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### EE 435 HW 6 Spring 2009

Course: EE 435, Fall 2009
School: Iowa State
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Word Count: 1198

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Assignment Homework EE 435 Homework 6 Spring 2009 Problem 1 An open-loop amplifier has a gain Ao and left-half plane poles at p1 and p2 where p2=kp1 If it is used as a feedback amplifier with a feedback factor , determine the pole ratio k so that the feedback amplifier has a 65 degree phase margin. Problem 2 Assume an amplifier has a dc gain of 40,000, a pole at -30 rad/sec and a pole at -500,000 rad/sec. Assume...

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Iowa State - EE - 435
Iowa State - EE - 435
%EE 435 Homework 7 %Problem 2 Ao = 1e4; %80 dB = 10000 p1 = 10; p2 = 1e4; n=1; for B = 0:0.01:1 D_FB = [1 (p1+p2) p1*p2*(1+Ao*B)]; poles(n,:) = roots(D_FB); n = n+1; end figure, plot(poles,'x-'),. title('Poles Locus Plot'), xlabel('Real Axis'), ylabe
Iowa State - EE - 435
EE 435 Assignment 9 Spring 2009Problem 1 a) Prove the following theorem. Theorem: If the INL of a DAC is less than LSB, then the DAC is monotone. b) The theorem of part a) was sufficient but not necessary. Give an example of a DAC that is monotone
Iowa State - EE - 435
EE 435 HW #9 solution Problem 1 (a) Proof: Vk+1-Vk = [(k+1)LSB+INLk+1]- (kLSB+INLk)=1LSB+INLk+1 - INLk &gt; 1 LSB-| INLk+1|-| INLk| &gt; 0 (b) ExampleProblem 2 12.1 of Johns and Martin Solution: The number of switches in the N-bit resistor-string DAC
Iowa State - EE - 435
EE 435 HW #9 solution for Problems 2, 4 and 6 Problem 2 12.1 of Johns and Martin Solution: The number of switches in the N-bit resistor-string DAC is: # of switches = 2*(1+2+2^2 2^ N1 2* 2^N1 N 1,2,3 Problem 4 If a 1024 p
Iowa State - EE - 435
Pelgrom Paper Assessment Most challenges and solutions are included in Lecture 35 of EE 435 as shown below. http:/class.ece.iastate.edu/ee435/lectures/EE%20435%20Lect%2035%20Spring%202009.pdf An example assessment from one of the students is also inc
Iowa State - EE - 435
1 EE 435 Final Exam Spring 2009 Name_Instructions: This is an open-book, open-notes exam but no collaboration with anyone except the course instructor is permitted. This exam is to be returned to Rm 2133 Coover by 12:00 noon on Thursday May 7. All
Iowa State - EE - 435
1 EE 435 Exam 1 Spring 2008 Name_Instructions: You may use up to 2 pages of notes for this exam. All work should be included on this exam attach additional pages only if more space is needed. There are 10 short questions worth 2 pts each. All prob
Iowa State - EE - 435
1 EE 435 Final Exam Spring 2006 Name_Instructions: This is an open-book, open-notes exam. All work should be included on this exam attach additional pages only if more space is needed. There are 10 short questions worth 2 pts each. All problems ar
Iowa State - EE - 435
EE435 Experiment 2: Amplifier Characterization Spring 2009Objective: The objective of this experiment is to develop measurementmethods for characterizing key properties of operational amplifiers1 IntroductionAmplifiers are one of the major compo
Iowa State - EE - 435
EE 435Lecture 4: Fully Differential Single-Stage Amplifier Design1Review from last lecture:Parameter Domains for Characterizing Amplifier Performance{gm,g0}W ,IDQ L Degrees of Freedom: 2 Small signal parameter domain : -g g A v0 = m GB
Iowa State - EE - 435
EE 435Lecture 7: High-Gain Single-Stage Op Amps1Review from last lecture:Signal SwingHow do the transfer characteristics relate to the signal swing ?ViNFor this circuit, high gain and large output signal swing for small VEB12Review f
Iowa State - EE - 435
Iowa State - EE - 435
EE 435 Lecture 11Other Gain Enhancement Strategies- Cascaded Amplifiers
Iowa State - EE - 435
EE 435Lecture 12 Cascaded Amplifier StructuresReview from last timeIncreasing Gain by CascadingProvided the stages are non-interactingXINA1A2XOUTX OUT = A 1A 2 X INX OUT = A 1A 2 A 3 X INn X OUT = Ai X IN i =1Gain can be easily
Iowa State - EE - 435
EE 435Lecture 15 Cascaded Amplifier Structures - the two-stage op ampReview from last lectureAnalysis of Internally Compensated TwoStage Op Amps gmF1 gmF2 A V0 = g + g g + g oP1 oF2 oP2 oF1p2 =(goF2 + goP2 )CLp1(goF1 + goP1
