Question

The free induction decay (FID) in the MRI of an object can be approximated as \[ s(t)=\iint m(x,y)\,e^{-j2\pi\,(K_x(t)x+K_y(t)y)}\,dx\,dy, \] where \(K_x(t)=\int_0^t G_x(\tau)\,d\tau\) and \(K_y(t)=\int_0^t G_y(\tau)\,d\tau\). Here \(G_x\) and \(G_y\) are pulses of identical period and are in–phase. By changing the amplitude of the pulses, one can obtain the two–dimensional Fourier transform of the object __________________.

• A. over radial lines in \((K_x,K_y)\) space
• B. over a parabolic contour in \((K_x,K_y)\) space
• C. along \(K_y\) only
• D. along \(K_x\) only

Hint:

In MRI, gradients scaled versions of the same waveform create a k–space trajectory with \(K_y/K_x=\) constant, i.e., straight radial lines through the origin.

Answer 1

The Correct Option is A

Step 1: With \(G_x\) and \(G_y\) in–phase and of the same period, we can write \(G_x(t)=a_x g(t)\) and \(G_y(t)=a_y g(t)\). Then \[ K_x(t)=a_x\!\int_0^t g(\tau)d\tau=a_x F(t),\qquad K_y(t)=a_y F(t), \] so \(K_y/K_x=a_y/a_x=\text{constant}\). Step 2: Hence the k–space trajectory \((K_x(t),K_y(t))\) lies on a straight line through the origin with slope \(a_y/a_x\); changing the amplitudes \((a_x,a_y)\) changes the slope, giving different straight lines through the origin—i.e., radial lines. Therefore the 2D Fourier data are obtained \emph{over radial lines} in \((K_x,K_y)\) space.