...Exercise 4.1.1 Because those same standards define how communication between different hardware and software involved, will take place. 4.1.2 The difference is between signal circuits and power circuits. Signal circuits operate at extremely low current levels and use digital logic voltage levels. Power circuits on the other hand require larger amounts of current to operate electrical devices such as motors, heating units 4.1.3 A very common application is the health care field. Shielded copper cable is often used in health care facilities for its efficiency and resistance to flame. The use of shielded cables in security systems provides some protection from power frequency and radio frequency interference, reducing the number of false alarms being generated. The best practice is to keep data or signal cables physically separated by at least 3 inches (75mm) from 'heavy' power circuits which are in parallel. 4.1.5 white/green stripe | white/orange stripe | green solid | orange solid | white/orange stripe | white/green strip | blue solid | blue solid | white/blue stripe | white/blue stripe | orange solid | green solid | white/brown stripe | white/brown stripe | brown solid | brown solid | 4.1.6 Coaxial cable conducts electrical signal using an inner conductor surrounded by an insulating layer and all enclosed by a shield, typically one to four layers of woven metallic braid and metallic tape. The cable is protected by an outer insulating jacket. Normally...
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...In: Computers and Technology Chapter 4 Lab Intro to Networking Lab Chapter 4 4.1.1 Exercise: Standards are important for NIC, Connectors and Media, because of the Stability, Consistency, and Minimization of packet errors. 4.1.2 Exercise: Why is it so low when the capacity for transmission electricity on the copper wire is so high? Because, due to truncation on the transmission of voltage with an electric current. 4,1,3 Exercise: It’s used in the Healthcare facilities, because it’s more fire resistance. 4.1.4 Exercise: Category | Maximum Speed | Application | 1 | 10 Mbps | Telephone Cabling (POTS) | 2 | 4 Mbps | Token Ring | 3 | 10 Mbps | Ethernet | 4 | 20 Mbps | Token Ring | 5 | 100 Mbps | Fast Ethernet | 5e | 1 Gbps | Gigabit Ethernet | 6 | 2500 Mbps | Gigabit Ethernet | 6a | 10,000 Mbps | Gigabit Ethernet | 4.1.5 Exercise: 4.1.6 Exercise: * , The central layer comprises of a conducting material. This layer transmits the baseband video signal. * The dielectric layer surrounds the centre copper core. The function of the dielectric is to separate the inner conductor from the shield and provide physical support. * The next layer is the metallic shield, generally composed of braided copper. It has two main purposes: to protect the conductor from noise or other unwanted signals, referred to as ingress, and to retain the transmitted signal in the conductor. * The last layer is the outside insulation which encloses...
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...information, streams of bytes representing characters, available to a wide audience. To do this that information has to be in a format which is as much as possible, "device independent". Text is a very lightweight way of conveying information. A small 143px by 143px image can easily run to 4500 bytes, which at a byte a character is the equivalent of about 900 words. A picture may indeed paint a thousand words in the real world, but in the virtual world a lot of users will thank you for not taking up all their bandwidth. But this is to digress from the main point, which is that the web began life as a medium for conveying information in as succinct and lightweight a manner as possible. During its period of explosive growth we lost sight of this a bit, but it's now an idea which is coming back. At lots of sites, for example intranets, news and information services and so on, it actually never went away. The important point to be made is that good web designers should be able to make text oriented pages appear enticing, interesting and most of all readable, without relying too heavily on bandwidth intensive images and animations, which also tend to reduce the usability and accessibility of pages. A significant challenge to designers and implementers of web sites is to present text based information in the most accessible...
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...INTEGRATED CIRCUITS 80C31/80C32 80C51 8-bit microcontroller family 128/256 byte RAM ROMless low voltage (2.7 V–5.5 V), low power, high speed (33 MHz) Product specification IC28 Data Handbook 2000 Aug 07 Philips Semiconductors Philips Semiconductors Product specification 80C51 8-bit microcontroller family 128/256 byte RAM ROMless low voltage (2.7V–5.5V), low power, high speed (33 MHz) 80C31/80C32 DESCRIPTION The Philips 80C31/32 is a high-performance static 80C51 design fabricated with Philips high-density CMOS technology with operation from 2.7 V to 5.5 V. The 80C31/32 ROMless devices contain a 128 × 8 RAM/256 × 8 RAM, 32 I/O lines, three 16-bit counter/timers, a six-source, four-priority level nested interrupt structure, a serial I/O port for either multi-processor communications, I/O expansion or full duplex UART, and on-chip oscillator and clock circuits. In addition, the device is a low power static design which offers a wide range of operating frequencies down to zero. Two software selectable modes of power reduction—idle mode and power-down mode are available. The idle mode freezes the CPU while allowing the RAM, timers, serial port, and interrupt system to continue functioning. The power-down mode saves the RAM contents but freezes the oscillator, causing all other chip functions to be inoperative. Since the design is static, the clock can be stopped without loss of user data and then the execution resumed from the point the clock was stopped. ...
