Computer Networking • CS 514

Test 5, 12 November 2014

This test has six questions. The test runs the entire class period, starting at 4:30 and ending at 6:20. When you finish the test, you may leave.

Always justify answers unless the question explicitly states it’s not necessary. Some questions require reasonable assumptions in order to be answered. In such cases the answer should clearly state the assumptions made. Answers should be no longer than 75 to 100 words.
  1. Keeping in mind that giving an example of X is not defining X, briefly (in one or two sentences, 25 to 30 words total) define:

    • Bandwidth (Hz)

    • ≈1 ft/nsec

    • Microwaves

    • Simplex Links

    • Harmonics

    Keeping in mind that giving an example of X is not defining X, briefly (in one or two sentences, 25 to 30 words) define:

    • Bandwidth (Hz):
    The range of frequencies over which a signal is transmitted. Bandwidth can be determined artificially, as in the 200 kHz bands used to transmit FM signals, or naturally, as in baseband communication.

    • ≈1 ft/nsec:
    The speed of light in a vacuum.

    • Microwaves:
    A band of the electromagnetic spectrum between radio waves on the low-frequency side and infrared on the high-frequency side.

    • Simplex links:
    A simplex link transmits data in one direction only.

    • Harmonics:
    The coefficients to the sin and cos terms of a Forier series expansion of a periodic signal.

  2. Describe all the signal transitions that can occur in a Manchester-encoded signal. Of the transitions described, which should be ignored by the receiver?

    The Manchester encoding has a transition in the middle of every bit time; these transitions correspond to bit data and can't be ignored at the receiver. A Manchester-encoded signal may also have a transition at the edge of a bit time if the signal is transmitting two consecutive bits of the same value. The edge transition resets the signal to the appropriate level (low or high) to transmit the next bit. The edge transitions should be ignored by the receiver.

  3. The MAC sublayer of the data-link layer can adjust framing to account for the error properties of the physical-layer medium. Assuming a frame structure for average error rates, suggest changes to the frame to account for high-error (say ×1000 larger than average) and low-error (say ×1000 smaller than average) media. You may assume an average error-rate frame looks like an Ethernet frame, or make up your own average error-rate frame.

    Assuming an Ethernet-like frame, in high error-rate environments, reduce the maximum frame size to reduce the probability that any particular frame would be damaged by a errors. The decrease in probability in any particular frame is offset by an increase in the number of frames, but losing smaller frames looses less payload and may provide easier recovery. With a smaller frame, the 32-bit CRC code can be changed to a error-correcting code of approximately the same size, depending on the code, to give the receiver the change to repair damaged frames.

    In low error-rate environments, do the opposite: increase the maximum frame size to reduce framing overhead, and reduce or eliminate the checksum over the frame. If errors are rare, it makes sense to pass the advantage up to the higher layers so they can take advantage of the low-error environment too.

  4. When designing an alert tone, such as that for a telephone, a human-factors guideline suggests that the alert tone be at a frequency that can be received with both ears in any head orientation relative to the tone generator. Under this guideline, what is the maximum frequency acceptable for an alert tone? Justify your answer. Assume 1) a perfectly spherical head of uniform density and 6-inch radius with ears on the equator exactly opposite each other and 2) tone amplitude is equal to tone wavelength. Note this question requires that the tone be received by both ears, which is not the same thing as being heard.

    A spherical head one foot in diameter can absorb any signal up to one foot in wavelength. From fλ = c with λ = 1 ft/cyc, get

    .
    f= (c mi/sec)/(1 ft/cyc)
    = (186,000 mi/sec)/(1 ft/cyc)
    = (186,000 mi/sec)(5280 ft/mi)/(1 ft/cyc)
    = 9.8×108 cyc/sec
    = 980 MHz
    Any frequency up to but not including 980 MHz will span a 1-ft diameter head and be received by both ears.
  5. A colleague of yours is designing a transmission system and misremembers Nyquist’s theorem as B log2 V. Describe the consequence of this mistake on your colleague’s design.

    My colleague mistake was to drop the 2 coefficient to B, the maximum line bandwidth in bits/sec. As a consequence, my colleague's design transmits at half its potential maximum rate.

  6. Researchers at Fumble Industries, Inc. have discovered that wires made with a certain alloy of unobtainium can transmit without loss over arbitrary distances. However, lossless transmission occurs only when the data being transmitted obeys the following property: any three consecutive bits being transmitted has at least one 1 and one 0. As chief communications engineer at Fumble Industries, Inc., you have been given the task of developing channel coding that ensures the three-bit property holds. What is your solution? Make sure you describe your solution with with enough detail so that your underlings, who did not study networking as hard or as well as you did in graduate school, can follow them.

    There are a number of possible solutions to this problem. The first is to define a 2B/1B encoding:

    1B—01—
    2B00011011
    In this encoding, a 1 bit is never more than two bits away from another 1 bit in either direction (left or right), and similarly for a 0 bit.

    The second solution uses bit stuffing. If the channel encoder sees two consecutive bits of the same value, it stuffs a bit of the opposite value after the pair. The trick to this encoding is that the stuffed bit becomes part of the bit stream to be examined by the encoder. For example, the bit string 0011 gets stuffed to 001101 where the green bits have been stuffed. When the decoder observes two consecutive bits with the same value, it tosses the next bit as a stuffed bit, remembering the value of the stuffed bit so it can continue to scan for identical consecutive bit values.

This page last modified on 2014 November 16.