# Rect to polar matlab torrent

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First we partition 7. Step 6 — Post-processing: In this step, we obtain the reactions at nodes 1 and 3, and the forces shears and moments in each beam element using MATLAB as follows. Next we set up the element nodal displacement vectors u1 and u2 , then we calculate the element force vectors f1 and f2 by making calls to the MATLAB function BeamElementForces. The Beam Element Element 1 has a shear force of Element 2 has a shear force of — 6.

Obviously the hinge at the right end has zero moment. This process is illustrated below. Example 7. Beam with Distributed Load for Example 7. We need first to replace the distributed loading on element 2 by equivalent nodal loads. This is performed as follows for element 2 with a uniformly distributed load. The resulting beam with equivalent nodal loads is shown in Fig. Equivalent nodal loads for example 7. Note that we only show the numbers to two decimal places although the MATLAB calculations are performed using at least four decimal places.

The moment at the fixed support node 4 is 4. Next we set up the ele- ment nodal displacement vectors u1 , u2 , and u3 , then we calculate the element force vectors f1 , f2 , and f3 by making calls to the MATLAB function BeamElement- Forces. In order to obtain the correct forces for element 2 we need to subtract from f2 the vector of equivalent nodal loads given in 7.

Element 1 has a shear force of —1. Element 2 has a shear force of Element 3 has a shear force of 6. Obviously the roller at the left end has zero moment. Problems: Problem 7. The Beam Element M 1 3 2 3. Beam with Two Elements for Problem 7. Do not put a node at the location of the concen- trated load but use the method of equivalent nodal forces for both the distributed load and the concentrated load. Beam with Distributed Load for Problem 7.

Beam with a Spring for Problem 7. The plane frame element has modulus of elasticity E, moment of inertia I, cross-sectional area A, and length L. In this case the element stiffness matrix is given by the following matrix including axial deformation see [1] and [18]. The sign convention used is that dis- placements are positive if they point upwards and rotations are positive if they are counterclockwise. Then the matrix 8. Inclined Support in a Plane Frame 8. PlaneFrameElementStiffness E, A, I, L, theta — This function calculates the element stiffness matrix for each plane frame element with modulus of elasticity E, cross- sectional area A, moment of inertia I, and length L.

PlaneFrameAssemble K, k, i, j — This function assembles the element stiffness matrix k of the plane frame element joining nodes i and j into the global stiffness matrix K. PlaneFrameElementAxialDiagram f , L — This function plots the axial force diagram for the element with nodal force vector f and length L. PlaneFrameElementShearDiagram f , L — This function plots the shear force dia- gram for the element with nodal force vector f and length L.

PlaneFrameElementMomentDiagram f , L — This function plots the bending mo- ment diagram for the element with nodal force vector f and length L. PlaneFrameInclinedSupport T , i, alpha — This function calculates the transforma- tion matrix of the inclined support using the node number i of the inclined support and the angle of inclination alpha in degrees.

Plane Frame with Three Elements for Example 8. Table 8. Element Connectivity for Example 8. The Plane Frame Element First we partition 8. Next we set up the element nodal displacement vectors u1 , u2 , and u3 , then we calculate the element force vectors f1 , f2 , and f3 by making calls to the MATLAB function PlaneFrameElementForces. Element 1 has an axial force of 8. Element 2 has an axial force of —7. Element 3 has an axial force of —8. Bending Moment Diagram for Element 3 Example 8. Plane Frame with Distributed Load for Example 8.

The resulting plane frame with equivalent nodal loads is shown in Fig. Equivalent Nodal Loads for Example 8. The Plane Frame Element Columns 1 through 7 0. Step 6 — Post-processing: In this step, we obtain the reactions at nodes 1 and 3, and the forces axial forces, shears and moments in each plane frame element using MATLAB as follows.

The moments at nodes 1 and 3 are 2. Next we set up the element nodal displacement vectors u1 and u2 , then we calcu- late the element force vectors f1 and f2 by making calls to the MATLAB function PlaneFrameElementForces. In order to obtain the correct forces for element 2 we need to subtract from f2 the vector of equivalent nodal loads given in 8. Plane Frame with Two Elements for Problem 8. Plane Frame with Distributed Load for Problem 8.

The Plane Frame Element 3. Hint: Use a plane frame element for the beam so as to include axial deformation effects. Also use a plane truss element for the spring so as to include the angle of inclination — in this case determine values for E and A for the spring using the value of k given and the length of the spring.

Plane Frame with a Spring for Problem 8. The grid element has modulus of elasticity E, shear modulus of elasticity G, moment of inertia I, torsional constant J, and length L. In this case the element stiffness matrix is given by the following matrix see [1] and [18]. The Grid Element It is clear that the grid element has six degrees of freedom — three at each node one displacement and two rotations.

Then the matrix 9. GridAssemble K, k, i, j — This function assembles the element stiffness matrix k of the grid element joining nodes i and j into the global stiffness matrix K. GridElementForces E, G, I, J, L, theta, u — This function calculates the element force vector using the modulus of elasticity E, shear modulus of elasticity G, moment of inertia I, torsional constant J, length L, and the element displacement vector u.

Grid with Three Elements for Example 9. Table 9. Element Connectivity for Example 9. The Grid Element 0. First we partition equation 9. The symmetry in the results in this problem is clear. Step 6 — Post-processing: In this step, we obtain the reactions at nodes 2, 3, and 4, and the forces and moments in each grid element using MATLAB as follows.

The vertical reaction and moments at node 3 are —1. The vertical reaction and moments at node 4 are Next we set up the element nodal displacement vectors u1 , u2 , and u3 , then we calculate the element force vectors f1 , f2 , and f3 by making calls to the MATLAB function GridElementForces. Element 1 has a force of — Element 2 has a force of 1.

Element 3 has a force of — Gird with Two Elements for Problem 9. The space frame element has modulus of elasticity E, shear modu- lus of elasticity G, cross-sectional area A, moments of inertia Ix and Iy , polar moment of inertia J, and length L. The Space Frame Element It is clear that the space frame element has twelve degrees of freedom — six at each node three displacements and three rotations.

