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LEARNING OBJECTIVESLEARNING OBJECTIVES
1. State four functions of a telephone set. 2. Label a block diagram of a telephone set. 3. Describe the electrical differences between pulse and tone dialing. 4. Describe the structure of the local telephone exchange. 5. Describe each of the BORSCHT functions. 6. Define “IDDD World Numbering Plan.” 7. Explain the differences between the telephone exchange classifications. 8. Describe the electrical characteristics of PAM, PDM, PCM, and DM. 9. Describe the advantage of companding.10. State the companding technique used in the United States and Europe.11. Analyze attenuation, delay, and line conditioning on a voice-channel.12. Interpret attenuation/ delay charts for conditioned lines.13. Prepare a brief definition of multiplexing and demultiplexing.14. Describe the characteristics of SDM, FDM, TDM, and STDM techniques.15. Describe the structure of the analog common carrier hierarchy.16. Describe the structure of the digital common carrier hierarchy.17. Explain the services offered through T1, T2, T3, and T4 connections.18. Label a block diagram of the T1 frame format.
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LEARNING OBJECTIVES (continued)LEARNING OBJECTIVES (continued)
19. Prepare a brief definition of CSU/DSU.20. State the advantages of superframes and extended superframes over standard T1 frames.21. Administer change controls by adding new telephone equipment.22. Design the network by identifying the availability of local T1 access.23. Describe the electrical characteristics of a T1 signal.24. State two types of acceptable T1 connectors.25. Ensure appropriate resources are available for implementing T-Carrier access.26. Describe the services available with a SONET connection.27. Describe the frame format of a SONET STS1.28. Define SONET terms such as virtual tributary, terminating multiplexer, regenerator, and add/drop multiplexer.29. While planning a customer’s job, study technology options, pro and con.30. Identify the technical capabilities of T-Carrier, SONET, and Multiplexing.
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Figure 9-1: Frequency Response of Voice Channel
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Figure 9-2: Producing Amplitude Modulation
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Figure 9-3: AM Modulation Envelope
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Figure 9-4: Frequency Modulation
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Using Carson’s rule, bandwidth is calculated by:
BW = 2(BW = 2(FFcc + + FFmm))
where Fc is the deviation produced by the modulating signal, and Fm is the frequency of the modulating signal.
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The bandwidth of the FM signal is:
BW = 2(BW = 2(FFcc + + FFmm))
BW = 2(1 kHz + 1 kHz)BW = 2(1 kHz + 1 kHz)BW = 2(2 kHz)BW = 2(2 kHz)
BW = 4 kHzBW = 4 kHz
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Figure 9-5: Phase Modulation
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Figure 9-6: Equivalent Two-Conductor Cable Circuit
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Table 9-1: Terminating Resistance Guide
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For any length of RG-58 coaxial cable, the characteristic impedance can be approximated by:
ZZoo = = LL//CC
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This means the RG-58 cable will behave as any circuit with a capacitor, and inductance, and a 50-ohm load.
It will exhibit phase-shifts and time constants as in any RCL circuit.
73 x 1073 x 10-9-9ZZoo = =
ZZoo = 49.745 ohms = 49.745 ohms295 x 10295 x 10-12-12__________________
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Figure 9-7: Effects Of Intersymbol Interference on Data Signals
(a)
(b)
(c)
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Figure 9-8: Evaluating the Eye Pattern with an Oscilloscope for ISI Distortion
(a)
(b)
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Figure 9-9: Equalizer Block Diagram
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The following illustrates examples of common logs:
101044 = 10,000 = a log of 4 = 10,000 = a log of 4
101033 = 1,000 = a log of 3 = 1,000 = a log of 3
101022 = 100 = a log of 2 = 100 = a log of 2
101011 = 10 = a log of 1 = 10 = a log of 1
101000 = 1 = a log of 0 = 1 = a log of 0
1010–1–1 = .1 = a log of –1 = .1 = a log of –1
1010–2–2 = .01 = a log of –2 = .01 = a log of –2
1010–3–3 = .001 = a log of –3 = .001 = a log of –3
1010–4–4 = .0001 = a log of –4 = .0001 = a log of –4
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Common logs are integrated with decibels in the following formula:
dB = 10 log (dB = 10 log (PPoo//PPii))
where Po = Power out and Pi = Power in.
