UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences. EE105 Lab Experiments

Similar documents
Solution: K m = R 1 = 10. From the original circuit, Z L1 = jωl 1 = j10 Ω. For the scaled circuit, L 1 = jk m ωl 1 = j10 10 = j100 Ω, Z L

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

Steady State Frequency Response Using Bode Plots

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

Single-Time-Constant (STC) Circuits This lecture is given as a background that will be needed to determine the frequency response of the amplifiers.

Homework 7 - Solutions

Response to a pure sinusoid

Frequency Response. Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). ve jφ

Frequency response. Pavel Máša - XE31EO2. XE31EO2 Lecture11. Pavel Máša - XE31EO2 - Frequency response

8.1.6 Quadratic pole response: resonance

Chapter 8: Converter Transfer Functions

CHAPTER ONE FUNCTIONS AND GRAPHS. In everyday life, many quantities depend on one or more changing variables eg:

2nd-order filters. EE 230 second-order filters 1

Asymptote. 2 Problems 2 Methods

Sophomore Physics Laboratory (PH005/105)

Chapter 8: Converter Transfer Functions

ESE319 Introduction to Microelectronics Bode Plot Review High Frequency BJT Model

Dynamic circuits: Frequency domain analysis

Definitions. Decade: A ten-to-one range of frequency. On a log scale, each 10X change in frequency requires the same distance on the scale.

9.7 Extension: Writing and Graphing the Equations

EKT 119 ELECTRIC CIRCUIT II. Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016 Dr. Mohd Rashidi Che Beson

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

EECE 301 Signals & Systems Prof. Mark Fowler

Poles and Zeros and Transfer Functions

(amperes) = (coulombs) (3.1) (seconds) Time varying current. (volts) =

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

'XNH8QLYHUVLW\ (GPXQG73UDWW-U6FKRRORI(QJLQHHULQJ. EGR 224 Spring Test II. Michael R. Gustafson II

EE1-01 IMPERIAL COLLEGE LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2013 ANALYSIS OF CIRCUITS. Tuesday, 28 May 10:00 am

Stability & Compensation

BIOEN 302, Section 3: AC electronics

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques

'XNH8QLYHUVLW\ (GPXQG73UDWW-U6FKRRORI(QJLQHHULQJ. EGR 224 Spring Test II. Michael R. Gustafson II

Experiment 3: Resonance in LRC Circuits Driven by Alternating Current

Test II Michael R. Gustafson II

Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros)

Time Varying Circuit Analysis

Second-order filters. EE 230 second-order filters 1

EE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 2

Homework 6 Solutions and Rubric

Frequency Dependent Aspects of Op-amps

New Mexico State University Klipsch School of Electrical Engineering EE312 - Signals and Systems I Fall 2015 Final Exam

ECE 350 Signals and Systems Spring 2011 Final Exam - Solutions. Three 8 ½ x 11 sheets of notes, and a calculator are allowed during the exam.

POLYTECHNIC UNIVERSITY Electrical Engineering Department. EE SOPHOMORE LABORATORY Experiment 2 DC circuits and network theorems

If you have completed your extra credit opportunity, please place it on your inbox.

Biasing BJTs CHAPTER OBJECTIVES 4.1 INTRODUCTION

16.30/31, Fall 2010 Recitation # 2

2.004 Dynamics and Control II Spring 2008

1 (20 pts) Nyquist Exercise

If we plot the position of a moving object at increasing time intervals, we get a position time graph. This is sometimes called a distance time graph.

Frequency Response part 2 (I&N Chap 12)

Control Systems I. Lecture 9: The Nyquist condition

College Algebra Through Problem Solving (2018 Edition)

DIGITAL SIGNAL PROCESSING UNIT III INFINITE IMPULSE RESPONSE DIGITAL FILTERS. 3.6 Design of Digital Filter using Digital to Digital

MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 3 PERIODIC FUNCTIONS

Geometry Summer Assignment 2018

Chapter 3 Data Acquisition and Manipulation

Transient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

Supplementary Reader. EECS 40 Introduction to Microelectronic Circuits

IMPERIAL COLLEGE OF SCIENCE, TECHNOLOGY AND MEDICINE UNIVERSITY OF LONDON DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2010

EE 16B Final, December 13, Name: SID #:

1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I

Chapter 2 Functions and Graphs

H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )

2. If the values for f(x) can be made as close as we like to L by choosing arbitrarily large. lim

Systems Analysis and Control

Designing Information Devices and Systems II Fall 2018 Elad Alon and Miki Lustig Discussion 5A

