Lesson 3 Microprocessors

科普版小学英语四年级下册Lesson3微课精讲知识点视频讲解第一课时第二课时第三课时知识点知识点Lesson 3说一说露露：看那个女孩。

东东：她的上衣多漂亮啊！露露：她是谁？东东：对不起，我不知道。

露露：你好，请问你来自哪里？女孩：我来自云南。

你来自哪里？东东：我来自北京。

露露：你是新生吗？女孩：是，我是。

东东：欢迎到我们学校来。

女孩：谢谢。

阅读我们是一个大家庭你来自哪里？我来自新疆。

你擅长什么？我擅长唱歌。

你来自吉林吗？是的。

我想你很擅长跳舞。

是的。

你们也来自吉林吗？不，我们来自黑龙江。

我们是赫哲族人。

你们擅长什么？我们擅长捕鱼。

他们来自哪里？我想他们是从西藏来的。

他们也擅长捕鱼吗？不，我想他们擅长骑马。

我来自四川，我喜欢爬树。

我们是一个大家庭。

让我们一起唱歌玩耍吧。

4.Are these elephants?1. 名词复数的规则变化情况构成方法读音例词一般情况加-s1. 清辅音后读/s/;2. 浊辅音和元音后读/z/;1. map-maps2. bag-bagscar-cars以s, x ch, sh,等结尾的词加-es读/iz/bus-buses watch-watchesbox-boxes以ce, se, ze, (d)ge等结尾的词加-s读/iz/blouse-blousesorange-oranges以辅音字母+y结尾的词变y 为i再加es读/z/baby-babiesfamily-families2. 其它名词复数的规则变化1) 以y结尾的专有名词，或元音字母+y结尾的名词变复数时，直接加s变复数：如： monkey---monkeysholiday---holidays比较：层楼：storey---storeys story---stories2) 以o结尾的名词，变复数时：a.加-s，如：photo---photospiano---pianos radio---radioszoo---zoos；b.加-es，如：potato--potatoestomato--tomatoesc.均可，如： zero---zeros / zeroes3) 以f或fe结尾的名词变复数时：a.加s，如：belief---beliefs roof---roofs safe---safes gulf---gulfs；b.去f, fe 加ves，如：half---halvesknife---knives leaf---leaves wolf---wolveswife---wives life---lives thief---thieves；c.均可,如：handkerchief---handkerchiefs / handkerchieves怎么获取配套课件教案习题试卷等资料？。

