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CEG3185 Quiz1 answers Computer Systems Design (University of Ottawa) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university CEG3185 Quiz1 answers Computer Systems Design (University of Ottawa) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Nic Lefebvre (niclefebvre252@gmail.com) lOMoARcPSD|19515774

Page 1/ 4 COURSE: CEG3585 / SEG3555 PROFESSOR: Miguel A. Garzón SEMESTER: Summer 2018 DATED: May 30, 2018 Quiz 1 - SOLUTIONS Out of 18 points N AME AND STUDENT NUMBER : / _ 1. [5 points] The standard spectrum of a regular telephone that is used, via modems, for transmitting digital data is 300 to 5000 Hz. The ratio of signal to noise is 32 dB. What is the channel capacity? If we assume that only 75% of the theoretical limit may be reached, derive from the Nyquist formula, the required number of signal levels? Bandwidth = 5000 Hz – 300 Hz = 4700 Hz (1 points) Convert SNRdb to SNR SNR db = 10log10(SNR) SNR = 1585 (1.5 points) Use Shannon Formula to find Capacity C = 4700 * Log 2 (1+1585) = 49966.53 bps (0.5 points) 75% of the Capacity 49966.53 * 0.75 = 37475 bps (2 points) Use Nyquist to derive number of signals 37475= 2 * 4700 * Log 2 L L = 15.85 (0.5 points) Conclusion We need 8 signal levels Downloaded by Nic Lefebvre (niclefebvre252@gmail.com) lOMoARcPSD|19515774

Page 2/ 4 2. [4 marks] The loss in a cable is defined in decibels per kilometer (dB / km). If the signal at the beginning of the cable with -0.2 dB / km has a power of 4 mW. What is the signal power at 7 km? (1.5 points) Total Power loss -0.2 db/km * 7 km = -1.4 db ( 2.5 points ) Calculate P 2 Db = 10 log 10 (P2/P1) -1.4 = 10 log 10 (P2/4 mW) P2 = 2.89 mW 3. [5 points ] A channel has a bandwidth of 8 kHz. If we want to send data at 150 kbps, what is the signal- to-noise ratio (SNR db ) in decibels? (3 points) Use Shannon formula to find SNR C = B * Log 2 (1+SNR) 150000 = 8000 * Log 2 (1 + SNR) 18.75 = Log 2 (1 + SNR) 1+SNR = 2 18.75 SNR = 440870.9 (2 point) Find SNRdb SNR db = 10log10(440870.9) SNR db = 56.4 dB Downloaded by Nic Lefebvre (niclefebvre252@gmail.com) lOMoARcPSD|19515774

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Page 3/ 4 4. [4 marks] Fill the tale in the figure below indicating the headers of the layers 2 and 4 at the sender and at the receiver when data is sent from Host 1 in the network N1 to Host 2 in the network N2. Networks are indicated by clouds and the routers are R1 and R2. The IP address of Host 1 is: IP1 The Host 1 MAC address is: MAC_H1 The IP address of host 2 is: IP2 The MAC address of Host 2 is: MAC_H2 The source port is 21 and the destination port is 80 The MAC addresses of R1 and R2 are respectively MAC_R1 AND MAC_R2 Layer 5 Data Data Layer 4 21 80 | Data 21 80 | Data Layer 3 21 80 | IP1 IP2 | Data 21 80 | IP1 IP2 | Data Layer 2 MAC_R1 MAC_H1 | 21 80 | IP1 IP2 | Data MAC_H2 MAC_R2 | 21 80 | IP1 IP2 | Data There was a mistake in Layer 3; Question 4 will be corrected out of 2 points only. Quiz will be then out of 18 points total. (1 point for layer 2; 0.5 at the source, 0.5 at the destination) (1 point for layer 4; 0.5 at the source, 0.5 at the destination) Note that the addresses at Layer 2 are inversed, destination MAC address comes first. Downloaded by Nic Lefebvre (niclefebvre252@gmail.com) lOMoARcPSD|19515774

Page 4/ 4 5. [2 points] Write an advantage and a disadvantage of the Star topology. (1 point for one correct advantage; only one is needed) Advantages of a Star Topology  Easy to install and wire.  No disruptions to the network when connecting or removing devices.  Easy to detect faults and to remove parts. (1 point for one correct disadvantage; only one is needed) Disadvantages of a Star Topology  Requires more cable length than a linear topology.  If the hub, switch, or concentrator fails, nodes attached are disabled.  More expensive than linear bus topologies because of the cost of the hubs, etc. (THIS IS NOT REQUIRED) Downloaded by Nic Lefebvre (niclefebvre252@gmail.com) lOMoARcPSD|19515774