Topic 7e Finite Difference Analysis of Transmission Lines

Transcription

1 Course Instructor r. Raymond C. Rumpf Office: A 337 Phone: (915) Mail: rcrumpf@utep.edu Topic 7e Finite ifference Analysis of Transmission Lines 4386/531 Computational Methods in 1 Outline Introduction to Transmission Lines High Level Solution Approach Formulation erivation of governing equations Finite difference approimation of governing equations Implementation Building the device on a grid grid technique Solving the matri equations Calculating the transmission line parameters ample Weird Coa 1

2 Introduction to Transmission Lines 3 What Are Transmission Lines? Homogeneous Has TM mode. Has T and TM modes. Inhomogeneous Supports only quasi (TM, T, & TM) modes. Single nded coaial stripline microstrip coplanar Transmission lines are essentially high frequency electrical cables composed of two or more metal wires. ifferential buried parallel plate coplanar strips shielded pair slotline 4

3 Transmission Line Parameters RLGC We can think transmission lines as being composed of millions of tiny little circuit elements that are distributed along the length of the line. In fact, these circuit element are not discrete, but continuous along the length of the transmission line. Slide 5 RLGC Circuit Model This model is accurate when the size z of the equivalent circuit is very small compared to the wavelength of the signal on the transmission line. lim z R z L z Gz Cz z z Slide 6 3

4 istributed TL Parameters istributed Circuit Parameters R (/m) Resistance per unit length. Arises due to resistivity in the conductors. L (H/m) Inductance per unit length. Arises due to stored magnetic energy around the line. G (1/m) Conductance per unit length. Arises due to conductivity in the dielectric separating the conductors. 1 G R C (F/m) Capacitance per unit length. Arises due to stored electric energy between the conductors. There are many possible circuit models for transmission lines, but most produce the same equations after analysis. R z L z Gz Cz Slide 7 z Characteristic Impedance, Z The characteristic impedance is the voltage divided the current at any point along the transmission line. Z V I If we perform a circuit analysis of the equivalent circuit, we get Z R jl G jc When we ignore loss (usually accurate to do this), we get Z L C Slide 8 4

5 ample RLGC Parameters RG 59 Coa CAT5 Twisted Pair Microstrip R 36 mω m L 43 nh m G 1 m C 69 pf m Z 75 R 176 mω m L 49 nh m G m C 49 pf m Z 1 R 15 mω m L 364 nh m G 3 m C 17 pf m Z 5 Surprisingly, almost all transmission lines have parameters very close to these same values. Slide 9 Reflections From Impedance iscontinuities Transmission Line 1 Transmission Line Z 1 Z z z 1% t r Z Z Z Z

6 Voltage Standing Wave Ratio 11 Impedance Transformation As we back away from the load, the impedance eperienced by the source changes! Short circuits can look like open circuits, open circuits can look like short circuits, inductors can look like capacitors, capacitors can look like inductors, its crazy! ZL jz tan Zin Z Z jz L tan Slide 1 6

7 Impedance Matching Similar to the anti reflection layer for waves, we can match a transmission line to a load impedance by inserting a quarter wave section of a second transmission line. Transmission Line Quarter Wave Transformer Load, Z Zin Z ar? Z ZZ ar L Z L z 4 We must perform an electromagnetic analysis of the transmission line to determine ar. z 4 ar z Multi Segment Transmission Lines Slide 13 Multi Segment Circuits We can design filters, pulse shaper, and much more. 14 7

8 High Level Solution Approach 15 Mawell s quations Mawell s equations describe classical electromagnetics. v B B t H J t Gauss law Gauss law for magnetic fields Faraday s law Ampere s circuit law electric field intensity V m electric flu density C m H magnetic field intensity A m B magnetic flu density Wb m v volume charge density C m J electric current density A m The constitutive relations describe how the electromagnetic fields interact with matter. lectric response permittivity F m permeability H m B H Magnetic response

