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University_of_Southern_California_-_EE_450_EE450-HW3-Fall-2012-solutions.pdf-EE450, Homework #3, Solutions 1. (
##### Page 1
EE450, Homework #3, Solutions
1.
(
5 points
) If we want to send data at a speed of 3000 bps through a channel of
bandwidth 1000 Hz, what is the minimum S/N ratio required?
2.
(
5 points
) What is the maximum bit rate that can be transmitted over a channel
with bandwidth of 500 Hz and a signal to noise ratio of 5 dB?
3.
(
20 points
) Encode 00110010 by (a) NRZ-L, (b) NRZI, (c) Manchester, and (d)
Differential Manchester.
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##### Page 2
4.
(
9 points
) Calculate the signaling rate for the following modulation schemes and
bit rates:
a)
36Kbps, 4-QAM
b)
8Kbps, 32-QAM
c)
3Kbps, 8-PSK
5.
(
9 points
) Calculate the bit rate for the following modulation schemes and
signaling rates
a)
1Kbaud, 32-QAM
b)
c)
5Kbaud, BFSK
6.
(
10 points
) Assume that you are to design a synchronous TDM carrier to support
30 voice channels using 6-bit samples and a structure similar to T1. Determine
the required bit rate.
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##### Page 3
7.
(
15 points
) Find the number of the following devices that could be
accommodated by a T1-type synchronous TDM line if 1% of the line capacity is
reserved for synchronization purposes.
a)
110-bps teleprinter terminals, (
2 points
)
b)
300-bps computer terminals, (
2 points
)
c)
1200-bps computer terminals, (
2 points
)
d)
9600-bps computer output ports, (
2 points
)
e)
64-kbps PCM voice frequency lines. (
2 points
)
How would these numbers change if each of the sources were operational an
average of 10% of the time? In this case, there will be statistical TDM usage
and the line should be utilized at most up to 80% of its capacity. (
5 points
)
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##### Page 4
8.
(
15 points
) Four 1 Kbps devices are to be multiplexed using synchronous TDM.
The multiplexor will take one bit from each source during each cycle. Find
a)
The duration of the bit before multiplexing
b)
The duration of the bit after multiplexing
c)
The duration of the multiplexed frame
d)
The multiplexer bit rate
e)
The multiplexer frame rate.
Duration of bit before MUX = 1/1K = 1 msec
Multiplexer bit rate is 4x1K = 4Kbps and hence the bit duration at output of
MUX is 1/4K = 0.25 msec
Duration of multiplexed frame is 4x0.25msec = 1msec
Frame rate is 1/1msec = 1000 frames/sec
9.
(
12 points
) We have 14 sources, each creating 500 8-bit characters per second.
Since only some of these devices are active at any moment, a statistical TDM,
using character interleaving, is used to aggregate these sources. Each frame
consists of 6 time slots (each time slot will support a character). Four bits of
overhead (address) are added to each character in each time slot.
a)
The number of bits in the multiplexed frame
b)
The multiplexer frame rate
c)
The duration of the multiplexed frame
d)
The multiplexer bit rate
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