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Why are standards so important for NICs, connectors, and media? Use your textbook and Internet research to justify your answer. The physical level determines how the 0 and 1 bits are identified. So both ends need to know the voltage range for a 1 and for a 0. If they used different values then the data would be lost, or the voltage at the destination may be so high that the current delivered to the destination could destroy the receiver. The NIC has the MAC address which has to be unique on each Local Area Network. When a frame is sent the format of the frame has to be understood by the destination NIC. If the format of the frames were different then the two devices could not communicate. Exercise 4.1.2 The voltage for registering a bit of 1 on Ethernet can be as low as 2.5 volts or even 1 volt. Why is this so low when the capacity for transmitting electricity on the copper wire is so high?

Use your textbook and Internet research to support your answer. Copper wire is used to power appliances which use more current. The difference is between signal circuits and power circuits. Signal circuits operate at extremely low current levels and use digital logic voltage levels (less than 5 VDC). Power circuits on the other hand require larger amounts of current to operate electrical devices such as motors, heating units, etc. Exercise 4.1.3 What are some applications of shielded copper cable? Why is the more expensive shielded cable used in these situations? Use your textbook or Internet research to support your answer. Microphones, PA systems, Audio cables, Power cables, and RG-6 Cables used for television. There are three reasons, that I can think of, to use shielded cable instead of unshielded. Radio Frequency Interference (RFI) was the earliest interference engineers had to deal with.

Early radio signals easily found their way into devices. Coaxial cables and shielded twisted pairs using copper can eliminate most of this type of interference. Electromagnetic Interference (EMI) comes more into play today with high power transmission lines, higher magnetic fields, such as MRI machines in hospitals, and other high power applications. Copper may not offer much resistance to higher magnetic fields; therefore the use of magnetic materials such as high permeable irons may be required. And lastly you have Electro Magnetic Pulse (EMP) is produced by the detonation of nuclear devices. When the Hydrogen Bombs were tested in the 1950s at Bikini Atoll in the south pacific, circuit breakers at power stations in Hawaii were tripped by the EMP wave from the detonation. Critical military and civilian circuits have to be protected from an EMP condition. These shields require the use of both high and low permeable materials to reduce the effect of an EMP.

Exercise 4.1.6 Briefly describe the layered construction of a coaxial cable from the inner core to the outer insulation. Use your textbook and Internet research to compose your answer. A coaxial cable starts with a copper wire in the core, which is surrounded by a dielectric insulator. These are then wrapped by a thin metallic shield. And finally it is protected by a plastic jacket. Exercise 4.1.7 Based solely on the distance limitation of UTP cable, what should you guarantee about any network you plan for UTP cable? Use your textbook and Internet research to justify your answer. You should always guarantee that all devices are under 100 meters away from each other so that you can prevent attenuation. Exercise 4.1.8 What type of cable is necessary for each connection in Figure 4- 2: straight or crossover? You can assume that S1 and S2 do not have the ability to resolve crossovers ( called Auto- MDIX).

Use your textbook and Internet research to compose your answer. S1-A(straight) S1-B (straight) S1-C (straight) C-D (crossover) S1-S2 (multimode fiber) S2-E (straight) S2-F (straight) S2-G (multimode fiber) F-G (crossover) Exercise 4.1.9 If all the connections in the network from Figure 4- 2 use UTP cable, will they all be able to communicate correctly? If not, which links will not function and what should be done to resolve the issue? Use your textbook and Internet research to justify your answer. No, Not all of them will be able to communicate properly. Everything on S-1 side will not communicate properly with all devices on S-2 side. The cable between the switches should be replaced with a multimode fiber optic cable.

