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Objectives To Understand; 1 . Sources of resolution loss

SPATIAL RESOLUTION. Objectives To Understand; 1 . Sources of resolution loss Point spread function, line spread function, edge response function 3. Convolutions and convolution theorem.

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Objectives To Understand; 1 . Sources of resolution loss

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  1. SPATIAL RESOLUTION Objectives To Understand; 1. Sources of resolution loss Point spread function, line spread function, edge response function 3. Convolutions and convolution theorem

  2. 4. Use of the modulation transfer function to describe the spatial frequency dependent resolution of imaging systems. 5. Calculation of the modulation transfer function 6. Magnification radiography 7. Image sampling and its effects on resolution

  3. The spatial resolution of a system is a measure of the system’s ability to see details of various sizes. In this section we will describe a method for quantitatively describing and measuring the spatial resolution of a system. Resolution can be limited by any of the elements in the imaging chain or, as in conventional photography, by motion of the subject. Let’s look at a few examples of effects which can limit resolution.

  4. Figure 1 illustrates the effect of x-ray focal spot size on spatial resolution. Due to the finite size of the focal spot, the image of a single point in the image can be no smaller than the image of the focal spot itself. The focal spot image is called the focal spot point spread function. As magnification, defined as (d1 + d2) /d1, is increased, focal spot blurring increases.

  5. t X-ray Intensifying screen Figure 2 illustrates the degradation of spatial resolution due to the finite thickness of a radiographic intensifying screen. film When an x-ray is absorbed in the screen it produces visible light. This light then diverges before reaching the film producing a point spread function due to the screen. This point spread function increases with screen thickness leading to a tradeoff between screen detection efficiency and spatial resolution.

  6. Another source of blurring is patient motion, shown below. In this case motion during the exposure leads to another point spread function ( 3A ) equivalent to that produced in the case of a stationary object and a finite focal spot ( 3B ) . B Motion Equivalent Focal spot A Blurred image Blurred image v Moving object Detector Detector

  7. A further example of resolution loss occurs when an image is digitized and represented by a picture element ( pixel ) matrix. This is illustrated below, which shows images of a chest phantom represented by pixel matrices of various sizes. 128 x 128 256 x 256 32 x 32 64 x 64

  8. 1024 x 1024

  9. 512 x 512

  10. Studies have been done to show that for diagnosis of clinical chest films information must be digitized to 4096 x 4096 to maintain diagnostic accuracy for the most demanding tasks. However, most diagnostic tasks are done acceptably well with 2048 x 2048 matrices. The smaller matrix is more convenient in terms of data storage and retrieval. There is also evidence that digital enhancement of contrast can compensate to some extent for slight losses in resolution.

  11. Modulation Transfer Function - MTF In order to quantitatively describe the effects of various system elements on the overall resolution of an imaging system the concept of modulation transfer function has been developed. In this description of the resolution properties of an imaging system components are described in terms of their ability to produce images of sinusoidally varying test objects of various spatial frequencies.

  12. The concept of spatial frequency is illustrated in Figure 5 which illustrates the variation in x-ray transmission through an object with a sinusoidally varying transmission in the x-direction of the form where k is the angular spatial frequency, related to the spatial frequency fx by (2)

  13. The contrast associated with the sinusoidal waveform is given by

  14. Note that many discussions of MTF use for the angular spatial frequency. We will use k to be consistent with our later discussions of magnetic resonance imaging where k is used for spatial angular frequency and  is used for temporal angular frequency. Spatial frequency fx is usually given in units of line pairs per mm, where one line pair or one cycle refers to a bright and dark band in the sinusoidal object transmission.

  15. Suppose the waveform of Figure 5 is sent into an arbitrary imaging system as shown in Figure 6. Imaging system Nin Nout

  16. The effect of the imaging system will in general be to multiply the signal variation by a complex system transfer function M(k) which reduces contrast and in general produces a phase shift  giving

  17. The factor MTF(k) modifying the magnitude of output signal variation is called the modulation transfer function and is given by the ratio of the output to input contrast for spatial frequency k. We will refer to both fx and k as spatial frequency. The distinction between linear and angular frequency should be clear from the context of the discussion.

  18. Physical objects can be represented as a weighted sum of spatial frequency components. Because M(k) varies with k, the imaging system will generally generate a distorted version of the actual x-ray transmission, typically failing to represent the higher spatial frequency content of the transmission.