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...Assignment 4.1 Chapter 5 pg 135-138 2. Identify the addressing mode for each of the following? (A) MOV B,#34H IMMEDIATE (B) MOV A 50H DIRECT (C) MOV R2,07 REGISTER MODE (D) R3,#0 IMMEDIATE (E) MOV R7,0 REGISTER MODE (F) MOV R6,#7FH IMMIDIATE (G) MOV R0,A REGISTER MODE (H) MOV B,A (I) MOV A,@R0 INDIRECT MODE (J) MOV R7,A REGISTER MODE (K) MOV A,@R1 INDIRECT MODE 9. Which register are allowed to be used for indicated addressing mode when accessing data in RAM? 20. Write a program to generate a square wave with 75% duty cycle on bit P1.5. 29. What bit addresses are assigned to the TCON register? 88H 30. What bit address are assigned to register A? 0E0H 31. What bit address are assigned to resister B? 0F0H 45. The byte address assigned to the SFR are REGISTERS to ADDRESSES 61. What addressing mode is used to access the upper 128 bytes of RAM in the 8052? DIRECT MODE 66. Which registers can be used to access the upper 128 bytes of RAM in the 8052? Chapter 6 pg. 174-178 1) a) AC=1 and CY= 0 b) AC= 2 and CY= 2 c) AC=0 and CY= 0 d) AC= 2 and CY= 1 e) AC=0 and CY=0 f) CY=1 and AC=1 9. Write a program to add 897F9AH to 34BC48H and save the result in RAM memory locations stating at 40H. CLR C MOV 12. Show how to perform 77 x 34 in the 8051. 77 x 34 in 8051 = MOV A, #77 MOV B, #34 MUL AB 13. Show how to perform 77 / 3 in the 8051. 77/3 in 8051 = MOV A, #77 MOV B, #3 DIV AB 14. True or False. The MUL and DIV instructions work on an...
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...Eddie Bond nt1210 Lab Review 1.1 1. 127 in binary is 01111111. I came up with tis answer by dividing by two till i couldnt anymore. 2.Both numbers represent the powers of 10. 3.16,32,64,128 4.It would be difficult becuase you would have to put multiple equations to find te rigt answer. Lab Review 1.3 1. It is important to know how many system words will fit in a primary device so you will know how big your storage device is. 2.Multiple bytes has more ones and zero as bytes with holds more information a single byte only has a series of 8 bit which has less data then a series of multiple sets of 8bit. 3. When converting a big number you cant just line up the digits, you need to make the number into bytes so that each one can easily convert into a unit of dex. Lab Review 1.4 1. My name in hexadecimal is 069 100 100 105 101 032 066 111 110 100. I would need 32bytes of storage. Lab Review 1.5 1. It is necessary to plan a file structure carefully because you want to have organization on the operating system . If it is too loosely organized or too detailed then it will be a chaotic environment and will be difficult to find anything. A nice organized file hierarchy would be easy to find things. Lab Review 1.6 1. To know what the system can handle and to know which operating system to use. 2. You must use the task manager, killing a program that is running your machine will cause your computer's operating sytem to stop working until the computer...