You will also have to run multimode fiber from S-2 to G to remove any chance of attenuation. Exercise 4.1.10 Briefly describe the purpose of each of the following network testing devices: Multimeter ¦ Tone generator ¦ Pair scanner ¦ Time domain reflectometer ( TDR) 1:Multimeter = A typical multimeter would include basic features such as the ability to measure voltage, current, and resistance. 2: Tone Generator = Can be used to send a sound or tone down a line so you may find the other end. 3: Pair Scanner = A device used to locate problems in LAN cabling systems 4: Time Domain Reflectometer = is an electronic instrument that uses time-domain reflectometry to characterize and locate faults in metallic cables (for example, twisted pair wire or coaxial cable). It can also be used to locate discontinuities in a connector, printed circuit board, or any other electrical path. For each device, use the Internet to find a vendor for the product and list its cost. 1: Multimeter-vendor is Transcat for $1499.95

2:Tone generator- vendor is Zoro for $160.00
3: Pair Scanner- vendor is CDW for $119.99
4: Time Domain Reflectometer- vendor is Mitchell Instruments for $314.95 Lab 4.1 Review
1: Modern routing devices have a software solution to detect crossover ( called Auto- MDIX) and correct for it, allowing the use of straight cables to connect these devices instead of crossover cables. What is the benefit of this functionality for wiring a network? This functionality allows you to use either crossover or straight through cabling. It eliminates the need to differentiate between the two types of cabling. 2: Signal repeaters will take the weak signal as input and retransmit it as a stronger signal. Why should these devices be placed before the end of the maximum distance for the cable? These devices should be placed before the end of the maximum distance for the cable because you want the repeater to retransmit the full signal. If it is placed after the maximum distance it would repeat the signal even if there is attenuation. 3: What is the limiting factor for the number of network connections a desktop or laptop computer can make?

The limit on network connections a desktop can make is dependent on acceptable speed. A vast number of connections can be established, however each connection is a drain on overall speed which can reduce working ability to a crawl. Exercise 4.2.1 The light in an SMF cable travels down the center of the fiber parallel to the direction of the fiber in a single path. Can this type of cable be used to transmit and receive bits simultaneously, or does it require one fiber for transmitting and one for receiving? Justify your answer using your textbook and Internet. Instead of using one fiber cable and enforcing half-duplex logic, most fiber links use a pair of cables, one for each direction. Each fiber NIC, port, interface, and so on has an interface that has two sockets: one for the transmit cable and one for the receive cable. When connecting the two devices, just make sure that each node’s transmit socket connects to the same cable as the other node’s receive socket, and vice versa.

Figure 4-23 shows the idea; note that the transmit (Tx) side on one end connects to the receive (Rx) side on the other, for both cables. Exercise 4.2.2 What characteristics of fiber- optic cables allow them to be used over longer distances with less risk than copper cable? Optical fiber can carry much higher frequency ranges, note that light is a very high frequency signal, while copper wire attenuates or loses signal strength at higher frequencies. Exercise 4.2.3 MMF distances can still far exceed copper cable, but MMF cannot match the distance available from the more expensive SMF. What is the reason for the distance limitation with the use of MMF? Multimode fiber gives you high bandwidth at high speeds over medium distances. Light waves are dispersed into numerous paths, or modes, as they travel through the cable’s core. However, in long cable runs (greater than 3000 feet), multiple paths of light can cause signal distortion at the receiving end, resulting in an unclear and incomplete data transmission. Lab 4.2 Review

1: What is the bend radius for a fiber- optic cable? Why is it important to adhere to the bend radius restrictions for fiber- optic cable? Fiber optic cable can be broken when kinked or bent too tightly, especially during pulling. If no specific recommendations are available from the cable manufacturer, the cable should not be pulled over a bend radius smaller than twenty (20) times the cable diameter. After completion of the pull, the cable should not have any bend radius smaller than ten (10) times the cable diameter. 2: Fiber- optic cables are often bundled together into thicker cables for wiring multiple devices in a location or to travel over long distances. What layers of the standard fiber- optic cable are necessary for each individual glass filament in order for them to be bundled together with other fibers? The cladding, the buffer, the strengthener, and the outer jacket.

