Chapter 03 INSY 3304

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University of Texas, Arlington *

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3304

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Electrical Engineering

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Apr 3, 2024

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Chapter 03 1. Describe the three types of data flows. - Simplex : Communication occurs in one direction only. Example: TV broadcasting. - Half-Duplex : Communication happens in both directions, but not simultaneously. Example: Walkie-talkies. - Full-Duplex : Communication occurs in both directions simultaneously. Example: Modern Ethernet networks. 2. How does bipolar signaling differ from unipolar signaling? Why is Manchester encoding more popular than either?  - Bipolar Signaling : n bipolar signaling, three voltage levels are used to represent binary data. It typically employs three voltage levels: positive, negative, and zero. The "zero" voltage level represents binary 0, while positive and negative levels represent binary 1 alternately. - Unipolar Signaling : In unipolar signaling, only one voltage level is used to represent binary data. It typically uses a single voltage level to represent both 0 and 1. For example, a positive voltage may represent binary 1, and zero voltage represents binary 0. Unipolar signaling is simpler but can suffer from DC offset issues since the average voltage is not zero. This can lead to problems with long- term signal integrity. - Manchester encoding is more popular than either because it overcomes the limitations of both bipolar and unipolar signaling by ensuring regular transitions in the signal, aiding clock recovery, and eliminating DC offset issues. It's widely used in digital communication protocols for its reliability and self-clocking properties. 3. Explain how pulse code modulation (PCM)works.  Pulse Code Modulation (PCM) works by sampling an analog signal at regular intervals, quantizing the samples into discrete amplitude levels, encoding them into binary numbers, transmitting or storing the binary representation, decoding it at the receiver, and reconstructing the original analog signal with the help of a reconstruction filter. 4. Draw how the bit pattern 01101100 would be sent. Using A. Single-bit FM B. Two-bit AM (i.e., four amplitude levels) C. Single-bit AM combined with single-bit PM

Chapter 04 1. Describe three approaches to detecting errors, including how they work, the probability of detecting an error, and any other benefits or limitations. Parity Bit: • How it works: Adds an extra bit to ensure even or odd number of set bits. • Probability of detecting an error: Can detect single-bit errors. • Benefits: Simple and adds minimal overhead. • Limitations: Cannot correct errors, only detect some of them. Checksums : • How it works: Involves summing data bytes and appending the result. • Probability of detecting an error: Can detect various errors, depending on the algorithm used. • Benefits: Relatively simple and can catch a range of errors. • Limitations: Cannot correct errors, and some error patterns may go undetected. Cyclic Redundancy Check (CRC): • How it works: Employs polynomial division to calculate and append a remainder (CRC value). • Probability of detecting an error: Highly effective, especially for burst errors. • Benefits: Very reliable and can detect a wide range of errors. • Limitations: More complex to implement than parity or checksums and cannot correct errors, only detect them. 2. Compare and contrast stop-and-wait ARQ and continuous ARQ. Stop-and-Wait ARQ :  Transmits one frame at a time and waits for acknowledgment.  Provides implicit flow control.  Simple but less efficient for high-speed channels.

Continuous ARQ (e.g., Go-Back-N or Selective Repeat ARQ) :  Allows multiple frames to be sent before waiting for acknowledgments.  Provides more efficient channel utilization.  Slightly more complex to implement than Stop-and-Wait. 3. How efficient would a 6-bit code be in asynchronous transmission if it had 1 parity bit, 1 start bit, and 2 stop bits? (Some old equipment uses 2 stop bits.) For Modern Equipment (2 Stop Bits): Total Bits per Character = Data Bits + Parity Bit + Start Bit + Stop Bits Total Bits per Character = 6 bits (data) + 1 bit (parity) + 1 bit (start) + 2 bits (stop) = 10 bits Now, use the efficiency formula: Efficiency (%) = (6 data bits / 10 total bits) * 100 = 60% For Older Equipment (1 Stop Bit): Total Bits per Character = Data Bits + Parity Bit + Start Bit + Stop Bits Total Bits per Character = 6 bits (data) + 1 bit (parity) + 1 bit (start) + 1 bit (stop) = 9 bits Now, use the efficiency formula: Efficiency (%) = (6 data bits / 9 total bits) * 100 ≈ 66.67%

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