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  • Slide 1
  • Slide 2
  • Jozef Goetz, 2009 1 expanded by Jozef Goetz, 2009 The McGraw-Hill Companies, Inc., 2006 expanded by Jozef Goetz, 2009 PART II
  • Slide 3
  • Jozef Goetz, 2009 2 Define transmission. Define transmission. Describe the concept of frequency, spectrum, and Describe the concept of frequency, spectrum, and bandwidth. bandwidth. Define analog and digital data transmission. Define analog and digital data transmission. Describe the concept of data, signals, and Describe the concept of data, signals, and transmission. transmission. List the various impairments to effective transmission. List the various impairments to effective transmission. Describe attenuation. Describe attenuation. Describe delay distortion. Describe delay distortion. Describe the concept of noise & channel capacity. Describe the concept of noise & channel capacity.
  • Slide 4
  • Jozef Goetz, 2009 3 3-3 DIGITAL SIGNALS In addition to being represented, information can also be represented by an analog signalby an analog signal by a digital signalby a digital signal For example,For example, a 1 can be encoded as a positive voltage anda 1 can be encoded as a positive voltage and a 0 as zero voltage.a 0 as zero voltage. A digital signal can have more than 2 levels.A digital signal can have more than 2 levels. In this case, we can send more than 1 bit for each level.In this case, we can send more than 1 bit for each level.
  • Slide 5
  • Jozef Goetz, 2009 4 3-3 DIGITAL SIGNALS Bit Rate vs Bit Length Digital Signal as a Composite Analog Signal Application Layer Topics discussed in this section:
  • Slide 6
  • Jozef Goetz, 2009 5 Figure 3.16 A digital signal The bit interval (bit duration) is the time required to send 1 bit. The bit rate is the # of bit intervals per sec
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  • Jozef Goetz, 2009 6 Example 6 A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) Solution The bit duration (interval) is the inverse of the bit rate. Bit interval = 1/ 2000 [bps] = 0.000500 [s/bit] = 0.000500 x [10 6 s/bit] = 500 [ s/bit]
  • Slide 8
  • Jozef Goetz, 2009 7 Figure 3.16 Bit rate and bit interval 1 s The bit duration BD (interval) is the time required to send 1 bit. The bit rate BR is the # of bit intervals per sec R = 1 / D
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  • Jozef Goetz, 2009 8 Bit length and bit duration p.73 and 64 The bit length BL is the distance 1 bit occupies on the transmission medium. The bit duration BD (interval) is the time required to send 1 bit. bit length = propagation speed * bit duration L = S * D Remember from your elementary school formula: D = S * T where: D distance, S speed, t - Time
  • Slide 10
  • Jozef Goetz, 2009 9 Figure 3.16 Two digital signals: one with two signal levels and the other with four signal levels
  • Slide 11
  • Jozef Goetz, 2009 10 Appendix C reviews information about exponential and logarithmic functions. Note Appendix C reviews information about exponential and logarithmic functions.
  • Slide 12
  • Jozef Goetz, 2009 11 A digital signal has 8 levels. How many bits are needed per level? Example 3.16 Each signal level is represented by 3 bits.
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  • Jozef Goetz, 2009 12 A digital signal has 9 levels. How many bits are needed per level? We calculate the number of bits by using the formula. Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level. Example 3.17
  • Slide 14
  • Jozef Goetz, 2009 13 Assume we need to download text documents at the rate of 100 pages per seconds. What is the required bit rate of the channel? Solution A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits, the bit rate is 100 x 24 [line/sec] = 100 x 24 x 80[chr/sec] = = 100 x 24 x 80 x 8 [bit/sec] = 1, 636, 000 [bit/sec] = 1.636 Mbps Example 3.18
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  • Jozef Goetz, 2009 14 A digitized voice channel, as we will see in Chapter 4, is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (2 samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate? Solution bit rate = 2 x 4000 [Hz] x 8 [bits] = 2 x 4000 [1/sec] x 8 [bits] = 64,000 [bits/sec] = 64 [kbps] Example 3.19
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  • Jozef Goetz, 2009 15 What is the bit rate for high-definition TV (HDTV)? Solution HDTV uses digital signals to broadcast high quality video signals. The HDTV screen is normally a ratio of 16 : 9. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second = 30 [1/sec] 24 [bits] represents one color pixel: 1920 x 1080 x 30 [1/sec] x 24 [bit] = = 1,492,992,000 [bit/sec] = 1.5 Gbps Example 3.20 The TV stations reduce this rate to 20 to 40 Mbps through compression.
