The Smith Chart provides a graphical tool for analyzing transmission lines and impedance matching circuits. It maps the reflection coefficient Γ onto a two-dimensional plane defined by the magnitude and phase of Γ. Transformations of Γ along a transmission line correspond to moving along a circular arc on the Smith Chart, allowing problems involving transmission line impedance transformations to be solved graphically. Special cases like quarter-wave and half-wave transmission lines correspond to rotations of 90 and 180 degrees on the Smith Chart.
P, NP, NP-Complete, and NP-Hard
Reductionism in Algorithms
NP-Completeness and Cooks Theorem
NP-Complete and NP-Hard Problems
Travelling Salesman Problem (TSP)
Travelling Salesman Problem (TSP) - Approximation Algorithms
PRIMES is in P - (A hope for NP problems in P)
Millennium Problems
Conclusions
Lab 9 D-Flip Flops: Shift Register and Sequence CounterKatrina Little
This document describes an experiment involving designing a 4-bit shift register and sequence counter using D-flip flops. It includes building the circuits in an FPGA tool, simulating their operation, and downloading them to a development board. A debouncing circuit is added to prevent erroneous output from noisy button inputs. The objectives of introducing sequential circuit design and implementing a shift register and sequence counter are met.
This document discusses methods for solving systems of linear equations. It covers direct methods like Gaussian elimination and LU factorization. Gaussian elimination reduces a system of equations to upper triangular form using elementary row operations. LU factorization expresses the coefficient matrix as the product of a lower triangular matrix and an upper triangular matrix. The document provides examples to demonstrate Gaussian elimination with partial pivoting and solving a system using LU factorization. Iterative methods are also introduced as an alternative to direct methods for large systems.
The document discusses several iterative methods for solving systems of equations, including Jacobi iteration, Gauss-Seidel method, and relaxation methods. The Gauss-Seidel method is commonly used and improves guesses for unknowns using values from the current iteration. Convergence is determined by calculating the relative percent change between iterations. Relaxation methods can enhance convergence by taking a weighted average of old and new values at each iteration.
The document defines the gamma and beta functions and provides examples of using them to evaluate integrals. The gamma function Γ(n) generalizes the factorial function to real and complex numbers. It satisfies properties like Γ(n+1)=nΓ(n). The beta function B(m,n) defines integrals over the interval [0,1]. It relates to the gamma function as B(m,n)=Γ(m)Γ(n)/Γ(m+n). Several integrals are evaluated using these functions, including changing variables to match their definitions. Proofs are also given for relationships between beta function integrals over [0,1] and [0,π/2].
Derivation of Laplace equation (2D Heat Equation)Manish Khose
1. The document describes heat transfer through a metal plate over time using a partial differential equation.
2. It derives an equation for the net heat gain of the plate based on the temperature distribution and thermal properties.
3. Taking the limit as the plate dimensions approach zero, it arrives at the heat equation, which relates the temperature change over time to conduction through the plate.
This document contains exercises involving functions and graphs. There are multiple choice and free response questions testing understanding of properties of functions including domain, range, intercepts, maxima/minima, and graphing functions. Solutions are provided for checking work.
P, NP, NP-Complete, and NP-Hard
Reductionism in Algorithms
NP-Completeness and Cooks Theorem
NP-Complete and NP-Hard Problems
Travelling Salesman Problem (TSP)
Travelling Salesman Problem (TSP) - Approximation Algorithms
PRIMES is in P - (A hope for NP problems in P)
Millennium Problems
Conclusions
Lab 9 D-Flip Flops: Shift Register and Sequence CounterKatrina Little
This document describes an experiment involving designing a 4-bit shift register and sequence counter using D-flip flops. It includes building the circuits in an FPGA tool, simulating their operation, and downloading them to a development board. A debouncing circuit is added to prevent erroneous output from noisy button inputs. The objectives of introducing sequential circuit design and implementing a shift register and sequence counter are met.
This document discusses methods for solving systems of linear equations. It covers direct methods like Gaussian elimination and LU factorization. Gaussian elimination reduces a system of equations to upper triangular form using elementary row operations. LU factorization expresses the coefficient matrix as the product of a lower triangular matrix and an upper triangular matrix. The document provides examples to demonstrate Gaussian elimination with partial pivoting and solving a system using LU factorization. Iterative methods are also introduced as an alternative to direct methods for large systems.
The document discusses several iterative methods for solving systems of equations, including Jacobi iteration, Gauss-Seidel method, and relaxation methods. The Gauss-Seidel method is commonly used and improves guesses for unknowns using values from the current iteration. Convergence is determined by calculating the relative percent change between iterations. Relaxation methods can enhance convergence by taking a weighted average of old and new values at each iteration.
The document defines the gamma and beta functions and provides examples of using them to evaluate integrals. The gamma function Γ(n) generalizes the factorial function to real and complex numbers. It satisfies properties like Γ(n+1)=nΓ(n). The beta function B(m,n) defines integrals over the interval [0,1]. It relates to the gamma function as B(m,n)=Γ(m)Γ(n)/Γ(m+n). Several integrals are evaluated using these functions, including changing variables to match their definitions. Proofs are also given for relationships between beta function integrals over [0,1] and [0,π/2].
Derivation of Laplace equation (2D Heat Equation)Manish Khose
1. The document describes heat transfer through a metal plate over time using a partial differential equation.
2. It derives an equation for the net heat gain of the plate based on the temperature distribution and thermal properties.
3. Taking the limit as the plate dimensions approach zero, it arrives at the heat equation, which relates the temperature change over time to conduction through the plate.
This document contains exercises involving functions and graphs. There are multiple choice and free response questions testing understanding of properties of functions including domain, range, intercepts, maxima/minima, and graphing functions. Solutions are provided for checking work.
This document discusses state diagrams and state tables for sequential circuits. It describes how state diagrams use circles for system states and arcs for transitions between states due to events. State tables have sections for present state, next state, and output. They show the state of flip-flops before and after a clock pulse and output values. Examples of SR, JK, D and T flip-flops are provided along with their characteristic equations and state diagrams. Characteristic equations define the next state in terms of the present state and inputs.
This document provides an overview of Python basics including:
1. How to print and comment in Python code
2. Taking input and manipulating strings
3. Converting between data types and working with dates and times
4. Making decisions with conditional statements and loops
5. Working with lists, tuples, and dictionaries
6. Handling exceptions
The document discusses designing state machines using state diagrams. It describes a state machine to control tail lights on a 1965 Ford Thunderbird with three lights on each side. The state machine has three inputs (left turn, right turn, hazard) and six outputs. It provides steps to design the state diagram, including ensuring it is mutually exclusive and all inclusive. It also describes techniques for synthesizing the state machine using a transition list, including writing transition equations and excitation equations. The document discusses variations in the scheme, such as output-coded state assignment and decomposing large state machines.
This document discusses digital logic circuits and binary logic. It begins with an overview of binary logic, logic gates like NAND, NOR and XOR, and Boolean algebra. It then covers analog vs digital signals, quantization, and converting between analog and digital formats. Various representations of digital designs are presented, including truth tables, Boolean algebra, and schematics. Common logic gates and their representations are described. The document discusses design methodologies and analyses, as well as simulation of logic circuits. It also covers elementary binary logic functions, basic identities of Boolean algebra, and converting between Boolean expressions and logic circuits.
Longest common subsequences in Algorithm AnalysisRajendran
The document discusses dynamic programming and the longest common subsequence (LCS) problem. It provides an example of using dynamic programming to find the LCS of two strings "ABCB" and "BDCAB" in multiple steps. The algorithm builds up the solution by filling a 2D array c[m,n] where m and n are the lengths of the two strings. Each entry c[i,j] represents the length of the LCS of the prefixes of length i and j of the two strings. The running time is O(mn).
The document discusses various optimization problems that can be solved using the greedy method. It begins by explaining that the greedy method involves making locally optimal choices at each step that combine to produce a globally optimal solution. Several examples are then provided to illustrate problems that can and cannot be solved with the greedy method. These include shortest path problems, minimum spanning trees, activity-on-edge networks, and Huffman coding. Specific greedy algorithms like Kruskal's algorithm, Prim's algorithm, and Dijkstra's algorithm are also covered. The document concludes by noting that the greedy method can only be applied to solve a small number of optimization problems.
This document provides an overview of pushdown automata (PDA). It defines a PDA as a finite automaton with an additional memory stack. This stack allows two operations - push, which adds a new symbol to the top of the stack, and pop, which removes and reads the top symbol. The document then discusses the formal definition of a PDA as a septuple and provides an example of a PDA that accepts the language of strings with an equal number of 0s and 1s. It concludes with an explanation of the state operations of replace, push, pop and no change and conditions for PDA acceptance and rejection.
Computer graphics - bresenham line drawing algorithmRuchi Maurya
You know that DDA algorithm is an incremental scan conversion method which performs calculations at each step using the results from the preceding step. Here we are going to discover an accurate and efficient raster line generating algorithm, the Bresenham's line-drawing algorithm.
