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Signal to noise ratio in a CT scanner

  1. Apr 27, 2016 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations


    3. 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
     
    Last edited: Apr 27, 2016
  2. jcsd
  3. Apr 29, 2016 #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.
     
  4. Apr 30, 2016 #3
    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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