Lesson 3 Analytical Chemistry Analytical chemistry is the science of making quantitative measurements. In practice, quantifying analytes in a complex sample becomes an exercise in problem solving. To be effective and efficient, analyzing samples requires expertise in: 1the chemistry that can occur in a sample2analysis and sample handling methods for a wide variety of problems (the tools-of-the-trade)3proper data analysis and record keepingTo meet these needs, Analytical Chemistry courses usually emphasize equilibrium, spectroscopic and electrochemical analysis, separations, and statistics.Analytical chemistry requires a broad background knowledge of chemical and physical concepts. With a fundamental understanding of analytical methods, a scientist faced with a difficult analytical problem can apply the most appropriate technique(s). A fundamental understanding also makes it easier to identify when a particular problem cannot be solved by traditional methods, and gives an analyst the knowledge that is needed to develop creative approaches or new analytical methods.1 GravimetryGravimetry is the quantitative measurement of an analyte by weighing a pure, solid form of the analyte. Obtaining pure solids from solutions containing an unknown amount of a metal ion is done by precipitation.Since gravimetric analysis is an absolute measurement, it is a principal method for analyzing and preparing primary standards. A typical experimental procedure to determine an unknown concentration of an analyte in solution is as follows: •quantitatively precipitate the analyte from solution•collect the precipitate by filtering and wash it to remove impurities•dry the solid in an oven to remove solvent•weigh the solid on an analytical balance•calculate the analyte concentration in the original solution based on the weight of the precipitateGravimetric Determination of Iron:•Determine constant weight of the crucibles•Oxidation of iron sample•Precipitation of iron hydroxide•Ignition of iron hydroxide to iron oxide•Determine constant weight of the crucibles plus iron oxide•Calculation of iron in the sample2 TitrationTitration is the quantitative measurement of an analyte in solution by completely reacting it with a reagent solution. The reagent is called the titrant and must either be prepared from a primary standard or be standardized versus a primary standard to know its exact concentration.The point at which all of the analyte is consumed is the equivalence point. The number of moles of analyte is calculated from the volume of reagent that is required to react with all of the analyte, the titrant concentration, and the reaction stoichiometry.The equivalence point is often determined by visual indicators are available for titrations based on acid-base neutralization, complexation, and redox reactions, and is determined by some type of indicator that is also present in the solution. For acid-base titrations, indicators are available that change color when the pH changes. When all of the analyte is neutralized, further addition of the titrant causes the pH of the solution to change causing the color of the indicator to change.If the pH of an acid solution is plotted against the amount of base added during a titration, the shape of the graph is called a titration curve. All acid titration curves follow the same basic shapes.Strong Acid Titration CurveAt the beginning, the solution has a low pH and climbs as thestrong base is added. As the solution nears the point where all ofthe H+are neutralized, the pH rises sharply and then levels outagain as the solution becomes more basic as more OH- ions areadded.Manual titration is done with a buret, which is a long graduated tube to accurately deliver amounts of titrant. The amount of titrant used in the titration is found by reading the volume of titrant in the buret before beginning the titration and after reaching the endpoint. The difference in these readings is the volume of titrant to reach the endpoint. The most important factor for making accurate titrations is to read the buret volumes reproducibly. The figure shows how to do so by using the bottom of the meniscus to read the reagent volume in the buret.The end point can be determined by an indicator as described above or by an instrumental method. The most common instrumental detection method is potentiometric detection. The equivalence point of an acid-base titration can bedetected with a pH electrode. Titrations, such as complexation or precipitation, involving other ions can use an ion-selective electrode (ISE). UV-vis absorption spectroscopy is also common, especially for complexometric titrations where a subtle color change occurs.For repetitive titrations, autotitrators with microprocessors are available that deliver the titrant, stop at the endpoint, and calculate the concentration of the analyte. The endpoint is usually detected by some type of electrochemical measurement. Some examples of titrations for which autotitrators are available include:•Acid or base determination by pH measurement with potentiometric detection.•Determination of water by Karl Fischer reagent (I2 and SO2 in methyl alcohol and pyridine) with coulometric detection.•Determination of Cl in aqueous solution with phenylarsene oxide using amperometric detection.3 ExtractionExtractions use two immiscible phases to separate a solute from one phase into the other. The distribution of a solute between two phases is an equilibrium condition described by partition theory. Boiling tea leaves in water extracts the tannins, theobromine, and caffeine (the good stuff) out of the leaves and into thewater. More typical lab extractions are of organic compounds out of an aqueous phase and into an organic phase.Analytical Extractions4 Precipitation (Insoluble Salts)Many metal ions form compounds that are insoluble in water. We call them insoluble salts or precipitates. Common precipitates are carbonates, hydroxides, sulfates, and sulfides. Ions that we consider spectator ions when discussing acid-base equilibria will form insoluble salts.An insoluble salt in contact with water maintains an equilibrium with the ions. In simple cases where there are no common ions or competing equilibria, the ion concentrations depend only on the equilibrium constant for the particular precipitate. When we talk about solubility equilibria we always write the equilibrium with the solid on the left. For example:Ba(IO3)2 (s) Ba2+(aq) + 2 IO3-(aq)The equilibrium constant expression for an insoluble salt is written following the same rules as for any other equilibrium. The equilibrium constant is called the solubility product, K sp. The K sp expression for the above equilibrium is:K sp = [Ba2+][IO3-]2K sp Values for Some PrecipitatesFormula Name K spAgCl silver chloride 1.8×10-10Al(OH)3aluminum hydroxide 2×10-32BaCO3barium carbonate 5×10-9Words & Phrasesamperometric [] adj. 测量电流的analyte [] n. （被）分析物buret [] n. 滴定管；量筒carbonate [] n. 碳酸盐complexometri c []n. 络合滴定（法）coulometric [] n. 库仑滴定crucible [] n. 坩埚endpoint [] n. 端点equilibrium [] n. 平衡（复数形式：equilibria）filtering [] n. 过滤gravimetry [] n. 重量测定法hydroxide [] n. 氢氧化物impurity [] n. 不纯，杂质insoluble [] adj. 不能溶解的，不能解决的neutralization []n. 中和（作用）reagent [] n. 反应物, 试剂solute [] n. 溶解物，溶质solvent [] n. 溶剂spectroscopic [] adj. 分光镜的，借助分光镜的sulfate [] n. 硫酸盐sulfide [] n. 硫化物tannin [] n. 单宁酸theobromin [] n. 可可碱titrant [] n. 滴定剂（滴定标准液）n. 分析天平analyticalbalanceaqueous phase n. 水相n. 等量点equivalencepointgraduated tube n. 刻度管n. 不混溶相immisciblephaseion-selective electrode n. 选择性离子电极organic phase n. 有机相partition theory n. 分配理论potentiometric adj. 电势测定的precipitation n. 沉淀(作用) quantitative measurement n. 定量测量solubility solution product n. 溶度积stoichiometry n. 化学计量法，化学计量学第3课分析化学分析化学是定量测量的科学。