This algorithm was developed by Jack E. Bresenham in 1962 at IBM.
The function given below handles all lines and implements the complete Bresenham's algorithm.
function line(x0, x1, y0, y1)
boolean steep := abs(y1 - y0) > abs(x1 - x0)
if steep then
swap(x0, y0)
swap(x1, y1)
if x0 > x1 then
swap(x0, x1)
swap(y0, y1)
int deltax := x1 - x0
int deltay := abs(y1 - y0)
real error := 0
real deltaerr := deltay / deltax
int y := y0
if y0 < y1 then ystep := 1 else ystep := -1
for x from x0 to x1
if steep then plot(y,x) else plot(x,y)
error := error + deltaerr
if error ? 0.5
y := y + ystep
error := error - 1.0
Synchronous down counter with full description.
All the flip-flop are clocked simultaneously.
Synchronous counters can operate at much higher frequencies than asynchronous counters.
As clock is simultaneously given to all flip-flops there is no problem of propagation delay. Hence they are high speed counters and are preferred when number of flip-flops increase's in the given design.
In this counter will counter
The one-dimensional heat equation describes heat flow along a rod. It can be solved using separation of variables. For a rod with insulated sides initially at uniform temperature u0 and ends suddenly cooled to 0°C:
1) The solution is a Fourier series involving eigenfunctions that satisfy the boundary conditions.
2) The temperature is the sum of the eigenfunctions weighted by Fourier coefficients involving u0.
3) As time increases, the temperature decreases towards the boundary values according to exponential decay governed by the eigenvalues.
The document discusses window to viewport transformation. It defines a window as a world coordinate area selected for display and a viewport as a rectangular region of the screen selected for displaying objects. Window to viewport mapping requires transforming coordinates from the window to the viewport. This involves translation, scaling and another translation. Steps include translating the window to the origin, resizing it based on the viewport size, and translating it to the viewport position. An example transforms a sample window to a viewport through these three steps.
This document describes the digital differential analyzer (DDA) algorithm for rasterizing lines, triangles, and polygons in computer graphics. It discusses implementing DDA using floating-point or integer arithmetic. The DDA line drawing algorithm works by incrementing either the x or y coordinate by 1 each step depending on whether the slope is less than or greater than 1. Pseudocode is provided to illustrate the algorithm. Potential drawbacks of DDA are also mentioned, such as the expense of rounding operations.
The document discusses Bram Cohen's view that Python is a good language for maintainability as it has clean syntax, object encapsulation, good library support, and optional parameters, and then provides details about the history and features of the Python programming language such as being dynamically typed, having a large standard library, and being cross-platform.
These slides have full understanding about Equivalent Moore Mealy... Having Moore to Mealy conversion and Mealy to Moore conversion...
These slides also describing the concept of Transducers as models of sequential circuits (both w.r.t Moore and Mealy)...
All these concepts are explained with easy examples...
This document outlines numerical methods for finding roots of nonlinear equations presented by Dr. Eng. Mohammad Tawfik. It introduces the fixed point, Newton-Raphson, and secant methods. The fixed point method rearranges the equation to an iterative form where the next estimate is a function of the previous. Newton-Raphson linearizes the function to get faster convergence. The secant method does not require derivatives by using the slope between previous points. Convergence conditions and algorithms are provided for each method. Students are assigned homework problems from the textbook.
This document outlines a multi-step process for designing T and pi matching networks using a Smith chart. It involves first completing the Smith chart with impedance curves, then taking two further steps which involve matching points on the chart and adding network components represented by symbols.
The 3D Smith Chart program is a new Java tool for visualizing and designing active and passive microwave circuits. It generalizes the traditional 2D Smith chart onto the surface of a sphere, addressing limitations of the 2D chart. Key features include reading measurement files, plotting reflection coefficients for complex impedances, and aiding oscillator and amplifier design by visualizing infinite mismatch and stability circles. The tool aims to provide a complete graphical solution for microwave circuit measurement and design.
This document discusses state diagrams and state tables for sequential circuits. It describes how state diagrams use circles for system states and arcs for transitions between states due to events. State tables have sections for present state, next state, and output. They show the state of flip-flops before and after a clock pulse and output values. Examples of SR, JK, D and T flip-flops are provided along with their characteristic equations and state diagrams. Characteristic equations define the next state in terms of the present state and inputs.
This document provides an overview of Python basics including:
1. How to print and comment in Python code
2. Taking input and manipulating strings
3. Converting between data types and working with dates and times
4. Making decisions with conditional statements and loops
5. Working with lists, tuples, and dictionaries
6. Handling exceptions
The document discusses designing state machines using state diagrams. It describes a state machine to control tail lights on a 1965 Ford Thunderbird with three lights on each side. The state machine has three inputs (left turn, right turn, hazard) and six outputs. It provides steps to design the state diagram, including ensuring it is mutually exclusive and all inclusive. It also describes techniques for synthesizing the state machine using a transition list, including writing transition equations and excitation equations. The document discusses variations in the scheme, such as output-coded state assignment and decomposing large state machines.
This document discusses digital logic circuits and binary logic. It begins with an overview of binary logic, logic gates like NAND, NOR and XOR, and Boolean algebra. It then covers analog vs digital signals, quantization, and converting between analog and digital formats. Various representations of digital designs are presented, including truth tables, Boolean algebra, and schematics. Common logic gates and their representations are described. The document discusses design methodologies and analyses, as well as simulation of logic circuits. It also covers elementary binary logic functions, basic identities of Boolean algebra, and converting between Boolean expressions and logic circuits.
Longest common subsequences in Algorithm AnalysisRajendran
The document discusses dynamic programming and the longest common subsequence (LCS) problem. It provides an example of using dynamic programming to find the LCS of two strings "ABCB" and "BDCAB" in multiple steps. The algorithm builds up the solution by filling a 2D array c[m,n] where m and n are the lengths of the two strings. Each entry c[i,j] represents the length of the LCS of the prefixes of length i and j of the two strings. The running time is O(mn).
The document discusses various optimization problems that can be solved using the greedy method. It begins by explaining that the greedy method involves making locally optimal choices at each step that combine to produce a globally optimal solution. Several examples are then provided to illustrate problems that can and cannot be solved with the greedy method. These include shortest path problems, minimum spanning trees, activity-on-edge networks, and Huffman coding. Specific greedy algorithms like Kruskal's algorithm, Prim's algorithm, and Dijkstra's algorithm are also covered. The document concludes by noting that the greedy method can only be applied to solve a small number of optimization problems.
This document provides an overview of pushdown automata (PDA). It defines a PDA as a finite automaton with an additional memory stack. This stack allows two operations - push, which adds a new symbol to the top of the stack, and pop, which removes and reads the top symbol. The document then discusses the formal definition of a PDA as a septuple and provides an example of a PDA that accepts the language of strings with an equal number of 0s and 1s. It concludes with an explanation of the state operations of replace, push, pop and no change and conditions for PDA acceptance and rejection.
Computer graphics - bresenham line drawing algorithmRuchi Maurya
You know that DDA algorithm is an incremental scan conversion method which performs calculations at each step using the results from the preceding step. Here we are going to discover an accurate and efficient raster line generating algorithm, the Bresenham's line-drawing algorithm.
This algorithm was developed by Jack E. Bresenham in 1962 at IBM.
The function given below handles all lines and implements the complete Bresenham's algorithm.
function line(x0, x1, y0, y1)
boolean steep := abs(y1 - y0) > abs(x1 - x0)
if steep then
swap(x0, y0)
swap(x1, y1)
if x0 > x1 then
swap(x0, x1)
swap(y0, y1)
int deltax := x1 - x0
int deltay := abs(y1 - y0)
real error := 0
real deltaerr := deltay / deltax
int y := y0
if y0 < y1 then ystep := 1 else ystep := -1
for x from x0 to x1
if steep then plot(y,x) else plot(x,y)
error := error + deltaerr
if error ? 0.5
y := y + ystep
error := error - 1.0
Synchronous down counter with full description.
All the flip-flop are clocked simultaneously.
Synchronous counters can operate at much higher frequencies than asynchronous counters.
As clock is simultaneously given to all flip-flops there is no problem of propagation delay. Hence they are high speed counters and are preferred when number of flip-flops increase's in the given design.
In this counter will counter
The one-dimensional heat equation describes heat flow along a rod. It can be solved using separation of variables. For a rod with insulated sides initially at uniform temperature u0 and ends suddenly cooled to 0°C:
1) The solution is a Fourier series involving eigenfunctions that satisfy the boundary conditions.
2) The temperature is the sum of the eigenfunctions weighted by Fourier coefficients involving u0.
3) As time increases, the temperature decreases towards the boundary values according to exponential decay governed by the eigenvalues.
The document discusses window to viewport transformation. It defines a window as a world coordinate area selected for display and a viewport as a rectangular region of the screen selected for displaying objects. Window to viewport mapping requires transforming coordinates from the window to the viewport. This involves translation, scaling and another translation. Steps include translating the window to the origin, resizing it based on the viewport size, and translating it to the viewport position. An example transforms a sample window to a viewport through these three steps.
