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Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 490490Chapter 15 Large-Scale Chromosomal ChangesAnother approach is to treat monoploid cells basically as a population of haploid organisms in a mutagenesis-and-selection procedure. A population of monoploid cells
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 49115.1 Changes in chromosome number491copies of each sex chromosome, such as XXY, XYY, XXX, or XO (the O stands for absence of a chromosome and is included to show that the single X symbol is not a typographical
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 492492Chapter 15 Large-Scale Chromosomal ChangesXO. Affected persons have a characteristic phenotype: they are sterile females, short in stature, and often have a web of skin extending between the neck and shoulde
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/22/04 4:08 PM Page 49315.1 Changes in chromosome number493One poleOther pole1 2 3A a a1 2A3aa A21 3aa a12 3AaFigure 15-15 Genotypes of the meiotic products of an A /a/atrisomic. The three segregations shown are equ
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 494494Chapter 15 Large-Scale Chromosomal ChangesIncidence of Down syndrome per number of births1/46faunlike ears, a small jaw, a narrow pelvis, and rockerbottom feet; almost all babies with trisomy 18 die within
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 49515.1 Changes in chromosome number495chromosome 1 and gene B on chromosome 2, the ratio is 4 A : 4 B, or 1 : 1. In contrast, in an aneuploid, the ratio of genes on the aneuploid chromosome to genes on the other
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 496496Chapter 15 Large-Scale Chromosomal ChangesXY males. In XXX females, on the other hand, the few transcribed genes are active at only 1.5 times the level that they are in XX females. This lower level of functi
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 49715.2 Changes in chromosome structure497mosomal rearrangements by breakage can be induced articially by using ionizing radiation. This kind of radiation, particularly X rays and gamma rays, is highly energetic a
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 498498Chapter 15 Large-Scale Chromosomal Changesthe two segments are adjacent to each other (a tandem duplication), as in this duplication of segment C: A B C D ENormal sequence Paracentric PericentricA A AB B
Aberystwyth University - BIOL - BIOL
44200_15_p481-520 3/12/04 1:06 PM Page 49915.2 Changes in chromosome structure499Chromosome rearrangements: formation of paracentric inversions(a)abacddeInversion loop c Normalb a c bdd e ebeInversiona(b)www. ANIMATED ARTable. Hence,
City University of Hong Kong - EE - 4115
Data EncryptionEfficient use of spectrumDigital VideoLess sensitive to noise & distortionsIntegration of digital servicesR: 83Mbit/s G: 83Mbit/s B: 83Mbit/s Total: 249Mbit/s 5.5MHz = 720 pixels Raw image R,G andB 576 lines 576 lines/frame 720 pixels/
City University of Hong Kong - EE - 4047
EE4047 Genetic Algorithms and Their ApplicationsK S Tang Department of Electronic Engineering City University of Hong KongRoom: G6354 Tel: 2788 7783 e-mail: eekstang@cityu.edu.hkCity University of Hong KongReading MaterialsGenetic Algorithms: Concept
City University of Hong Kong - EE - 4047
Part 1: Optimization Problems and Classical MethodsCity University of Hong KongOptimization ProblemsFind the best solution from all feasible solutions for a problem Best for a function, cost, Examples:Dimensions of Antenna Wireless network layout Para
City University of Hong Kong - EE - 4047
Part 2: Basic Genetic AlgorithmCity University of Hong KongGenetic AlgorithmNatural selection: Survival of the fittest DNA structuresA C T G Nucleotides AAA CGA Codons . A A A C G A A T C . Genes ATCAAA C GAAT CDNA1Biological DNA Double HelixFrom