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An amplifier has an input power of 10 mW (milliwatts) and a power out of 20 mW. Calculate the gain in decibels.
dB = 10 log (dB = 10 log (PPoo//PPii))
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An amplifier has an input power of 10 mW (milliwatts) and a power out of 20 mW. Calculate the gain in decibels.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (20dB = 10 log (20
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An amplifier has an input power of 10 mW (milliwatts) and a power out of 20 mW. Calculate the gain in decibels.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (20 dB = 10 log (20 mW/10 mW)mW/10 mW)
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An amplifier has an input power of 10 mW (milliwatts) and a power out of 20 mW. Calculate the gain in decibels.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (20 dB = 10 log (20 mW/10 mW)mW/10 mW)
dB = 10 log (2)dB = 10 log (2)
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An amplifier has an input power of 10 mW (milliwatts) and a power out of 20 mW. Calculate the gain in decibels.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (20 dB = 10 log (20 mW/10 mW)mW/10 mW)
dB = 10 log (2)dB = 10 log (2)dB = 10 (.3)dB = 10 (.3)
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An amplifier has an input power of 10 mW (milliwatts) and a power out of 20 mW. Calculate the gain in decibels.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (20 dB = 10 log (20 mW/10 mW)mW/10 mW)
dB = 10 log (2)dB = 10 log (2)dB = 10 (.3)dB = 10 (.3)
dB = 3dB = 3
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The power received through a transmission line is .25 mW. Using decibels, evaluate the performance of the line if the power at the transmitter is 1 mW.
dB = 10 log (dB = 10 log (PPoo//PPii))
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The power received through a transmission line is .25 mW. Using decibels, evaluate the performance of the line if the power at the transmitter is 1 mW.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (.25 mW/1 mW)dB = 10 log (.25 mW/1 mW)
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The power received through a transmission line is .25 mW. Using decibels, evaluate the performance of the line if the power at the transmitter is 1 mW.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (.25 mW/1 mW)dB = 10 log (.25 mW/1 mW)dB = 10 (–.6)dB = 10 (–.6)
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The power received through a transmission line is .25 mW. Using decibels, evaluate the performance of the line if the power at the transmitter is 1 mW.
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (.25 mW/1 mW)dB = 10 log (.25 mW/1 mW)dB = 10 (–.6)dB = 10 (–.6)
dB = –6dB = –6
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What is the value of the signal strength?
dB = 10 log (dB = 10 log (PPoo//PPii))
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What is the value of the signal strength?
dB = 10 log (dB = 10 log (PPoo//PPii))
––6 dB = 10 log (6 dB = 10 log (xx/1 mW)/1 mW)
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What is the value of the signal strength?
dB = 10 log (dB = 10 log (PPoo//PPii))
––6 dB = 10 log (6 dB = 10 log (xx/1 mW)/1 mW)––.6 dB = log (.6 dB = log (xx/1 mW)/1 mW)
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What is the value of the signal strength?
dB = 10 log (dB = 10 log (PPoo//PPii))
––6 dB = 10 log (6 dB = 10 log (xx/1 mW)/1 mW)––.6 dB = log (.6 dB = log (xx/1 mW)/1 mW)
.25 = .25 = xx/1 mW/1 mW
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What is the value of the signal strength?
dB = 10 log (dB = 10 log (PPoo//PPii))
––6 dB = 10 log (6 dB = 10 log (xx/1 mW)/1 mW)––.6 dB = log (.6 dB = log (xx/1 mW)/1 mW)
.25 = .25 = xx/1 mW/1 mW.25 mW = .25 mW = xx = = PPoo
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A power amplifier has 5-dB gain referenced to 6 mW. What is the output power?
dB = 10 log (dB = 10 log (PPoo//PPii))
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A power amplifier has 5-dB gain referenced to 6 mW. What is the output power?