This document describes the digital differential analyzer (DDA) algorithm for rasterizing lines, triangles, and polygons in computer graphics. It discusses implementing DDA using floating-point or integer arithmetic. The DDA line drawing algorithm works by incrementing either the x or y coordinate by 1 each step depending on whether the slope is less than or greater than 1. Pseudocode is provided to illustrate the algorithm. Potential drawbacks of DDA are also mentioned, such as the expense of rounding operations.
The document discusses Bram Cohen's view that Python is a good language for maintainability as it has clean syntax, object encapsulation, good library support, and optional parameters, and then provides details about the history and features of the Python programming language such as being dynamically typed, having a large standard library, and being cross-platform.
These slides have full understanding about Equivalent Moore Mealy... Having Moore to Mealy conversion and Mealy to Moore conversion...
These slides also describing the concept of Transducers as models of sequential circuits (both w.r.t Moore and Mealy)...
All these concepts are explained with easy examples...
This document outlines numerical methods for finding roots of nonlinear equations presented by Dr. Eng. Mohammad Tawfik. It introduces the fixed point, Newton-Raphson, and secant methods. The fixed point method rearranges the equation to an iterative form where the next estimate is a function of the previous. Newton-Raphson linearizes the function to get faster convergence. The secant method does not require derivatives by using the slope between previous points. Convergence conditions and algorithms are provided for each method. Students are assigned homework problems from the textbook.
This document outlines a multi-step process for designing T and pi matching networks using a Smith chart. It involves first completing the Smith chart with impedance curves, then taking two further steps which involve matching points on the chart and adding network components represented by symbols.
The 3D Smith Chart program is a new Java tool for visualizing and designing active and passive microwave circuits. It generalizes the traditional 2D Smith chart onto the surface of a sphere, addressing limitations of the 2D chart. Key features include reading measurement files, plotting reflection coefficients for complex impedances, and aiding oscillator and amplifier design by visualizing infinite mismatch and stability circles. The tool aims to provide a complete graphical solution for microwave circuit measurement and design.
The Smith Chart is a graphical tool developed in the 1930s to help solve radio frequency circuit problems. It displays impedances on a chart using a system of circles and arcs. Impedances are plotted based on their resistance (circle) and reactance (arc) values after normalizing to the system characteristic impedance. In this example, the Smith Chart is used to calculate the input impedance seen at the transmitter given the load impedance of an antenna and the electrical length of the transmission line connecting them.
The document discusses the Smith chart, a tool for analyzing transmission lines. It describes how the Smith chart allows transmission line impedances, voltages, currents, and other parameters to be normalized and plotted. Examples are provided to illustrate how the Smith chart can be used to locate a load impedance and draw reflection coefficient circles, as well as to solve problems like matching a transmission line to a load using a stub. The document aims to provide supplemental information on using the Smith chart for transmission line analysis.
Smith chart:A graphical representation.amitmeghanani
The document discusses the Smith chart, which is a graphical tool used to solve transmission line problems. Some key points:
- The Smith chart was developed in 1939 and allows tedious transmission line calculations to be done graphically.
- It provides a mapping between the normalized impedance plane and the reflection coefficient plane. Circles of constant resistance and reactance are plotted, along with the reflection coefficient.
- Parameters like impedance, admittance, reflection coefficient, VSWR can all be plotted and derived from locations on the chart.
- Examples are given of using the Smith chart to determine input impedance, reflection coefficient, and stub matching of transmission lines with various termination impedances.
The Smith chart is a graphical method that is essential for microwave engineering. It allows microwave engineers to represent normalized impedances on a chart. Developed by Philip Smith in 1939, the chart maps the reflection coefficient Γ which relates the load and source impedances. It uses constant resistance and reactance circles to plot impedance points. Microwave engineers can use the Smith chart and vector network analyzers to measure reflection coefficients over frequency sweeps.
This document provides an overview of transmission lines and matching techniques for high-frequency circuit design. It begins with basic transmission line models and the Telegrapher's equations that describe voltage and current propagation along a transmission line. It then discusses transmission line properties such as characteristic impedance and reflection coefficient. The Smith chart is introduced as a tool for visualizing impedance transformations along a transmission line of varying lengths. Finally, the document covers matching conditions and techniques for minimizing reflections, including using a single stub to transform the impedance to match the characteristic impedance of the transmission line.
The document describes how the Smith Chart maps impedances on the normalized complex impedance plane to the complex Γ plane. Vertical lines on the impedance plane representing constant resistance values r map to circles on the Γ plane, with centers located along the Γi = 0 line. Horizontal lines representing constant reactance values x also map to circles, with centers located along the Γr = 1 line. By mapping many such lines, the rectilinear grid of the impedance plane is distorted into the curvilinear grid of circles that make up the Smith Chart.
Brief 5AC RL and RC CircuitsElectrical Circuits Lab VannaSchrader3
Brief 5
AC RL and RC Circuits
Electrical Circuits Lab I
(ENGR 2105)
Dr. Kory Goldammer
Review of Complex Numbers and Transforms
Transforms
The Polar Coordinates / Rectangular Coordinates Transform
The Complex Plane
We can use complex numbers to solve for the phase shift in AC Circuits
Instead of (x,y) coordinates, we define a point in the Complex plane by (real, imaginary) coordinates
Real numbers are on the horizontal axis
Imaginary numbers are on the vertical axis
The Complex Plane (cont.)
Imaginary numbers are multiplied by j
By definition,
(Mathematicians use i instead of j, but that would confuse us since i stands for current in this class)
Imaginary Plane: Rectangular Coordinates
We can identify any point in the 2D plane using (real, imaginary) coordinates
Complex Plane Using Rectangular Coordinates
Imaginary Plane: Transform to Polar Coordinates
We can identify any point in the complex plane using (r,) coordinates.
The arrow is called a Phasor.
r is the length of the Phasor, and is the angle between the Positive Real Axis and the Phasor
is the Phase Angle we want to calculate
Complex Plane Using Rectangular Coordinates
r
Imaginary Plane: Transform to Polar Coordinates
Complex Plane Using Rectangular Coordinates
(we will discuss the meaning of r later)
Use either the sin or cos term to find :
But we need in radians:
r
=7.07
Complex Math
For addition or subtraction, add or subtract the real and j terms separately.
(3 + j4) + (2 – j2) = 5 + j2
To multiply or divide a j term by a real number, multiply or divide the numbers. The answer is still a j term.
5 * j6 = j30
-2 * j3 = -j6
j10 / 2 = j5
Complex Math (cont. 1)
To divide a j term by a j term, divide the j coefficients to produce a real number; the j factors cancel.
j10 / j2 = 5
-j6 / j3 = -2
To multiply complex numbers, follow the rules of algebra, noting that j2 = -1
Complex Math (cont. 2)
To divide by a complex number: Can’t be done!
The denominator must first be converted to a Real number!
Complex Conjugation
Converting the denominator to a real number without any j term is called rationalization.
To rationalize the denominator, we need to multiply the numerator and denominator by the complex conjugate
Complex Number Complex Conjugate
5 + j3 5 – j3
–5 + j3 –5 – j3
5 – j3 5 + j3
–5 – j3 –5 + j3
Complex Math (cont. 2)
Multiply the original equation by the complex conjugate divided by itself (again, j2 = -1):
Phase Shift
Time Domain - ω Domain Transforms
Transforming from the Time (Real World) Domain to the (Problem Solving Domain
Note that in the ω Domain, Resistance, Inductance and Capacitance all of units of Ohms!ElementTime Domainω Domain TransformApplied Sinusoidal AC Voltage
(Volts) (ω=2πf)Vp
(Volts)Series Current(Amps) (ω=2πf)(Amps)ResistanceR
(Ohms)R
(Ohms)InductanceL
(Henry’s)(Ohms) (ω=2πf)CapacitanceC
(Farads)(Ohms) (ω=2πf)
Solving For Current
...
This document provides an overview of Smith charts and how they can be used for impedance and admittance calculations. It defines the generalized reflection coefficient Γ and shows how impedance/admittance values map to points on the Smith chart complex Γ plane. Circles on the chart correspond to constant values of normalized impedance/admittance. The document discusses four possible Smith chart variants depending on whether impedance/admittance calculations are being performed in the Γ or Γ' planes.
The document describes results from graph theory, specifically matching theory, flows, coloring, and combinatorial nullstellensatz. It investigates classical theorems like Hall's theorem, Tutte's theorem, Dilworth's theorem, and Nash-Williams theorem on matchings in bipartite and general graphs. It also discusses network flows and Baranyai's theorem. The problems were taken from exercises in a book on modern graph theory.