SAMPLE EXAMINATION PAPER

Moment Distribution Method

EE40 Lecture 11 Josh Hug 7/19/2010

Transfer func+ons, block diagram algebra, and Bode plots. by Ania- Ariadna Bae+ca CDS Caltech 11/05/15

Summer Review Packet for Students Entering AP Calculus BC. Complex Fractions

Review of Linear Time-Invariant Network Analysis

Testing Newton s 2nd Law

Refinements to Incremental Transistor Model

Draft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 1: Complex numbers 1

Frequency domain analysis

( ) 2. 1) Bode plots/transfer functions. a. Draw magnitude and phase bode plots for the transfer function

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain

Asymptotic Bode Plot & Lead-Lag Compensator

Modeling and System Identification for a DC Servo

Basic Electronics. Introductory Lecture Course for. Technology and Instrumentation in Particle Physics Chicago, Illinois June 9-14, 2011

Revision of Basic A.C. Theory

v t 2 2t 8. Fig. 7 (i) Write down the velocity of the insect when t 0. (ii) Show that the insect is instantaneously at rest when t 2and when t 4.

Calculus Summer TUTORIAL

Horizontal asymptotes

Kinematics. Vector solutions. Vectors

Introduction Assignment

Calculus I Practice Test Problems for Chapter 2 Page 1 of 7

Frequency Response Analysis

Linear Equations. Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)}

SOLUTIONS FOR PROBLEMS 1-30

STUDY KNOWHOW PROGRAM STUDY AND LEARNING CENTRE. Functions & Graphs

Learn2Control Laboratory

Systems Analysis and Control

ECE2210 Final given: Fall 13

Analysis and Design of Analog Integrated Circuits Lecture 12. Feedback

(Continued on next page)

Designing Information Devices and Systems I Fall 2018 Lecture Notes Note Introduction to Capacitive Touchscreen

Exercises for lectures 13 Design using frequency methods

Transcription:

UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EE15 Lab Experiments Bode Plot Tutorial Contents 1 Introduction 1 2 Bode Plots Basics 1 2.1 Magnitude............................................... 1 2.2 Phase................................................. 2 3 Combining Poles and Zeroes 4 1 Introduction Although you should have learned about Bode plots in previous courses, this tutorial will give you a brief review of the material in case your memory is fuzzy. 2 Bode Plots Basics Making the Bode plots for a transfer function involves drawing both the magnitude and phase plots. The magnitude is plotted in decibels (db) while the phase is plotted in degrees ( ). For both plots, the horizontal axis is either frequency (f) or angular frequency (ω), measured in Hz and rad/s respectively. In addition, the horizontal axis should be logarithmic (i.e. increasing by factors of 1). Most of the transfer functions we will encounter in this lab manual can be rearranged into the general form: H(jω) = A jω/ω z1 (1 + jω/ω z2 )(1 + jω/ω z3 )... jω/ω p1 (1 + jω/ω p2 )(1 + jω/ω p3 )..., (1) where A is an arbitrary constant and j is 1. Besides the exception of, the basic component of this transfer function is 1 +, where ω c is some numerical constant. Let us analyze this basic component first before we analyze the transfer function as a whole. 2.1 Magnitude Recall the definition of magnitude (measured in db): H(jω) db = 2 log H(jω) = 2 log (R [H(jω)]) 2 + (I [H(jω)]) 2 (2) Let us apply this definition to our basic component (1+ ), which is also called a zero when it appears in the numerator of the transfer function: 1 + db = 2 log 1 + = 2 log 1 + (ω/ω c ) 2 (3) For small values of ω, we have 2 log 1 + db. For large values of ω, 2 log 1 +. When ω = ω c, the magnitude of the transfer function is approximately 3 db. 1