9 Physical Constants The permittivity and permeability are usually epressed in terms of the relative parameters. permittivity F m free space permittivity r Permittivity F m r relative permittivity no units r Permeability permeability H m 1 6 free space permeability H m F m H m relative permeability no units 1 1 r r is commonly called the dielectric constant. r r The speed of light in free space can be calculated from the permittivity and permeability as follows. 1 c 99,79,458 m s speed of light in vacuum m s c 17 lectrostatic Approimation Transmission lines are waveguides. To be rigorous, they should be modeled as such. This can be rather computationally intensive. An alternative is to analyze transmission lines in the electrostatic approimation using Laplace s equation. The dimensions of a transmission are typically much smaller than the operating wavelength so the wave nature is less important to consider. Therefore, we can solve Mawell s equations assuming static fields. t B v H J t B t B v H J 18 9

10 lectric Potential V The vector electric field and the scalar electric potential V describe the same physical phenomenon. This allows equations to be solved in terms of a scalar quantity instead of a vector quantity. In electrostatics, we have. In vector calculus, the curl of a gradient is always zero, V. Based on this, if a vector field does not have any curl, it must be possible to epress it as the gradient of a scalar field. V The negative sign is incorporated to be consistent with the sign conventions used with charges and fields. V V electric potential volts 19 Inhomogeneous Laplace s quation Away from charges v, the divergence condition for the electric field is q. (1) But, we know that, so q. (1) can be written in terms of. r q. () The electric field is related to the scalar potential V as follows. V q. (3) The inhomogeneous Laplace s equation is derived by substituting q. (3) into q. (). r V 1

11 Homogeneous Laplace s quation When the transmission line is embedded in a homogeneous dielectric, the permittivity is not a function of position and drops out of Laplace s equation. r V r V V V V 1 istributed Capacitance In the electrostatic approimation, the transmission line is a capacitor. The total energy stored in a capacitor is the total energy in the fields: The integral is taken over the entire cross section of device. For open devices like microstrips, this is an infinite area. In practice, we integrate over a large enough area to U da 1 incorporate as much of of the electric field as possible. A From circuit theory, the capacitance is related to the total stored energy through CV U V is the voltage across the capacitor. If we set the above equations equal, we derive the equation to calculate the distributed capacitance from the electric field functions and. 1 C da V A 11

12 istributed Inductance The voltage along the transmission line embedded in a homogeneous material travels at the same velocity as the electric field so we can write 1 1 r,h r,h vv v LC LC c Solving this equation for L, we get L r,h r,h cc h h This means that for transmission lines embedded in homogeneous materials, we can calculate the distributed inductance L directly from the distributed capacitance C. ielectric materials should not alter the inductance. However if we use the value of C calculated previously, it will. This is incorrect. The solution is to calculate distributed capacitance with air dielectric C h and then calculate the distributed inductance L from this. L cc r,h h For most materials rh = 1 3 Calculating Transmission Line Parameters The characteristic impedance Z c is calculated from the distributed inductance L and distributed capacitance C through Z c L C The velocity of a signal travelling along the transmission line is v 1 LC The effective refractive inde is therefore c Both Z c and n are needed to analyze transmission line circuits. v n c LC 4 1

13 Two Step Modeling Approach Step 1 Homogeneous Case air outer conductor inner conductor distributed inductance L air Step Inhomogeneous Case outer conductor 1 inner conductor distributed capacitance C air Slide 5 How We Will Analyze Transmission Lines Step 1 Construct homogeneous TL air outer conductor inner conductor air Step Construct and solve matri equation Vh Lv h h v v L v h src 1 h src Step 3 Calculate distributed inductance h Vh C da h h h V A L cc rh h Step 4 Construct inhomogeneous TL 1 outer conductor inner conductor air Step 5 Construct and solve matri equation r V Lv vsrc v 1 L vsrc Step 6 Calculate distributed capacitance V C da V Z A Step 7 Calculate TL parameters L C n c LC c Slide 6 13