Iowa State - EE - 435
EE 435Lecture 17 A Design Flow For Two-Stage Op AmpsReview from last lectureBasic Two-Stage Op Amp gmd (gm0 - sCc ) A FB (s) 2 s CCCL + sCC (gmo - gmd ) + gmd gmoIt can be shown thatgmogmd CL Q= CC gmo - gmdgmo gmd CL CC = 2 Q (gmo - gm
Iowa State - EE - 435
EE 435Lecture 19 Loop Gain Calculations Other Gain Enhancement Strategies Linearity of the Op AmpReview from last lectureWhy does the RHP zero limit performance ?Gain Magnitude in dB 100 80 60 40 20 0 -20 -40 -60 -801 Phase in Degrees0.00E+
Iowa State - EE - 435
EE 435Lecture 20 Linearity of the Op AmpReview from last lectureOpen-loop gain simulationsXX OUT IN icLoad with Termination TerminationBias Must first adjust VXX to trim out any systematic offset Always verify all devices are operating in
Iowa State - EE - 435
EE 435Lecture 21 Linearity of the Op Amp Offset VoltageReview from last lectureSignal Swing and LinearityVOUTOutput RangeVINInput RangeReview from last lectureLinearity of AmplifiersSingle-Stage Linearity of differential pair of maj
Iowa State - EE - 435
EE 435Lecture 22 Offset Voltage Common-Centroid Layouts Common-Mode Feedback CircuitsReview from last lectureLinearity of Common-Source Amplifier VVOS 1 2 2 ViS + ViS VEB 2VEB ViSDDIB VOUTIs this a linear or nonlinear relationship?
Iowa State - EE - 435
EE 435Lecture 23 Data Converters
Iowa State - EE - 435
EE 435Lecture 24 Data Converter Specifications
Iowa State - EE - 435
EE 435Lecture 26 Exam 1 Given in this Time Slot
Iowa State - EE - 435
EE 435Lecture 27 Data Converter CharacterizationREVIEW (slides 2-45)D/A ConvertersXINXINXREFXOUTXOUTREVIEWD/A ConvertersXIN =&lt;bn-1,bn-1,.b1,b0 &gt;An Ideal DAC transfer characteristic (3-bits)XOUT X-REFXINXOUT&lt;0 0 0&gt;&lt;0 0 1&gt;
Iowa State - EE - 435
EE 435Lecture 29 Data Converter CharacterizationSpectral PerformancePerformance Characterization of Data Converters Static characteristics Resolution Least Significant Bit (LSB) Offset and Gain Errors Absolute Accuracy Relative Accu
Iowa State - EE - 435
EE 435Lecture 31 Quantization Noise in Data ConvertersData Converter Design Strategies There are many different DAC and ADC architectures that have been proposed and that are in widespread use today Almost all work perfectly if all componen
Iowa State - EE - 435
EE 435Lecture 38 ADC DesignReview from Last Time:ADC TypesOver-Sampled Single-bit Multi-bit First-order Higher-order Continuous-timeNyquist Rate Flash Pipeline Two-Step Flash Multi-Step Flash Cyclic (algorithmic) Successive Appro
Iowa State - EE - 435
EE 435 Lecture 42Switched-Capacitor Filters and AmplifiersTypical Filter ImplementationBiquads often LP or BPVINI1(s)Integrator a1I2(s)IntegratorI3(s)IntegratorI4(s)IntegratorIk-1(s)IntegratorIk(s)Integrator a2VOUTTypical
Iowa State - EE - 435
EE 435 Homework 3 Spring 2008 In the following problems, if reference to a semiconductor process is needed, assume processes with the following characteristics: CMOS Process - nCOX=100A/v2 ,pCOX=nCOX/3 ,VTNO=0.5V, VTPO= - 0.5V, COX=2fF/2, n= p= 0.01V
Iowa State - EE - 435
Homework Assignment EE 435 Spring 2008 (Due Wednesday March 12 turn in to Vaibhav) Problem 1 Assume the area for the n-channel and the p-channel devices in the amplifier shown are identical and that each transistor has an gate area of 1002. Determin
Michigan State University - HAMIL - 199
J. Ross Hamilton ATL 110 Sect. 009 Frank Manista 9/19/02 Prenatal Structuring of Offspring and its Ethical Dilemmas One of the questions that has recently plagued me has been that of which is about the genetic structuring of our offspring. I ask myse
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Chapter1: Asymptotic Tests for Binomial and Multinomial Models Lecture 2Three Asymptotically equivalent tests (Walds test, Likelihood ratio test and Score test): ^ Let be the MLE of a parameter of a distribution which satisfies the regularity con
Texas San Antonio - CHAPTER - 7853
Chapter1: More on Exact Confidence Intervals for Discrete Distributions Lecture 3 Confidence Interval by Pivoting: Theorem:(Pivoting a discrete cdf ): Let T be a discrete statistic with the cdf FT (t|) Let 1 + 2 = with 0 &lt; &lt; 1 be fixed values. Supp