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...HT12A/E: 18-pin DIP/20-pin SOP package Operating voltage - 2.4V~5V for the HT12A - 2.4V~12V for the HT12E Low power and high noise immunity CMOS technology Low standby current: 0.1mA (typ.) at VDD=5V HT12A with a 38kHz carrier for infrared transmission medium · · · · · Applications · · · · Burglar alarm system Smoke and fire alarm system Garage door controllers Car door controllers · · · · Car alarm system Security system Cordless telephones Other remote control systems General Description The 212 encoders are a series of CMOS LSIs for remote control system applications. They are capable of encoding information which consists of N address bits and 12-N data bits. Each address/data input can be set to one of the two logic states. The programmed addresses/data are transmitted together with the header bits via an RF or an infrared transmission medium upon receipt of a trigger signal. The capability to select a TE trigger on the HT12E or a DATA trigger on the HT12A further enhances the application flexibility of the 212 series of encoders. The HT12A additionally provides a 38kHz carrier for infrared systems. Selection Table Function Address Address/ Data Oscillator No. Data No. No. Part No. HT12A HT12E 8 8 0 4 4 0 455kHz resonator RC oscillator Trigger D8~D11 TE Package 18 DIP 20 SOP 18 DIP 20 SOP Carrier Output 38kHz No Negative Polarity No No Note: Address/Data represents pins that can be address or data according to the decoder requirement. 1 April...
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...DAC de 10 bits tendrá una resolución menor que un DAC de 12 bits para la misma salida a plena escala. Cierto o falso: un DAC de 10 bits con salida a plena escala de 10 V tiene un porcentaje de resolución menor que un DAC de 10 bits con salida a plena escala de 12 V. 2. Un DAC de ocho bits produce un voltaje de salida de 2.0 v para un código de entrada de 01100100. ¿Cuál será el valor de Vsal para un código de entrada de 10110011? 3. Determine los factores de ponderación para cada bit de entrada del DAC en el problema 10.2 4. ¿Cuál es la resolución del DAC del problema 10.2? Exprésela en volts y como porcentaje. 5. ¿Cuál es la resolución en volts de un DAC de 10 bits cuya salida F.S es 5 V.? 6. ¿Cuántos bits se requieren para un DAC de modo que su salida F.S sea 10 mA y su resolución menor que 40 µA? 7. ¿Cuál es el porcentaje de resolución de un DAC de 3 bits y cuál es el tamaño del escalón si el valor máximo es 2 V.? 8. Un DAC de 12 bits (3 dígitos) que usa el código de entrada BCD tiene una salida a plena escala de 9.99 V. Determine el tamaño del escalón, el porcentaje de resolución y el valor de Vsal para un código de entrada de 011010010101 9. Compare el tamaño del escalón y el porcentaje de resolución de un DAC con una entrada binaria de ocho bits con los de un DAC que tiene una entrada BCD de ocho bits. Suponga un valor a plena escala de 990 mV para cada DAC. |DAC de 8 bits con entrada...
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...are the same regardless, so the only change is the addition of 103 as the leftmost digit value. 2 | 9 | 3 | 1 | 2,000 | 900 | 30 | 1 | Exercise 1.1.2 The mapping of 1102 will have 1, 1, and 0 in the lower bit values and 4, 2, and 0 in the bottom bits, for a sum of 6. 1 | 1 | 0 | 4 | 2 | 0 | Exercise1.1.3 The mapping of 112 will have a leading 0 to fit the mapping, which does not affect the results. The lower bits will have leading 0s and then 1 and 1 in the rightmost bits. This results in all 0s at the bottom, with 2 and 1 in the rightmost bits, for a sum of 3. 0 | 0 | 1 | 1 | 0 | 0 | 2 | 1 | Exercise 1.1.4 The placeholder 24 = 16 is necessary for this mapping because it is five digits. The result is 16 + 2 = 18 ...
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...NIKHILREDDY, ID: 10000126161 Problem 3.1 Design a 1-bit memory circuit and provide a truth table for the circuit. Answer: 1-bit memory circuit is made of flip-flop (or) RS trigger. It may be NAND gate (or) NOR gate. The below figure shows the NOR gate RS trigger. S R State 0 0 Keep state(memory mode) 0 1 Q=0 1 0 Q=1 1 1 Unstable/race condition Problem 3.2 MDR is used to decode address signals from CPU. For a 16-bit address bus, how much memory can be attached to the...
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...understand and take advantage of newly emerging careers because the seven fastest growing occupations are computer related. 5. It helps you deal with challenges associated with technology because mostly everything deals with computers having to know how to use them in order to get your jobs done. Chapter Two Questions 1. Computers are data processing devices. The four main functions of a computer are that gathers data (users input data), processes data into information, outputs data or information, stores data and information. 2. The difference between data and information is that data is representation of a fact, figure, or idea and information is organized, meaningful data. 3. Bits and Bytes are the language of computers. Bit is Binary digit (0 or 1). Byte is 8 bits, each letter, number, and character = a string of eight 0’s and 1’s. 4. Devices that you would use to get information into a computer would be keyboard, mouse/pointing device, stylus, scanner, digital cameras, and microphones. 5. Devices you would use to get information out of a computer...