14 Formulation: erivation of Governing quations 7 quations of lectrostatics Any electromagnetic analysis begins with Mawell s equations. We start with the following for electrostatic problems. q. (1) q. () V q. (3) 8 14

15 pand quations y y z z y yy y z zz z V y V y z z 9 Analysis of Cross Section We wish to analyze a transmission line that is uniform in the direction that the wave propagates. Let this direction be z. Since the device is uniform in z, nothing changes in the z direction. z 3 15

16 Reduced quations for y y y y z z y yy y y yy y z zz z z z V y y V V y y z z z This implies TM. z 31 Final Governing quations y y y y y yy y V y y y yy y y V V y 3 16

17 Formulation: Finite ifference Approimation of Governing quations 33 Grid Strategy (1 of ) The manner in which we stagger the functions across the grid comes from the governing equations themselves. From the constitutive equations, we see that and will lie on the same points as. i, j i, j i, j Similarly, y and yy will lie on the same points as y. y yy y i, j i, j i, j y yy y Thus, the constitutive relations do not require any staggering

18 Grid Strategy ( of ) Net, we inspect the equations relating and V. We see that V will need to be staggered around in the direction. i1, j i, j V i, j V V i, j V i, j i 1, j V We see that V will need to be staggered around y in the y direction. y i, j 1, V i, j V V y y y i j y V i, j 1 i, j y i, j V 35 Grid Strategy (3 of 3) Putting all of this together, we arrive at our staggered grid Actual staggered grid showing the true position of the function values. An alternative view of the staggered grid that more clearly conveys which cell each function value resides in

19 Matri Form of Governing quations We have five coupled equations. We can immediately write these in matri form as: y y i, j i1, j i, j i, j1 y y y d d e e y y y yy y i, j i, j i, j i, j i, j i, j y yy y d d ε e ε e y yy y V V y y i, j i, j y V V V V y i1, j i, j i, j1 i, j e e y v v v v y 37 Block Matri Form We write our matri equations in block matri form. y y e e d y d y y yy y d ε e dy εyyey V y y v e v v e y y 38 19

20 liminate Field We can eliminate the field by substituting the second equation into the first. d e e y d y e e ε e y ε yy ey e e ε e y εyy ey d ε e dy εyy ey 39 liminate Field We can similarly eliminate the field e e ε e y εyy ey v e e ε y v v εyy y v e e ε y v v εyy y v e v v e y y We now have a single scalar differential equation in matri form. 4

21 Final Matri quations Our final matri equation is the matri form of the inhomogeneous Laplace s equation. Inhomogeneous Laplace s quation r V v e e ε y v v εyy y Homogeneous Laplace s quation We can easily write the block matri form of the homogeneous Laplace s equation from this result because r = 1 everywhere. V h v e e y v v y 41 Implementation: Build the evice on a Grid 4 1

22 Grid Representation of a Microstrip Transmission Line N unit cells N y unit cells The spacer regions are needed so that the transmission line is not affected by the edges of the grid. 43 Four Arrays to escribe a Transmission Line ( r = 6) Note: If you wish to simulate a transmission line with more than two conductors, you will need one array for each conductor in addition to the two permittivity arrays. 44

23 Building the iagonal Matrices and yy 1. Build R and Ryy using the grid technique.. Reshape these arrays into 1 arrays (column vectors). R = R(:); Ryy = Ryy(:); 3. eclare the 1 arrays as sparse. R = sparse(r); Ryy = sparse(ryy); 4. Convert the 1 sparse arrays into sparse diagonal matrices. R = diag(r); Ryy = diag(ryy); Actually, we can do all of this in a single line: R = diag(sparse(r(:))); Ryy = diag(sparse(ryy(:))); 45 The Forced Potentials Grid vf, y v f Reshaped to a column vector. vf = vf(:); 46 3