2 BODE PLOTS BASICS 2 Since there is little change in the magnitude of the transfer function from ω = to ω = ω c, we can approximate the magnitude expression as equal to db within this interval. As for the ω > ω c interval, the (ω/ω c ) term dominates in the expression; thus, we can approximate the magnitude as 2 log(ω/ω c ). Also, notice that, within the ω > ω c interval, the magnitude increases by 2 db when ω increases by a factor of 1. Therefore, the overall Bode plot approximation for a zero is the following: db for ω < ω c and a 2 db/decade line for ω > ω c. Please see Figure 1 for an illustration of this approximation. Figure 1 also shows the magnitude Bode plot for a DC zero, which has the form. Because the DC zero lacks the 1 term, its Bode plot approximation differs from the normal zero: instead of two different regions, the DC zero consists of only a 2 db/decade line. Its intersection with the frequency axis is located at ω c because 2 log(ω c /ω c ) = db. Note that ω c is 1 5 rad/s in this example. 4 Magnitude (db) 2 2 2 log 1 + 2 log 4 Figure 1: Magnitude Bode plots of a normal zero and a DC zero (with both having ω c = 1 5 rad/s). Notice the plots overlap for ω > ω c. The basic transfer function component, 1 +, can also appear in the denominator (in which case it is called a pole). Even though this may seem like an entirely different problem, the same principles apply. Recall that we took the logarithm of our transfer function when we expressed our results in decibels. Also, recall that taking the logarithm of the inverse of a function simply gives the negated logarithm of the function. In other words, we simply have to negate the results of our zero analysis to get the appropriate expressions for poles. The same argument applies with DC poles, which has the form. In general, a normal pole will have a constant db value for ω < ω c and will drop by 2 db/decade for ω > ω c. A DC pole will drop by 2 db/decade for any ω and will intersect the frequency axis ( db) at ω = ω c. The results are shown in Figure 2. 2.2 Phase Now, let us take a look at the respective phases of a zero, DC zero, pole, and DC pole. Recall the definition of phase: ( ) I[H(jω)] Arg (H(jω)) = arctan (4) R[H(jω)]

3 COMBINING POLES AND ZEROES 3 4 Magnitude (db) 2 2 2 log 1 1+ 1 2 log 4 Figure 2: Magnitude Bode plots of a normal pole and a DC pole (with both having ω c = 1 5 rad/s). Notice the plots overlap for ω > ω c. Applying this definition to the normal zero: ( ) ω Arg(1 + ) = arctan ω c (5) For ω =, Arg(1 + ) =. For ω, Arg(1 + ) 9. For ω = ω c, Arg(1 + ) 45. Thus, our approximation for the phase of a zero is the following: for ω <.1ω c, 45 for ω = ω c, and 9 for ω > 1ω c. The phase Bode plot is then constructed using straight lines that join these regions together. As for the DC zero, its phase stays constant at 9. Please refer to Figure 3 for an illustration of these plots. The phase analysis of poles and DC poles can be conducted in a similar fashion. Let us begin by applying the definition of phase to the pole and DC pole, respectively: ( ) 1 Arg = Arg (1) Arg (1 + ) = Arg (1 + ) = Arg (1 + ) (6) 1 + ( ) 1 Arg = Arg (1) Arg ( ) = Arg ( ) = Arg ( ) (7) As you might notice, our phase plots for poles and DC poles will simply be the negated versions of the zero plots. Please see Figure 4 for an illustration. 3 Combining Poles and Zeroes Generally, a transfer function may involve many poles and zeroes (as well as their DC counterparts). In order to simplify the task of drawing Bode plots, your first step should be to factor the transfer function into the canonical form as shown in Equation 1. This makes it easy to identify all of the poles and zeroes. Then, you ll have to handle the constant coefficient, A (if it is present). The magnitude of A will affect your magnitude plot, and the sign of A will affect your phase plot. More specifically, your magnitude plot must be offset by 2 log A. For example, if A = 1, then your magnitude plot must be shifted up by 2 db.

3 COMBINING POLES AND ZEROES 4 1 8 Phase (degrees) 6 4 2 Arg (1 + ) Arg ( ) Figure 3: Phase Bode plots of a normal zero and a DC zero (with both having ω c = 1 5 rad/s). Notice the plots overlap for ω > 1ω c. 2 Phase (degrees) 4 6 8 Arg ( 1 1+ ) Arg ( 1 ) 1 Figure 4: Phase Bode plots of a normal pole and a DC pole (with both having ω c = 1 5 rad/s). Notice the plots overlap for ω > 1ω c. Similarly, if A = 1/1, then your magnitude plot must be shifted down by 2 db. If A <, then your phase plot must be shifted up (or down it s the same in this case) by 18. Next, you need to draw each pole and zero plot individually on the same graph (whether you re making a magnitude or phase plot). Finally, add together all the curves that you have drawn to obtain the final Bode plot. Remember to shift your plots accordingly based on the constant A as mentioned previously. This superposition principle is possible

3 COMBINING POLES AND ZEROES 5 because of the decomposition of the transfer function into zeroes and poles. When adding the poles and zeroes in the final plot, remember that in areas where two curves are constant, the result will just be the sum of the constant values. When one is a constant and the other is linear, the result will be a line that starts at that constant value and progress with a slope equal to the linear curve. Finally, when both are linear, the sum will be a line that has a slope equal to the sum of the slopes.