Lecture 1 -Introduction to MicroprocessorsObjective : 1.General Architecture of a Microcomputer System2. Types of Microprocessors3. Number Systems------------------------------------------------------------------------------1. General Architecture of a Microcomputer SystemThe hardware of a microcomputer system can be divided into four functional sections: the Input unit , Microprocessing Unit , Memory Unit , and Output Unit . See Fig. 1Figure 1∙M icro P rocessor U nit (MPU ) is the heart of a microcomputer. A microprocessor is a general purpose processing unit built into a single integrated circuit (IC).The Microprocessor is the part of the microcomputer that executes instructions of the program and processes data. It is responsible for performing all arithmetic operations and making the logical decisions initiated by the computer’s program. In addition to arithmetic and logic functions, the MPU controls overall system operation.∙Input and Output units are the means by which the MPU communicates with the outside world.o Input unit: keyboard, mouse, scanner, etc.o Output unit: monitor, printer, etc.∙Memory unit:o Primary: is normally smaller in size and is used for temporary storage of active information. Typically ROM, RAM.o Secondary: is normally larger in size and used for long-term storage of information. Like Hard disk, Floppy, CD, etc.Memory UnitPrimary Storage Unit Secondar y Storage UnitOutp ut UnitInput UnitData Storage MemoryProgram Storage Memory2. Types of MicroprocessorsMicroprocessors generally is categorized in terms of the maximum number of binary bits in the data they process – that I, their word length. Over time, five standard data widths have evolved for microprocessors: 4-bit, 8-bit, 16-bit, 32-bit, 64-bit.There are so many manufacturers of Microprocessors, but only two companies have been produces popular microprocessors: Intel and Motorola. Table 1 lists some of types that belong to these companies (families) of microprocessors.Table 1: Some Types of Microprocessors:Memory sizeType Data buswidthIntel family:8085864K8086161M802861616M80386EX , 80386DX 16 , 3264M , 4G80486DX4324G + 16K cachePentium 644G + 16K cachePentiumIII , Pentium46464G+32K L1 cache +256 L2cacheMotorola family:6800864K68060644G + 16K cacheNote that the 8086 has data bus width of 16-bit, and it is able to address 1Megabyte of memory.It is important to note that 80286, 80386,80486, and Pentium-Pentium4 microprocessors are upward compatible with the 8086 Architecture. This mean that 8086/8088 code will run on the 80286, 80386, 80486, and Pentium Processors, but the reverse in not true if any of the new instructions are in use.Beside to the general-purpose microprocessors, these families involve another type called special-purpose microprocessors that used in embedded control applications. This type of embedded microprocessors is called microcontroller.The 8080, 8051, 8048, 80186, 80C186XL are some examples of microcontroller.3. Number SystemsFor Microprocessors, information such as instruction, data and addresses are described with numbers. The types of numbers are not normally the decimal numbers we are familiar with; instead, binary and hexadecimal numbers are used. Table 2 shows Binary and Hexadecimal representations for some decimal numbers.Table 1: Binary, and Hexadecimal representation of some numbers:Decimal B0001112102311341004510156110671117810008910019101010A111011B121100C131101D141110E151111FExample 1: Evaluate the decimal equivalent of binary number 101.012 Solution:101.012= 1(22) + 0(21) + 1(20) + 0(2-1) + 1(2-2)= 1(4) + 0(2) + 1(1) + 0(0.5) + 1(0.25)= 4 + 0 +1 + 0 + 0.25= 5.25Example2: Evaluate the binary representation of decimal number 8.875 Solution:Integer Fraction8/2=04/2=02/2=01/2=10/2=00/2=01000.1111000.111Generally, Binary numbers are expressed in fixed length either:8-bit called Byte16-bit called Word32-bit called Double WordExample3: Evaluate the 16-bit binary representation of decimal number 10210, then evaluate its hexadecimal representationSolution:10210 = 011010112 = 6BHLecture 2-Software Architecture of 80861. Internal Architecture of the 8086The internal architecture of the 8086 contains two processing units: the bus interface unit(BIU) and the execution unit(EU). Each unit has dedicated functions and both operate at the same time. This parallel processing makes the fetch and execution of instructions independent operations. See Fig. 1The BIU is responsible for performing all external bus operations, such