  19. Mathematical Background for MTF slit a x N(x) b x L/2 Fourier Series Representation of the Transmitted Image Let’s consider the case of a one dimensional transmitted fluence N(x) representing a slit of width L as shown in Figure 7a. The image profile is shown in Figure 7b. Detector -L/2

  20. ( dx Since the transmitted image is symmetric about x=0, it can be represented as a cosine series(in general sines and cosines are required) of the form, (6) where (7) and, for any n, (8)

  21. multiply both sides of Eq. 6 by cos(kn’x) and integrate from -L/2 to L/2 0 =Lan/2 if k= k’ = 0 k not = k’ Orthogonality

  22. This leads to

  23. Assuming that N(x) has a value of unity in some appropriate unit and doing the integrals, we obtain and giving

  24. The graph of an which will be compared later to the Fourier transform of the transmitted image is shown below. Note that the points go through zero at k = 2/L or fx = 1/L. (See equations 2 and 7)

  25. Fourier Integral Representation of the Transmitted Image See Bracewell-Chapter 2 Usually, it is more convenient to express the image as an integral over the various spatial frequency components required to represent the image. Any distribution in x can be written as a sum of integrals of the form

  26. Since and (12)

  27. This is just a sum over spatial frequencies similar to the discrete example above, but now with a continuous distribution of frequency components. The quantities Ñ+ and Ñ- are in general complex numbers and are basically the weighting coefficients for the various frequencies analogous to the discrete points in Figure 8.

  28. For convenience, equation 11 is usually written in compact form as (13) The weighting function for the various sinusoidal frequencies Ñ(k) is called the Fourier transform of N(x), or FT(N(x)). It should be realized that the appearance of negative spatial frequencies in equation 13 is a reflection of the fact that there must, in general, be terms of the form e-ikx. These terms are of course associated with positive, physically realizable spatial frequencies. The artificial concept of negative spatial frequencies just arises in association with writing the integral in compact form.

  29. The Fourier transform or expansion coefficient of the image is related to the image through the Fourier transform relationship, (14) Equation (13) expressing N(x) in terms of Ñ(k) is called the inverse Fourier transform. N(x) and Ñ(k) are called a Fourier transform pair. It is interesting to mention at this point that in magnetic resonance imaging, data is obtained in the form of Ñ(k) in “k space”. The image is then obtained through an two dimensional transform equation analogous to equation 13.

  30. slit a x N(x) b x L/2 Following a brief introduction to the Dirac delta function, which will be presented below, the path from equation 13 to 14 may become more clear. However ,at this point lets continue our discussion of the single slit experiment of Figure 7.

  31. The Fourier transform of the detected image is given by equation 14 as

  32. where we have used the definition of the sinc function, namely (16)

  33. This sinc function is shown in Figure 9 and, aside from the normalization, has the same shape as the expansion coefficient distribution shown in Figure 8. Note that in the integral representation the weighting of the complex exponentials is symmetric about fx =o which is necessary to reconstruct the cosine behavior corresponding to the expansion of equation 6. You can think of a point at each positive frequency having an equally weighted point at negative frequency in order to construct a cosine function in accordance with the last of equations 12.

  34. Note that the weighting function (Fourier transform) goes to zero at fx = 1/L and 2/L as in the discrete case.

  35. The frequency space representation of images will be helpful in the understanding of spatial resolution as described by the modulation transfer function. A physical realization of this representation is to consider the fact that it would be possible in the laboratory to construct an image on film of any one dimensional image by making x-ray exposures through a series of carefully positioned sinusoidal bar patterns of appropriately selected transmissions and spatial frequencies. In fact the idea can be generalized to two dimensions by using bar patterns rotated by ninety degrees. In thisway it would be possible, in principle, to make an image of the Mona Lisa on film, at least a black and white version.

  36. Jean Baptiste Joseph Fourier 1768-1830

  37. ky kx

  38. The Mona Lisa in k-space

  39. ky kx k - space Low frequency Mona

  40. ky kx High frequency Mona k - space

  41. Spock in k-space

  42. The frequency content of a given image is related to its size and shape. In general, for example, the Fourier transform a broad slit will contain mostly low spatial frequencies, while a narrow slit will require much higher frequencies. This comparison is shown in Figure 10 for slits of width L and L/4. The curve for the slit of width L corresponds to the sinc function of Figure 9. The first zero of the Fourier transform of the slit of width L/4 occurs at fx = 1/(L/4) = 4/L. Figure 10

  43. Note regarding area of integrals in x-space and k-space: Since so the signal at the origin of k-space is equal to the integral over all image space signal.

  44. It can also be seen that since i.e. the value at the center of x-space is the integral of the k-space signal.

  45. In Figure 10 it is assumed that the intensity has been kept the same as the slit width has been decreased. The Fourier transform goes from Since N(0) is the same in each case (fixed intensity through the slit), the integrals in k-space are the same.

  46. One further example will serve to introduce a mathematical function which will be useful in several subsequent sections. Figure 11 shows the Dirac delta function and the magnitude of its Fourier transform. •   •  •  Magnitude of Fourier transform of delta function  (x-a) a x k 0 Figure 11

  47. The delta function, located for example at x=a, is an infinitely narrow, infinitely intense signal distribution defined by the following properties

  48. The integral property of the delta function whereby it selects out the value of the integrand at the point where the argument of the delta function is zero is very convenient in a number of applications. For example in calculating the Fourier transform of x-a) we obtain (18) For points away from x=o the complex Fourier transform is modulated by the phase factor e-ika which shuffles signal between the real and imaginary parts. However, the magnitude of the Fourier transform is constant at all frequencies.

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