LGW2eCh4Solutions

LGW2eCh4Solutions - Communication Networks (2nd Edition)...

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Communication Networks (2 nd Edition) Chapter 4 Solutions Leon-Garcia/Widjaja 1 Solutions to Chapter 4 1. A television transmission channel occupies a bandwidth of 6 MHz. Solutions follow questions: a. How many two-way 30 kHz analog voice channels can be frequency-division multiplexed in a single television channel? 6 × 10 6 / 30 × 10 3 = 200 channels b. How many two-way 200 kHz GSM channels can be frequency-division multiplexed in a single television channel? 6 × 10 6 / 200 × 10 3 = 30 channels c. Discuss the tradeoffs involved in converting existing television channels to cellular telephony channels? The biggest advantage of using existing television channels to provide cellular telephony channels is the very large bandwidth that they occupy. In theory, one could divide each television channel into many cellular telephone channels, as shown in parts (a) and (b). The frequency reuse aspect of cellular networks would multiply these channels many times. This would be at the cost of television channels that cover a broad region and presumably provide service to a large audience. 2. A cable sheath has an inner diameter of 2.5 cm. Solutions follow questions: a. Estimate the number of wires that can be contained in the cable if the wire has a diameter of 5 mm. Ignoring empty space between the wires: 2 2 ) 2 5 ( ) 2 25 ( π = 25 wires b. Estimate the diameter of a cable that holds 2700 wire pairs. Ignoring empty space between the wires: 2 2 ) 2 5 ( ) 2 ( d = 2x2700 wires Æ d 368 mm
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Communication Networks (2 nd Edition) Chapter 4 Solutions Leon-Garcia/Widjaja 2 3. Suppose that a frequency band W Hz wide is divided into M channels of equal bandwidth. a. What bit rate is achievable in each channel? Assume all channels have the same SNR. Each user uses W/M bandwidth. Using Shannon’s Channel Capacity formula: Bit rate = ) 1 ( log ) ( 2 SNR M W + bps b. What bit rate is available to each of M users if the entire frequency band is used as a single channel and TDM is applied? In this case, the total bit rate afforded by the W Hz is divided equally among all users: Bit rate = M SNR W ) 1 ( log 2 + bps c. How does the comparison of (a) and (b) change if we suppose that FDM requires a guard band between adjacent channels? Assume the guard band is 10% of the channel bandwidth. Because of the guard band we expect that the scheme in (b) will be better since the bit rate in (a) will be reduced. In (a), the bandwidth usable by each channel is 0.9 W/M . Thus, we have: Bit rate = ) 1 ( log ) 9 . 0 ( 2 SNR M W + bps 4. In a cable television system (see Section 3.8.2), the frequency band from 5 MHz to 42 MHz is allocated to upstream signals from the user to the network, and the band from 550 MHz to 750 MHz is allocated for downstream signals from the network to the users. Solutions follow questions:
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This note was uploaded on 03/09/2009 for the course ECE 410 taught by Professor Black during the Spring '09 term at Rose-Hulman.

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LGW2eCh4Solutions - Communication Networks (2nd Edition)...

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