as instruction fetching, reading and writing of data operands for memory, address generating, and inputting or outputting data for input/output peripherals. These operations are take place over the system bus. This bus includes 16-bit bidirectional data bus, a 20-bit address bus, and the signals needed to control transfer over the bus.Fig 1: Execution and bus interface unitsThe BIU uses a mechanism known as instruction queue. This queue permits the 8086 to prefetch up to 6 bytes of instruction code.The EU is responsible for decoding and executing instructors. It contains arithmetic logic unit (ALU), status and control flags, general-purpose register, and temporary-operand registers.2. Memory address space and data organization8086 can supports 1Mbyte of external memory that organized as individual bytes of data stored at consecutive addresses over the address range 0000016 to FFFFF 16. The 8086 can access any two consecutive bytes as a word of data. The lower-addressed byte is the least significant byte of the word, and the higher- addressed byte is its most significant byte.Example 1: For the 1Mbyte memory shown in Fig 2, storage location of address 0000916contains the value 000001112=716 , while the location of address 0001016 contains the value 01111101= 7D 16 . The 16-bit word 225A 16 is stored in the locations 0000C 16 to 0000D 16 .The word of data is at an even-address boundary if its least significant byteis in even address. It’s also called aligned word . The word of data is at an odd-address boundary if its least significant byte is in odd address. It’s also called misaligned word , as shown in Fig 3.To store double word four locations are needed. The double word that it’s least significant byte store at an address that is a multiple of 4 (e.g. 016, 416, 816,....) as shown in Fig 4.000090000A 0000B 0000C 0000D 0000E 0000F 00010Fig 2:Part of 1Mbyte3. Segment registers and memory segmentation Even though the 8086 has a 1Mbyte address space, not all this memory is active at one time. Actually, the 1Mbytes of memory are partitioned into 64Kbyte (65,536) segments . Each segment is assigned a Base Address that identifies its starting point (identify its lowest address byte-storage location).Only four of these 64Kbyte segments are active a time: the code segment, stack segment, data segment, and extra segment. The addresses of these four segments are held in four segment registers: CS (code segment), SS (stack segment), DS (data segment), and ES (extra segment). These registers contain a 16-bit base address that points to the lowest addressed byte of the segment (see Fig 5).Note that the segment registers are user accessible. This means that the programmer can change their contents through software.There is one restriction on the value assigned to a segment as base address: it must reside on a 16-byte address boundary. This is because the memory address is 20 bits while the segment register width is 16 bits. Four bits (0000) must be added to the segment register content to evaluate the segment starting address.Fig 3 Aligned and misaligned word Fig 4 Aligned and misaligned doublewordFig 5: Software model of 8086 microprocessorExample 2:Let the segment registers be assigned as follow:CS = 0009H, DS = 0FFFH, SS = 10E0, and ES = 3281H. We note here that code segment and data segment are overlapped while other segments are disjointed (see Fig 6).Fig 6: Overlapped and disjointed segments4. Instruction PointerInstruction pointer (IP ): is a 16 bits in length and identifies the location of the next word of instruction code to be fetched from the current codesegment of memory, it contains the offset of the next word of instruction code instead of its actual address.The offset in IP is combined with the current value in CS to generate the address of the instruction code (CS:IP). 5. Data RegistersThe 8086 has four general-purpose data register, which can be used as the source or destination of an operand during arithmetic and logic operations (see Fig 5).Notice that they are referred to as the accumulator register (A), the base register (B), the count register (C), and the data register (D). Each one of these registers can be accessed either as a whole (16 bits) for word data operations or as two 8-bit registers for byte-wide data operations.Code segment (64kbyte)Data segment (64kbyte)Stack segment (64kbyte)Extra segment (64kbyte)CS DS SS ES0009H 0FFFH 10E0H 3281H00090These two segmentsare overlappeSegment registers1Mbyte memory unitFig 7: (a) General purpose data Registers, (b) dedicated register functions6. Pointer and Index RegistersThe 8086 has four other general-purpose registers, two pointer registers SP and BP, and two index registers DI and SI. These are used to store what are called offset addresses.An offset address represents the displacement of a storage location in memory from the segment base address in a segment register.Unlike the general-purpose data registers, the pointer and index registers are only accessed as words (16 bits).