporal resolution without a loss of spatial resolution by limited data acquisition.