Lecture Notes - EEEC6430310 Electromagnetic Fields and Waves - Smith ChartAIMST University
The Smith chart is a graphical tool used to analyze high frequency circuits. It represents all possible complex impedances in terms of the reflection coefficient. Circles of constant resistance and arcs of constant reactance intersect on the chart to indicate impedance values. The chart can be used to determine impedances, reflection coefficients, standing wave ratios, and more from various circuit parameters. It provides a clever way to visualize complex impedance functions that continues to be popular for high frequency applications.
1. The document provides examples of using a Smith chart to solve transmission line problems involving impedance matching and locating voltage/current minima.
2. Key information extracted from measurements include VSWR, shift in voltage minima when terminating a line with a short, and load impedance.
3. The Smith chart is used to determine normalized load impedances, line impedances at various distances from the load, and stub parameters for matching a given load.
The document discusses transmission line theory and the propagation of waves on transmission lines. It introduces the lumped element circuit model of a transmission line and derives the telegrapher's equations that describe wave propagation on the line. It then shows how a transmission line can be modeled as a two-port network and discusses wave propagation on lossless transmission lines, including when the line is terminated by different impedances.
1. The document discusses the relationship between multicommodity flow problems and polyhedra related to Seymour's conjecture on binary clutters.
2. Seymour's conjecture states that a binary clutter is weakly bipartite if and only if it does not contain a forbidden minor like K5.
3. Weak bipartiteness of a signed graph is related to the cut condition being sufficient for the existence of multicommodity flows. If the signed graph is weakly bipartite, the cut condition is sufficient.
This document summarizes research on Casimir torque in the weak coupling approximation. It examines manifestations of Casimir torque between planar objects characterized by delta function potentials. The key findings are:
1) An exact calculation of the Casimir torque between a finite rectangular plate above a semi-infinite plate is presented and agrees well with the proximity force approximation when the plate separation is small compared to their sizes.
2) Cusps in the torque arise when the corners of the finite plate pass over the edge of the semi-infinite plate.
3) A similar calculation is done for a disk above a semi-infinite plate, again finding good agreement with the proximity force approximation.
This document discusses transmission line models used in power system analysis. It begins with an overview of the distributed parameter model that represents an infinitesimal length of transmission line using series impedance and shunt admittance. It then derives the telegrapher's equations and uses them to develop a single second-order differential equation for the voltage along the line. The document presents solutions for this equation that allow determining the voltage and current at any point on the line given conditions at one end. It introduces transmission line parameters including characteristic impedance and propagation constant and shows how they relate the sending and receiving end quantities. Equivalent lumped-parameter π models are also derived in terms of the transmission line parameters. Finally, it discusses short
This page is intentionally left blank. The following document discusses algorithms for finding minimum spanning trees (MSTs) in graphs, including:
1) Methods for constructing graphs containing the MST generalize to higher dimensions, though some results are less informative in higher dimensions.
2) Yao's algorithm finds MSTs in higher dimensions in time proportional to the number of points raised to a power that depends on the dimension.
3) Agarwal et al. presented a more efficient method using bichromatic nearest neighbors, showing that finding MST edges is no harder than computing bichromatic nearest neighbors.
1. The document analyzes the forces acting on a rope as it falls over a support. It derives an expression for the force N from the support as a function of time t until the free end of the rope reaches a distance of 2L.
2. It examines the motion of a mass attached to a string winding around a thin pole, and derives an expression for the angle θ at which the string becomes fully unwound using conservation of energy and the horizontal component of tension.
3. It considers electrical resistance in a circuit composed of squares connected at the corners. By simplifying the circuit in stages, it derives an expression for the desired resistance factor x between opposite corners of the original circuit.
1. The document analyzes the forces acting on a rope as it falls over a support. It derives an expression for the force N from the support as a function of time t until the free end of the rope falls a distance of 2L.
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2 4 the_smith_chart_package
1. 2/7/2005
2_4 The Smith Chart
1/2
2.4 – The Smith Chart
Reading Assignment: pp. 64-73
The Smith Chart
The Smith Chart provides:
1)
2)
The most important fact about the Smith Chart is:
HO: The Complex Γ plane
Q: But how is the complex Γ plane useful?
A:
HO: Transformations on the Complex Γ Plane
Q: But transformations of Γ are relatively easy—
transformations of line impedance Z is the difficult one.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2. 2/7/2005
2_4 The Smith Chart
2/2
A:
HO: Mapping Z to Γ
HO: The Smith Chart
HO: Zin Calculations using the Smith Chart
Example: The Input Impedance of a Shorted Transmission
Line
Example: Determining the Load Impedance of a
Transmission Line
Example: Determining the Length of a Transmission Line
Expressing a load or line impedance in terms of its admittance
is sometimes helpful. Additionally, we can easily map
admittance onto the Smith Chart
HO: Impedance and Admittance
Example: Admittance Calculations with the Smith Chart
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3. 2/7/2005
The Complex Gamma Plane
1/6
The Complex Γ Plane
Resistance R is a real value, thus we can indicate specific resistor values
as points on the real line:
R =20 Ω
R =0
R =50 Ω
R
R =5 Ω
Likewise, since impedance Z is a complex value, we can indicate specific
impedance values as point on a two dimensional complex plane:
Im {Z }
Z =30 +j 40 Ω
Re {Z }
Z =60 -j 30 Ω
Note each dimension is defined by a single real line: the horizontal line
(axis) indicating the real component of Z (i.e., Re {Z } ), and the vertical
line (axis) indicating the imaginary component of impedance Z (i.e.,
Im {Z } ). The intersection of these two lines is the point denoting the
impedance Z = 0.
* Note then that a vertical line is formed by the locus of all points
(impedances) whose resistive (i.e., real) component is equal to, say, 75.
* Likewise, a horizontal line is formed by the locus of all points
(impedances) whose reactive (i.e., imaginary) component is equal to -30.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4. 2/7/2005
The Complex Gamma Plane
2/6
Im {Z }
R =75
Re {Z }
X =-30
If we assume that the real component of every impedance is positive,
then we find that only the right side of the plane will be useful for
plotting impedance Z—points on the left side indicate impedances with
negative resistances!
Im {Z }
Invalid
Region
(R<0)
Valid
Region
(R>0)
Re {Z }
Moreover, we find that common impedances such as Z = ∞ (an open
circuit!) cannot be plotted, as their points appear an infinite distance
from the origin.
Im {Z }
Z =0
(short)
Z =Z0
Z = ∞ (open)
(matched)
Re {Z }
Jim Stiles
The Univ. of Kansas
Somewhere way the
heck over there !!
Dept. of EECS
5. 2/7/2005
The Complex Gamma Plane
3/6
Q: Yikes! The complex Z plane does not appear to be a very helpful.
Is there some graphical tool that is more useful?
A: Yes! Recall that impedance Z and reflection coefficient Γ are
equivalent complex values—if you know one, you know the other.
We can therefore define a complex Γ plane in the same manner
that we defined a complex impedance plane. We will find that there
are many advantages to plotting on the complex Γ plane, as opposed
to the complex Z plane!
Im {Γ }
Γ =0.3 +j 0.4
Γ =-0.5 +j 0.1
Re {Γ }
Γ =0.6 -j 0.3
Note that the horizontal axis indicates the real component of Γ ( Re {Γ } ),
while the vertical axis indicates the imaginary component of Γ ( Im {Γ } ).
We could plot points and lines on this plane exactly as before:
Im {Γ }
Re {Γ}=0.5
Re {Γ }
Im {Γ} =-0.3
Jim Stiles
The Univ. of Kansas
Dept. of EECS
6. 2/7/2005
The Complex Gamma Plane
4/6
However, we will find that the utility of the complex Γ pane as a graphical
tool becomes apparent only when we represent a complex reflection
coefficient in terms of its magnitude ( Γ ) and phase (θ Γ ):
Γ = Γ e jθΓ
In other words, we express Γ using polar coordinates:
Γ = 0.6 e
j 3π 4
Im {Γ }
Γ
Γ
θΓ
Re {Γ }
Γ = 0.7 e j 300
Note then that a circle is formed by the locus of all points whose
magnitude Γ equal to, say, 0.7. Likewise, a radial line is formed by the
locus of all points whose phase
θ Γ is equal to 135 .
θ Γ = 135
Im {Γ }
Γ = 0.7
Re {Γ }
Jim Stiles
The Univ. of Kansas
Dept. of EECS
7. 2/7/2005
The Complex Gamma Plane
5/6
Perhaps the most important aspect of the complex Γ plane is its validity
region. Recall for the complex Z plane that this validity region was the
right-half plane, where Re {Z } > 0 (i.e., positive resistance).
The problem was that this validity region was unbounded and infinite in
extent, such that many important impedances (e.g., open-circuits) could
not be plotted.
Q: What is the validity region for the complex Γ plane?
A:
Recall that we found that for Re {Z } > 0 (i.e., positive
resistance), the magnitude of the reflection coefficient was
limited:
0 < Γ <1
Therefore, the validity region for the complex Γ plane consists of
all points inside the circle Γ = 1 --a finite and bounded area!