∙The stack pointer (SP) and base pointer (BP) are used with the stack segment register (SS) to access memory locations within the stack segment.∙The source index (SI) and destination index (DI) are used with DS or ES to generate addresses for instructions that access data stored in the data segment of memory.7. Status RegisterThe status register also called flag register: is16-bit register with only nine bits that are implemented (see Fig 8). Six of theses are status flags:1.The carry flag (CF): CF is set if there is a carry-out or a borrow-in for themost significant bit of the result during the execution of an instruction.Otherwise FF is reset.2.The parity flag(PF): PF is set if the result produced by the instructionhas even parity- that is, if it contains an even number of bits at the 1 logic level. If parity is odd, PF is reset.3.The auxiliary flag(AF): AF is set if there is a carry-out from the lownibble into the high nibble or a borrow-in from the high nibble into the low nibble of the lower byte in a 16-bit word. Otherwise, AF is reset.4.The zero flag (ZF): ZF is set if the result produced by an instruction iszero. Otherwise, ZF is reset.5.The sign flag (SF): The MSB of the result is copied into SF. Thus, SF is set ifthe result is a negative number of reset if it is positive.6.The overflow flag (OF): When OF is set, it indicates that the signedresult is out of range. If the result is not out of range, OF remains reset.The other there implemented flag bits are called control flags:1.The trap flag(TF): if TF is set, the 8086 goes into the single-step modeof operation. When in the single-step mode, it executes an instruction and then jumps to a special service routine that may determine the effect of executing the instruction. This type of operation is very useful for debugging programs.2.The interrupt flag(IF): For the 8086 to recognize maskable interruptrequests at its interrupt (INT) input, the IF flag must be set. When IF is reset, requests at INT are ignored and the maskable interrupt interface is disabled.3.The direction flag (DF): The logic level of DF determines the directionin which string operations will occur. When set, the string instructions automatically decrement the address; therefore the string data transfers proceed from high address to low address.15141312119420D IF TF D ZFFig 8: Flag registerThe 8086 provides instructions within its instruction set that are able to use status flags to alter the sequence in which the program is executed. Also it contains instructions for saving, loading, or manipulation flags.8. Generating a memory address∙In 8086, logical address is described by combining two parts: Segment address and offset .∙Segment address is 16-bit data from one of the segment registers (CS, SS, DS and ES).∙Offset address is 16-bit data from one of the index and pointer registers (DI, SI, SP and BP). Also it could be base register BX.∙To express the 20-bit Physical Address of memory1 Multiply Segment register by 10H ( or shift it to left by four bit)2Add it to the offset(see Fig 9)Fig 9: Generating a Memory AddressExample 3: if CS = 002AH, and IP = 0023H, write the logical address that they represent, then map it to Physical address.Solution:Logical address =CS :IP 002A : 0023Physical address = ( CS X 10H ) + IP = 002A0 +0023 = 002C3Example 4: if CS = 002BH, and IP = 0013H, write the logical address that they represent, then map it to Physical address.Solution: Logical address =CS :IP 002B : 0013 Physical address = ( CS X 10H ) + IP = 002B0 +0013 = 002C3Offset value:IP BP DI SI or BXSegment Register:CS SS DSPhysicaladdressesare identicalhere !Actually, many different logical addresses map to the same physical address location in memory.9. The stackThe stack is implemented in the memory and it is used for temporary storage of information such as data and addresses. The stack is 64Kbytes long and is organized from a software point of view as 32Kwords (see Fig 10).