City University of Hong Kong - EE - 4047
Part 4: Theory and HypothesisCity University of Hong KongHyperplanesAssume that we have a problem, where the solution can be encoded in 3 bits (X1,X2,X3)X2010 110011 000 001111 100X1101X3*11TerminologySchema: binary string may contain * (don
City University of Hong Kong - EE - 4047
Part 5: Modification on Simple GACity University of Hong KongSimple Genetic Algorithm Binary representation Roulette Wheel Selection Single Point Crossover Bit Mutation High crossover rate and low mutation rate Generational Replacement Policy1Chromo
City University of Hong Kong - EE - 4047
Part 6: AdvantagesCity University of Hong KongStrength of GAHandle multi-modal problems Handle multi-objective problems Parallelism Handle constrainted problems1Multi-objective ProblemsLinear combination Nonlinear combination Pareto-based approachP
City University of Hong Kong - EE - 4047
Part 7: Problems and DifficultiesCity University of Hong KongProblem 1:Premature Convergence and Genetic DriftStochastic errors in sampling caused by small population sizes Genetic Drift: Population converges on a single peak without differential adva
City University of Hong Kong - EE - 4047
Part 8: Advanced GA Designs(I) Hierarchical Chromosome StructureCity University of Hong KongBiological InspirationRegulatory Sequences and Structural Genestran-acting factor transcription initiation siteA promoterBCstructural genesmRNAtranscrip
City University of Hong Kong - EE - 4047
Test Questions in Past Few Years Question 5 The yellow region is for the goals. Those meet all the goals will fall inside.To build the table using Fonseca and Fleming approach with goal attainment, you should consider three cases (see the notes). Case 1:
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
Assignment 3, Oscillator Design 4 The transistor LAE2001R is unconditionally stable at 1.9GHz in both the common emitter and common base configuration. Connecting a 1pF between the base and the collector makes the transistor potentially unstable and the r
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
SuggestedRevisionforEE5602Handouts,Notes,Tutorial,Assignments,Test,ExamplesandReviewofTransmissionLines1 Topic AllexceptSmithchart Equationswhichwillbegiven Whatyouareexpectedtoknow Principleanduse ConversionfromtoZ andZto AdmittanceorImpedancechart Ad
City University of Hong Kong - EE - 5602
EE5602 Active Radio Frequency & Microwave TechniquesLT11Course Lecturer: Dr. Wing Shing Chan Room: FYW6322 Email: eeej2710@cityu.edu.hkSome material are reproduced with permission from Q. XUEs lecture notes.General Aims This course aims to provide st
City University of Hong Kong - EE - 5602
Traveling WavesZSV + e - z Ve = + jzZ z R e fe re n c e p la n eL Voltage CurrentV ( z ) = V + e z + V ez I ( z ) = I + e z I ezReflection Coefficient at LoadV V ZS + -V + e - z VezZ zL Load reflection coefficient In polar formV L = + V L
City University of Hong Kong - EE - 5602
Ell Matching Using Smith Chart (CCT A)C L Y y10y 1 lo c u szinz 1 lo c u s z in a r e aEll Matching Using Smith Chart (CCT B)L C Z y in y 1 lo c u s y in a r e a z0z 1 lo c u s1Ell Matching Using Smith Chart (CCT C)L C Y y 1 lo c u s0y inz
City University of Hong Kong - EE - 5602
Scattering Parameters [S]I1 VV1 V21I2 [ 2 -p o r t], I1 = = =V2= V1+ + V1 = V2+ + V2, I21+ (V1 V1 ) Z0 1+ (V2 V2 ) Z0 s11 V1+ V2+ + s12 Z0 Z0 V1+ V2+ + s22 Z0 Z0V V 1 2V1 = s11V1+ + s12V2+ Z0 + + V2 = s21V1 + s22V2 Z0= s21Defining New Trave
City University of Hong Kong - EE - 5602
[Y] Short-Circuit Admittance Parametersi1 v1i1 y11 = v1 i1 y12 = v2 i2 y21 = v1 y22 = i2 v2i2 [y] v2i1 = y11v1 + y12 v2 i2 = y21v1 + y22 v2s/c input admittancev2 = 0s/c reverse transfer admittancev1 = 0s/c forward transfer admittancev2 = 0s/c o