Then the matrix SpaceFrameElementStiffness E, G, A, Iy , Iz , J, x1 , y1 , z1 , x2 , y2 , z2 — This function calculates the element stiffness matrix for each space frame element with modulus of elasticity E, shear modulus of elasticity G, cross-sectional area A, moments of inertia Ix and Iy , polar moment of inertia J, coordinates x1 , y1 , z1 for the first node, and coordinates x2 , y2 , z2 for the second node.

SpaceFrameAssemble K, k, i, j — This function assembles the element stiffness matrix k of the space frame element joining nodes i and j into the global stiffness matrix K. SpaceFrameElementForces E, G, A, Iy , Iz , J, x1 , y1 , z1 , x2 , y2 , z2 , u — This function calculates the element force vector using the modulus of elasticity E, shear modulus of elasticity G, cross-sectional area A, moments of inertia Ix and Iy , polar moment of inertia J, coordinates x1 , y1 , z1 for the first node, coordinates x2 , y2 , z2 for the second node, and the element displacement vector u.

SpaceFrameElementAxialDiagram f , L — This function plots the axial force diagram for the element with nodal force vector f and length L. SpaceFrameElementTorsionDiagram f , L — This function plots the torsional moment diagram for the element with nodal force vector f and length L. Space Frame with Three Elements for Example Table Element Connectivity for Example The Space Frame Element The Space Frame Element Columns 1 through 7 1. The Space Frame Element Columns 22 through 24 0 0 0.

We will proceed directly to the boundary conditions. First we partition the equation by extracting the submatrix in rows 1 to 6 and columns 1 to 6. Also the three rotations at node 1 are 0. Step 6 — Post-processing: In this step, we obtain the reactions at nodes 2, 3, and 4, and the forces and moments in each space frame element using MATLAB as follows.

Space Frame with Eight Elements for Problem This element can be used for plane stress or plane strain problems in elasticity. It is also called the constant strain triangle. Each linear triangle has three nodes with two in- plane degrees of freedom at each node as shown in Fig. The global coordinates of the three nodes are denoted by xi , yi , xj , yj , and xm , ym.

The order of the nodes for each element is important — they should be listed in a counterclockwise direction starting from any node. LinearTriangleAssemble K, k, i, j, m — This function assembles the element stiffness matrix k of the linear triangle joining nodes i, j, and m into the global stiffness matrix K.

It returns the stress vector for the element. LinearTriangleElementPStresses sigma — This function calculates the element prin- cipal stresses using the element stress vector sigma. The plate is discretized using two linear triangular elements as shown in Fig.

Thin Plate for Example Step 1 — Discretizing the Domain: We subdivide the plate into two elements only for illustration purposes. More elements must be used in order to obtain reliable results. Thus the domain is subdivided into two elements and four nodes as shown in Fig. The total force due to the distributed load is divided equally between nodes 2 and 3.

Since the plate is thin, a case of plane stress is assumed. First we partition When a larger number of elements is used we expect to get the same result for the horizontal displacements at nodes 2 and 3. The horizontal and vertical reactions at node 4 are forces of 9. Next we set up the element nodal displacement vectors u1 and u2 then we calculate the element stresses sigma1 and sigma2 by making calls to the MATLAB function LinearTrian- gleElementStresses.

The Linear Triangular Element 0. It is clear that the stresses in the x-direction approach closely the correct value of 3 MPa tensile. Example The plate is discretized using twelve linear triangles as shown in Fig. Solution: Use the six steps outlined in Chap.

The total force due to the distributed load is divided equally between nodes 5 and However, the resultant applied force at node 10 cancels out and we are left with a concentrated force of The Linear Triangular Element 1. The Linear Triangular Element 2. The following are the MATLAB commands — note that the result for the global stiffness matrix is not shown at each step except at the last step. The Linear Triangular Element 0 0. First we partition the matrix equation by extracting the submatrices in rows 3 to 6, 9 to 12, 15 to 22, and columns 3 to 6, 9 to 12, 15 to We set up the global nodal displacement vector U , then we calculate the global nodal force vector F.

The stresses and principal stresses in each element are not required in this problem but they can be easily obtained for each one of the twelve elements by making successive calls to the MATLAB functions LinearTriangleElementStresses and LinearTriangleElementPStresses. Problems: Problem Solve the problem again us- ing four linear triangular elements instead of two elements as shown in Fig.

Compare your answers for the displacements at nodes 2 and 3 with the answers ob- tained in the example. Compare also the stresses obtained for the four elements with those obtained for the two elements in the example. The Linear Triangular Element P 0.

Thin Plate with a Hole 20 kN 13 14 15 16 0. It is also called the linear strain triangle. Each quadratic triangle has six nodes with two in-plane degrees of freedom at each node as shown in Fig. The global coordinates of the six nodes are denoted by x1 , y1 , x2 , y2 , x3 , y3 , x4 , y4 , x5 , y5 , and x6 , y6.

The order of the nodes for each element is important — they should be listed in a counterclockwise direction starting from the corner nodes then the mid- side nodes. In this case the element stiffness matrix is not written explicitly but calculated through symbolic differentiation and integration with the aid of the MATLAB Symbolic Math Toolbox.

The six shape functions for this element are listed explicitly as follows see [1] and [14]. The partial differentiation of The reader should note the calculation of this matrix will be somewhat slow due to the symbolic computations involved. It is clear that the quadratic triangular element has twelve degrees of freedom — two at each node. It should be noted that in this case this vector is a linear function of x and y.

Usually numerical results are obtained at the centroid of the element. The MATLAB function QuadTriangleElementStresses gives two results — the general linear stress functions in x and y, and the numerical values of the stresses at the centroid of the element. The Quadratic Triangular Element node, x2 , y2 for the second node, and x3 , y3 for the third node. QuadTriangleAssemble K, k, i, j, m, p, q, r — This function assembles the element stiffness matrix k of the linear triangle joining nodes i, j, m, p, q, and r into the global stiffness matrix K.

QuadTriangleElementPStresses sigma — This function calculates the element prin- cipal stresses using the element stress vector sigma. This is the problem solved in Example It will be solved here using quadratic triangles. The plate is discretized using quadratic triangles as shown in Fig. Thus the domain is subdivided into two elements and nine nodes as shown in Fig. The Quadratic Triangular Element 0. The Quadratic Triangular Element Columns 15 through 18 0 0 0 0.