∙SS register points to the lowest address word in the stack∙SP and BP points to the address within stack∙Data transferred to and from the stack are word-wide, not byte-wide.∙The first address in the Stack segment (SS : 0000) is called End of Stack.∙The last address in the Stack segment (SS : FFFE) is called Bottom of Stack.∙The address (SS:SP) is called Top of Stack.∙POP instruction is used to read word from the stack.∙PUSH instruction is used to write word to the stack.∙When a word is to be pushed onto the top of the stack:o the value of SP is first automatically decremented by twoo and then the contents of the register written into the stack.∙When a word is to be popped from the top of the stack the o the contents are first moved out the stack to the specific registero then the value of SP is first automatically incremented by two.Fig 10: Stack segment of memoryExample 5: let AX =1234H , SS =0105H and SP =0006H. Fig 11 shows the state of stack prior and after the execution of next program instructions:PUSH AX POP BX POP AXFig 11PUSH and POP instruction10. Input and Output address space010540105501056010570105001051010520105301058010590105A 0105B SSSP BX 010500065D00551F DF DD 0200C0529068A255010540105501056010570105001051010520105301058010590105A 0105B SSSP BX 010500045D003412DF DD 0200C0529068A255010540105501056010570105001051010520105301058010590105A 0105B SSSP BX 0105000612343412DF DD 0200C0529068A255010540105501056010570105001051010520105301058010590105A 0105B SSSP BX 0105000812343412DF DD 0200C0529068A255(a) Initial state(b) After execution of PUSH AX(d) After execution of POP AX(c) After execution of POP BXAX 1234AX 1234AX 1234AX DDDFThe 8086 has separate memory and input/output (I/O) address spaces. The I/O address space is the place where I/O interfaces, such as printer and monitor ports, are implemented. Notice that this address range is form 0000H to FFFFH. This represents just 64Kbyte addresses; therefore only 16 bits of address are needed to address I/O space.Problems1.What are the length of the 8086’s address bus and data bus?2.How large is the instruction queue of the 8086?3.List the elements of the executionunit.4.What is the maximum amount ofmemory that can be active at a given time in the 8086?5.Which part of the 8086’s memoryaddress space can be used to store the instruction of a program? two dedicated operationsassigned to the CX register.7.Calculate the value of each ofthe physical addresses that follows. Assume all numbers are hexadecimal numbers.a)A000 : ? =A0123b)? : 14DA =235DAc)D765 : ? =DABC0d)? : CD21 =322D218.If the current values in the codesegment register and the instruction pointer are 020016 AND 01AC16 , respectively, what physical address is used in the next instruction fetch?.9.If the current values in the stacksegment register and stack pointer are C00016and FF0016, respectively, what is the address of the current top of the stack?Lecture 3-Addressing MODES1. Introduction to assembly language programming∙Program is a sequence of commands used to tell a microcomputer what to do.∙Each command in a program is an instruction∙Programs must always be coded in machine language before they can be executed by the microprocessor.∙ A program written in machine language is often referred to as machine code.∙Machine code is encoded using 0s and 1s∙ A single machine language instruction can take up one or more bytes of code∙In assembly language, each instruction is described with alphanumeric symbols instead of with 0s and 1s∙Instruction can be divided into two parts : its opcode and operands ∙Opcode identify the operation that is to be performed.∙Each opcode is assigned a unique letter combination called a mnemonic.∙Operands describe the data that are to be processed as the microprocessor carried out the operation specified by the opcode.∙Instruction set includes1.Data transfer instructions2.Arithmetic instructions3.Logic instructions4.String manipulation instructions5.control transfer instructions6.Processor control instructions.∙As an example for instructions, next section discusses the MOV instruction.2. The MOV instruction∙The move instruction is one of the instructions in the data transfer group of the 8086 instruction set.∙Execution of this instruction transfers a byte or a word of data from a source location to a destination location.Fig 1 shows the general format of MOVinstruction and the valid source anddestination variations.Fig 1The MOV instruction and the valid source and destinationvariations。