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...n this new age of Information, the Internet has made all types of information readily available. Some of this information can be very useful, some can be malicious. Child pornography, also known as Paedophilia is one of these problems. Any one person can find child pornography on the internet with just a few clicks of the mouse using any search engine. Despite webmaster's and law enforcement officials' efforts to control child pornography and shut down illegal sites, new sites are posted using several ways to mask their identity. The Internet provides a new world for curious children. It offers entertainment, opportunities for education, information and communication. The Internet is a tool that opens a window of opportunities. As Internet use grows, so do the risks of children being exposed to inappropriate material, in particular, criminal activity by paedophiles and child pornographers. Many children first come in contact with the Internet at a very young age. Some children become victims of child pornography through close relatives who may have abused them. Some children become involved with chat services or newsgroup threads. It is usually through these sites that they meet child pornographers. Children may be asked to send explicit pictures of themselves taken either by a digital camera or scanned from a polaroid. The pornographer will then post the pictures on their web site, sometimes hiding...
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...Timers and Counters Exercise 2.2 1. Define the following timer bits: timer enable, timer timing, and timer done. ANSWER: Timer enable bit: The enable bit is true when the rung input logic is true, and the enable bit is false when the rung input logic is false. When the EN bit is true the timer accumulator is incrementing at the rate set by the timer time base. Timer timing bit: indicates when timing action is occurring and can be used to control timed events in automation applications. Timer done bit: the end of the timing process by changing states from false to true or from true to false depending on the type of timer instruction used. 2. Compare and contrast the true and false states of the timer timing bit for the on-delay timer, the off-delay timer, and the retentive timer. ANSWER: In the true states for all three have the accumulator value is less than the preset value, but for off-delay timer the timer rung is false unlike the retentive timer and on-delay that the timer rung is true. In the false states for all three have the accumulator is equal to or greater than the preset value. Like the true state retentive timer and on-delay timer both have the same condition and that are false if the timer rung is false, but the on-delay timer has other condition that are false if the timer done bit is true. The off-delay has a condition that are false if the timer done is false. 3. What is the difference between a retentive timer and a non-retentive timer? ...
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...utlo |[Drum Roll Production / How I Met Your Father] | |MS Word Struggle | |Script | |By | |Mia Narcelles | | | | | | | | | | | | | |October 4...
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...shorter. 13. Compiler 14. true 15. lst file 23. (a) MOV A, #55H - 2 bytes (b) MOV R3, #3 2 bytes (c) INC R2- 1 byte (d) ADD A, #0 -2 bytes (e) MOV A, R1 -1 byte (f) MOV R3, A -1 byte (g) ADD A, R2 -1 byte 37. (a) MOV A, #54H , ADD A, #0C4h CY=1 (b) MOV A, #00, ADD A, #0FFh CY=0 (c) MOV A, #250, ADD A, #05 CY=0 38. Add 25H and 70H and find the contents of the AC, CY, and P flags. 25h 70h 95h AC=0 CY=0 P=0 42. 8 bits 43. bank 0 48. MOV SP, #52H stack starts at MOV A, #04 move 4 into acc MOV R0, #05H move value to r0 ADD A, R0 add and store to acc PUSH 0E0H push the value of acc.on the stack PUSH 0 push value of r0 on the stack SET PSW.4 select reg bank 2 POP 10H pop value at r0 0E0H pop the value.of 00 ------------------------------------------------- Chapter 3 4. Short jump; 2 byte 5. long jump; 3 bytes 6. The LJMP instruction is 3-bytes in which the first byte is the opcode, and the second and third bytes represent the 16-bit address of the target location from 0000 to FFFFH. The SJMP is a 2-byte instruction where the first byte is the opcode and the second byte is the address location FFH 7. true 8. false 9. (c)LJMP 10. A short jump is a 2-byte instruction. Why? The first byte is the opcode (e.g MOV, ADD, etc.) and the second byte is the relative address of the target location. 0-FFH MAX 11. true 14. The loop is executed 200x100 = 20,000 times 17. 3 18. 2 19. 2 ...
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