dB = 10 log (dB = 10 log (PPoo//PPii))
5 dB = 10 log (5 dB = 10 log (PPoo/6 mW)/6 mW)
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A power amplifier has 5-dB gain referenced to 6 mW. What is the output power?
dB = 10 log (dB = 10 log (PPoo//PPii))
5 dB = 10 log (5 dB = 10 log (PPoo/6 mW)/6 mW)
.5 dB = log (.5 dB = log (PPoo/6 mW)/6 mW)
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A power amplifier has 5-dB gain referenced to 6 mW. What is the output power?
dB = 10 log (dB = 10 log (PPoo//PPii))
5 dB = 10 log (5 dB = 10 log (PPoo/6 mW)/6 mW)
.5 dB = log (.5 dB = log (PPoo/6 mW)/6 mW)
Inverse log .5 = Inverse log .5 = PPoo/6 mW/6 mW
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A power amplifier has 5-dB gain referenced to 6 mW. What is the output power?
dB = 10 log (dB = 10 log (PPoo//PPii))
5 dB = 10 log (5 dB = 10 log (PPoo/6 mW)/6 mW)
.5 dB = log (.5 dB = log (PPoo/6 mW)/6 mW)
Inverse log .5 = Inverse log .5 = PPoo/6 mW/6 mW
3.16 = 3.16 = PPoo/6 mW/6 mW
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A power amplifier has 5-dB gain referenced to 6 mW. What is the output power?
dB = 10 log (dB = 10 log (PPoo//PPii))
5 dB = 10 log (5 dB = 10 log (PPoo/6 mW)/6 mW)
.5 dB = log (.5 dB = log (PPoo/6 mW)/6 mW)
Inverse log .5 = Inverse log .5 = PPoo/6 mW/6 mW
3.16 = 3.16 = PPoo/6 mW/6 mW
18.96 mW = 18.96 mW = PPoo
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A data signal is permitted no more than 1 dB of attenuation, or loss, from transmitter to receiver. You measure the transmitter power at 24.2 mW and the power received at 19.5 mW. Is this an acceptable amount of attenuation?
dB = 10 log (dB = 10 log (PPoo//PPii))
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A data signal is permitted no more than 1 dB of attenuation, or loss, from transmitter to receiver. You measure the transmitter power at 24.2 mW and the power received at 19.5 mW. Is this an acceptable amount of attenuation?
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (19.5 mW/24.2 mW)
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A data signal is permitted no more than 1 dB of attenuation, or loss, from transmitter to receiver. You measure the transmitter power at 24.2 mW and the power received at 19.5 mW. Is this an acceptable amount of attenuation?
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (.806)dB = 10 log (.806)
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A data signal is permitted no more than 1 dB of attenuation, or loss, from transmitter to receiver. You measure the transmitter power at 24.2 mW and the power received at 19.5 mW. Is this an acceptable amount of attenuation?
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (.806)dB = 10 log (.806)dB = 10 (–.0936)dB = 10 (–.0936)
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A data signal is permitted no more than 1 dB of attenuation, or loss, from transmitter to receiver. You measure the transmitter power at 24.2 mW and the power received at 19.5 mW. Is this an acceptable amount of attenuation?
dB = 10 log (dB = 10 log (PPoo//PPii))
dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (19.5 mW/24.2 mW)dB = 10 log (.806)dB = 10 log (.806)dB = 10 (–.0936)dB = 10 (–.0936)
dB = –.936dB = –.936
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Initiates use of the telephone system when the handset is lifted.
11
At a minimum, the telephone set:
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Table 9-2: Typical SNRs of Communication Channels
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The signal-to-noise ratio is a measure of the desired signal power to the noise signal power at the same point in a circuit. It’s expressed mathematically as:
SNR = SNR = PPss//PPnn
where Ps = the power of the desired signal and Pn = the power of the noise.