24 Implementation: Grid Technique 47 Grid Technique (1 of 9) We define our ordinary 1 grid as usual. The output of this step is the number of cells in the grid, N and Ny, and the size of the cells in the grid, d and dy. % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; 48 4

25 Grid Technique ( of 9) Recall how the various functions overlay onto the grid. Functions assigned to the same grid cell are in physically different positions and may reside in different materials as a result. % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; 49 Grid Technique (3 of 9) It is like we are getting twice the resolution due to the staggering of the functions. In order to sort out what values go where, we construct a grid at twice the resolution of the original grid. The grid occupies the same physical amount of space as the original grid. % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; 5 5

26 Grid Technique (4 of 9) Let s say we wish to construct a cylinder of radius on our grid. % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; % CRAT CYLINR r = ; 51 Grid Technique (5 of 9) We start by building our object on the grid, ignoring anything about or original grid for now. % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; R % CRAT CIRCL r = ; a = [:N-1]*d; ya = [:Ny-1]*dy; a = a - mean(a); ya = ya - mean(ya); [Y,X] = meshgrid(ya,a); R = (X.^ + Y.^) <= r^; 5 6

27 Grid Technique (6 of 9) Given the object on the grid, we etract R by grabbing values from R that correspond to the locations of R. R R % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; % CRAT CYLINR r = ; a = [:N-1]*d; ya = [:Ny-1]*dy; a = a - mean(a); ya = ya - mean(ya); [Y,X] = meshgrid(ya,a); R = (X.^ + Y.^) <= r^; % XTRACT 1X GRI PARAMTRS R = R(::N,1::Ny); 53 Grid Technique (7 of 9) We then etract Ryy by grabbing values from R that correspond to the locations of Ryy. Ryy R R % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; % CRAT CYLINR r = ; a = [:N-1]*d; ya = [:Ny-1]*dy; a = a - mean(a); ya = ya - mean(ya); [Y,X] = meshgrid(ya,a); R = (X.^ + Y.^) <= r^; % XTRACT 1X GRI PARAMTRS R = R(::N,1::Ny); Ryy = R(1::N,::Ny); 54 7

28 Grid Technique (8 of 9) R % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; Ryy R and Ryy are the outputs of the grid technique. They are defined on the original 1 grid. % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; % CRAT CYLINR r = ; a = [:N-1]*d; ya = [:Ny-1]*dy; a = a - mean(a); ya = ya - mean(ya); [Y,X] = meshgrid(ya,a); R = (X.^ + Y.^) <= r^; % XTRACT 1X GRI PARAMTRS R = R(::N,1::Ny); Ryy = R(1::N,::Ny); 55 Grid Technique (9 of 9) After building R and Ryy, the grid is no longer used anywhere. % FIN GRI N = 5; Ny = 6; d = 1; dy = 1; All of the grid parameters may be deleted at this point because they are no longer needed. R R Ryy % X GRI N = *N; Ny = *Ny; d = d/; dy = dy/; % CRAT CYLINR r = ; a = [:N-1]*d; ya = [:Ny-1]*dy; a = a - mean(a); ya = ya - mean(ya); [Y,X] = meshgrid(ya,a); R = (X.^ + Y.^) <= r^; % XTRACT 1X GRI PARAMTRS R = R(::N,1::Ny); Ryy = R(1::N,::Ny); No more grid!!! 56 8

29 Implementation: Solving the Matri quation 57 Are We Ready to Solve? So far we have discussed: (1) Construction of the permittivity matrices, and yy. () Construction of the derivative operators, e and e y. (3) Construction of the matri problem. That puts us here: v e e ε y v v εyy y How do we solve this for v? v d d ε L y v εyy y Lv 1 v L We can only find a trivial solution! 58 9

