Signal to noise ratio in a CT scanner

  • Thread starter BobP
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  • #1
BobP
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Homework Statement


If the relative noise (noise divided by signal) in a region of a CT slice is measured to be N what would be the value if the tube current was halved, the pitch increased from 0.9 to 1.2, the rotation time increased from 0.75s to 1s and the slice width increased from 1mm to 2mm.


Homework Equations




The Attempt at a Solution


I think the answer is N
Working: sqrt(0.5*(1/0.75)*(0.9/1.2)*2)
but am not certain
 
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Answers and Replies

  • #2
FactChecker
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Doesn't your textbook have any discussion of the influence of any of those changes? There must be some relevant equations. You should learn them.
 
  • #3
BobP
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Doesn't your textbook have any discussion of the influence of any of those changes? There must be some relevant equations. You should learn them.
My course is entire lecture-based and we were told we didn't need books. as they are so expensive I didn't buy any. I did a google search but couldn't find anything


Having said that I realised I probably got two things in the working the wrong way around but the answer is still N? right?
sqrt(0.5*(0.75/1)*(1.2/0.9)*2)
 

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