1 / 29

Physical Layer

Physical Layer. Cont. Outline. Digital Signal Digital versus Analog Data Rate Limits Transmission Impairment More about signals. Digital Signals. Bit Rate and Bit Interval. Bit Interval: the time required to send one single bit

drucilla
Download Presentation

Physical Layer

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. Physical Layer Cont.

  2. Outline • Digital Signal • Digital versus Analog • Data Rate Limits • Transmission Impairment • More about signals

  3. Digital Signals

  4. Bit Rate and Bit Interval • Bit Interval: the time required to send one single bit • Bit rate: the number of bit interval per second, bits per second (bps)

  5. Example • Question: A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) • Solution: The bit interval is the inverse of the bit rate. Bit interval = 1/ 2000 s = 0.000500 s = 500 μs

  6. Digital Signals as a composite Analog Signals • A Digital Signal is a composite signal with an infinite bandwidth • Digital Signal Through a Wide-BW medium • A dedicated medium such as a coaxial cable to send digital data through LAN • Digital Signal Through a Band-Limited Medium • What is the minimum required bandwidth B in hertz if we want to send n bps?

  7. Digital versus Analog • Using only One Harmonic • B = n/2 • Using More Harmonics • B>=n/2 or n<=2B

  8. Digital versus Analog

  9. Bandwidth Requirement • The bit rate and the bandwidth are proportional to each other

  10. Digital versus Analog Bandwidth • The analog bandwidth of a medium is expressed in hertz, the digital bandwidth, in bits per second • Higher bit rate : modulation technique

  11. Analog Versus Digital • Low pass channel: bandwidth frequencies between 0 and f • Band pass channel: bandwidth frequencies between f1 and f2 • Digital transmission needs a low pass channel • Analog transmission can use a band pass channel

  12. Low pass & Band pass

  13. Data Rate Limits • Noiseless Channel: Nyquist Bit Rate • Bit rate = 2* Bandwidth * log2L • L = numbers of signal level • Noisy Channel: Shannon Capacity • Capacity = Bandwidth * log2 (1+SNR) • SNR = signal-to-noise ratio

  14. Example • Question: Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as • Solution: BitRate = 2  3000  log2 2 = 6000 bps

  15. Example • Question: Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). The maximum bit rate can be calculated as: • Solution: Bit Rate = 2 x 3000 x log2 4 = 12,000 bps

  16. Example • Question: We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level? • Solution: First, we use the Shannon formula to find our upper limit. • Then we use the Nyquist formula to find the number of signal levels C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps 4 Mbps = 2  1 MHz  log2L L = 4

  17. Transmission Impairment • Attenuation: • Loss of energy • need amplifier • Distortion: • Change of form or shape • Noise • Thermal noise, induced noise, cross talk, etc.

  18. Attenuation

  19. Decibel • Showing that a signal has lost or gained strength • dB = 10 log10(P2/P1)

  20. Example • Question: Imagine a signal travels through a transmission medium and its power is reduced to half. • Solution: This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as 10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(–0.3) = –3 dB

  21. Example • Question: Imagine a signal travels through an amplifier and its power is increased ten times. • Solution: This means that P2 = 10 * P1. In this case, the amplification (gain of power) can be calculated as 10 log10 (P2/P1) = 10 log10 (10P1/P1) = 10 log10 (10) = 10 (1) = 10 dB

  22. Example In the following figure, a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the signal just by adding the decibel measurements between each set of points.

  23. Example dB = –3 + 7 – 3 = +1

  24. Distortion

  25. Noise

  26. More about the signal • Throughput • The measurement of how fast the data can pass through an entity (n per second) • Propagation Speed • The measurement of the distance a bit can travel through a medium in one second • Propagation Time • The measurement of time required for a bit to travel from one point to another point of the medium • Propagation time = Distance/Propagation speed • Wavelength • Wave length = Propagation speed * period

  27. Throughput

  28. Propagation Time

  29. Wavelength

More Related