30 What is Missing? Lv v We are missing the ecitation b The forced potentials. The metals of the transmission line must be set to some known voltage. Lv b vl 1 b How do we construct b? 59 Constructing b (1 of ) It turns out we must also modify L in addition to constructing b. For each point where there is a forced potential, we do the following to the corresponding row in the matri equation: 1. Replace entire row in L with all zeros.. Place a 1 along the diagonal element. 3. Place the forced potential in b. # # # # # # 1 V # # # # # # 1 V m Vapplied # # # # # # V NNy1 # # # # # # V N Ny L' v b V The applies voltages are usually just and V, but you may use other values if you are doing something special. Vm V applied 6 3

31 Constructing b ( of ) efine two functions: F Force matri v f forced potentials iagonal matri containing 1 s in the diagonal positions corresponding to values we wish to force (i.e. metals). s otherwise. Column vector containing the forced potentials. Numbers in positions not being forced are ignored. 1. Replace rows in L with all zeros that correspond to metals on the grid. L IF L. Then place a 1 in the diagonal element by adding F to L. LF IF L FL 3. Place the forced potentials in b. The remaining elements in b must be in order to be consistent with Laplace s equation.. b Fv f 61 The Force Matri The force matri F starts off as an array F(,y) containing 1 s in the positions we wish to force to some potential and s everywhere else. We already have two arrays defined this same way that describe the two conductors of the transmission line. We simply combine all of the conductor arrays to construct the force array. SIG GN F = SIG GN + = Last, we construct the force matri by diagonalizing the force array. F = diag(sparse(f(:))); 6 31

32 Complete Matri Solution We are now ready to solve the problem. The scalar potential is calculated as 1 v L b We can then calculate the field from V. v e v v e y y We can then calculate the field from the field and the permittivity tensor. d ε e d ε e y yy y There should be an here. We have removed it from this equation and will incorporate it in our final calculation of distributed capacitance in order to keep our functions normalized. 63 tract the Vector Components We must etract the terms e and e y from column vectors. v = L\b; e = -[VX;VY]*v; M = N*Ny; e = e(1:m); ey = e(m+1:*m); v e v v y e e e y 64 3

33 Reshape Back to Arrays The terms v, e, e y, d, and d y are column vectors, or 1 arrays. These need to be reshaped back to arrays before they can be visualized. v = reshape( v,n,ny); e = reshape(e,n,ny); ey = reshape(ey,n,ny); 65 Implementation: Calculating the Transmission Line Parameters 66 33

34 istributed Capacitance, C The distributed capacitance C is calculated by numerical integration of C da V A This is where we incorporate the constant. Assuming V = 1, this integral is easily evaluated as C T de y C = d. *e*(e*d*dy); 67 istributed Inductance, L First, the distributed capacitance C h for the homogeneous case is calculated. C de T h h h y Ch = dh. *eh*(e*d*dy); Second, the distributed inductance L is calculated from C h. L cc r,h h L = urh/(c^*ch); 68 34

35 Characteristic Impedance, Z Given the distributed inductance L and distributed capacitance C, the characteristic impedance of the transmission line is Z L C Z = sqrt(l/c); 69 ffective Refractive Inde, n eff Given the distributed inductance L and distributed capacitance C, the effective refractive inde n eff of the transmission line is n c LC neff = c*sqrt(l*c); 7 35

36 ample Weird Coa 71 Step 1: Choose a Coaial Transmission Line Choose a transmission line. Look only at the cross section. r 3 r r.15 mm 1 r.5 mm r.7 mm 3.3 r1.3 r r 1 1 Slide 7 36