（科普版）三年级英语上册 Lesson 3 教案教学内容：本课教学内容为科普版三年级英语上册Lesson 3。

课程内容以“动物”为主题，学生将学习认识不同的动物，并掌握相关的英语词汇和句型。

课程内容丰富，旨在激发学生对英语学习的兴趣，同时提高他们的英语听说能力。

教学目标：1. 知识与技能：学生能够掌握本课的动物词汇和句型，并能运用所学知识进行简单的英语交流。

2. 过程与方法：通过图片展示、游戏互动等方式，培养学生观察、思考和表达能力。

3. 情感态度价值观：培养学生关爱动物、保护环境的意识，激发他们对英语学习的兴趣。

教学难点：1. 动物词汇的发音和记忆。

2. 句型的正确运用。

教具学具准备：1. 教具：PPT、图片、录音机、磁带。

2. 学具：英语课本、练习册、文具。

教学过程：1. 导入a. 跟唱英文歌曲《Old MacDonald Had a Farm》，让学生在轻松愉快的氛围中进入英语学习状态。

b. 引导学生观察教室内的动物图片，复习已学的动物词汇。

2. 新课展示a. 利用PPT展示本课的动物词汇，教授单词的发音和拼写。

b. 通过图片和实物，让学生了解动物的特点和生活习性。

c. 引导学生用英语描述动物，如“I have a cat. It's white and black.”3. 练习与互动a. 组织学生进行“动物猜谜”游戏，巩固所学词汇。

b. 分组进行角色扮演，模拟动物的声音和动作，提高学生的听说能力。

b. 布置作业，让学生课后收集其他动物的英语词汇。

板书设计：1. 动物词汇：cat, dog, fish, bird, elephant, tiger, panda, rabbit2. 句型：I have a/an… It's…作业设计：1. 抄写本课所学动物词汇。

2. 仿照课文，用英语描述自己喜爱的动物。

3. 收集其他动物的英语词汇，准备下一课的分享。

课后反思：本节课通过图片展示、游戏互动等方式，让学生在轻松愉快的氛围中学习英语。

第三课时教学目标1、知识与技能：学习单词ruler eraser sharpener knife，学习句型Is this(that)your …?Yes,it is./No, it i snˊt.2.过程与方法：通过实物练习，使学生能够熟练运用句型：“Is this(that)your …?Yes,it is./No,it isnˊt. 于日常交际。

3.情感态度与价值观：培养学生运用所学内容进行日常交际的能力和积极英语表达的习惯。

教学重难点1、能听懂，认读单词ruler eraser sharpener knife2、能用主题句型“Is this(that)your …?Yes,it is./No,it isnˊt. 于日常交际。