Im {Γ }
Invalid
Region
( Γ > 1)
Valid
Region
( Γ < 1)
Re {Γ }
Γ =1
Jim Stiles
The Univ. of Kansas
Dept. of EECS
8. 2/7/2005
The Complex Gamma Plane
6/6
Note that we can plot all valid impedances (i.e., R >0) within this finite
region!
Im {Γ }
Γ = e j π = −1.0
Γ=0
(matched)
(short)
Re {Γ }
Γ = e j 0 = 1.0
(open)
Γ =1
(Z = jX → purely reactive)
Jim Stiles
The Univ. of Kansas
Dept. of EECS
9. 2/7/2005
Transformations on the Complex
1/7
Transformations on the
Complex Γ Plane
The usefulness of the complex Γ plane is apparent when we
consider again the terminated, lossless transmission line:
z =0
z =−
Z0, β
Γ in
Z0, β
Γ
L
Recall that the reflection coefficient function for any location z
along the transmission line can be expressed as (since z L = 0 ):
Γ ( z ) = ΓL e j 2 β z
= ΓL e
j (θ Γ +2 β z )
And thus, as we would expect:
Γ(z = 0) = ΓL
and
Γ(z = − ) = Γ Le - j 2 β = Γin
Recall this result “says” that adding a transmission line of length
to a load results in a phase shift in θ Γ by −2 β radians, while
the magnitude Γ remains unchanged.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10. 2/7/2005
Transformations on the Complex
2/7
Q: Magnitude Γ and phase θ Γ --aren’t those the
values used when plotting on the complex Γ plane?
A: Precisely! In fact, plotting the
transformation of ΓL to Γin along a transmission
line length has an interesting graphical
interpretation. Let’s parametrically plot Γ ( z )
from z = z L (i.e., z = 0 ) to z = z L −
(i.e., z = − ):
Im {Γ }
Γ (z )
θL
Γ (z = 0 ) = ΓL
ΓL
Γ (z = − ) = Γ
Re {Γ }
θin = θL − 2β
in
Γ =1
Since adding a length of transmission line to a load ΓL modifies
the phase θ Γ but not the magnitude ΓL , we trace a circular arc
as we parametrically plot Γ ( z ) ! This arc has a radius ΓL and an
arc angle 2β radians.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11. 2/7/2005
Transformations on the Complex
3/7
With this knowledge, we can easily solve many interesting
transmission line problems graphically—using the complex Γ
plane! For example, say we wish to determine Γin for a
transmission line length = λ 8 and terminated with a short
circuit.
Z0, β
z =0
z =−
Γ in
Z0 , β
ΓL = −1
=λ 8
The reflection coefficient of a short circuit is ΓL = −1 = 1 e j π ,
and therefore we begin at that point on the complex Γ plane.
We then move along a circular arc −2 β = −2 (π 4 ) = − π 2
radians (i.e., rotate clockwise 90 ).
Im {Γ }
Γ (z )
Γin = 1 e
+jπ2
Re {Γ }
ΓL = 1 e
+ jπ
Γ =1
Jim Stiles
The Univ. of Kansas
Dept. of EECS
12. 2/7/2005
Transformations on the Complex
4/7
When we stop, we find we are at the point for Γin ; in this case
Γin = 1 e j π 2 (i.e., magnitude is one, phase is 90 o ).
Now, let’s repeat this same problem, only with a new
transmission line length of = λ 4 . Now we rotate clockwise
2β = π radians (180 ).
Im {Γ }
Γ (z )
Γin = 1 e
+j π2
Re {Γ }
ΓL = 1 e
+ jπ
Γ =1
For this case, the input reflection coefficient is Γin = 1 e j 0 = 1 :
the reflection coefficient of an open circuit!
Our short-circuit load has been transformed into an open
circuit with a quarter-wavelength transmission line!
But, you knew this would happen—right?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
13. 2/7/2005
Transformations on the Complex
z =0
z =−
Z0, β
Γin = 1
(open)
5/7
Γ L = −1
Z0, β
(short)
=λ 4
Recall that a quarter-wave transmission line was one of the
special cases we considered earlier. Recall we found that the
input impedance was proportional to the inverse of the load
impedance.
Thus, a quarter-wave transmission line transforms a short into
an open. Conversely, a quarter-wave transmission can also
transform an open into a short:
Im {Γ }
Γ =1
ΓL = 1 e
+jπ2
Re {Γ }
Γin = 1 e
+ jπ
Γ (z )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
14. 2/7/2005
Transformations on the Complex
6/7
Finally, let’s again consider the problem where ΓL = −1 (i.e.,
short), only this time with a transmission line length
=λ 2 (a
half wavelength!). We rotate clockwise 2β = 2π radians (360 ).
Hey look! We came clear
around to where we started!
Im {Γ }
Γ (z )
ΓL = 1 e
+ jπ
Re {Γ }
Γin = 1 e
+ jπ
Γ =1
Thus, we find that Γin = ΓL if
= λ 2 --but you knew this too!
Recall that the half-wavelength transmission line is likewise a
special case, where we found that Zin = Z L . This result, of
course, likewise means that Γin = ΓL .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
15. 2/7/2005
Transformations on the Complex
7/7
Now, let’s consider the opposite problem. Say we know that the
input impedance at the beginning of a transmission line with
length = λ 8 is:
Γin = 0.5 e j 60
Q: What is the reflection coefficient of the load?
A: In this case, we begin at Γin and rotate COUNTERCLOCKWISE along a circular arc (radius 0.5) 2β = π 2 radians
(i.e., 60 ). Essentially, we are removing the phase shift
associated with the transmission line!
Im {Γ }
Γ (z )
θ
in
θL = θin + 2β
0.5
Γ in
ΓL
= 0.5 e j 60
Re {Γ }
= 0.5 e j 150
Γ =1
The reflection coefficient of the load is therefore:
ΓL = 0.5 e j 150
Jim Stiles
The Univ. of Kansas
Dept. of EECS
16. 2/7/2005
Mapping Z to Gamma
1/8
Mapping Z to Γ
Recall that line impedance and reflection coefficient are
equivalent—either one can be expressed in terms of the other:
Γ (z ) =
Z (z ) − Z 0
Z (z ) + Z 0
and
⎛ 1 + Γ (z ) ⎞
⎟
1 − Γ (z ) ⎠
⎝
Z (z ) = Z 0 ⎜
Note this relationship also depends on the characteristic
impedance Z0 of the transmission line. To make this relationship
more direct, we first define a normalized impedance value z ′
(an impedance coefficient!):
z ′ (z ) =
Z (z ) R (z )
X (z )
=
+j
= r (z ) + j x (z )
Z0
Z0
Z0
Using this definition, we find:
Γ (z ) =
Z (z ) − Z 0
Z (z ) + Z 0
=
=
Jim Stiles
Z (z ) Z 0 − 1
Z (z ) Z 0 + 1
z ′ (z ) − 1
z ′ (z ) + 1
The Univ. of Kansas
Dept. of EECS
17. 2/7/2005
Mapping Z to Gamma
2/8
Thus, we can express Γ ( z ) explicitly in terms of normalized
impedance z ′ --and vice versa!
z ′ (z ) − 1
Γ (z ) =
z ′ (z ) + 1
z ′ (z ) =
1 + Γ (z )
1 − Γ (z )
The equations above describe a mapping between coefficients
z ′ and Γ . This means that each and every normalized impedance
value likewise corresponds to one specific point on the complex
Γ plane!
For example, say we wish to mark or somehow indicate the
values of normalized impedance z’ that correspond to the
various points on the complex Γ plane.
Some values we already know specifically:
case
z′
Γ
1
∞
∞
1
2
0
0
-1
3
Z0
1
0
4
j Z0
j
j
5
Jim Stiles
Z
−j Z0
j
−j
The Univ. of Kansas
Dept. of EECS
18. 2/7/2005
Mapping Z to Gamma
3/8
Therefore, we find that these five normalized impedances map
onto five specific points on the complex Γ plane:
Γi
Γ =1
z ′ = −j
z′ = 1
( Γ=− j )
z′ = ∞
( Γ=0 )
( Γ=1)
Γr
z′ = 0
( Γ=−1)
z ′ = −j
( Γ=− j )
Or, the five complex Γ map onto five points on the normalized
impedance plane:
x
z ′ = −j
z′ = 0
z′ = ∞
( Γ=− j )
( Γ=1)
( Γ=−1)
z ′ = −j
( Γ=− j )
Jim Stiles
r
z′ = 1
( Γ=0 )
The Univ. of Kansas
Dept. of EECS
19. 2/7/2005
Mapping Z to Gamma
4/8
Now, the preceding provided examples of the mapping of points
between the complex (normalized) impedance plane, and the
complex Γ plane.
We can likewise map whole contours (i.e., sets of points)
between these two complex planes. We shall first look at two
familiar cases.
Z =R
In other words, the case where impedance is purely real, with no
reactive component (i.e., X = 0 ).
Meaning that normalized impedance is:
z′ =r + j0
(i .e., x
= 0)
where we recall that r = R Z 0 .