It’s often expressed in decibels as:
SNR = 10 log (SNR = 10 log (PPss//PPnn))
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For example, if the SNR is to be determined at the output of a transmitter, and the signal level is determined to be 2 mW and the noise is 500 mW, the SNR is:
SNR = 2 mW/.5 mWSNR = 2 mW/.5 mWSNR = 4SNR = 4
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or, as a decibel:
SNR = 10 log (SNR = 10 log (PPss//PPnn))
SNR = 10 log (2 mW/.5 mW)SNR = 10 log (2 mW/.5 mW)SNR = 10 log (4)SNR = 10 log (4)
SNR = 10 (.602) dBSNR = 10 (.602) dBSNR = 6 dBSNR = 6 dB
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Noise factor is calculated as:
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
where Si/Ni is the signal-to-noise ratio at the input, and So/No is the signal-to-noise ratio at the output.
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For example, the NF of a transmission line is to be calculated. The signal level at the transmitter is 22 mW and the noise level is 1 mW. At the receiver, the signal level has dropped to 18.5 mW and the noise increased to 2.25 mW.
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
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For example, the NF of a transmission line is to be calculated. The signal level at the transmitter is 22 mW and the noise level is 1 mW. At the receiver, the signal level has dropped to 18.5 mW and the noise increased to 2.25 mW.
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))
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For example, the NF of a transmission line is to be calculated. The signal level at the transmitter is 22 mW and the noise level is 1 mW. At the receiver, the signal level has dropped to 18.5 mW and the noise increased to 2.25 mW.
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log (22/8.22)NF = 10 log (22/8.22)
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For example, the NF of a transmission line is to be calculated. The signal level at the transmitter is 22 mW and the noise level is 1 mW. At the receiver, the signal level has dropped to 18.5 mW and the noise increased to 2.25 mW.
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log (22/8.22)NF = 10 log (22/8.22)NF = 10 log (2.676)NF = 10 log (2.676)
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For example, the NF of a transmission line is to be calculated. The signal level at the transmitter is 22 mW and the noise level is 1 mW. At the receiver, the signal level has dropped to 18.5 mW and the noise increased to 2.25 mW.
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log (22/8.22)NF = 10 log (22/8.22)NF = 10 log (2.676)NF = 10 log (2.676)NF = 10 (.427) dBNF = 10 (.427) dB
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For example, the NF of a transmission line is to be calculated. The signal level at the transmitter is 22 mW and the noise level is 1 mW. At the receiver, the signal level has dropped to 18.5 mW and the noise increased to 2.25 mW.
NF = 10 log ((NF = 10 log ((SSii//NNii)/()/(SSoo//NNoo))))
NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log ((22 mW/1 mW)/(18.5 mW/2.25 mW))NF = 10 log (22/8.22)NF = 10 log (22/8.22)NF = 10 log (2.676)NF = 10 log (2.676)NF = 10 (.427) dBNF = 10 (.427) dB
NF = 4.27 dBNF = 4.27 dB
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Figure 9-10: Noise and Distortion Contribute to Data Errors
(a)
(b)
(c)
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Initiates use of the telephone system when the handset is lifted.
11
At a minimum, the telephone set:
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22 Receives a dial tone indicating thesystem is ready to be used.
At a minimum, the telephone set: (continued)
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33 Transmits the number to be called to the telephone system.
At a minimum, the telephone set: (continued)
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44 Supervises the status of the set by indicating to a caller if the set is in use (busy), or available to receive a call (ring).
At a minimum, the telephone set: (continued)
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55 Acknowledges an incoming call by ringing.
At a minimum, the telephone set: (continued)
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66 Performs the duties of a transducer by converting audio signals into electrical signals, and electrical signals into audio signals.
At a minimum, the telephone set: (continued)
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77 Compensates for varying power levels supplied to it.
At a minimum, the telephone set: (continued)
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88 Provides an indication to the telephone system that a call is finished when the caller hangs up.
At a minimum, the telephone set: (continued)
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Figure 9-11: The Telephone Set
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When working on telephone lines in the premises, if unable to disconnect the phone service, take the phone off the hook to avoid electrical shock by reducing the continuous voltage.