教具单词卡片，录音机及相关磁带、尺子、卷笔刀、橡皮、小刀等文具教学过程Step 1 Warming up1 Free talk2 Play a gameListen and doT says and Ss doT:Touch your pen. Touch your pencil.Show me your pencil-box. Show me your book.Show me your bag.Step 2 Presentation1. T: Now look at the cards.教师分别出示单词卡片ruler eraser sharpenerknife并配合实物反复教生读这些词2.师手持这些文具，以Chant形式教学生学这些词ruler ruler ruler This is a ruler.eraser eraser eraser This is an eraser.sharpener sharpener sharpener This is a sharpener.knife knife knife This is a knife.3. 师拿着一名学生的橡皮T：Hi,xxx！Is this your eraser?S: Yes,it is.师再拿一名学生的尺子T：Hi,xxx！Is this your ruler?S: Yes,it is.师再拿一名学生的卷笔刀T：Hi,xxx！Is this your sharpener?S: Yes,it is.师再拿一名学生的小刀T：Hi,xxx！Is this your knife?S: Yes,it is.Step 3 Drills1.以“Chant”形式操练Let 's learn内容。

《专业英语》课程教学大纲一、课程基本情况屮文名称：专业英语英文名称：Specialty English课程号：1020046授课对象：电了信息工程专业、电了信息科学与技术专业木科生开课学期：秋学时数：36学分数：2. 5课程性质：选修专业课考核方式：考查先修课程：大学英语后续课程：无开课教研室：电了信息教研室执笔人：李红二、课程教学目标1.任务和地位木课稈是对于电了专业学生专业英语能力训练和培养的一门重要课程，是对大学高年级学生继公共英语课程Z后的一个重要补充和提高。

2.知识要求木课稈着重从实用角度出发，使学生掌握电子信息类专业中大量的专业词汇、英语术语及用法, 提高学生阅读和撰写英文科技文章的能力。

3.能力要求通过教师讲解，结合学生课后杳阅英文资料，培养学生听、说、写的综合能力，掌握本专业的当前动态和前沿发展。

三、教学内容的基本要求和学时分配1.教学内容及要求(1)Unit 1 Electronic DevicesLesson 1 VLSI Technology； Lesson 2 Memory Devices；Lesson 3 Microprocessors； Exercises基木要求：掌握超大规模集成技术、存储器件和微处理器的基木知识。

(2)Unit 2 Electronic CircuitsLesson 4 Operaticnal Amplifiers； Lesson 5 Low pass Filters；Lesson 6 Analog to Digital Converters； Exercises 基木要求：掌握常用的放大器、低通滤波器、模数转换器的基木结构及原理。

(3)Unit 3 Electronic System ComponentsLesson 7 Switching Power Supply； Lesson 8 Clock Sources；Lesson 9 Intercorinect； Exercises 基木要求：掌握常用的开关电源、时钟信号源和互连器件的基木结构。

Lesson3 An unknown goddess无名女神课文Some time ago,an interesting discovery was made by archaeologists on the Aegean is land of Kea.An American team explored a temple which stands in an ancient city on t he promontory of Ayia Irini.The city at one time must have been prosperous,for it en joyed a high level of civilization.Houses -- often three storeys high -- were built of st one.They had large rooms with beautifully decorated walls.The city was even equipped with a drainage system,for a great many clay pipes were found beneath the narrow streets.The temple which the archaeologists explored was used as a place of worship from the fifteenth century B.C.until Roman times.In the most sacred room of the temple,clay fragments of fifteen statues were found.Each of these represented a godd ess and had,at one time,been painted.The body of one statue was found among rema ins dating from the fifteenth century B.C.Its missing head happened to be among rem ains of the fifth century B.C.This head must have been found in classical times and ca refully preserved.It was very old and precious even then.When the archaeologists rec onstructed the fragments,they were amazed to find that the goddess turned out to be a very modern-looking woman.She stood three feet high and her hands rested on her hi ps.She was wearing a full-length skirt which swept the ground.Despite her great age, she was very graceful indeed,but,so far,the archaeologists have been unable to disc over her identity.译文不久之前，在爱琴海的基亚岛上，考古工作者有一项有趣的发现。