The equation is not written out explicitly below because it is too large. First we partition the resulting equation by extracting the submatrix in rows 3 to 6, rows 9 to 12, rows 15 to 18, and columns 3 to 6, columns 9 to 12, columns 15 to Therefore we obtain the following matrix equation showing the numbers to two decimal places only although the MATLAB calculations are carried out using at least four decimal places. The horizontal and vertical displacements at node 6 are 0.

The horizontal and vertical displacements at node 9 are 0. When a larger number of elements is used we expect to get the same result for the horizontal displacements at nodes 3, 6, and 9. The Quadratic Triangular Element The horizontal and vertical reactions at node 4 are forces of The horizontal and vertical reactions at node 7 are forces of 3.

It is noted that the results for the reactions are different than those obtained in Example We need to use more elements to get reliable results for the reactions and stresses. Next we set up the element nodal displacement vectors u1 and u2 then we calculate the element stresses sigma1 and sigma2 by making calls to the MATLAB function QuadTriangleElementStresses.

Then numerical values for the stresses are computed at the centroid of each element. It is clear that the stresses in the x-direction approach closely the correct value of 3 MPa tensile at the centroid of element 2 because this element is located away from the supports at the left end of the plate. Solve the problem again us- ing four quadratic triangular elements instead of two elements as shown in Fig.

Compare your answers for the displacements at nodes 3, 8, and 13 with the answers obtained in the example. Compare also your answers with those obtained in Example It is characterized by linear shape functions in each of the x and y directions.

It is a generalization of the 4-node rectangular element. This is the first isoparametric element we deal with in this book. Each bilinear quadrilateral element has four nodes with two in-plane degrees of freedom at each node as shown in Fig. The global coordinates of the four nodes are denoted by x1 , y1 , x2 , y2 , x3 , y3 , and x4 , y4. The double integration of It is clear that the bilinear quadrilateral element has eight degrees of freedom — two at each node.

This function uses a different form of the element equations but produces exactly the same result as the function BilinearQuadElementStiffness. BilinearQuadAssemble K, k, i, j, m, n — This function assembles the element stiff- ness matrix k of the bilinear quadrilateral element joining nodes i, j, m, and n into the global stiffness matrix K. BilinearQuadElementPStresses sigma — This function calculates the element prin- cipal stresses using the element stress vector sigma.

This problem was solved in Example Solve this problem again using two bilinear quadrilateral elements as shown in Fig. The Bilinear Quadrilateral Element Step 1 — Discretizing the Domain: We subdivide the plate into two elements only for illustration purposes. Thus the domain is subdivided into two elements and six nodes as shown in Fig. The total force due to the distributed load is divided equally between nodes 3 and 6. The Bilinear Quadrilateral Element Column 8 0.

The Bilinear Quadrilateral Element 0 0 The Bilinear Quadrilateral Element 0. Next we set up the element nodal displacement vectors u1 and u2 then we calculate the element stresses sigma1 and sigma2 by making calls to the MATLAB function BilinearQuadElementStresses. The Bilinear Quadrilateral Element 3.

It is clear that the stresses in the x-direction approach closely the correct value of 3 MPa tensile at the centroids of both elements. This problem was solved previously in Example Solve this problem again using three bilinear quadri- lateral elements as shown in Fig. The Bilinear Quadrilateral Element 7 8 0. Step 1 — Discretizing the Domain: We subdivide the plate into three elements and eight nodes as shown in Fig. The total force due to the distributed load is divided equally between nodes 5 and 6.

However, the resultant applied force at node 6 cancels out and we are left with a concentrated force of First we partition the matrix equation by extracting the submatrices in rows 3 to 6, rows 9 to 12, rows 15 to 16, and columns 3 to 6, columns 9 to 12, columns 15 to Therefore we obtain: 5. These results can be compared with those obtained in Example We note that the vertical displacement at node 5 is obtained here as —0. It is clear that these results are very close to each other indicating that the two types of elements and meshes used give similar results in this case.

The Bilinear Quadrilateral Element These results can also be compared to those obtained in Example We note that the horizontal and vertical reactions at node 1 are 6. It is also seen that the results obtained using the two types of elements and meshes give very similar results.

We cannot compare these results with Example However, accurate results for the stresses can be obtained by refining the mesh and using more elements. Solve this problem again using eight bilinear quadrilateral elements as shown in Fig. Compare your answers with those obtained in Example Compare the results obtained for the displacements and reactions with those ob- tained in Problem Thin Plate with a Hole for Problem The Bilinear Quadrilateral Element Problem Use two bilinear quadrilaterals to solve this problem.

Compare your answers with those obtained for Problem It is characterized by quadratic shape functions in each of the x and y directions. This is the second isoparametric element we deal with in this book. Each quadratic quadrilateral element has eight nodes with two in-plane degrees of freedom at each node as shown in Fig. The global coordinates of the eight nodes are denoted by x1 , y1 , x2 , y2 , x3 , y3 , x4 , y4 , x5 , y5 , x6 , y6 , x7 , y7 , and x8 , y8. The order of the nodes for each element is important — they should be listed in a counterclockwise direction starting from the corner nodes followed by the midside nodes.

It is clear that the quadratic quadrilateral element has sixteen degrees of freedom — two at each node. QuadraticQuadElementPStresses sigma — This function calculates the element prin- cipal stresses using the element stress vector sigma. Solve this problem again using one quadratic quadrilateral element as shown in Fig.

The Quadratic Quadrilateral Element 0. Step 1 — Discretizing the Domain: We use one quadratic quadrilateral element to model the plate only for illustration purposes. Thus the domain is subdivided into one element and eight nodes as shown in Fig. The Quadratic Quadrilateral Element Columns 8 through 14 Only the final result is shown after the element is assembled.

The Quadratic Quadrilateral Element The Quadratic Quadrilateral Element Columns 8 through 10 0 The horizon- tal and vertical displacements at node 8 are 0. The horizontal and vertical displacements at node 5 are 0. Thus the horizontal and vertical reactions at node 6 are forces of 3. Obviously force equilibrium is satis- fied for this problem. Then numerical values for the stresses are computed at the centroid of the element.

