Name: __Jeff Absher_____________
TDC462
DATA COMMUNICATIONS
Mid-Term Assignment
July 3, 2002
Wednesday, 9:15pm
Total
Overall Grade Points
Total Points
100
20
Points
Remarks:
1. Write clearly and eligibly, unreadable answers will be marked wrong.
2. Clearly state any assumptions that you make.
3. You must show all of your work, including all the intermediate steps of a calculation,
to get ANY credit.
4. Submit your answers via DLWeb only.
5. It is recommended that you print a copy of your answers before turning it in as this is
what the grader is going to do. Check to make sure that it looks exactly how you want
it to. This will help you minimize formatting errors.
(10 Points) 1. Explain how the standardization process of the Internet-related
protocols function. Draw a diagram to show the steps involved and include the names
of the various documents used at different stages of that diagram.
An Internet draft is submitted to IESG the Internet Engineering Steering Group.
They will approve it as an RFC (request for comments) and place it as either an
experimental standard or a proposed standard. If it is an experimental standard, it
can be reworked and resubmitted to become a proposed standard, or it may move
directly into the Historic state if it is decided that it is not a good idea.
The RFC will be a proposed standard for a minimum length of time (6 months) as
the community reviews it and implements it. After 2 parties have independently
implemented the proposed standard and have shown that it works, it can be
promoted to a Draft Standard if the IETF Internet Engineering Task Force
recommends promotion and the IESG ratifies the promotion.
The proposal must remain as a Draft Standard for 4 months. Again this is time for
the community to review it. If the IETF and IESG recommend and agree on
promotion, after this time the proposal will be promoted to an actual Internet
Standard and will receive a STD number.
The standard may become obsolete and then moves into the Historic state.
(10 Points) 2. A PCM encoder accepts a signal with a full range of values from –4V
to +4V and generates 4-bit codes using uniform quantization.
1) What is the quantization step size?
+4 - -4 = 8V difference.
24=16 quanta levels
8 V/16 steps = 0.5 V/step
2) What is the greatest possible quantization noise in volts that can occur on any
single PAM pulse?
0.5 V/ 2 = 0.25 V
3) What is the quantization noise of SNR?
6.02 (4) + 1.76 db = 25.84dB
(10 Points) 3. Consider the following diagram:
A
12
B
5
C
A power source is located at Point A. The power of the signal received at point C is
0.0378W. Point B is located 12 miles from Point A. At Point B, there is an amplifier that
increases the signal strength by 35dB. Point C is located 5 miles from Point B. What is
the power of the signal produced at Point A? Assume a line loss of 3dB per mile.
-15 = 10 log(0.0378/PBout)
35 = 10 log(PBout/PBin)
-36 = 10 log(PBin/PAout)
10-1.5=0.0378/x … PBout = 1.195
103.5=1.195/x …PBout= 0.000378
10-3.6=0.000378/x …PAout = 1.505 W
A channel has a data rate of 64Kbps and a propagation delay of 10ms. What is the
minimum frame size that gives an efficiency of at least 35% for a sliding window
flow control with a window size of 5?
.35 <= 5/(2a + 1)
.35 <= 5/(2(1e-3/tframe) + 1)
tframe = 1.51e-4 secs
1.51e-4secs/frame* 64000 bits/sec= 9.6344 bits/frame
10 bits/frame
(10 Points) 4. Define quantization noise and slope overload noise in delta
modulation.
In delta modulation, slope overload noise is the noise (difference between actual
signal and the transmitted signal) that occurs when the magnitude of the rate of
change of the actual signal (|Δx/Δt|) is greater than the rate of change allowed by
the step function (|δ/Ts|).
Quantization noise is due to the fact that at each new Ts a step (up or down) in
the transmitted value must occur and the transmitted value can only be one of
two possible values “around” the actual value of the sample signal. Quantization
noise occurs when the input signal and the output signal are less than δ different.
Explain the relationship between the following terms in delta modulation:
a) Quantization step size and quantization noise
quantization noise is always < quantization step size and the mean
of quantization noise should always be < ½ the quantization stepsize. Hence the larger the step-size, the larger the quantization
noise
b) Quantization step size and the slope overload noise.
The larger the quantization step size, the less slope overload noise
there will be on the system because the transmitted wave can more
quickly adapt to large changes in the sampled wave. But due to a)
increasing the step-size has its drawbacks as well.
(10 Points) 5. Suppose that a modem can operate at 30.42Kbps over telephone lines
whose baud rate is 3380. What is the minimum number of signal levels that this
modem must have been designed to handle?
30420 bits/1 sec * 1 sec/3380 signals = 9 bits/signal
log2(9) = 3.1 = 4 levels
(10 Points) 7. A data source generates 7-bit IRA characters. If the data rate on the
channel is 45Mbps, give the actual user data rate for the following systems:
a) Asynchronous transmission, with a 2-unit stop bits and a parity bit
Asynch always implies a start bit!
therefore (1+ 7 + 1 + 2) bits/char
45000 bits/1 second * 1 char/11 bits = 4090.1 chars/sec
4090.1 chars/1 sec * 7 user bits/ 1 char = 28636.4 user bits/sec
b) Synchronous transmission, with a frame consisting of 88 control bits and 2048 bits of
payload. The payload contains 8-bit (parity included) IRA characters. Flags are assumed
to be part of the control bits and included in the 88 bits.
45000 bits/1 second * 1 frame/ 2136 bits * 256 chars /1 frame* 7 user bits/1 char
= 37752.8 user bits/sec
(10 Points) 8. Compare and contrast NRZI and Differential Manchester encoding
schemes. Include a comparison of their signal transition rates under worst-case bit
sequence, best–case bit sequence and a bit sequence of alternating binary values
(101010) for the data to be transmitted.
NRZI Best-case = all zeros = 0 transitions/bit
NRZI Worst-case = all ones = 1 transition/bit
NRZI alternating = HHLLHHLL = 0.5 transitions/bit
Diff Man Best-case = all ones = 1 transition/bit
Diff Man Worst-case = all zeros = 2 transitions/bit
Diff Man alternating = 1.5 transitions/bit
The key is that Manchester and Differential Manchester have a transition in
the middle of the bit for each bit which helps with synchronization while
NRZ/NRZI do not have that transition. The clocking can also be used for error
detection and there is a net DC component of 0 in biphase. The negative is that
bandwidth for the biphase family of encoding schemes needs to be higher
because of the excess transitions.
(10 Points) 9. An original message of M = 11100011 is to be transmitted by using
CRC error detection scheme with a pattern of P = 110011. Find the bit string that
will be transmitted, i.e. message + FCS.
11100011 * 25 = 1110001100000
1110001100000/110011 get remainder..
Remainder = 10110
1110001110110 will be transmitted.
(10 Points) 10.
a) Assuming the
phase as zero, write the function for the solid curve in the figure below
in the form A.sin(2ft+).
2.5 sin(25t/2+0)
b) Write the dotted curve below in the form A.sin(2ft+) by considering the solid line
as the reference signal for the phase.
2 sin(21t/2+)