1 / 11

Test 1

Test 1. Solution. I/. OSI and Layered Architecture. Sergeants can talk indirectly with other sergeants and directly with privates and lientenants B. General->Colonel->Major…->Private-> Private->Sergeant->Lieut…->General C. Link Layer Error-free point-to-point connection

estrella-gomez
Download Presentation

Test 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Test 1 Solution

  2. I/. OSI and Layered Architecture • Sergeants can talk indirectly with other • sergeants and directly with privates and • lientenants • B. General->Colonel->Major…->Private-> • Private->Sergeant->Lieut…->General • C. Link Layer • Error-free point-to-point connection • Network Layer • Connectivity anywhere in the net by Routing • and universal naming (IP name mapping) • Transport Layer • Error-free end-to-end connection General Colonel Major Captain Lieut Sergeant Private General Colonel Major Captain Lieut Sergeant Private

  3. Basic Concepts

  4. Frequency and Bandwidth (A) cos(A)cos(B)= ½ (cos (A+B) + cos(A-B)) Lowest frequency cos(1240k)cos(0)= ½ (cos (1240k+0) + cos(1240-0)) ½ (cos (1240k) + cos(1240k)) Highest frequency cos(1240k)cos(5k)= ½ (cos (1240k+5k) + cos(1240-5k)) ½ (cos (1235k) + cos(1245k)) Keeping higher subbands -> 1240k to1245k I’m being sloppy with the frequency. (should be 2piF not F)

  5. Frequency and Bandwidth (B) Encoding a digitized talk show over an analog medium would use a modem-type encoding technique ASK PSK FSK or some combination of these

  6. Frequency and Bandwidth (C)

  7. T, duration of 1 bit IV. Encoding (A) high value Time low value 1 1 1 0 1 0 0 Part 1 Manchester Encoding

  8. FSK 1 1 1 0 1 0 0 Frequency (FSK)

  9. ASK 2 bits per baud 01 00 11 10 transmission 11 10 01 00

  10. IV. Encoding B 64 levels -> 6 bits per sample 10kHz -> 20kHz sampling rate 6 bits/sample x 20k samples/sec = 120k bits/sec 120 secs in 2 minutes -> 120 x 120k bits = 14,400,000 bits

  11. Analog to Digital 0.3 = 0000 0000 0.25 0001 0.5 0010 0.75 0011 1.0 0100 1.25 0101 1.5 0110 1.75 0111 2.00 1000 2.25 1001 2.5 1010 2.75 1011 3.0 1100 3.25 1101 3.5 1110 3.75 1111 4.0 1.3 = 0100 1.8 = 0110 2.05 = 0111 3.55 = 1101 0100 0111 0110 0000 1101

More Related