It is clear that the stress in the x- direction approaches closely the correct value of 3 MPa ten- sile at the centroid of the element. It is seen that only one quadratic quadrilat- eral element gives accurate results in this problem. Solve this problem using one quadratic quadri- lateral element as shown in the figure.

It is also called the constant strain tetrahedron. Each linear tetrahedron has four nodes with three degrees of freedom at each node as shown in Fig. The global coordinates of the four nodes are denoted by x1 , y1 , z1 , x2 , y2 , z2 , x3 , y3 , z3 , and x4 , y4 , z4. The numbering of the nodes for each element is very important — you should number the nodes such that the volume of the element is positive. In this case the element stiffness matrix is given by see [1] and [8].

TetrahedronAssemble K, k, i, j, m, n — This function assembles the element stiffness matrix k of the linear tetrahedron joining nodes i, j, m, and n into the global stiffness matrix K. TetrahedronElementPStresses sigma — This function calculates the three principal stresses for the element using the element stress vector sigma.

This function does not return the principal angles. Use five linear tetrahedral elements to solve this problem as shown in Fig. Step 1 — Discretizing the Domain: We subdivide the plate into five linear tetrahedral elements only for illustration pur- poses. Thus the domain is subdivided into five elements and eight nodes as shown in Fig. The total force due to the distributed load is divided equally between nodes 3, 4, 7, and 8 in the ratio 1 : 2 : 2 : 1.

This ratio is obtained considering that nodes 4 and 7 take loads from two elements each while nodes 3 and 8 take loads from one element each. The Linear Tetrahedral Solid Element 2. The Linear Tetrahedral Solid Element 1. The final result is shown only after the fifth element has been assembled. The Linear Tetrahedral Solid Element 0 0 1. First we partition the resulting equation by extracting the submatrces in rows 7 to 12, rows 19 to 24, and columns 7 to 12, columns 19 to Therefore we obtain the following equation noting that the numbers are shown to only two decimal places although MATLAB carries out the calculations using at least four decimal places.

The Linear Tetrahedral Solid Element 0. These results are compared with the result of approximately 0. Next we set up the element nodal displacement vectors u1 , u2 , u3 , u4 , and u5 then we calculate the element stresses sigma1, sigma2, sigma3, sigma4, and sigma5 by making calls to the MATLAB func- tion TetrahedronElementStresses. The Linear Tetrahedral Solid Element 0 It is clear that the stresses in the y-direction approach closely the correct value of 3 MPa tensile.

Solve the problem again us- ing six linear tetrahedral elements instead of five elements as shown in Fig. Compare your answers for the displacements at nodes 3, 4, 7, and 8 with the answers obtained in the example. Compare also the stresses obtained for the six elements with those obtained for the five elements in the example.

Compare also your answers with those obtained in the related examples and problems in Chaps. Hint: Table Element Connectivity for Problem It is characterized by linear shape functions in each of the x, y, and z directions. It is also called a trilinear hexahedron.

This is the third isoparametric element we deal with in this book. Each linear brick element has eight nodes with three degrees of freedom at each node as shown in Fig. The order of the nodes for each element is important — they should be numbered such that the volume of the element is positive. It is clear that the linear brick element has twenty-four degrees of freedom — three at each node. This process will be illustrated in detail in the example.

LinearBrickAssemble K, k, i, j, m, n, p, q, r, s — This function assembles the element stiffness matrix k of the linear brick element joining nodes i, j, m, n, p, q, r, and s into the global stiffness matrix K. LinearBrickElementPStresses sigma — This function calculates the element principal stresses using the element stress vector sigma. Use one linear brick element to solve this problem as shown in Fig.

Step 1 — Discretizing the Domain: We subdivide the plate into one linear brick element only for illustration purposes. The total force due to the distributed load is divided equally between nodes 5, 6, 7, and 8. In this case, it is clear that the element stiffness matrix is the same as the global stiffness matrix. The Linear Brick Solid Element 0.

First we partition the re- sulting equation by extracting the submatrix in rows 13 to 24, and columns 13 to Results may be inaccurate. This is due to using one element only in this example. This is also due to the element having a bad aspect ratio. Therefore, the results obtained above for the displacements are not reliable, even totally wrong. It is anticipated that using more elements will result in reliable answers to the displacements. This will be done by the reader in Problem Solve the problem again us- ing two linear brick elements instead of one element as shown in Fig.

Compare your answers for the displacements at nodes 9, 10, 11, and 12 with the answers ob- tained in the example. Element connectivity for problem This is intended to be a concluding chapter to the book. The first sixteen chapters were mainly concerned with finite elements in structural analysis and mechanics.

In this chapter, I show how the same methodology in terms of consistency and simplicity can be used to formulate and code other types of elements in MATLAB. First, we review the applications of finite elements in other areas, then we provide a sample example with its MATLAB code, namely the fluid flow one-dimensional element.

The reader may then be able to code other types of elements more complicated or in other areas following the same format and procedure. However, finite elements are also used in other diverse areas like fluid flow, heat transfer, mass transport, electro-magnetics, geotechnical engineering, structural dynamics, plasticity and visco-plasticity, etc. In this section, we review the use of finite elements in these areas.

Finite elements are used to model membrane, plate and shell problems in structural mechanics. We have shown in this book different types of membrane elements plane stress and plane strain elements but have not shown the plate and shell elements. Different types of triangular, rectangular, and quadrilateral plate bending elements can be formulated using both the Kirchoff and Reissner plate theories.

Shell elements of different types can also be used. For example, flat shell elements can be developed by superimposing the stiffness of the membrane element and plate bending element. Other types of structural elements can still be developed. Axisymmetric elements may be formulated to be used in different types of problems like pressure vessels. Finite elements may also be used in fracture mechanics and damage mechanics to simulate cracks, micro-cracks, voids and other types of defects.

Other Elements dynamics and earthquake engineering are also important areas for finite element analysis. Finally, in structural mechanics, nonlinear finite elements may be used in problems in plasticity, visco-plasticity and stability of structures. In geotechnical engineering, finite elements are used extensively in solving geotechnical problems.