37 Step : Build Materials Arrays % BUIL INNR CONUCTOR CIN = (RSQ<=r1^); inner conductor % BUIL OUTR CONUCTOR COUT = (RSQ<r3^ & RSQ>=r^); outer conductor R dielectric 1 Grid 1 Grid % BUIL ILCTRIC r3 = (r + r3)/; %middle of outer conductor R = ones(n,ny); %air R = R + (er1-1)*(rsq<r3^ & X<); %dielectric 1 R = R + (er - 1)*(RSQ<r3^ & X>=); %dielectric Grid Slide 73 Step 3: Form iagonal Materials Matrices This task reshapes the dielectric functions into 1 arrays and then places them along the diagonal of two matrices. ε εyy % XTRACT R AN Ryy FROM R R = R(::N,1::Ny); Ryy = R(1::N,::Ny); % FORM IAGONAL PRMITTIVITY MATRICS R = diag(sparse(r(:))); Ryy = diag(sparse(ryy(:))); % FORM PRMITTIVITY TNSOR Z = sparse(n*ny,n*ny); R = [ R, Z ; Z, Ryy ]; Slide 74 37

38 Step 4: Construct erivative Operators This task will be performed by the function tlder(). % CALL FUNCTION TO CONSTRUCT RIVATIV OPRATORS NS = [N Ny]; %grid size RS = [d dy]; %grid resolution [VX,VY,X,Y] = tlder(ns,rs); %build matrices d d 1 1 y y v v y 1 1 y Slide 75 Step 5: Build Matri quations Inhomogeneous Laplace s quation v e e ε L y v εyy y L = [X Y] * R * [VX ; VY]; Homogeneous Laplace s quation v e e L h y v y Lh = [X Y] * [VX ; VY]; L Slide 76 38

39 Step 6: Force Known Potentials This same procedure should be done for both the inhomogeneous and homogeneous matri equations. V 1 V= everywhere in the outer conductor. V=1 everywhere in the inner conductor. V % FORC MATRIX F = CIN COUT; F = diag(sparse(f(:))); % FORC POTNTIALS vf = 1*CIN + *COUT; % FORC KNOWN POTNTIALS I = speye(m,m); L = (I - F)*L + F; Lh = (I - F)*Lh + F; b = F*vf(:); Slide 77 Step 7: Compute the Fields Compute the Potentials h 1 v L b 1 h v L b WARNING: O NOT compute the inverse of L. % COMPUT POTNTIALS v = L\b; vh = Lh\b; Compute the Fields e v v v e y y v h, v e yh, y e v % COMPUT FILS e = - [ VX ; VY ] * v; eh = - [ VX ; VY ] * vh; h Compute the Fields d ε e dy εyyey dh, eh, dyh, eyh, Note: We dropped the in these equations and will incorporate that when we calculate C. % COMPUT FILS d = R*e; dh = eh; Slide 78 39

40 Step 8: Calculate TL Parameters istributed Capacitance C da V A Note: We added the constant here. % ISTRIBUT CAPACITANC C = d. *e*(e*d*dy); We have assumed V = 1. Characteristic Impedance L Z C % CHARACTRISTIC IMPANC Z = sqrt(l/c); istributed Inductance C da h h h V A 1 L cc h % ISTRIBUT INUCTANC Ch = dh.'*eh*(e*d*dy); L = 1/(c^*Ch); ffective Refractive Inde n c LC % FFCTIV RFRACTIV INX neff = c*sqrt(l*c); Recall that we have 1 rh We have assumed V = 1. rh Slide 79 Step 9: Reshape the ata We need to reshape the column vectors back to a grid. V, y # # v # # % RSHAP TH FUNCTIONS BACK TO A GRI Vh = reshape(vh,n,ny); V = reshape(v,n,ny); Slide 8 4

41 Step 1: Show the results! Problem definition: Problem results: r.3 mm 1 r.5 mm r.7 mm Standard RG 59 C 7.73 pf m L nh m Z 71.5 n 1.5 Slide 81 Step 1: Another Case Problem definition: r.1 mm 1 r 1.5 mm r 1.6 mm C 11.1 pf m L 551. nh m Z 67.5 n.45 Problem results: Slide 8 41