Remember, this real-valued impedance results in a real-valued
reflection coefficient:
r −1
Γ=
r +1
I.E.,:
Γr
Jim Stiles
Re {Γ} =
r −1
r +1
The Univ. of Kansas
Γi
Im {Γ} = 0
Dept. of EECS
20. 2/7/2005
Mapping Z to Gamma
5/8
Thus, we can determine a mapping between two contours—one
contour ( x = 0 ) on the normalized impedance plane, the other
( Γi = 0 ) on the complex Γ plane:
x =0
Γi = 0
⇔
Γi
Γ =1
x =0
( Γi =0 )
Γr
x
r
x =0
( Γi =0 )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
21. 2/7/2005
Mapping Z to Gamma
6/8
Z = jX
In other words, the case where impedance is purely imaginary,
with no resistive component (i.e., R = 0 ).
Meaning that normalized impedance is:
(i .e., r
z ′ = 0 + jx
= 0)
where we recall that x = X Z 0 .
Remember, this imaginary impedance results in a reflection
coefficient with unity magnitude:
Γ =1
Thus, we can determine a mapping between two contours—one
contour ( r = 0 ) on the normalized impedance plane, the other
( Γ = 1 ) on the complex Γ plane:
r =0
Jim Stiles
⇔
The Univ. of Kansas
Γ =1
Dept. of EECS
22. 2/7/2005
Mapping Z to Gamma
7/8
Γi
Γ =1
r =0
( Γ =1)
Γr
x
r =0
( Γ =1)
r
Jim Stiles
The Univ. of Kansas
Dept. of EECS
23. 2/7/2005
Mapping Z to Gamma
8/8
Q: These two “mappings” may
very well be fascinating in an
academic sense, but they seem
not particularly relevant, since
actual values of impedance
generally have both a real and
imaginary component.
Sure, mappings of more general
impedance contours (e.g., r = 0.5
or x = −1.5 ) onto the complex Γ
would be useful—but it seems
clear that those mappings are
impossible to achieve!?!
A: Actually, not only are mappings of more general impedance
contours (such as r = 0.5 and x = −1.5 ) onto the complex Γ
plane possible, these mappings have already been achieved—
thanks to Dr. Smith and his famous chart!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
24. 2/7/2005
The Smith Chart
1/11
The Smith Chart
Say we wish to map a line on the normalized complex impedance
plane onto the complex Γ plane.
For example, we could map the vertical line r =2 ( Re{z ′} = 2 ) or
the horizontal line x =-1 ( Im{z ′} = −1 ).
Im {z ′}
r =2
Re {z ′}
x =-1
Recall we know how to map the vertical line r =0; it simply maps
to the circle Γ = 1 on the complex Γ plane.
Likewise, we know how to map the horizontal line x = 0; it simply
maps to the line Γi = 0 on the complex Γ plane.
But for the examples given above, the mapping is not so straight
forward. The contours will in general be functions of both
Γr and Γi (e.g., Γ2 + Γi2 = 0.5 ), and thus the mapping cannot be
r
stated with simple functions such as Γ = 1 or Γi = 0 .
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
25. 2/7/2005
The Smith Chart
2/11
As a matter of fact, a vertical line on the normalized impedance
plane of the form:
r = cr ,
where cr is some constant (e.g. r = 2 or r = 0.5 ), is mapped onto
the complex Γ plane as:
2
2
⎛
⎛ 1 ⎞
c ⎞
Γr − r ⎟ + Γi2 = ⎜
⎜
⎟
1 + cr ⎠
⎝
⎝ 1 + cr ⎠
Note this equation is of the same form as that of a circle:
2
2
( x − x c ) + ( y − yc ) = a 2
where:
a = the radius of the circle
Pc ( x = xc , y = yc )
⇒ point located at the center of the circle
Thus, the vertical line r = cr maps into a circle on the complex Γ
plane!
By inspection, it is apparent that the center of this circle is
located at this point on the complex Γ plane:
⎛
Pc ⎜ Γr =
⎝
Jim Stiles
⎞
cr
, Γi = 0 ⎟
1 + cr
⎠
The Univ. of Kansas
Dept. Of EECS
26. 2/7/2005
The Smith Chart
3/11
In other words, the center of this circle always lies somewhere
along the Γi = 0 line.
Likewise, by inspection, we find the radius of this circle is:
a=
1
1 + cr
We perform a few of these mappings and see where these
circles lie on the complex Γ plane:
Γi
r = −0.3
Γ =1
r = 0. 0
r = 1.0
r = 0.3
Jim Stiles
The Univ. of Kansas
Γr
r = 3. 0
Dept. Of EECS
27. 2/7/2005
The Smith Chart
4/11
We see that as the constant cr increases, the radius of the
circle decreases, and its center moves to the right.
Note:
1. If cr > 0 then the circle lies entirely within the circle
Γ = 1.
2. If cr < 0 then the circle lies entirely outside the circle
Γ = 1.
3. If cr = 0 (i.e., a reactive impedance), the circle lies on
circle Γ = 1 .
4. If cr = ∞ , then the radius of the circle is zero, and its
center is at the point Γr = 1, Γi = 0 (i.e., Γ = 1 e j 0 ). In
other words, the entire vertical line r = ∞ on the
normalized impedance plane is mapped onto just a single
point on the complex Γ plane!
But of course, this makes sense! If r = ∞ , the impedance is
infinite (an open circuit), regardless of what the value of
the reactive component x is.
Now, let’s turn our attention to the mapping of horizontal lines
in the normalized impedance plane, i.e., lines of the form:
x = ci
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
28. 2/7/2005
The Smith Chart
5/11
where ci is some constant (e.g. x = −2 or x = 0.5 ).
We can show that this horizontal line in the normalized
impedance plane is mapped onto the complex Γ plane as:
( Γr
− 1)
2
2
⎛
1⎞
1
+ ⎜ Γi − ⎟ = 2
ci ⎠ ci
⎝
Note this equation is also that of a circle! Thus, the horizontal
line x = ci maps into a circle on the complex Γ plane!
By inspection, we find that the center of this circle lies at the
point:
⎛
Pc ⎜ Γr = 1, Γi =
⎝
1⎞
⎟
ci ⎠
in other words, the center of this circle always lies somewhere
along the vertical Γr = 1 line.
Likewise, by inspection, the radius of this circle is:
a=
Jim Stiles
1
ci
The Univ. of Kansas
Dept. Of EECS
29. 2/7/2005
The Smith Chart
6/11
We perform a few of these mappings and see where these
circles lie on the complex Γ plane:
Γr = 1
Γi
x = 0 .5
x = 1 .0
x = 2 .0
x = 3. 0
Γ =1
Γr
x = −3 . 0
x = −0 . 5
x = −1.0
x = −2.0
We see that as the magnitude of constant ci increases, the
radius of the circle decreases, and its center moves toward the
point ( Γr = 1, Γi = 0 ) .
Note:
1. If ci > 0 (i.e., reactance is inductive) then the circle lies
entirely in the upper half of the complex Γ plane (i.e.,
where Γi > 0 )—the upper half-plane is known as the
inductive region.
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
30. 2/7/2005
The Smith Chart
7/11
2. If ci < 0 (i.e., reactance is capacitive) then the circle
lies entirely in the lower half of the complex Γ plane (i.e.,
where Γi < 0 )—the lower half-plane is known as the
capacitive region.
3. If ci = 0 (i.e., a purely resistive impedance), the circle
has an infinite radius, such that it lies entirely on the line
Γi = 0 .
4. If ci = ±∞ , then the radius of the circle is zero, and its
center is at the point Γr = 1, Γi = 0 (i.e., Γ = 1 e j 0 ). In other
words, the entire vertical line x = ∞ or x = −∞ on the
normalized impedance plane is mapped onto just a single
point on the complex Γ plane!
But of course, this makes sense! If x = ∞ , the impedance
is infinite (an open circuit), regardless of what the value of
the resistive component r is.
5. Note also that much of the circle formed by mapping
x = ci onto the complex Γ plane lies outside the circle
Γ = 1.
This makes sense! The portions of the circles laying
outside Γ = 1 circle correspond to impedances where the
real (resistive) part is negative (i.e., r < 0).
Thus, we typically can completely ignore the portions of the
circles that lie outside the Γ = 1 circle !
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
31. 2/7/2005
The Smith Chart
8/11
Mapping many lines of the form r = cr and x = ci onto circles on
the complex Γ plane results in tool called the Smith Chart.
Im{ Γ }
Re{ Γ }
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
32. 2/7/2005
The Smith Chart
9/11
Note that around the outside of the Smith Chart there is a
scale indicating the phase angle θ Γ , from −180 < θ Γ < 180 .
However, there is another scale that also directly indicates the
equivalent transmission line distance ∆z associated with phase
shift ∆θ Γ = 2β ∆z , in terms of λ (i.e., the electrical distance).
