Medical Imaging Systems

Drexel University

ECE Department

S684 Image Modalities

Fall 2005

Homework 1&2

Due October 19, 2005

Problem 1. Let r and s be homogeneous coordinates (four-dimensional vectors). Show that the set of points specified by ar +bswhere a and b are scalar constants represent a straight line in Cartesian space.

Answer: One of the definitions of a straight line is

,

where psand peare the starting and end points of the line segment, and is a scalar parameter. Let the coordinates of the point be ps = (ps1, ps2, ps3). The value of p1, the first dimension of the point is

.

We now examine the coordinates of t = ar +bs, the proposed equation in homogenous coordinates. Let t = (t0, t1,t2,t3), with similar notation for r, s. It follows that

, etc.

The Cartesian points are defined as p1 = t1/t0, p2 = t2/t0, p3 = t3/t0. Writing this in terms of r and s, we get

, etc.

This does not look like a straight line: for that matter, it could be a two-dimensional surface. However, if we define

then it turns out that

,

so that

.

The same applies to p2 and p3, QED.

Problem 2.. An X-ray source is located at the origin of the coordinate system, and the film in the x-y plane at z = 1 meter. An object is placed between the source and the film.

(a) Find a formula for the detector coordinates (xd, yd) of an object point (x, y, z).

.

(b) What is the lateral (xy) magnification as a function of z? In what plane is the magnification equal to 3?

Answer: The magnification is given by m = xd/x = 1/z. For m = 3, z = 0.333 = 33.3 cm.

(c) A square 0.2 meters on the side is placed with it's center at z = 0.5 and the plane of the square makes angle of 45° with the z axis. Two sides of the square are parallel to the y-axis, and two sides are at an angle of 45° to the x-axis. Sketch the shadow cast by the square on the film plane. Answer: The geometry is shown below. We draw the x-z projection: the y coordinate is treated the same way as the x coordinate

The four corners of the square are at

Using the above formula, the projected coordinates are:

Problem 3. A Gaussian random variable X has mean 3 and standard deviation 0.5. What is the probability that X>4? Specify an interval such that X will be in this range 95% of the time.

Answer: P[X>4] = 1 - Normdist(4,3,0.5,1) = 0.023.

We wish to find a, b such that P[a<X<b] = 0.95. Since the Gaussian distribution is symmetric about the mean, a = mean - c, b = mean + c. P[X<b] = 0.975. We see that mean + c is pretty close to 4. To do it more exactly, we use

Norminv(0.975, 3, 0.5) = 3.98.

We find that 3 + c = 0.98, c = 0.98, a = 2.02, b = 3.98

Problem 4. X-ray intensity without “contrast material” is I1(x, y), and with contrast material it is

I2(x, y) = I1(x, y)e-C(x, y).

Write the equations for the processing to be performed that will produce an image that will be proportional to C(x, y).

We measure f1(х, y) = I1(x, y), the intensity when no object is present. We then compute, for each (x, y)

g(x, y) = - ln(I2(x, y) + ln(f1(x, y)).

Problem 5. Assume that the dose in an abdominal X-ray is 550 mR (millirads) The exposure, in Röntgens, is approximately equal to the dose. At 60 keV one Röntgen produces about 3x1010 photons/cm2. Assume that the thickens of the abdomen is 25 cm, the beam is monoenergetic, and the attenuation coefficients is the same as for water. How many photons fall on a 0.1x0.1 mm pixel? Assume the signal-to-noise ratio is N0.5. What is the signal-to-noise ratio in the image?

Answer: At 60 keV,  = 0.20. The attenuation through the abdomen is exp(-25x0.2) = exp(-5) = 0.0067. The dose is 0.55 rads, so that the exposure is 0.55 Röntgens. This produces an incident flux of 1.65 e10 photons/cm2, and a transmitted flux of 1.11e8 photons per cm2. (0.1mm)2 = 10e-4 (cm)2. Therefore the recorded photons per pixel is 1.11e4, and the signal-to-noise ratio is 1.05e2 = 100.

Problem 6. Assume that air has the same mass attenuation coefficient of water. Estimate the mass (grams per square centimeter) of air between see level and interplanetary space. From this, estimate the attenuation of 100 KeV X-rays by the atmosphere. Find the equivalent thickness of lead to produce the same attenuation.

Answer: The pressure of the atmosphere is equivalent to about 10 m of water. Therefore the atmospheric attenuation is approximately the same as that of 10 m of water. The linear attenuation coefficient of water at 100 KeV is 1.5070. The linear attenuation coefficient of lead at the same energy is 52.9. Lead is 52.9/1.51 = 35 times more attenuator. Therefore the atmosphere is equivalet to 10/35 = 28 cm of lead (about 11 inches).

Data on attenuation is available from