TIPTIP
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Figure 9-12: Pulse Dialing 3-4-7
(a)
(b)
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The total time required to pulse dial the three numbers is:
(3 x 100) + 500 + (4 x 100) + 500 + (7 x 100)(3 x 100) + 500 + (4 x 100) + 500 + (7 x 100)= 300 + 500 + 400 + 500 + 700= 300 + 500 + 400 + 500 + 700
= 2,400 ms= 2,400 ms
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Figure 9-13: DTMF Keypad
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Figure 9-14: Central Office Exchange Loop
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Figure 9-15: Structure of the Central Office
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Figure 9-16: Overvoltage Protection
(a)
(b)
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Figure 9-17: Central Office Supervisory Signals
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Figure 9-18: Two-Wire to Four-Wire Hybrid
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Figure 9-19: IDDD World Numbering System
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Figure 9-20: Telephone Exchange Classification
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Figure 9-21: Possible Long Distance Routings
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Figure 9-22: Interconnection of Central Office Exchanges
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Figure 9-23: Pulse Amplitude Modulation
(b)(a)
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Figure 9-24: Pulse Duration Modulation
(a)
(b)
(c)
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Figure 9-25: Pulse Code Modulation
(a)
(b)
(c)
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Figure 9-26: The -Law Companding Code
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Figure 9-27(a): Delta Modulation Transmitter Block Diagram
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Figure 9-27(b): Delta Modulation Transmitter
Waveforms
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Figure 9-28: Interconnection of Central Office Exchanges
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The signal strength ratio is expressed as the logarithmic function of the ratio:
dB = 10 log dB = 10 log PP11//PP22
where P1 = signal power and P2 = the 1mW reference.
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For example, a signal delivers 50 mW of power at a specified frequency to a load.
The power gain is found by:
dBm = 10 log dBm = 10 log PP11/1 mW/1 mW
dBm = 10 log 50 mW/1 mWdBm = 10 log 50 mW/1 mWdBm = 17dBm = 17
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For example: a signal delivering .5 mW to a load has a loss of:
dBm = 10 log .5 mW/1 mWdBm = 10 log .5 mW/1 mWdBm = –3dBm = –3
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Figure 9-29: Voice-Channel Attenuation Response Curve
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Figure 9-30: Voice-Channel Delays
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Figure 9-31: 3002 Leased Line (b)
(a)
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Figure 9-32: C1, C2, and C3 Conditioned Lines
(a) C1
(c) C4
(b) C2
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Figure 9-32(a): C1 Conditioned Line (continued)
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Figure 9-32(b): C2 Conditioned Line (continued)
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Figure 9-32(c): C3 Conditioned Line (continued)
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LAB 25 OBJECTIVELAB 25 OBJECTIVEMeasuring Data Throughput of the Local
Subscriber Loop
Use an Internet download speed Use an Internet download speed measurement site to evaluate the measurement site to evaluate the current condition of your local current condition of your local loop.loop.
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Figure 9-33: MSN Connection Speed Test Results
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Table 9-3: Speed Test
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LAB 25 QUESTIONSLAB 25 QUESTIONS
Summarize the results of the download test in Table 9-3.
11
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LAB 25 QUESTIONSLAB 25 QUESTIONS
Why should you choose a compressed file when measuring modem data rates?
22
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LAB 25 QUESTIONSLAB 25 QUESTIONS
How close was your speed to the theoretical maximum for a conditioned line of 53.3 kbps?
33
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Figure 9-34: Multiplexing Reduces Hardware
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Figure 9-35: Space-Division Multiplexing
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Figure 9-36: FDM Frame Formatting
(a)
(b)
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Figure 9-37: FDM Analog Common-Carrier Hierarchy
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Figure 9-38: TDM Frame Format
(a)
(b)
(c)
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Figure 9-39: T-Carrier Multiplexing Scheme
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Figure 9-40: T1 Frame Format
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Figure 9-41: Simplified STDM
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Figure 9-42: STDM Frame Format
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Consider the phrase:
THAT CATTHAT CAT
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Another compression technique is the dictionary method. The dictionary is composed of the words contained in the data to be transmitted.
For example:
THAT CAT ATE THAT RATTHAT CAT ATE THAT RAT
may be filed alphabetically in the dictionary.
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Once all of the redundancies in the phrase are deleted, the above phrase would look like:
ATE, CAT, RAT, THATATE, CAT, RAT, THAT
1 2 3 41 2 3 4
THAT CAT ATE THAT RATTHAT CAT ATE THAT RAT
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The numbers are encoded into a 3- or 4-bit format and transmitted in the sequence shown.