The two scales are related by the equation:
∆z =
∆θ Γ ⎛ ∆θ Γ ⎞
=⎜
⎟λ
2β ⎝ 4π ⎠
Note the Smith Chart is simply the vertical lines r = cr and
horizontal lines x = ci of the normalized impedance plane,
mapped onto the two types of circles on the complex Γ plane.
Note for the normalized impedance plane, a vertical line r = cr
and a horizontal line x = ci are always perpendicular to each
other when they intersect. We say these lines form a
rectilinear grid.
However, a similar thing is true for the Smith Chart! When a
mapped circle r = cr intersects a mapped circle x = ci , the two
circles are perpendicular at that intersection point. We say
these circles form a curvilinear grid.
In fact, the Smith Chart is formed by distorting the rectilinear
grid of the normalized impedance plane into the curvilinear grid
of the Smith Chart!
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
34. 2/7/2005
The Smith Chart
11/11
And then distorting some more—we have the curvilinear grid of
the Smith Chart!
x
r
Jim Stiles
The Univ. of Kansas
Dept. Of EECS
35. 2/8/2005
Zin Calculations using the Smith Chart.doc
1/7
Zin Calculations using
the Smith Chart
′
z0 = 1
′
zin
z = −
z L′
z = 0
′
The normalized input impedance zin of a transmission line length
, when terminated in normalized load z L′ , can be determined as:
′
zin =
Zin
Z0
⎛ Z L + j Z 0 tan β ⎞
⎟
Z 0 ⎝ Z 0 + j Z L tan β ⎠
Z Z + j tan β
= L 0
1 + j Z L Z 0 tan β
=
=
1
Z0 ⎜
z L′ + j tan β
1 + j z L′ tan β
Q: Evaluating this unattractive expression
looks not the least bit pleasant. Isn’t there a
′
less disagreeable method to determine zin ?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
36. 2/8/2005
Zin Calculations using the Smith Chart.doc
2/7
A: Yes there is! Instead, we could determine this normalized
input impedance by following these three steps:
1. Convert z L′ to ΓL , using the equation:
Zin − Z 0
Zin + Z 0
Z Z −1
= in 0
Zin Z 0 + 1
z ′ −1
= in
′
zin + 1
Γin =
2. Convert ΓL to Γin , using the equation:
Γin = Γ L e − j 2 β
′
3. Convert Γin to zin , using the equation:
z L′ =
Z L 1 + ΓL
=
Z 0 1 − ΓL
Q: But performing these three
calculations would be even more
difficult than the single step
you described earlier. What
short of dimwit would ever use
(or recommend) this approach?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
37. 2/8/2005
Zin Calculations using the Smith Chart.doc
3/7
A: The benefit in this last approach is that each of the three
steps can be executed using a Smith Chart—no complex
calculations are required!
1. Convert z L′ to ΓL
Find the point z L′ from the impedance mappings on your
Smith Chart. Place you pencil at that point—you have now
located the correct ΓL on your complex Γ plane!
For example, say z L′ = 0.6 − j 1.4 . We find on the Smith
Chart the circle for r =0.6 and the circle for x =-1.4. The
intersection of these two circles is the point on the
complex Γ plane corresponding to normalized impedance
z L′ = 0.6 − j 1.4 .
This point is a distance of 0.685 units from the origin, and
is located at angle of –65 degrees. Thus the value of ΓL is:
ΓL = 0.685 e − j 65
2. Convert ΓL to Γin
Since we have correctly located the point ΓL on the
complex Γ plane, we merely need to rotate that point
clockwise around a circle ( Γ = 0.685 ) by an angle 2β .
When we stop, we are located at the point on the complex
Γ plane where Γ = Γin !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
38. 2/8/2005
Zin Calculations using the Smith Chart.doc
4/7
For example, if the length of the transmission line
terminated in z L′ = 0.6 − j 1.4 is = 0.307 λ , we should
rotate around the Smith Chart a total of 2β = 1.228π
radians, or 221 . We are now at the point on the complex
Γ plane:
Γ = 0.685 e + j 74
This is the value of Γin !
′
3. Convert Γin to zin
When you get finished rotating, and your pencil is located
at the point Γ = Γin , simply lift your pencil and determine
the values r and x to which the point corresponds!
For example, we can determine directly from the Smith
Chart that the point Γin = 0.685 e + j 74 is located at the
intersection of circles r =0.5 and x =1.2. In other words:
′
zin = 0.5 + j 1.2
Jim Stiles
The Univ. of Kansas
Dept. of EECS
39. 2/8/2005
Zin Calculations using the Smith Chart.doc
5/7
Step 1
Γ = 0.685
ΓL = 0.685 e − j 65
θ Γ = −65
Jim Stiles
The Univ. of Kansas
Dept. of EECS
40. 2/8/2005
Zin Calculations using the Smith Chart.doc
6/7
Step 2
2
= 0.147 λ
Γin = 0.685 e − j 74
Γ = 0.685
ΓL = 0.685 e − j 65
1
= 0.16λ
=
1
+
2
= 0.160λ + 0.147 λ = 0.307 λ
2β = 221
Jim Stiles
The Univ. of Kansas
Dept. of EECS
41. 2/8/2005
Zin Calculations using the Smith Chart.doc
7/7
Step 3
′
zin = 0.5 + j 1.2
Jim Stiles
The Univ. of Kansas
Dept. of EECS
42. 2/8/2005
Example Shorted Transmission Line.doc
1/3
Example: The Input
Impedance of a Shorted
Transmission Line
Let’s determine the input impedance of a transmission line that
is terminated in a short circuit, and whose length is:
a)
= λ 8 = 0.125λ
b)
= 3λ 8 = 0.375λ
′
zin
2β = 90
⇒
⇒
2β = 270
′
z0 = 1
z = −
Jim Stiles
z L′ = 0
z = 0
The Univ. of Kansas
Dept. of EECS
43. 2/8/2005
a)
= λ 8 = 0.125λ
Example Shorted Transmission Line.doc
⇒
2/3
2β = 90
′
Rotate clockwise 90 from Γ = −1.0 = e j 180 and find zin = j .
zin = j
ΓL = −1 = e j 180
Jim Stiles
The Univ. of Kansas
Dept. of EECS
44. 2/8/2005
b)
Example Shorted Transmission Line.doc
= 3λ 8 = 0.375λ
⇒
3/3
2β = 270
′
Rotate clockwise 270 from Γ = −1.0 = e j 180 and find zin = − j .
ΓL = −1 = e j 180
zin = j
Jim Stiles
The Univ. of Kansas
Dept. of EECS
45. 2/8/2005
Example The Load Impedance.doc
1/2
Example: Determining the
Load Impedance of a
Transmission Line
Say that we know that the input impedance of a transmission
line length = 0.134λ is:
′
zin = 1.0 + j 1.4
Let’s determine the impedance of the load that is terminating
this line.
′
zin =
1 + j 1. 4
′
z0 = 1
z L′ = ??
= 0.134λ
z = −
z = 0
′
Locate zin on the Smith Chart, and then rotate counter-
clockwise (yes, I said counter-clockwise) 2β = 96.5 .
Essentially, you are removing the phase shift associated with
the transmission line. When you stop, lift your pencil and find
z L′ !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
46. 2/8/2005
Example The Load Impedance.doc
2/2
= 0.134λ
2β = 96.5
′
zin = 1 + j 1.4
z L′ = 0.29 + j 0.24
Jim Stiles
The Univ. of Kansas
Dept. of EECS
47. 2/8/2005
Example Determining the tl length.doc
1/7
Example: Determining
Transmission Line Length
A load terminating at transmission line has a normalized
impedance z L′ = 2.0 + j 2.0 . What should the length of
transmission line be in order for its input impedance to be:
a) purely real (i.e., xin = 0 )?
b)
have a real (resistive) part equal to one (i.e., rin = 1.0 )?
Solution:
a) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate
clockwise until you “bump into” the contour x = 0 (recall this is
contour lies on the Γr axis!).
When you reach the x = 0 contour—stop! Lift your pencil and
note that the impedance value of this location is purely real
(after all, x = 0 !).
Now, measure the rotation angle that was required to move
clockwise from z L′ = 2.0 + j 2.0 to an impedance on the x = 0
contour—this angle is equal to 2β !
You can now solve for , or alternatively use the electrical
length scale surrounding the Smith Chart.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
48. 2/8/2005
Example Determining the tl length.doc
2/7
One more important point—there are two possible solutions!
Solution 1:
2β = 30
= 0.042λ
z L′ = 2 + j 2
Γ (z )
′
zin = 4.2 + j 0
x =0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
49. 2/8/2005
Example Determining the tl length.doc
3/7
Solution 2:
z L′ = 2 + j 2
′
zin = 0.24 + j 0
x =0
Γ (z )
2β = 210
= 0.292λ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
50. 2/8/2005
Example Determining the tl length.doc
4/7
b) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate
clockwise until you “bump into” the circle r = 1 (recall this circle
intersects the center point or the Smith Chart!).
When you reach the r = 1 circle—stop! Lift your pencil and note
that the impedance value of this location has a real value equal
to one (after all, r = 1 !).