City University of Hong Kong - EE - 5602
RF/Microwave Transistor Types Bipolar Junction Transistor Ge-BJT, Si-BJT, SiGe-BJT Heterojunction Bipolar Transistors (HBT) Field Effect Transistor (FET) Metal Insulator Semiconductor FET (MISFET) Metal Oxide Semiconductor FET (MOSFET) Junction FET
City University of Hong Kong - EE - 5602
Power Amplifier DesignClass of Operation (conduction angle ) Class A ( = 360o), MAX = 50% Class AB (180o < < 360o), MAX = 50-78.5% Class B ( = 180o), MAX = 78.5% Class C (0o < < 180o), MAX = 78.5-100% Class E ( = 0o), MAX = 100%Difference Between Avera
City University of Hong Kong - EE - 5602
3-Port Scattering Parametersa1b 33 b1 s11 b = s 2 21 b3 s31 b1 s11 = a1 b1 s12 = a2 b1 s13 = a3s12 s22 s32s13 a1 s23 a2 s33 a3 b11a33 -p o rt N /W2 a2 b2= input with ports 2 and 3 matcheda 2 , a3 = 0= input with ports 1 and 3 matcheda1 ,
City University of Hong Kong - EE - 5602
Oscillators Oscillation Conditions I Nyquist Criterion Barkhausen Criterion Oscillation Conditions II Concept of negative resistance Use of s-parameters Oscillator Configurations Basic oscillator Matrix conversion Pierce, Colpitts and Clapp family R
City University of Hong Kong - EE - 5602
Diode Circuits Types of Diodes Detector Square law Envelope Mixer Single ended Doubly balanced Control Switches Attenuators Phase shiftersDiode Detector Diode equation qvd id = I S exp 1 kT Integrating over one RF cycle gives the average current 1
City University of Hong Kong - EE - 5602
EE5602 Active RF and Microwave Techniques, Assignment 1.D Z 0 = 50 Z 0 =5 0Z L =50 Z O U T = 1 5+ j 25 S/CLUsing a Smith Chart show clearly how you would design a single stub matching network to match the transistor output impedance ZOUT to 50.0 .
City University of Hong Kong - EE - 5602
EE5602 Active RF and Microwave Techniques, Assignment 2. A transistor is biased with a collector emitter voltage of 3V and a collector current of 120mA. The s-parameters under this bias condition are given in table 1. Frequency s11 s12 o 2.0GHz 0.737 163
City University of Hong Kong - EE - 5602
Assignment 4, Diode Detector DesignDue on Tuesday 20 April at 7:30pm before test. A diode detector is used to monitor the level of power from the output of a power amplifier at 915MHz. Diode HSMS2800 made by Avago is used as the diode detector biased wit
City University of Hong Kong - EE - 5602
Traveling WavesZSV + e - z Ve = + jzZ z R e fe re n c e p la n eL Voltage CurrentV ( z ) = V + e z + V ez I ( z ) = I + e z I ezReflection Coefficient at LoadV V ZS + -V + e - z VezZ zL Load reflection coefficient In polar formV L = + V L
City University of Hong Kong - EE - 5602
Chapter 1Introduction to Wireless TechniquesScope of Applications1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Wireless Communications Radar Navigation Remote Sensing RFID Broadcasting Automobile and Highway Sensors Surveillance and Electronic Warfare Medical
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 5602
City University of Hong Kong - EE - 4015
Design of FIR filter Design of FIR filters by windowing FIR filter can only be implemented in digital form. Its filter design is based on the approximation of the desired frequency response in the digital domain. Most design methods assume the filter stru
City University of Hong Kong - EE - 4015
Discrete-time SystemsThe function of a discrete-time system is to process a given input sequence to produce an output sequence. In practice, the input is a digital signal and the output is also a digital signal. Such a discrete-time system is called as a
City University of Hong Kong - EE - 4015
Discrete-time Fourier Analysis (DTFT) Discrete-time Fourier analysis (transform) is a powerful tool to analyze the frequency characteristics of discrete-time signals and systems. If the discrete-time signal is periodic, discrete-time Fourier series will b