4 2 1 4 34 2 1 4 3
(THAT) (CAT) (ATE) (THAT) (RAT)(THAT) (CAT) (ATE) (THAT) (RAT)
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Figure 9-43(a): FDM Common-Carrier Analog Hierarchy
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Figure 9-43(b): FDM Typical Modulation Method
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Figure 9-44: T-Carrier Digital Hierarchy
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Table 9-4: T-Carrier Digital Hierarchy
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Figure 9-45: T-Simplified D4 Channel Bank
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Table 9-5: Superframe Format
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Each synchronization technique uses a specific 6-bit code that’s interleaved within the superframe.
The bit codes are as follows:
Terminal: 101010Terminal: 101010Superframe: 001110Superframe: 001110
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Once the two codes are interleaved at bit position 193 in each of the T1 frames, they generate the following 12-bit code:
100011011100100011011100
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Table 9-6: Extended Superframe Format
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The ESF contains 24, 12-channel T1 frames (288 total channels). Bit 193 is multiplexed to serve three distinct purposes:
The Fe bit provides frame synchronization at every fourth frame using the bit pattern 001011. Fe serves the same purpose as the S and T bits in the D4 format.
The Data Link (DL) bit carries line performance information at every other frame.
The CRC-6 is a 6-bit cyclic redundancy check that inspects all 4,632 bits of the frame for errors.
11
22
33
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Table 9-7: DS1 Signal Characteristics
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Table 9-8: AT&T T1 Connector Pinout
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Table 9-9: ANSI T1 Connector Pinout
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Figure 9-46: Word and Bit Interleaving
(a)
(b)
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The four levels of service are:
The customer may change the location of terminating equipment with carrier’s assistance.
11
22
33
44
The use of multiplexing to allow the customer to connect up to 24 channels to switched or dial-up services
The customer controls all configurations. This allows dynamic allocation of circuits—all without assistance from the carrier.
The use of multiplexing to allow two T1 lines that carry up to 22 channels on a single T1 connection
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Figure 9-47: Typical T1 Interconnection
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SONET networks are tasked to provide the following services:
Reduction in copper-based equipment costs and increased reliability
Frame lengths of sufficient size to carry management information about the link and the pay-load carried in the frame
The establishment of accepted standards that permit networks to be built that are vendor independent
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SONET networks are tasked to provide the following services (continued) :
The ability to format lower speed frames such as DS1, and multiplex these using a synchronous structure
The creation of an architecture that promotes future development at varyingtransmission rates
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Table 9-10: SONET Signaling Hierarchy
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Figure 9-48: SONET Frame Format
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Figure 9-49: SONET Terminating Multiplexer
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There are three basic configurations used with SONET:
Point-to-Point
Hub (Star)
Ring
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33
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Figure 9-50: SONET Configuration Topologies
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REVIEW QUESTIONSREVIEW QUESTIONS
What are the components of the local exchange loop?
11
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REVIEW QUESTIONSREVIEW QUESTIONS
22 What characteristic of a telephone set allows for full-duplex operations?
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REVIEW QUESTIONSREVIEW QUESTIONS
33 What are the BORSCHT functions?
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REVIEW QUESTIONSREVIEW QUESTIONS
44 State the voltages present on the line when a telephone is on-hook and off-hook.
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REVIEW QUESTIONSREVIEW QUESTIONS
55 What is the purpose of companding a PCM-encoded signal?
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REVIEW QUESTIONSREVIEW QUESTIONS
66 What signal characteristics are controlled through line conditioning?
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REVIEW QUESTIONSREVIEW QUESTIONS
77 What is the data rate of a T1 channel?
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REVIEW QUESTIONSREVIEW QUESTIONS
88 Describe the difference between FDM and TDM.
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REVIEW QUESTIONSREVIEW QUESTIONS
99 What is the data rate of an OC1 channel?
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REVIEW QUESTIONSREVIEW QUESTIONS
1010 How many PCM-encoded channels comprise a T1 frame?