They are used in modeling foundations, slopes, soil-structure interaction, consolidation, seepage and flow nets, excavation, earth pressures, slope stability and other complex boundary value problems in geotechnical engineering. Finite elements are also used in mechanical design especially in machines and components. In addition, they are used in heat transfer and mass transport to solve problems in conduction, radiation, etc. They can be used to solve both linear and nonlinear problems, and both steady-state and transient processes.

All types of thermal boundary conditions may be modeled including isotropic and orthotropic properties, heat loads, etc. The area of electrical engineering is also important for finite elements. Different types of elements may be used to in electromagnetic analysis. They may be used in studying antennas, radar, microwaves engineering, high-speed and high-frequency circuits, wireless communications, electro-optical engineering, remote sensing, bio- electromagnetics and geo-electromagnetics.

The finite element method provides a very powerful technique for solving problems in circuits and circuit analysis. A very important area for the use of finite elements is the area of water resources and the global atmosphere. Finite elements can be used in groundwater hydrology, turbulent flow problems, fluid mechanics, hydrodynamic flows, flow separation pat- terns, incompressible Navier-Stokes equations in three dimensions, oceanic general circulation modeling, wave propagation, typhoon surge analysis, mesh generation of groundwater flow, non-steady seepage, coastal aquifers, soil-moisture flow, contam- inant transport, and sediment transport.

Finally, the finite element method is a recognized method used in the numerical analysis branch of mathematics and mathematical physics. In addition to finite dif- ferences, finite elements are used extensively in solving partial differential equations of all types. In the next two sections, we provide as an example the full formulation and MAT- LAB code for the one-dimensional fluid flow element.

It is characterized by linear shape functions and is iden- tical to the spring and linear bar elements. In general, if the time-sequence is real-valued, then the DFT will have real components which are even and imaginary components that are odd. Simi- larly, for an imaginary valued time sequence, the DFT values will have an odd real component and an even imaginary component.

The FFT can be used to a obtain the power spectrum of a signal, b do digi- tal filtering, and c obtain the correlation between two signals. The vector x is truncated or zeros are added to N, if necessary. The sampling interval is ts. Its default value is 1. The spectra are plotted versus the digital frequency F. Solution a From Equation 8. With the sampling interval being 0.

The duration of g t is 0. The am- plitude of the noise and the sinusoidal signal can be changed to observe their effects on the spectrum. Math Works Inc. Using the FFT algorithm, generate and plot the frequency content of g t. Assume a sampling rate of Hz. Find the power spectrum. Diode circuit analysis techniques will be discussed. The electronic symbol of a diode is shown in Figure 9. Ideally, the diode conducts current in one direction.

The cur- rent versus voltage characteristics of an ideal diode are shown in Figure 9. The characteristic is divided into three regions: forward-biased, reversed- biased, and the breakdown. If we assume that the voltage across the diode is greater than 0. The following example illustrates how to find n and I S from an experimental data. Example 9. Figure 9. The thermal voltage is directly propor- tional to temperature. This is expressed in Equation 9.

The reverse satura- tion current I S increases approximately 7. T1 and T2 are two different temperatures. Assuming that the emission constant of the diode is 1. We want to determine the diode current I D and the diode volt- age VD. There are several approaches for solving I D and VD. In one approach, Equations 9. This is illustrated by the following example.

Assume a temperature of 25 oC. Then, from Equation 9. Using Equation 9. The iteration technique is particularly facilitated by using computers. It consists of an alternat- ing current ac source, a diode and a resistor. The battery charging circuit, explored in the following example, consists of a source connected to a battery through a resistor and a diode.

Use MATLAB a to sketch the input voltage, b to plot the current flowing through the diode, c to calculate the conduction angle of the diode, and d calculate the peak current. Assume that the diode is ideal. The output of the half-wave rectifier circuit of Figure 9. The smoothing circuit is shown in Figure 9. When the amplitude of the source voltage VS is greater than the output volt- age, the diode conducts and the capacitor is charged.

When the source voltage becomes less than the output voltage, the diode is cut-off and the capacitor discharges with the time constant CR. The output voltage and the diode cur- rent waveforms are shown in Figure 9. Therefore, the output waveform of Figure 9.

When v S t is negative, diode D1 is cut-off but diode D2 conducts. The current flowing through the load R enters it through node A. The current entering the load resistance R enters it through node A. The output voltage of a full-wave rectifier circuit can be smoothed by connect- ing a capacitor across the load.

The resulting circuit is shown in Figure 9. The output voltage and the current waveforms for the full-wave rectifier with RC filter are shown in Figure 9. The capacitor in Figure 9. Solution Peak-to-peak ripple voltage and dc output voltage can be calculated using Equations 9.

I ZM is the maximum current that can flow through the zener without being destroyed. A zener diode shunt voltage regulator circuit is shown in Fig- ure 9. Con- versely, if R is constant and VS decreases, the current flowing through the zener will decrease since the breakdown voltage is nearly constant; the output voltage will remain almost constant with changes in the source voltage VS.

Now assuming the source voltage is held constant and the load resistance is decreased, then the current I L will increase and I Z will decrease. Con- versely, ifVS is held constant and the load resistance increases, the current through the load resistance I L will decrease and the zener current I Z will increase. In the design of zener voltage regulator circuits, it is important that the zener diode remains in the breakdown region irrespective of the changes in the load or the source voltage.

From condition 1 and Equation 9. I Z ,max 9. Solution Using Thevenin Theorem, Figure 9. In addition, when the source voltage is 35 V, the output voltage is The zener breakdown characteristics and the loadlines are shown in Figure 9. Lexton, R. Shah, M. Angelo, Jr.

Sedra, A. Beards, P. Savant, Jr. Ferris, C. Ghausi, M. Warner Jr. Assume a temperature of 25 oC, emission coef- ficient, n , of 1. Both intrinsic and extrinsic semicon- ductors are discussed. The characteristics of depletion and diffusion capaci- tance are explored through the use of example problems solved with MATLAB. The effect of doping concentration on the breakdown voltage of pn junctions is examined. Electrons surround the nucleus in specific orbits.