Exercise 4.3.2 Rank the three main types of cable ( UTP, SMF, and MMF) in a hierarchy of use for connecting network devices based on your completed comparison table. Use your textbook and Internet research to justify your ranking. 1: UTP is most common. It is cheaper, and is easy to install. 2: MMF is good for long distances between buildings, more expensive, and gives faster data transmission. 3: SMF is good for really long distances, is the most expensive, and gives the fastest data transmission rates. Exercise 4.3.3 Using the description of Network A and Figure 4- 4, create a cabling solution for this network that meets the owner’s needs. Justify your choices. Switch 1 is not needed(Router has 6 ethernet ports)

Router 1 using UTP to connect A,B,C,D, and E= 205 meters for $51.25 Add a coaxial NIC to Router 1 for $40
Router 1 using coaxial thicknet to switch 2 = 210 meters for $56.70 Add a coaxial NIC to switch 2 for $40
Switch 2 using UTP to connect F,G, and H= 150 meters for $37.50 Total network cost= $225.45
Exercise 4.3.4 Using the description of Network B and Figure 4- 5, create a cabling solution for this network that meets the owner’s needs. Choose whether you want to connect the three routers in full mesh and explain your choice. Justify your choices. Add 3 Coaxial NIC’s for A,B, and C = $180

Router 1 connects to A,B, and C using Coaxial Thinnet for 30 meters= $8.10 Add SMF NIC to Router 1 and Router 2 for $300
Router 1 connects to Router 2 using SMF for 410 meters = $184.50 Add 3 Coaxial NIC’s for D,E, and F for $180
Router 2 connects to D,E, and F using coaxial thinnet for 15 meters = $4.05 Add MMF NIC’s to Router 2 and Router 3 for$200
Router 2 connects to Router 3 for 170 meters using MMF = $59.50 Add 2 Coaxial NIC’s to G and H for $120
Router 3 connects to G and H using Coaxial Thinnet for 10 meters = $2.70 Add one more SMF NIC to Router 1 for $150
Router 3 connects connects to Router 1 using SMF for 502 meters = $225.90 Total network Cost = $1614.75
No, I don’t want full mesh because the scenario states that the 3 Routers do not need full mesh to function. Exercise 4.3.5 Using the description of Network C and Figure 4- 6, create a cabling solution for this network that meets the owner’s needs. Justify your choices. Router 1 connects to A,B, and C using UTP for 15 meters =$ 3.75 Add 2 Coaxial NIC’s to Router 1 for $120

Router 1 connects to I using Coaxial Thinnet for 120 meters =$32.40 Add 2 Coaxial NIC’s to Router 2 for $120
Router 1 connects to Router 2 using Coaxial Thinnet for 80 meters =$21.60 Router 2 connects to D,E, and F using UTP for 15 meters =$ 3.75 Add a Coaxial NIC to Router 3 for $60
Router 2 connects to Router 3 using Coaxial Thinnet for 120 meters = $32.40 Router 3 connects to G and H using UTP at 10 meters = $2.50
Total network cost = $ 396.40
Lab 4.3 Review
1: As the cost of copper increases and the cost of fiber- optic cable decreases, eventually fiber might be less expensive for local network connections than copper cable. What are some barriers to widespread adoption of fiber in place of UTP? One barrier is the difficulty of installing fiber. Another is that fiber is more fragile than an ordinary cat 5 cable. And lastly most computers in consumer homes are pre-built for UTP connections.

The consumer would have to purchase additional hardware to make a fiber connection in the house. 2: Most wireless devices for home and small office use have an effective broadcast radius of 35 meters ( or approximately 115 feet) and can share up to approximately 25 devices ( though for voice and video, it is recommended to reduce this to 10– 15 devices) per access point ( AP). How would this fit into your cabling hierarchy as a networking solution? If your focus is speed, it wouldn’t change anything because hardline is best. This would be my last resort option. However, if speed is not your concern, then using wireless to clean up cabling messes would make it more essential. It is basically up to the owners discretion.

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