Data com ch#3 (part 2)

Transcription

1 Data com ch#3 (part 2) ENG. IBRAHEEM LUBBAD

2 TRANSMISSION IMPAIRMENT

3 Attenuation a loss of energy db =20log 10 V2 V1 db < 0 db > 0 db = 0 attenuated amplified not changed

4 Decibel numbers can be added or subtracted (cascading) db= = +1

5 P3-15. A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the attenuation in decibels?

6 P3-15. A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the attenuation in decibels? db = 10 log 10 ( 90 ) = db 100

7 P3-16. The attenuation of a signal is 10 db. What is the final signal power if it was originally 5 W?

8 P3-16. The attenuation of a signal is 10 db. What is the final signal power if it was originally 5 W? db=10log 10 P2 p1-10=10log 10 P2 5 log 10 P2 5 =-1 P2 5 =10 1 P2 = 0.5 W

9 P3-17. A signal has passed through three cascaded amplifiers, each with a 4 db gain. What is the total gain? How much is the signal amplified?

10 P3-17. A signal has passed through three cascaded amplifiers, each with a 4 db gain. What is the total gain? How much is the signal amplified? Total gain = 4dB + 4dB + 4dB = 12dB P1 P2 P3 P4 For power gain of the first stage: P2 4dB=10 log 10 P1 P2 = 10 ( 4 10 ) = P1 For power gain of three stages: 2.512*2.512*2.512= or 12dB =log 10 P4 P4 P1 P1 (12 = ) = 15.85

11 P3-18. If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of 100,000 bits out of this device?

12 P3-18. If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of 100,000 bits out of this device? Given,bandwidth 5000 bps, Frame 100,000 bit -> b 5000 bps = 20sec

13 P3-20. A signal has a wavelength of 1 μm in air. How far can the front of the wave travel during 1000 periods?

14 P3-20. A signal has a wavelength of 1 μm in air. How far can the front of the wave travel during 1000 periods? Given, wavelength = 1 μm, periods= 1000 Wavelength = propagation speed period (For one period ) Distance = 1 μm *1000= 1000μm =1 mm

15 Distortion means that the signal changes its form or shape.

16 Noise: Several types of noise Thermal noise :random motion of electrons in a wire Induced :noise from sources such as motors Crosstalk :effect of one wire on the other Impulse noise : a signal with high energy in a very short time

17 Signal-to-Noise Ratio (SNR) SNR= average signal power average noise power SNR db = 10 log 10 (SNR)

18 DATA RATE LIMITS: Data rate depends on three factors: 1. The bandwidth available 2. The level of the signals we use 3. The quality of the channel (the level of noise) To calculate the data rate: Nyquist for a noiseless channel, Shannon for a noisy channel.

19 Noiseless Channel: Nyquist Bit Rate BitRate =2* BW* log 2 l Bandwidth is the bandwidth of the channel L : is the number of signal levels used to represent data BitRate: is the bit rate in bits per second. Note: Increasing the levels of a signal may reduce the reliability of the system.

20 Noisy Channel: Shannon Capacity Capacity =BW* log 2 ( 1 + SNR ) CAPACITY is the capacity of the channel in bits per second the formula defines a characteristic of the channel, not the method of transmission we cannot achieve a data rate higher than the capacity of the channel When the SNR is very high then Capacity =0 the theoretical channel capacity can be simplified to C = BW * SNR db 3

21 Using Both Limits We need to use both methods to find the limits and signal levels 1) The Shannon capacity gives us the upper limit. 2) The Nyquist formula tells us how many signal levels we need Example : We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level? Capacity =BW* log 2 ( 1 + SNR )= 10 6 * log 2 ( )=6Mbps 6Mbps=2*10 6 *log 2 ( l) l = 4

22 P3-21. A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is the maximum data rate supported by this line?

23 P3-21. A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is the maximum data rate supported by this line? Given Bandwidth =4000KHz, SNR=1000 So, We Know the Shannon capacity Capacity =BW* log 2 ( 1 + SNR ) Capacity =4000K* log 2 ( )=40 Kbps

24 P3-22. We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 10 V, the noise is 5 mv. What is the maximum data rate supported by this telephone line?

25 P3-22. We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 10 V, the noise is 5 mv. What is the maximum data rate supported by this telephone line? Given, BW= 4 KHz, Signal=10V, Noise= 5mV, SNR= = 2000 Capacity =BW* log 2 ( 1 + SNR ) Capacity =4000* log 2 ( )=43866 bps

26 P3-23. A file contains 2 million bytes. How long does it take to download this file using a 56-Kbps channel? 1-Mbps channel?