The electrons are negatively charged and the nucleus is positively charged. If an electron absorbs energy in the form of a photon , it moves to orbits further from the nucleus. An electron transition from a higher energy orbit to a lower energy orbit emits a photon for a direct band gap semiconductor. The energy levels of the outer electrons form energy bands. In insulators, the lower energy band valence band is completely filled and the next energy band conduction band is completely empty.

The valence and conduction bands are separated by a forbidden energy gap. In semicon- ductors the forbidden gap is less than 1. Some semiconductor materials are silicon Si , germanium Ge , and gallium arsenide GaAs. Figure Silicon has four valence electrons and its atoms are bound to- gether by covalent bonds.

At absolute zero temperature the valence band is completely filled with electrons and no current flow can take place. As the temperature of a silicon crystal is raised, there is increased probability of breaking covalent bonds and freeing electrons. The vacancies left by the freed electrons are holes. The process of creating free electron-hole pairs is called ionization.

The free electrons move in the conduction band. Since electron mobility is about three times that of hole mobility in silicon, the electron current is considerably more than the hole current. The following ex- ample illustrates the dependence of electron concentration on temperature. Solution From Equation The width of energy gap with temperature is given as [1]. An n-type semiconductor is formed by doping the silicon crystal with elements of group V of the periodic table antimony, arse- nic, and phosphorus.

The impurity atom is called a donor. The majority car- riers are electrons and the minority carriers are holes. A p-type semiconductor is formed by doping the silicon crystal with elements of group III of the peri- odic table aluminum, boron, gallium, and indium. The impurity atoms are called acceptor atoms.

The majority carriers are holes and minority carriers are electrons. The law of mass action enables us to calculate the majority and minority car- rier density in an extrinsic semiconductor material. In an n-type semiconductor, the donor concentration is greater than the intrin- sic electron concentration, i.

Example It is used to describe the energy level of the electronic state at which an electron has the probability of 0. Equation In addition, the Fermi energy can be thought of as the average energy of mobile carriers in a semiconductor mate- rial. In an n-type semiconductor, there is a shift of the Fermi level towards the edge of the conduction band.

The upward shift is dependent on how much the doped electron density has exceeded the intrinsic value. Drift current is caused by the application of an elec- tric field, whereas diffusion current is obtained when there is a net flow of car- riers from a region of high concentration to a region of low concentration.

This is shown in Figure Practical pn junctions are formed by diffusing into an n-type semiconductor a p-type impurity atom, or vice versa. Because the p-type semiconductor has many free holes and the n-type semiconductor has many free electrons, there is a strong tendency for the holes to diffuse from the p-type to the n-type semi- conductors.

Similarly, electrons diffuse from the n-type to the p-type material. When holes cross the junction into the n-type material, they recombine with the free electrons in the n-type. Similarly, when electrons cross the junction into the p-type region, they recombine with free holes. In the junction a transition region or depletion region is created. In the depletion region, the free holes and electrons are many magnitudes lower than holes in p-type material and electrons in the n-type material.

As electrons and holes recombine in the transition region, the region near the junc- tion within the n-type semiconductor is left with a net positive charge. The re- gion near the junction within the p-type material will be left with a net negative charge. This is illustrated in Figure Because of the positive and negative fixed ions at the transition region, an elec- tric field is established across the junction.

The electric field creates a poten- tial difference across the junction, the potential barrier. The potential barrier pre- vents the flow of majority carriers across the junction under equilibrium condi- tions. That is, from Figure Typically, VC is from 0. For germanium, VC is ap- proximately 0. When a positive voltage VS is applied to the p-side of the junction and n-side is grounded, holes are pushed from the p-type material into the transition re- gion.

In addition, electrons are attracted to transition region. The depletion region decreases, and the effective contact potential is reduced. This allows majority carriers to flow through the depletion region. The depletion region increases and it become more difficult for the majority carriers to flow across the junction.

The current flow is mainly due to the flow of minority carriers. Using Equations The following example shows how I S is affected by tem- perature. During device fabrication, a p-n junction can be formed using process such as ion-implantation diffusion or epitaxy. The dop- ing profile at the junction can take several shapes. Two popular doping pro- files are abrupt step junction and linearly graded junction.

In the abrupt junction, the doping of the depletion region on either side of the metallurgical junction is a constant. This gives rise to constant charge densi- ties on either side of the junction. If the doping density on one side of the metallurgical junction is greater than that on the other side i.

This condition is termed the one-sided step junction approximation. This is the practical model for shallow junctions formed by a heavily doped diffusion into a lightly doped region of opposite polarity [7]. In a linearly graded junction, the ionized doping charge density varies linearly across the depletion region. The charge density passes through zero at the metallurgical junction.

It can be obtained from Equations Equations The positive voltage, VC , is the contact potential of the pn junction. However, when the pn junction becomes slightly forward biased, the capacitance increases rapidly. It is also used to plot the depletion ca- pacitance. The holes are momentarily stored in the n-type material before they recombine with the majority carriers electrons in the n-type material.

Similarly, electrons are injected into and temporarily stored in the p-type material. The electrons then recombine with the majority carriers holes in the p-type material. The diffusion capacitance is usually larger than the depletion capacitance [1, 6]. Typical values of Cd ranges from 80 to pF. A small signal model of the diode is shown in Figure Cd rd Rs Cj Figure RS is the semiconductor bulk and contact resistance.

The model of the diode is shown in Figure Cj Rs Rd Figure The diffusion capacitance is zero. The resistance Rd is reverse resistance of the pn junction normally in the mega-ohms range. At a critical field, E crit , the accelerated carriers in the depletion region have sufficient energy to create new electron-hole pairs as they collide with other atoms. The secondary electrons can in turn create more carriers in the depletion region. This is termed the avalanche breakdown process.

For silicon with an impurity concentration of cm-3, the critical electric field is about 2. This high electric field is able to strip electrons away from the outer orbit of the silicon atoms, thus cre- ating hole-electron pairs in the depletion region. This mechanism of break- down is called zener breakdown. This breakdown mechanism does not involve any multiplication effect. Normally, when the breakdown voltage is less than 6V, the mechanism is zener breakdown process.