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  • Jozef Goetz, 2009 16 Figure 3.17 The time and frequency domains of periodic and nonperiodic digital signals Based on Fourier analysis, a digital signal is a composite analog signal with an infinite bandwidth B The periodic signal has discrete frequencies The nonperiodic signal has continues frequencies
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  • Jozef Goetz, 2009 17 Figure 3.18 Baseband transmission A digital signal can be transmitted by using 2 approaches: Baseband Sending a digital signal without changing to an analog signal Broadband transmission => slide 33
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  • Jozef Goetz, 2009 18 Figure 3.19 Bandwidths of two low-pass channels Baseband transmission requires a low-pass channel with bandwidth that starts from 0
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  • Jozef Goetz, 2009 19 Figure 3.20 Baseband transmission using a dedicated medium with a. wide-bandwidth Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low- pass channel with an infinite or very wide bandwidth.
  • Slide 21
  • Jozef Goetz, 2009 20 An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN. Almost every wired LAN today uses a dedicated channel for 2 stations communicating with each other. In a bus topology LAN with multipoint connections, only two stations can communicate with each other at each moment in time (timesharing); the other stations need to refrain from sending data. In a star topology LAN, the entire channel between each station and the hub is used for communication between these two entities. Example 3.21
  • Slide 22
  • Jozef Goetz, 2009 21 b. Low-pass channels with Limited Bandwidth Assume digital bit rate = N, the worst case, max changes happens when the signal carries. The sequence 01010101010101.or 1010101010101010. To simulate this we need an analog signal of f = N / 2. Low-pass channels with Limited Bandwidth, we approximate the digital signal with an analog signal. The level of approximation depends on the bandwidth available.
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  • Jozef Goetz, 2009 22 Low-pass channels with Limited Bandwidth Figure 3.21 Rough approximation of a digital signal using the first harmonic N/2 and N/4 for worst case Where: bit rate = N 1 is represented by positive amplitude, 0 is represented by negative amplitude Here is a sequence of all 3-bit combinations:
  • Jozef Goetz, 2009 28 Table 3.2 Bandwidth Requirement Bit Rate Harmonic 1 Harmonics 1, 3 Harmonics 1, 3, 5 Harmonics 1, 3, 5, 7 1 Kbps500 Hz1.5 KHz2.5 KHz3.5 KHz 10 Kbps5 KHz15 KHz25 KHz35 KHz 100 Kbps50 KHz150 KHz250 KHz350 KHz How is it calculated: B = n / 2 is related to the first harmonic B = 3 n / 2 is related to the 2 nd column etc. B = 5 n / 2 is related to the 3rd column etc. So B >= n / 2 => n
  • Jozef Goetz, 2009 29 Note In baseband transmission, the required bandwidth is proportional to the bit rate; if we need to send bits faster, we need more bandwidth. The bit rate and the bandwidth are proportional to each other and B >= n / 2 from the previous slide A typical voice signal has a bandwidth of approximately 4 kHz, so we should be able send about 8000 bps if we use a single sine (with the 1 st harmonic)
  • Slide 31
  • Jozef Goetz, 2009 30 What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission? Solution The answer depends on the accuracy desired. a. The minimum bandwidth, is B = bit rate /2, or 500 kHz. b. A better solution is to use the first and the third harmonics with B = 3 500 kHz = 1.5 MHz. c. Still a better solution is to use the first, third, and fifth harmonics with B = 5 500 kHz = 2.5 MHz. Example 3.22
  • Slide 32
  • Jozef Goetz, 2009 31 We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel? Solution See note under table 3.2 From B >= bit rate / 2 => bit rate

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