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X-ray SNR in 3 steps. I ∆I. X-ray transmission SNR Review Let N = average number of transmitted x-rays N = N 0 exp [ - ∫  dz ] Emission and transmission.

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Presentation on theme: "X-ray SNR in 3 steps. I ∆I. X-ray transmission SNR Review Let N = average number of transmitted x-rays N = N 0 exp [ - ∫  dz ] Emission and transmission."— Presentation transcript:

1 X-ray SNR in 3 steps. I ∆I

2 X-ray transmission SNR Review Let N = average number of transmitted x-rays N = N 0 exp [ - ∫  dz ] Emission and transmission are Poisson processes where  = 2.5 x 10 10 photon 2 / cm 2 / Roentgen A = pixel area in cm 2,  screen efficiency R = dose t = transmission

3 Detection M  X  Z  W Transmitted g 1 g 2 developed grains photonslight grains / photons light photon /x-ray

4 Let’s consider N again as the average number of transmitted, not captured, photons per unit area. g 1 = hv x-ray / hv light = light / x-ray = 5000 A 0 / 0.2 A 0 ≈ 25,000 Typically efficiency gives 500 to 1000 light photons per x-ray For worst case, g 1 = 500 g 2 = 1/200

5 -Image catheters -Image injections -Old Method  is the solid angle subtended from a point on the detector to the pupil Subject Fluorescent screen Source

6  is the solid angle subtended from a point on the detector to the pupil Subject Fluorescent screen Source Instead of film, a fluorescent screen was first used. The problem is that the eye will only capture a portion of the light rays generated by the screen. How could the eye’s efficiency be increased?

7  is the solid angle subtended from a point on the detector to the pupil Fluorescent screen Let’s calculate the equivalent of the screen efficiency in this system, the eye’s efficiency capturing light   Where r = viewing distance ( minimum 20 cm) T e  retina efficiency ( approx..1) A = pupil area ≈ 0.5 cm 2 8 mm pupil diameter(usually 2-3mm)

8 At best   ≈ 10 -5 Typically   ≈ 10 -7 Recall With g 1 =10 3 light photons / x-ray g 1 g 2 = 10 3 10 -5 = 10 -2 at best Loss in SNR of 10 Have to up dose a factor of 100 ! (more likely to compromise resolution rather than dose) At each stage, we want to keep the gain product >> 1 or quantum effects will harm SNR.

9 g2g2 g3g3 phosphor output screen phosphor Photo cathode X-rays Light photons converted into electrons with g 2 Electrons accelerated and turned back into light photons with g 3 Electrostatic lenses g 4 eye efficiency

10 g2g2 g3g3 phosphor output screen phosphor Photo cathode X-rays Electrostatic lenses g1g1

11 g2g2 g3g3 phosphor Photo cathode X-rays For each captured x-ray  g 1 =10 3 light photons g 2 = electrons / light photons = 0.1 g 1 g 2 = 100 g 3 = emitted light photons / electron = 10 3 g 4 =   eye efficiency = 10 -5 optimum g 1 g 2 g 3 g 4 = 10 3 10 -1 10 3 10 -5 = 1 g 4 eye efficiency g1g1 Anode Section 5.2.6

12 For each captured x-ray  g 1 =10 3 light photons g 2 = electrons / light photons = 0.1 g 1 g 2 = 100 g 3 = emitted light photons / electron = 10 3 g 4 =   eye efficiency = 10 -5 optimum g 1 g 2 g 3 g 4 = 10 3 10 -1 10 3 10 -5 = 1 SNR loss =

13 g2g2 g3g3 Electronic focusing And amplification g1g1 TV cathode g4g4 Lens efficiency  0 ≈ 0.04 Much better than eye. g 1 = 10 3 g 2 = 10 -1 g 3 = 10 3  0 g 4 = 0.1 electrons / light photon g 1 g 2 g 3 g 4 = 4 x 10 2

14 Up to now, if g 1 g 2 g 3 g 4 >> 1 and all the intermediate gain products >> 1 But TV has an additive electrical noise component. Let’s say the noise power (variance) is N a 2.

15 NN N a =  = k  N In X-ray, the number of photons is modeled as our source of signal. We could consider modeling or expressing, N a ( which is actually a voltage), as its equivalent number of photons. Electrical noise then occupies some fraction of the signal’s dynamic range. Let’s use k to represent the portion of the dynamic range that is occupied by additive noise.

16 For k 2  N >> 1 SNR ≈ C/kpoor, not making use of radiation k 2  N << 1 k = 10 -2 to 10 -3 typically  N = 10 5 photons / pixel Regular k = 10 -2 k 2  N = 10 Better design k = 10 -3 k 2  N = 10 -1 Much Better!

17 I s + I b }∆I 0 IsIs We will view scattered photons as increasing the background intensity. Scatter increases the level of the background and the lesion. Let the ratio of scattered photons to desired photons be Section 5.4.3

18 I s + I b }∆I 0 IsIs Scatter increases the level of the background and the lesion. Conclusion: So does scatter affect contrast?

19  N 2 =  N +  N s where N s is mean number of scattered photons SNR Effects of Scatter Scatter doesn’t change the amount of signal. It is still  I. The variance of the background now depends on the variance of trans- mitted and scattered photons, however. Both are Poisson and independent so we can sum the variances. Here C is the scatter free contrast. So does SNR affect scatter? Grids to reduce scatter are described in Sec. 5.2.4


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