For breakdown voltages be- yond 6V, the mechanism is generally an avalanche breakdown process. For an abrupt junction, where one side is heavily doped, the electrical proper- ties of the junction are determined by the lightly doped side. The following example shows the effect of doping concentration on breakdown voltage. Solution Using Equation Impurity Concentration' axis [1.

Singh, J. Jacoboni, C. Mousty, F. Caughey, D. IEEE, Vol. Hodges, D. Neudeck, G. II, Addison-Wesley, Beadle, W. McFarlane, G. Sze, S. Plot the difference between of ni for Equations Assume donor concentrations from to Use impurity gradient values from to It can be used to perform the basic mathematical operations: addition, subtrac- tion, multiplication, and division. They can also be used to do integration and differentiation.

There are several electronic circuits that use an op amp as an integral element. Some of these circuits are amplifiers, filters, oscillators, and flip-flops. In this chapter, the basic properties of op amps will be discussed. The non-ideal characteristics of the op amp will be illustrated, whenever possi- ble, with example problems solved using MATLAB. Its symbol is shown in Figure It is a differ- ence amplifier, with output equal to the amplified difference of the two inputs.

It also has a very large input resistance to ohms. The out- put resistance might be in the range of 50 to ohms. The offset voltage is small but finite and the frequency response will deviate considerably from the infinite frequency response. The common-mode rejection ratio is not infinite but finite. Table This condi- tion is termed the concept of the virtual short circuit. In addition, because of the large input resistance of the op amp, the latter is assumed to take no cur- rent for most calculations.

Thus, Equation R2 R1 Vin Vo Figure With the assumptions of very large open-loop gain and high input resistance, the closed-loop gain of the inverting amplifier depends on the external com- ponents R1 , R2 , and is independent of the open-loop gain.

The integrating time con- stant is CR1. It behaves as a lowpass filter, passing low frequencies and at- tenuating high frequencies. However, at dc the capacitor becomes open cir- cuited and there is no longer a negative feedback from the output to the input. The output voltage then saturates. To provide finite closed-loop gain at dc, a resistance R2 is connected in parallel with the capacitor. The circuit is shown in Figure The resistance R2 is chosen such that R2 is greater than R.

From Equation This circuit is shown in Figure For Figure The input impedance of the amplifier Z IN approaches infinity, since the current that flows into the posi- tive input of the op-amp is almost zero. R2 R1 Vo Vin Figure In addition, from Equation Plot the closed-loop gain as the open-loop gain increases from to The pole of the voltage amplifier and level shifter is KHz and has a gain of The pole of the output stage is KHz and the gain is 0.

Sketch the magnitude response of the operational amplifier open-loop gain. This causes the op amp to have a single pole lowpass response. The process of making one pole dominant in the open-loop gain characteristics is called frequency compensation, and the latter is done to ensure the stability of the op amp. For an op amp connected in an inverting configuration Figure Slew rate is important when an output signal must follow a large input signal that is rapidly changing.

If the slew rate is lower than the rate of change of the input signal, then the output voltage will be distorted. The output voltage will become triangular, and attenuated. However, if the slew rate is higher than the rate of change of the input signal, no distortion occurs and input and output of the op amp circuit will have similar wave shapes. As mentioned in the Section In addition, the op amp has a limited output current capability, due to the saturation of the input stage.

The latter is the frequency at which a sinusoidal rated output signal begins to show distortion due to slew rate limiting. The fol- lowing example illustrates the relationship between the rated output voltage and the full-power bandwidth. CMRR decreases as frequency increases.

For an inverting amplifier as shown in Figure Thus, the common-mode input voltage is approximately zero and Equation A method normally used to take into account the effect of finite CMRR in cal- culating the closed-loop gain is as follows: The contribution of the output voltage due to the common-mode input is AcmVi ,cm.

This output voltage con- tribution can be obtained if a differential input signal, Verror , is applied to the input of an op amp with zero common-mode gain. Schilling, D. Wait, J. Irvine, R. The resistance values are in kilohms. A square wave of zero dc voltage and a peak voltage of 1 V and a frequency of KHz is connected to the input of the unity gain follower. Assume the following values of the open-loop 3 5 7 9 gain: 10 , 10 , 10 and The operation of the BJT depends on the flow of both majority and minority carriers.

There are two types of BJT: npn and pnp transistors. The electronic symbols of the two types of transistors are shown in Figure The model is shown in Figure In the case of a pnp transistor, the directions of the diodes in Figure In addition, the voltage polarities of Equations The four regions of operations are forward-active, reverse-active, saturation and cut-off.

Forward-Active Region The forward-active region corresponds to forward biasing the emitter-base junction and reverse biasing the base-collector junction. It is the normal operational region of transistors employed for amplifications. The cut-off region corresponds to reverse biasing the base-emitter and base- collector junctions. The collector and base currents are very small compared to those that flow when transistors are in the active-forward and saturation regions. Solution From Equations Assume a temperature of oK.

The variation on V BE with temperature is similar to the changes of the pn junction diode voltage with temperature. The collector-to-base leakage current, I CBO , approximately doubles every 10o temperature rise. As discussed in Section 9. The change in collector current can be obtained using partial derivatives. The derivation is assisted by referring to Figure Calculate the collector current at 25 oC and plot the change in collector current for temperatures between 25 and oC.

At each temperature, the stability factors are calculated using Equations The change in I C for each temperature is calculated using Equation It is uneconomical to fabricate IC resistors since they take a disproportionately large area on an IC chip. In addition, it is almost impossible to fabricate IC inductors.

Biasing of ICs is done using mostly transistors that are connected to create constant current sources. Examples of integrated circuit biasing schemes are discussed in this section. The current mirror consists of two matched transistors Q1 and Q2 with their bases and emitters connected.

The transistor Q1 is connected as a diode by shorting the base to its collector. In the latter mode of transistor operation, the device Q2 behaves as a current source. To obtain an expression for the output current, we assume that all three transistors are identical. Solution We use Equation Similarly, we use Equation The input resistance is medium and is essentially independent of the load resistance RL. The output resistance is relatively high and is essentially independent of the source resistance.

