Theorems, Detailed explanations

then the image on the detector plane is the reversed and magnified image of the source.

With this formalism the source, the object and the detector transfer 'effects' can be modeled as multiplication in the Fourier domain. If planar imaging is projections of 3D objects we should integrate along the magnification too as the magnification changes from point to point.

2.6.9 Appendix (Mathematics)

2.6.9.1 Theorems, Detailed explanations

2.6.9.1.1 Description of Fourier series by complex expression

The following comment shows, how is possible to convert the trigonometric expression of Fourier series into complex formula. The conversion will be carried by means of Euler‟s formula.

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Describe in the following way:

Write down the trigonometric expression of Fourier series:

Substitute the above written expressions into the Fourier series formula in order to obtain the Euler‟s description way:

Apply the following denoting:

Use common for separate summing operation of Fourier series:

135 2.6.9.1.2 Parzeval theorem

136

137 2.6.9.1.3 Response function in general case

The following description shows, how is possible to give a system response for a general input function (non-step function) by a linear invariant system having weight function

and under the condition: .

Let‟s consider the next two figures:

Let‟s approximate the function by the series of such Dirac-impulses, where , for

Let infinitesimal small, i.e. represents much more smaller value comparing to the corresponded system characteristic parameter value. The system response according to the approximated input impulse with the known weighted function can be expressed as follow:

Let‟s determine the following limit in order to get the response function for the complete real parameter domain:

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2.6.9.1.4 Deriving of Duhamel Theorem

Deriving of Duhamel-theorem will be executed in the followings.

Let‟s start from the weak derivative of convolution

Apply the following denoting:

Let‟s describe the convolution formula in the argument of weak derivative:

If , and

, where Now, it is possible to get the final expression of Duhamel-theorem

, where means the conventional derivative, which is denoted by “ ” as usual.

139 2.6.9.2 Solution of problems

2.6.9.2.1 S1 Solution:

As we know, the Fourier transformation may be written by the following way:

Let‟s apply the Euler‟s formula now!

; where we have to use the available condition with the following substitutions:

; ; .

Consequently the final solution is:

2.6.9.2.2 S2

Solution:

Let‟s attempt to derive the Laplace transformation of function instead of .

; as it is known Then, let‟s make the following operations!

Consequently the obtained results:

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2.6.9.2.3 S3

Solution:

Let‟s start to write the F(s) function by the following way:

Let‟s execute the partial fraction operation as follow!

Let's substitute in the equation value! Parameter will be then determined.

Next step: let‟s substitute into the same equation for obtaining parameter :

Let‟s substitute into the same equation in order to get parameter !

Let‟s write A3, B and C into the equation

141 where only and are unknown. and will be determined by the following way.

First , then will be substituted in the equation obtaining the following set of two linear equations for and .

Let‟ substitute the two equation : Consequently

Now all the coefficients of the partial fraction are known, and Laplace transformation can be executed.

2.6.9.2.4 S4 Solution:

Let‟s try to find the complementary function of the second ordered linear differential equation with constant coefficients at first.

Let‟s find the complementary function by the following formula

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where the auxiliary equation is obtained. Roots of the second ordered polynomial equation:

i.e. and .

Consequently the solutions are and

Linear combination of the two independent solution is also solution of the homogenous equation, meaning the complementary function will be the following:

Let‟s find the particular solution as , where B is a constant value, due to the constant input function.

General solution of the inhomogeneous differential equation is:

Let‟s use the initial conditions, in order to determine coefficients.

Now the following set of two linear equations has to solve:

Let‟s substitute the A1 coefficient into the II. equation

as well as .

Consequently the general solution of the differential equation is:

143 The following expression can be written by Euler‟s formula:

The solution is

Let‟s do some conversion at the final formula: it is known, that

, where , and

2.6.9.2.5 S5 Solution:

Let‟s apply another way to solve the differential equation, i.e. do the Laplace transformation of both side of the equation!

Let‟s substitute the initial conditions, i.e.

function can be determined by the inverse Laplace transformation!

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, where the method of partial fraction is working, and very useful, but in the particular case would be very long calculation with many traps. Let‟s try another way now, i.e. let‟s apply some algebraic conversions.

Let‟s execute the inverse Laplace transformation now!

2.6.9.2.6 S6

Solution:

Let‟s execute the Laplace transformation both side of the equation!

Let‟s substitute the available initial conditions

Result of the inverse Laplace transformation will give , as the general solution:

First step is to find the roots of denominator:

145 One real root of the third order polynomial is , which means the polynomial can be expressed by fully-factored form as follow

Next task is to determine the coefficients of the partial function Solution of this equation will give the coefficients.

Let‟s see the following cases:

If will be substituted, then C1 can be obtained.

Consequently, the solution is

2.6.9.2.7 S7

Solution:

The problem can be solved by applying Duhamel-theorem.

, or let‟s apply the other formula:

The latter formula will be used in order to determine the output response:

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Let‟s execute the following integral .

Partial integral method will be used.

Let‟s apply again the partial integral method.

Now the integral element can be expressed as follow:

The unknown integral element:

Completely same way can be written the other integral element:

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After substituting both integral elements, the response will be the following!

2.6.9.2.8 S8 Solution:

It is known, the relation between the transfer characteristic and the weighting function is:

It is possible to see from the given figures and

Consequently, the formula of the transfer characteristic is the following:

, for , otherwise .

The weighting function can be derived by the inverse Fourier transformation of the transfer characteristic:

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2.6.9.2.9 S9

Solution:

It is known from the relation between the step response function and the weighting function of a linear system:

, i.e. the weak derivative of the step response function gives the weighting function.

Consequently,

2.6.9.2.10 S10

Solution:

Let‟s use the relation between the weighting function and transfer characteristic at first:

Let‟s use, is general step function.

Step response function can be derived by the weighting function as follow:

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2.6.9.2.11 S11

Solution:

The partial differential equation can be written by expressing the Laplace operator as follow:

Let‟s apply the Fourier transformation first by x and then y.

It is known from the Fourier transformation rules:

Let‟s execute Fourier transformation on both side of partial differential equation:

The second ordered partial differential equation has been transformed to second order linear differential equation with constant coefficient.

It is known, the solution of that differential equation is:

like harmonic oscillator Let‟s use the available initial condition in Fourier frequency space:

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Consequently:

Consequently is expressed by the initial condition in frequency space (i.e. by Fourier transformed expression):

Then can be obtained by the p and q variables inverse Fourier transformation .

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In document Development of an electronic educational material for the analysing of the structures and the bio-chemical processes of living organisms by imaging technology (Pldal 134-152)