Now, measure the rotation angle that was required to move
clockwise from z L′ = 2.0 + j 2.0 to an impedance on the r = 1
circle—this angle is equal to 2β !
You can now solve for , or alternatively use the electrical
length scale surrounding the Smith Chart.
Again, we find that there are two solutions!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
51. 2/8/2005
Example Determining the tl length.doc
5/7
Solution 1:
z L′ = 2 + j 2
Γ (z )
r =1
′
zin = 1.0 − j 1.6
2β = 82
= 0.114λ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
52. 2/8/2005
Example Determining the tl length.doc
6/7
Solution 2:
′
zin = 1.0 + j 1.6
Γ (z )
z L′ = 2 + j 2
r =1
2β = 339
= 0.471λ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
53. 2/8/2005
Example Determining the tl length.doc
7/7
′
Q: Hey! For part b), the solutions resulted in zin = 1 − j 1.6 and
′
zin = 1 + j 1.6 --the imaginary parts are equal but opposite! Is
this just a coincidence?
A: Hardly! Remember, the two impedance solutions must result
in the same magnitude for Γ --for this example we find
Γ ( z ) = 0.625 .
Thus, for impedances where r =1 (i.e., z ′ = 1 + j x ):
Γ=
jx
z ′ − 1 (1 + jx ) − 1
=
=
z ′ + 1 (1 + jx ) + 1 2 + j x
and therefore:
2
Γ =
jx
2
2+ j x
2
x2
=
4 +x2
Meaning:
x =
2
4 Γ
2
1− Γ
2
of which there are two equal by opposite solutions!
x = ±
2 Γ
1− Γ
2
Which for this example gives us our solutions x = ±1.6 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
54. 2/10/2005
Admittance.doc
1/4
Admittance
As an alternative to impedance Z, we can define a complex
parameter called admittance Y:
Y =
I
V
where V and I are complex voltage and current, respectively.
Clearly, admittance and impedance are not independent
parameters, and are in fact simply geometric inverses of each
other:
1
1
Y =
Z =
Z
Y
Thus, all the impedance parameters that we have studied can
be likewise expressed in terms of admittance, e.g.:
Y (z ) =
1
Z (z )
YL =
1
ZL
Yin =
1
Zin
Moreover, we can define the characteristic admittance Y0 of a
transmission line as:
I + (z )
Y0 = +
V (z )
And thus it is similarly evident that characteristic impedance
and characteristic admittance are geometric inverses:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
55. 2/10/2005
Admittance.doc
Y0 =
1
Z0 =
Z0
2/4
1
Y0
As a result, we can define a normalized admittance value y ′ :
y′ =
Y
Y0
An therefore (not surprisingly) we find:
y′ =
Y Z0 1
=
=
Y0 Z
z′
Note that we can express normalized impedance and
admittance more compactly as:
y ′ = Y Z0
and
z ′ = Z Y0
Now since admittance is a complex value, it has both a real and
imaginary component:
Y = G + jB
where:
Re {Y } G = Conductance
Im {Z }
Jim Stiles
B = Susceptance
The Univ. of Kansas
Dept. of EECS
56. 2/10/2005
Admittance.doc
3/4
Now, since Z = R + jX , we can state that:
G + jB =
1
R + jX
Q: Yes yes, I see, and from this
we can conclude:
G =
1
R
and
B=
−1
X
and so forth. Please speed this up
and quit wasting my valuable time
making such obvious statements!
A: NOOOO! We find that G ≠ 1 R and B ≠ 1 X
(generally). Do not make this mistake!
In fact, we find that
G + jB =
R − jX
1
R + jX R − jX
R − jX
R2 + X 2
R
X
= 2
−j 2
R +X2
R +X2
=
Jim Stiles
The Univ. of Kansas
Dept. of EECS
57. 2/10/2005
Admittance.doc
4/4
Thus, equating the real and imaginary parts we find:
G =
R
and
R2 + X 2
B=
−X
R2 + X 2
Note then that IF X = 0 (i.e., Z = R ), we get, as expected:
G =
1
R
and
B =0
And that IF R = 0 (i.e., Z = R ), we get, as expected:
G =0
and
B=
−1
X
I wish I had a
nickel for every
time my software
has crashed—oh
wait, I do!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
58. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
1/9
Example: Admittance
Calculations with the
Smith Chart
Say we wish to determine the normalized admittance y1′ of the
network below:
z 2′ =
1 .7 − j 1 . 7
y1′
z L′ =
1.6 + j 2.6
′
z0 = 1
= 0.37 λ
z = 0
z = −
First, we need to determine the normalized input admittance of
the transmission line:
′
yin
z L′ =
1.6 + j 2.6
′
z0 = 1
= 0.37 λ
z = −
Jim Stiles
z = 0
The Univ. of Kansas
Dept. of EECS
59. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
2/9
There are two ways to determine this value!
Method 1
First, we express the load z L = 1.6 + j 2.6 in terms of its
admittance y L′ = 1 z L . We can calculate this complex value—or
we can use a Smith Chart!
z L = 1.6 + j 2.6
y L = 0.17 − j 0.28
Jim Stiles
The Univ. of Kansas
Dept. of EECS
60. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
3/9
The Smith Chart above shows both the impedance mapping
(red) and admittance mapping (blue). Thus, we can locate the
impedance z L = 1.6 + j 2.6 on the impedance (red) mapping, and
then determine the value of that same ΓL point using the
admittance (blue) mapping.
From the chart above, we find this admittance value is
approximately y L = 0.17 − j 0.28 .
Now, you may have noticed that the Smith Chart above, with
both impedance and admittance mappings, is very busy and
complicated. Unless the two mappings are printed in different
colors, this Smith Chart can be very confusing to use!
But remember, the two mappings are precisely identical—they’re
just rotated 180 with respect to each other. Thus, we can
alternatively determine y L by again first locating z L = 1.6 + j 2.6
on the impedance mapping :
z L = 1.6 + j 2.6
Jim Stiles
The Univ. of Kansas
Dept. of EECS
61. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
4/9
Then, we can rotate the entire Smith Chart 180 --while keeping
the point ΓL location on the complex Γ plane fixed.
y L = 0.17 − j 0.28
Thus, use the admittance mapping at that point to determine
the admittance value of ΓL .
Note that rotating the entire Smith Chart, while keeping the
point ΓL fixed on the complex Γ plane, is a difficult maneuver to
successfully—as well as accurately—execute.
But, realize that rotating the entire Smith Chart 180 with
respect to point ΓL is equivalent to rotating 180 the point ΓL
with respect to the entire Smith Chart!
This maneuver (rotating the point ΓL ) is much simpler, and the
typical method for determining admittance.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
62. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
5/9
z L = 1.6 + j 2.6
y L = 0.17 − j 0.28
′
Now, we can determine the value of yin by simply rotating
clockwise 2β from y L′ , where = 0.37 λ :
Jim Stiles
The Univ. of Kansas
Dept. of EECS
63. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
6/9
2β
yin = 0.7 − j 1.7
y L = 0.17 − j 0.28
Transforming the load admittance to the beginning of the
′
transmission line, we have determined that yin = 0.7 − j 1.7 .
Method 2
Alternatively, we could have first transformed impedance z L′ to
′
the end of the line (finding zin ), and then determined the value
′
of yin from the admittance mapping (i.e., rotate 180 around the
Smith Chart).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
64. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
7/9
′
zin = 0.2 + j 0.5
z L = 1.6 + j 2.6
2β
The input impedance is determined after rotating clockwise
′
2β , and is zin = 0.2 + j 0.5 .
Now, we can rotate this point 180 to determine the input
′
admittance value yin :
Jim Stiles
The Univ. of Kansas
Dept. of EECS
65. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
8/9
2β
′
zin = 0.2 + j 0.5
yin = 0.7 − j 1.7
The result is the same as with the earlier method-′
yin = 0.7 − j 1.7 .
Hopefully it is evident that the two methods are equivalent. In
method 1 we first rotate 180 , and then rotate 2β . In the
second method we first rotate 2β , and then rotate 180 --the
result is thus the same!
Now, the remaining equivalent circuit is:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
66. 2/17/2005
Example Admittance Calculations with the Smith Chart.doc
z 2′ =
1 .7 − j 1 . 7
y1′
9/9
′
yin =
0.7 − j 1.7
Determining y1′ is just basic circuit theory. We first express
z 2′ in terms of its admittance y2′ = 1 z 2′ .
Note that we could do this using a calculator, but could likewise
′
use a Smith Chart (locate z 2 and then rotate 180 ) to
accomplish this calculation! Either way, we find that
y2′ = 0.3 + j 0.3 .
y2′ =
0.3 + j 0.3
y1′
′
yin =
0.7 − j 1.7
Thus, y1′ is simply:
′
y1′ = y2′ + yin
= ( 0.3 + j 0.3) + ( 0.7 − j 1.7 )
= 1 .0 − j 1 . 4
Jim Stiles
The Univ. of Kansas
Dept. of EECS