The bypass capacitance C E is used to increase the midband gain, since it effectively short circuits the emitter resistance R E at midband frequencies. The resistance R E is needed for bias stability. The internal capacitances of the transistor will influence the high frequency cut-off. Solution Using Equations The zero of the overall amplifier gain is calculated using Equation The terminals of the device are the gate, source, drain, and substrate.

There are two types of mosfets: the enhancement type and the depletion type. In the enhancement type MOSFET, the channel between the source and drain has to be induced by applying a voltage on the gate. In the depletion type mosfet, the structure of the device is such that there exists a channel between the source and drain. Because of the oxide insulation between the gate and the channel, mosfets have high input resistance.

The electronic symbol of a mosfet is shown in Figure Because the enhancement mode mosfet is widely used, the presentation in this section will be done using an enhancement-type mosfet. In the latter device, the channel between the drain and source has to be induced by applying a voltage between the gate and source.

The voltage needed to create the channel is called the threshold voltage, VT. For an n-channel enhancement-type mosfet , VT is positive and for a p-channel device it is negative. This implies that the drain current is zero for all values of the drain-to-source voltage.

In the latter region, the device behaves as a non-linear voltage-controlled resistance. The resistances RG1 and RG 2 will define the gate voltage. The resistance RS improves operating point stability. To obtain the drain current, it is initially assumed that the device is in saturation and Equation If Equation The method is illustrated by the following example.

Solution Substituting Equation Two solutions of I D are obtained. However, only one is sensible and possible. If the device is not in saturation, then substituting Equation The circuit is normally referred to as diode-connected enhancement transistor. A circuit for generating dc currents that are constant multiples of a reference current is shown in Figure Assuming the threshold voltages of the transistors of Figure In practice, because of the finite output resistance of transistor T2, I 0 will be a function of the output voltage v 0.

Neglect channel length modulation. Using equation However, the common- source amplifier has higher input resistance than that of the common-emitter amplifier. The circuit for the common source amplifier is shown in Figure The internal capacitances of the FET will affect the high frequency response of the amplifier. The overall gain of the common-source amplifier can be written in a form similar to Equation The midband gain, Am , is obtained from the midband equivalent circuit of the common-source amplifier.

The equivalent circuit is obtained by short-circuiting all the external capacitors and open- circuiting all the internal capacitances of the FET. The high frequency equivalent circuit of a common-source amplifier is shown in Figure The external capacitors of the common of common- source amplifier are short-circuited at high frequencies. Determine a midband gain, b the low frequency cut-off, c high frequency cut-off, and d bandwidth of the amplifier. Geiger, R. Savant, C. Belanger, P. Wildlar R.

Do not neglect the effect of V AF on the collector current. Indicate the region of operation for each value of RS. Introduction to By Friend Name.

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You have to describe the mapping you want in a well defined manner For one you need to think about where the origin is located before converting to polar coordinates. The previous example assume the origin to be the axes base at 0,0. To better illustrate the effect, consider a source image with concentric circles drawn in Cartesian coordinates, and notice how they map to straight lines in polar coordinates when using the center of the circles as origin:.

Here is another example of how to display an image in polar coordinates as requested in the comments. Note that we perform the mapping in the inverse direction pol2cart :. Again the effect is better show if you feed it an input image with straight lines, and see how they map in polar coordinates vertical lines become circles, and horizontal lines become rays emanating from the origin :.

Stack Overflow for Teams — Start collaborating and sharing organizational knowledge. Create a free Team Why Teams? Learn more. How to change an image from Cartesian to Polar coordinates in Matlab? Ask Question. Asked 10 years, 8 months ago. Modified 4 years, 4 months ago. Viewed 28k times. Improve this question. It's not clear what you're after. Can you add a clearer text description of what you want. For example: "for each pixel i,j in the new image, then I want to Add a comment.

Sorted by: Reset to default. Highest score default Date modified newest first Date created oldest first. It is not quite clear what you are trying to do, which is why I am making my own example To better illustrate the effect, consider a source image with concentric circles drawn in Cartesian coordinates, and notice how they map to straight lines in polar coordinates when using the center of the circles as origin: EDIT Here is another example of how to display an image in polar coordinates as requested in the comments.

Improve this answer. Community Bot 1 1 1 silver badge. I tried but couldn't upload. The app interface is in Russian, which I don't speak or understand. The website says "there will be no registration in our service". Couldn't figure out what to do. If the text description of the problem is enough for you, I prefer continuing to do it that way. If the text description of the problem is enough for you, I prefer continuing to do it that way Hope this explains the problem better Better than previously given "nothing"!

But you have many important questions which you simply ignored! I missed that part of your reply. However, I think I found the problem: Some packages are in the torrent, while others act just as links to download location. For your questions: My platform is win I installed Rb from your torrent, it is working perfectly. Next time I'll use a picture sharing site for providing you with better info.

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On the left alerts every time there is a answers A steel copy or open a fully independent number or jumble. You can even content and collaborate around the technologies. Pros : We is used in manage remote computers. It is for Streamer will automatically be updated in online assistance, or easy to search.Sorted by: Reset to default. Highest score default Date modified newest first Date created oldest first. It is not quite clear what you are trying to do, which is why I am making my own example To better illustrate the effect, consider a source image with concentric circles drawn in Cartesian coordinates, and notice how they map to straight lines in polar coordinates when using the center of the circles as origin: EDIT Here is another example of how to display an image in polar coordinates as requested in the comments.

Improve this answer. Community Bot 1 1 1 silver badge. Amro Amro k 25 25 gold badges silver badges bronze badges. Masi: did you even try it? The example is complete as is Both X and map come from clown. As for the rotation-mapped-image, I added another example, see the edit.. I updated the example so you can use any image you want. You can read more about the different image types here indexed, grayscale, truecolor : mathworks.

Masi here are the pattern images if you want to try it yourself: circles , vlines , hlines. The variable s here controls the max radius in rho , it's not the center, you can really use any other value, but the size of the matrix produced by meshgrid will depend on it. Also you don't necessarily have to start at 0 in that linspace. Show 6 more comments. Sign up or log in Sign up using Google.

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