File Exchange

image thumbnail

mriphantom

version 1.0.0.0 (2.53 KB) by Ronald Ouwerkerk
Simulation of raw MRI data from k-space coordinates using the Shepp and Logan head phantom function.
4.7
12 Ratings

9 Downloads

Updated 10 Jun 2002

No License

Raw MRI data are generated analytically from k-space coordinates using the (continuous) Shepp and Logan head phantom function. K -space coordinates are typically entered as 2D matrices (frequency and phase encodes).

This tool is ideal for testing reconstruction of non-linear k-space trajectory MRI data such as projection reconstruction (PR) and spiral PR MRI.

The default k-space trajectory is a simple Carthesian trajectory. The reconstruction of the image in this case is a simple 2D FFT. Running the scripts without arguments results in plots of the raw and reconstructed data for this k-spce sampling path.

Cite As

Ronald Ouwerkerk (2021). mriphantom (https://www.mathworks.com/matlabcentral/fileexchange/1759-mriphantom), MATLAB Central File Exchange. Retrieved .

Comments and Ratings (15)

Ke Shi

Lauren Hirt

zhao li

hi,do you have codes of Variable density sampling,thank you

YUER LI

HUANG CHUAN

Hoang Tung NGUYEN

Dear Ronald Ouwerkerk,
I have a different images (not MRI image): a simple images bmp 8bits, a line or a point. I want to input this image to your *.m. Please show me the way (full code download). Thanks you so much.^^

Minh Chinh

Thanks

Irvin

Abdulrahman Sharaf Sharaf

THANKS

Julia Woody

I could not run this program, are there anybody who can write a small program which uses this function ? (or can you given example input parameters)

thanks..

Wen-Tung Wang

Very nice work. However, there are some issues. The contrast in the image reconstructed from feeding a Cartesian trajectory to mriphantom is different from feeding it a spiral trajectory.

jianfei ge

Thanks! Very good!

Ken Johnson

This nice work has been implemented throughout the MR imaging community, and is visible in many publications. I noticed one small bug. At k = 0 care was taken to avoid a nan by setting it to 1. However this is slightly off, and any cartesian sampling of k-space will result in an image that looks like it has been subtracted by .31. The part under consideration is J1(2pi*a*k)/(a*k).

Using l'Hopital's rule
J1(2pi*a*k)/(a*k) = J1'(2pi*a*k)/a

J1'(x) = J0(x)-J1(x)/x = -J2(x)+J1(x)/x
= [J0(x)-J2(x)]/2

Therefore
J1(2pi*a*0)/(a*0) = pi

In short change the following line from
bess = ones(size(akphi.*kr));
to
bess = pi*ones(size(akphi.*kr));

Naoyuki Takei

- -

like the image processing toolbox phantom function?

MATLAB Release Compatibility
Created with R12.1
Compatible with any release
Platform Compatibility
Windows macOS Linux

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!