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Chapter2:Probability Structure and Descriptive Measures in Contingency Tables: Lecture 2: Types of Studies and Measures of AssociationTypes of Studies: There are two types of studies: 1. Observational Study: In this study, subjects are simply obser
Texas San Antonio - CHAPTER - 7853
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Texas San Antonio - CHAPTER - 7853
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Texas San Antonio - CHAPTER - 7853
Chapter 3: Inference for Contingency Tables Lecture 1 CI for the Association Parameters: 1) Odds Ratio For a 2 2 table, =n11 n22 n21 n12 .It is 0 or if any nij = 0, and is undened if all the entries in any row or column are zeros. In this ca
Texas San Antonio - CHAPTER - 7853
Chapter 3: Two Way Tables with Ordinal Classification Lecture 2 Linear Trend Alternative to Independence: When the categories in X and Y are ordinal, a positive or negative trend may exist in the association. Assign scores to summarize the linear tre
Texas San Antonio - CHAPTER - 7853
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Texas San Antonio - CHAPTER - 7853
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Texas San Antonio - CHAPTER - 7853
Chapter 5: Logistic Regression (LR) Lecture 3 Multiple Logistic Regression and Special Cases: Multiple Logistic Regression: Suppose there are p regressor variables x = (x1 , x2 , , xp ), then multiple logistic linear model is logit(x) = 0 + 1 x1 +
Texas San Antonio - CHAPTER - 7853
Chapter 8: Loglinear Models for Contingency Tables Lecture 1 Two-way tables: Consider an I J two-way table based on a sample of size n with 1) cell probabilities {ij } 2) cell counts {nij } 3) expected cell count {ij = nij } The loglinear models use
Texas San Antonio - CHAPTER - 7853
Chapter 10: Measuring Agreement Between Observers Lecture 3 Consider n subjects which are classied or rated by two observers, A and B. The rating categories are ordinal. Agreement: Departure from Independent: Let ab denote the probability that A clas
Texas San Antonio - CHAPTER - 7853
Solutions to HW #1 STA 7853, SP 2007 1. (1.2). Let Y : No. of correct answers out 100. Then, if the student is guessing, =Prob(correct answer)=.25 a) Y Bin(100, .25)b) = n = 25, and = n(1 - ) = 4.33 Since P (Y 50) P (Z 49.5-25 4.33 )= P (Z
Texas San Antonio - STA - 5313
Chapter 3: Ratio and Regression Estimators Lecture 2 Properties of Ratio Estimators: Recall that yS ^ ^ ^ ^ ^ ^ B= , y r = B xU , ty = Btx xS Bias: ^ 1) Bias in B:^ Bias(B) = - ^ 2) Bias in tyr :^ Cov(B, xS ) xU ^ ^ ^ Bias(tyr ) = -Cov(B, tx
Texas San Antonio - STA - 5313
Chapter 8: Nonresponse Nonrespone Failure to observe the response form a unit in the sample Best ways to deal with Nonrespone: 1) Prevent it from happening 2) Predict by modeling missing data.Two Types of Nonresponses: 1) Unit Nonrespone: Entire ob
Trinity U - CS - 1300
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Trinity U - CS - 1300
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Ill. Chicago - PHY - 101
Statistics of the 1st Midterm Exam of Keung's section, September 25, 1997Number of students in the exam: 54Mean: 49; Median: 4415-19 X 120-24 XXXXX 525-29 XXXX 430-34 XX 2 35-3
Ill. Chicago - PHY - 101
Statistics (Keung's section ONLY) 1st Midterm Exam September 25, 1997,2nd Midterm Exam October 21, 1997. # of students | # of students in the 1st Midterm: 54 | in the 2nd Midterm: 47 Mean: 49; Median: 44
Penn State - GNM - 5015
Curr iculum VitaeGenevieve N. Miller825 S. Allen Street, Apt. 9 State College, PA 16801 412-418-9547 gnm5015@psu.eduEDUCATION The Pennsylvania State University, University Park, PA Anticipated Graduation, May 2010 The Schreyer Honors College B
Midwestern State University - MATH - 105
Math 105 Summer 2009 Exam 1BName: Section 1You must show your work and/or briefly explain your answers for all problems. Credit will not be given for correct answers that are not justified. Unless otherwise specified, use eight decimal places in
Midwestern State University - MATH - 105
Table for the Standard Normal Distributionz=x-z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.40.00 0.0000 0.0398 0.0793 0.1179 0.1554 0.1915 0.2257
Midwestern State University - MATH - 206
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UCSB - PSYCH - 168
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