ECE 6382 Functions of a Complex Variable as Mappings David R. Jackson Notes are adapted from D. R. Wilton, Dept. of ECE 1

A Function of a Complex Variable as a Mapping 2

Simple Mappings: Translations 3

Simple Mappings: Rotations 4

Simple Mappings: Dilations 5 Note:

A General Linear Transformation (Mapping) is a Combination of Translation, Rotation, and Dilation 6 Shapes do not change!

Simple Mappings: Inversions 7

Inversion: Line-to-Circle Property 8

Inversion: Line-to-Circle Property (cont.) 9

Simple Mappings: Inversions (cont.) 10

Circle Property of Inversion Mapping 11 Consider: This is in the form Hence (This maps circles into circles.) J. W. Brown and R. V. Churchill, Complex Variables and Applications, 9 th Ed., McGraw-Hill, 2013.

Circle Property of Inversion Mapping (cont.) 12 Multiply by u 2 + v 2 : This is in the form of a circle: or

A General Bilinear Transformation (Mapping) Is a Succession of Translations, Rotations, Dilations, and Inversions 13

Bilinear Transformation Example: The Smith Chart Bilinear Transformation Example: The Smith Chart For an interpretation of mobius transformations as projections on a sphere, see 14

The Squaring Transformation 15

Another Representation of the Squaring Transformation o 180 o 270 o 360 o -180 o -270 o -360 o 0o0o -90 o Re Im 16

The Square Root Transformation Principal branch Second branch The principal branch is the choice in MATLAB and most programming languages! 17 We say that there are two “branches” (i.e., values) of the square root function. Note that for a given branch (e.g., the principal branch), the square root function is not continuous on the negative real axis. (There is a “branch cut” there.)

o 45 o 67.5 o 90 o -45 o o -90 o 0o0o o Top sheet, k = o 225 o o 270 o 135 o 90 o 180 o o Bottom sheet, k = o 18 The Square Root Transformation (cont.)

Constant u and v Contours are Orthogonal 19

Constant u and v Contours are Orthogonal (cont.) Example: so Also, recall that 20

Mappings of Analytic Functions are Conformal (Angle-Preserving) 21 Hence

Constant u and v Contours are Orthogonal (Revisited) Since the contours u = constant and v = constant are (obviously) orthogonal in the w plane, they must remain orthogonal in the z plane. 22

Constant | w | and Arg( w) Contours are also Orthogonal 23

The Logarithm Function 24 There are an infinite number of branches (values) for the ln function!

Arbitrary Powers of Complex Numbers 25 ( a may be complex)

Arbitrary Powers of Complex Numbers (cont.) For z p/q the repetition period is k=q. For irrational powers, the repetition period is infinite; i.e., values never repeat! 26

Conformal Mapping 27 This is a method for solving 2D problems involving Laplace’s equation. J. W. Brown and R. V. Churchill, Complex Variables and Applications, 9 th Ed., McGraw-Hill, 2013.

Conformal Mapping (cont.) 28 z plane w plane C  A complicated boundary in the z plane is mapped into a simple one in the w plane -- for which we know how to solve the Laplace equation. The function f(z) is assumed to be analytic in the region of interest.

Conformal Mapping (cont.) 29 Theorem: If  ( u,v ) satisfies the Laplace equation in the ( u, v ) plane, then  ( x,y ) satisfies the Laplace equation in the ( x, y ) plane. Proof: Assume that or We want to prove that

Conformal Mapping (cont.) 30

Conformal Mapping (cont.) 31

Conformal Mapping (cont.) 32 Use Cauchy-Riemann equations (red and blue terms):

Conformal Mapping (cont.) 33 Hence, we have (proof complete) Recall that

Conformal Mapping (cont.) 34 Special cases: Clearly, in either case we have or Hence, we have The functions u and v, the real and imaginary parts of an analytic function, satisfy Laplace’s equation.

Conformal Mapping (cont.) 35 Theorem: If  ( u,v ) satisfies Dirichlet or Neumann boundary conditions in the ( u, v ) plane, then  ( x,y ) satisfies the same boundary conditions in the ( x, y ) plane. Proof: Assume that Then we immediately have that since

Conformal Mapping (cont.) 36 Next, assume that Because of the angle-preserving (conformal) property of analytic functions, we have: Hence,

Conformal Mapping (cont.) 37 Theorem: The capacitance (per unit length) between two conductive objects remains unchanged between the z and w planes. Proof:

Conformal Mapping (cont.) 38 Similarly,

Conformal Mapping (cont.) 39 We have: so that or Hence, we have Therefore,

Example 40 Solve for the potential inside and outside of a semi-infinite parallel-plate capacitor.

41 Example (cont.)

42 Example (cont.) The corresponding colored points show the mapping along the top plate.

Example (cont.) 43 This is an ideal parallel-plate capacitor, whose solution is simple:

Example (cont.) 44 For any given ( x, y ), these two equations have to be solved numerically to find ( u,v ). where The solution is:

Example 45 Solve for the potential surrounding a metal strip, and the surface charge density on the strip. Note: The potential goes to -  as   .

46 The outside of the circle gets mapped into the entire z plane. Example (cont.)

47 Outside the circle, we have (from simple electrostatic theory): Example (cont.) Note: To be more general, we could use Changing the constant A 1 changes the voltage on the strip. Changing the constant A 2 changes the total charge on the strip.

48 Hence, we have: For any given ( x, y ), these two equations can be solved numerically to find ( R,  ). Example (cont.)

49 One the upper surface of the strip, we have: Example (cont.) Charge density on strip Also,

Hence, we have: Example (cont.) Hence, 50

Hence, we have: Example (cont.) Use so or 51

Example (cont.) On the circle: so 52

Example (cont.) Hence, we have so This result was first derived by Maxwell! Note that the surface charge density goes to infinity as we approach the edges. On the strip: 53

Example (cont.) w x y Note: The normalization of 1/  corresponds to a unity total line charge density: The strip now has a width of w, and the total line charge density is assumed to be  l [C/m]. 54

Example (cont.) w h x y Microstrip line Note: The increased current density near the edges causes increased conductor loss. 55

Example 56

Example (cont.) 57 We therefore have