27 P3-23. A file contains 2 million bytes. How long does it take to download this file using a 56-Kbps channel? 1-Mbps channel? Given, file contains= 2 000,000 bytes or 16000,000 bit With a 56-Kbps channel,it s take = bit bit per sec = sec = 4.76 min 1-Mbps channel is the same

28 P3-24. A computer monitor has a resolution of 1200 by 1000 pixels. If each pixel uses 1024 colors, how many bits are needed to send the complete contents of a screen?

29 P3-24. A computer monitor has a resolution of 1200 by 1000 pixels. If each pixel uses 1024 colors, how many bits are needed to send the complete contents of a screen? To represent 1024 color we need = log = 10bits Total number of bit = 1200*1000*10 = 12000,000 bit = 12 million bit

30 P3-25. A signal with 200 milliwatts power passes through 10 devices, each with an average noise of 2 microwatts. What is the SNR? What is the SNRdB?

31 P3-25. A signal with 200 milliwatts power passes through 10 devices, each with an average noise of 2 microwatts. What is the SNR? What is the SNR db? Given, Signal =200 mw= 0.2 W Total Noise = 10*2 *10 6 =2*10 5 We Know, SNR= average signal power average noise power = = SNR db = 10 log 10 (SNR) SNR db = 10 log = 40

32 P3-26. If the peak voltage value of a signal is 20 times the peak voltage value of the noise, what is the SNR? What is the SNRdB?

33 P3-26. If the peak voltage value of a signal is 20 times the peak voltage value of the noise, what is the SNR? What is the SNRdB? We Know SNR= SNR= signal power noise power signal voltage 2 noise voltage 2 =202 =400 SNR db = 10 log 10 (SNR) SNR db = 10 log =

34 P3-27. What is the theoretical capacity of a channel in each of the following cases? We can approximately calculation the capacity as a. Bandwidth: 20 KHz SNR db = 40 b. Bandwidth: 200 KHz SNR db = 4 c. Bandwidth: 1 MHz SNR db = 20

35 P3-27. What is the theoretical capacity of a channel in each of the following cases? We can approximately calculation the capacity as a. Bandwidth: 20 KHz SNR db = 40 Capacity =BW * SNR db 3 =2000* 40 3 =267Kbps b. Bandwidth: 200 KHz SNR db = 4 c. Bandwidth: 1 MHz SNR db = 20

36 P3-28. We need to upgrade a channel to a higher bandwidth. Answer the following questions: 1) How is the rate improved if we double the bandwidth? 2)How is the rate improved if we double the SNR?

37 P3-28. We need to upgrade a channel to a higher bandwidth. Answer the following questions: 1) How is the rate improved if we double the bandwidth? C2=2*BW1* log 2 ( 1 + SNR ) The data rate is doubled (C2 =2*C1) 2)How is the rate improved if we double the SNR? C2=BW1* log 2 ( SNR ) C2=BW1* log 2 ( 2 SNR ) C2=BW1* log 2 ( 2) + BW1 log 2 ( SNR) C2=BW1+ C1

38 P3-29. We have a channel with 4 KHz bandwidth. If we want to send data at 100 Kbps, what is the minimum SNR db? What is the SNR?

39 P3-29. We have a channel with 4 KHz bandwidth. If we want to send data at 100 Kbps, what is the minimum SNR db? What is the SNR? We know, SNR db Capacity =BW * 3 3 Capacity so minimum SNR db= = BW 3 100,000 = SNR db Minimum SNR =10 10 =

40 P3-30. What is the transmission time of a packet sent by a station if the length of the packet is 1 million bytes and the bandwidth of the channel is 200 Kbps? Given, Packet length =1 million byte=8 million bit Transmission time = packet length bandwidth =8,000, ,000 =40 sec

41 P3-31. What is the length of a bit in a channel with a propagation speed of m/s if the channel bandwidth is a. 1 Mbps? Bit length= propagation speed Bandwith Bit length= m/s 10 6 b/s =200 m b. 10 Mbps? c. 100 Mbps?

42 PERFORMANCE Bandwidth Hertz Bits per Seconds the range of frequencies contained in a composite signal or the range of frequencies a channel can pass he number of bits per second that a channel, a link, or even a network can transmit

43 PERFORMANCE Throughput A measure of how fast we can actually send data through a network Bandwidth in bits per second and throughput seem the same, they are different. A link may have a bandwidth of B bps, but we can only send T bps through this link with T always less than B. Example : A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network? Throughput = (12,000 * 10,000) / 60 = 2 Mbps

44 Bandwidth-Delay Product Define the maximum number of bits that can fill the link

45 P3-32. How many bits can fit on a link with a 2 ms delay if the bandwidth of the link is a. 1 Mbps? b. 10 Mbps? c. 100 Mbps?

46 P3-32. How many bits can fit on a link with a 2 ms delay if the bandwidth of the link is a. 1 Mbps? Bandwidth-Delay = Bandwith*delay Bandwidth Delay = ( ) ( )=2000 bits b. 10 Mbps? c. 100 Mbps?

47 END