Download Solutions to End of Chapter Problems

Solutions to End of Chapter Problems
1.
For an ASCII ‘Z,’ sketch both the On-Off Keying (OOK) binary waveform (voltage pulses) and the modulated signal,
where the amplitude of the carrier is modulated to either 10 V or 0 V and Tb = 100 ms. Hint: use the ASCII table from
Chapter 1 of the course notes to determine the bits that represent ‘Z’.
‘Z’ is 0x5A in hex, or in binary, 0101 1010
2.
Given this FSK transmission where individual symbols are denoted by vertical lines:
a. Draw the corresponding binary transmission (voltage pulses),
assuming that the higher frequency represents a 1-bit:
b.
Determine the bit rate.
3 bits take 100 μsec, so Tb = 33.333 μsec, and Rb = 1/Tb =
30 kbps
c.
How many bits per symbol could be conveyed if four different
frequencies were used to transmit data instead of two (that is, if 4 symbols were used vice 2 symbols)?
N = log2M = log24 = 2
3.
The following is a BPSK transmission. The dashed vertical lines separate the bits.
On this plot, a binary ‘1’ is represented by this signal:
a.
Determine the transmitted bits. As shown above, 11010010
b.
Determine the bit rate.
4 bits take 10 msec, so Tb = 2.5 msec, and Rb = 1/Tb = 400 bps
c.
What is the bandwidth for this transmission?
BW = 2 Rb = 800 Hz
4.
QAM is a combination of which two types of modulation?
Amplitude Shift Keying and Phase Shift Keying
5.
The “forward” signal transmitted to control a remotely-controlled (RC) car is captured on an oscilloscope and displayed
below.
Answer the following questions with regards to this signal:
a.
This modulation is binary, meaning that there are two possible symbols. What type of digital modulation is
being used?
You might think it is ASK because it looks like there are 2 different amplitudes of the carrier…but
actually it is OOK with noise in the signal
b.
What is the bit rate? (Hint: Two time cursors are shown on the display as the two dashed vertical lines…these
cursors isolate a single bit. Also shown are some measurements about the time cursors below the plot and to the
μs is the difference in time between the two cursors).
Tb = 496 μsec, and Rb = 1/Tb = 2016 bps
c.
What bit sequence is represented by the O-scope display?
0 111 0 111 0 111 0 111 01 01 01 01 01 01 01 01 01 01 0 111 0
6.
16-QAM can be used for higher data rate transmissions.
a.
How many bits are transmitted with each symbol?
N = log2M = log216 = 4
b.
If 4 different phases and 4 different amplitudes are used in a 16-QAM modulation system, sketch a constellation
diagram that could be associated with the system (you do not need to label the bits for each symbol, just show
the symbols).
Two possible constellations:
c.
If 8 different phases and 2 different amplitudes are used in a 16-QAM
modulation system, sketch a constellation diagram that could be
associated with the system (you do not need to label the bits for each
symbol, just show the symbols).
One possible constellation:
d.
If the bit rate associated with either of these 16-QAM systems was 1.2 Mbps, what is the bandwidth of the
transmission?
BW = 2 Rb / N = 2 (1.2x106)/4 = 600 kHz
7.
For a given bandwidth system, what is the advantage and disadvantage of using a multi-symbol encoding scheme (that
is, using more than 2 symbols)?
Advantage: lower bandwidth required to transmit a given bit rate, or if bandwidth is fixed, you can transmit
a higher bit rate
Disadvantage: Symbols are closer together making errors in decoding more likely. Also, system is more
complex.
8.
9.
A communication system transmits 100 kbps. For each of the following modulation types, determine the bandwidth of
the transmission.
a.
FSK, with frequency deviation 200 kHz. BW = 2 (Δf + Rb) = 2 (200x103 + 100x103) = 600 kHz
b.
OOK.
BW = 2 Rb / N = 2 (100x103)/1 = 200 kHz
c.
QPSK.
BW = 2 Rb / N = 2 (100x103)/2 = 100 kHz
d.
16-PSK.
BW = 2 Rb / N = 2 (100x103)/4 = 50 kHz
e.
16-QAM.
BW = 2 Rb / N = 2 (100x103)/4 = 50 kHz
f.
512-QAM.
BW = 2 Rb / N = 2 (100x103)/9 = 22.2 kHz (Note: N = log2M = log2512 = 9)
Suppose the FCC has leased you the portion of the frequency spectrum from 1.2 MHz to 1.3 MHz for your free-space
communication system. What is the maximum bitrate you could obtain if you used the following modulation schemes:
For this problem, bandwidth = 1.3 MHz – 1.2 MHz = 100 kHz
a.
FSK, with fmark = 1.27 MHz and fspace = 1.23 MHz.
For FSK, BW = fmark – fspace + 2 Rb, so Rb = (BW – (fmark – fspace)) / 2
Rb = (100x103 – (1.27 x106 –1.23 x106)) / 2 = 30 kbps
For the following modulation types, BW = 2 Rb / N, so Rb = N x BW / 2
b.
ASK.
Rb = N x BW / 2 = 1 x 100x103/2 = 50 kbps
c.
BPSK.
Rb = N x BW / 2 = 1 x 100x103/2 = 50 kbps
d.
8-PSK.
Rb = N x BW / 2 = 3 x 100x103/2 = 150 kbps (Note: N = log2M = log28 = 3)
e.
32-QAM.
Rb = N x BW / 2 = 5 x 100x103/2 = 250 kbps (Note: N = log2M = log232 = 5)
f.
256-QAM.
Rb = N x BW / 2 = 8 x 100x103/2 = 400 kbps (Note: N = log2M = log2256 = 8)