City University of Hong Kong - EE - 4015
Structures of Digital Filters The difference equation of a general infinite impulse response (IIR) digital filter is given by y[n] = al y[n l ] +l =1 Nk =0 bk x[n k ] .NThe filter consists of two sets of coefficients: feedback coefficients cfw_al an
City University of Hong Kong - EE - 4015
IIR Filter Design A common method of designing IIR filter is to transform a continuous-time filter into a discrete-time filter. The reasons for this approach are (1) many useful continuous-time IIR filter design methods have relatively simple closed form
City University of Hong Kong - EE - 4015
Digital Signal Processing in CommunicationsDr. Shu-hung LEUNG Office: G6509 Tel: 2788-7784 Email: EEEUGSHL@cityu.edu.hk1Information Textbook: An Introduction to Digital Signal Processing by Edmund M K Lai Assessment: Assignments (No marking)Quizzes:
City University of Hong Kong - EE - 4015
Signals Signals are represented mathematically as functions of one or more independent variables. A signal is called as continuous time signal when it is a function of continuous time. It is a discrete-time signal when the time takes a discrete set of val
City University of Hong Kong - EE - 4015
Problem (1)(a) X ( e j ) =n= (u[n] u[n 5])e jn = e jn =n=04 1 e j 5 e j 5 / 2 (e j 5 / 2 e j 5 / 2 ) = j / 2 j / 2 1 e j e ( e e j / 2 ) = e j 2 sin( 5 / 2) sin( / 2)(b) X ( e j ) =n= n (u[n] u[n 8])e jn = ne jn = (e j )n =n=0 n=0 n77 1
City University of Hong Kong - EE - 4015
Problem 1: It is easily obtained that + s c = p = 0.525 , 2 = p s = 0.15 , = mincfw_ p , s = 0.005 . Therefore we have A = 20 log10 0.005 = 46.02dB . For Hamming window, we can calculate that M = 6.64 / 0.15 = 44.3 44 . Therefore length of the filter wil
City University of Hong Kong - EE - 4015
(1) Answer:It is easily calculated from the specification of the digital lowpass filter that: p = p T = 2f p T = 8k / 24k = / 3 . s = s T = 2f s T = 12k / 24k = / 2 And the value of the filters frequency response at the above edge frequency can be calcul
City University of Hong Kong - EE - 4015
Problem (1)(a). Use phasors to simplify the sum x[n] = cos(0.4n + / 3) + 2 cos(0.4n + 2 / 3) = A cos(0.4n + ) The values of A and are obtained as 2 2 Ae j = e j / 3 + 2e j 2 / 3 = cos( ) + 2 cos( ) + j[sin( ) + 2 sin( )] 3 3 3 3 = 0.5 1 + j[0.866 + 1.732
City University of Hong Kong - EE - 4015
Problem 1 Solution:(a) 1 + z 1 1 2 H (z) = 1 1.2 z 1 0.32 z 2 1 + 2 z + z 1 1 2 1 + z 1 + 2z + z = 1 1.2 z 1 0.32 z 2 1 + 3 z 1 + 3 z 2 + z 3 Y ( z ) = = 1 1.2 z 1 0.32 z 2 X ( z)()()()So the difference equation is y[n] 1.2 y[n 1] 0.32 y[n 2] = x[n
City University of Hong Kong - EE - 4015
(1)Simplify 3e j / 3 + 4e j / 6Answer: 3e j / 3 + 4e j / 6 = 3(cos + j sin ) + 4(cos j sin ) 3 3 6 6 = 3(cos + cos ) + j (3 sin 4 sin ) 3 6 3 6 3+ 4 3 3 34 = +j 2 2 = 5ej arctan 3 3 4 3+ 4 3= 5e j 0.12(2)Simplify x(t ) = 5 cos(t + 900 ) + 5 cos(t 30
City University of Hong Kong - EE - 4015
Problem (1) Solution:3 (a) The system described by y[n] = n x[n] is linear, causal, unstable and time-variant. 3 For the input x1[n] , the output is y1[n] = n x1[n] ; 3 For the input x2 [n] , the output is y2 [n] = n x2 [n] ;For the input x3 [n] = ax1[n
City University of Hong Kong - EE - 4015
Problem (1)(a) Y ( z ) 2 z 1Y ( z ) + z 2Y ( z ) = X ( z ) + 3z 3 X ( z ) H ( z) = Y ( z) 1 + 3 z 3 = X ( z ) 1 2 z 1 + z 2(b) x[ n] = (0.4) n u[n] + ( 0.6)(0.6) n 1 u[n 1] According to the time shit property, the z-transform is X ( z) = 1 0.6 z 1 + 1 0
City University of Hong Kong - EE - 4015
Problem 1Suppose that a discrete-time signal x[ n] is given by the formula: x[n] = 2.2 cos(0.3 n / 3) and that it was obtained by sampling a continuous-time signal x(t ) = A cos(2 f 0t + ) at a sampling rate of f s = 6000samples/sec . Determine three dif