ELEC166 Tutorial Week 10 Solutions
Problem 1.
As the length of an optical fibre is increased, the total loss increases, while the
bandwidth decreases. Explain the physical reasons why these effects occur.
Answer:
The attenuation or loss in the fibre occurs because some light is
absorbed as it travels through the fibre; that is, the fibre is not perfectly
transparent. As the light travels further, a greater fraction of it will have been
absorbed, so the loss is greater.
The bandwidth decreases due to dispersion in the fibre, which means that a
single pulse of light entering the fibre will be “smeared out” in time when it
arrives at the other end. This can be due to a number of factors, including the
presence of multiple modes of propagation (which can be thought of as slightly
different paths) in the fibre, or the speed of propagation being different for
slightly different wavelengths or colours (chromatic dispersion). Higher
frequencies (that is, faster variations in light intensity) will suffer more than
lower frequencies, so the bandwidth will be limited. For a longer fibre, the
amount of time smearing will be greater, meaning that the same degree of
impairment will occur at even lower frequencies, so the bandwidth will be
reduced.
Problem 2.
An optical fibre with a length of 1 km has a bandwidth of 200 MHz. What
bandwidth would you expect for the same type of fibre over a 15 km length? What
would be the maximum length of fibre you could use and still get a bandwidth in
excess of 30 MHz?
Answer:
The product of the bandwidth and the distance should be constant,
which in this case is 1 km × 200 MHz = 200 MHz-km. Hence for a distance of 15
km, the bandwidth should be 200/15 = 13.3 MHz. For a bandwidth of ≥30 MHz,
the maximum fibre length would be 200/30 = 6.7 km.
Problem 3.
Very briefly explain, with the aid of a diagram, the basic principle of amplitude
modulation, and give two examples of applications which use AM.
Answer:
Refer to the chapter on modulation in the lecture notes. There are
plenty of examples, including AM radio, IR remote controls, TV picture (but not
sound). You should have a diagram which looks something like figure 13-2 in the
notes.
Problem 4.
Sketch as accurately as you can a graph of voltage as a function of time showing
a carrier at a frequency of 100 kHz, amplitude modulated by a 10 kHz sine wave,
with a modulation index (m) of 0.4. On the graph indicate how the value of m
may be measured.
Answer:
Given that m = 0.4, we need to work out maximum and minimum
amplitudes for the carrier. The equation for AM modulation index
V max − V min
V min 1 − m
can be rearranged to give
=
, so that for m = 0.4,
m=
V max + V min
V max 1 + m
V min 1 − 0.4
=
= 0.43 , so that the minimum carrier amplitude will be 0.43 of its
V max 1 + 0.4
1
Vmax
Vmin
0.5
0
-0.5
-1
maximum. Vmax and Vmin should be indicated on the graph, and the equation
for m shown. The graph could look something like this:
Problem 5.
Explain why an audio signal with a bandwidth of 9 kHz requires a total
bandwidth of 18 kHz to be transmitted via an amplitude modulated (AM) carrier.
Include a carefully labelled diagram of the frequency spectrum of an AM signal in
your answer.
Answer:
The spectrum of a carrier amplitude modulated by a 9 kHz sine
wave (the highest frequency present in the audio signal) looks like this:
The total range of frequencies occupied by the signal is thus 9 + 9 = 18 kHz.
Clearly, this will also cover all audio frequencies up to 9 kHz. The spectrum of
the carrier modulated by an audio signal with 9 kHz bandwidth might look like
this:
carrier
18 kHz
Problem 6.
A carrier at 200 MHz is frequency modulated by a sine wave signal at a
frequency of 15 kHz. The peak carrier deviation is (±) 50 kHz.
•
•
•
What is the value of the modulation index, m?
Approximately what total bandwidth is needed to transmit the signal?
Draw a sketch to show the general idea of the spectrum of the FM sidebands
for the modulated carrier. Indicate (and give a value for) the spacing between
sidebands in frequency.
Answer:
(i) The modulation index is given by
modulation index (m) =
peak carrier deviation ( ∆f) 50kHz
=
= 3.33
modulating frequency (f m ) 15kHz
(ii) The approximate total bandwidth required is:
2 [ m + 1 ] fm Hz = 2 × (3.33 + 1) × 15kHz = 130 kHz.
(iii) The diagram should look something like this, with a spacing between
adjacent sidebands of 15 kHz.
Total bandwidth
required ~ 130 kHz
Carrier at 200 MHz
15 kHz spacing
Problem 7.
Sketch the frequency spectrum (say from about 90 kHz to 110 kHz) of a 100 kHz
carrier which is amplitude modulated by a 5 kHz sine wave.
• What total bandwidth is required to transmit this signal?
• If the carrier had been frequency modulated instead, very briefly explain
why the total bandwidth required would be different.
• To calculate the total bandwidth required in the FM case, what additional
information would you need?
Answer: The spectrum will look like that of problem 5, but centred at 100 kHz,
and with sidebands spaced 5 kHz from the carrier:
• The total bandwidth required is just twice the modulating frequency (fm) = 2 ×
5 kHz = 10 kHz.
• Frequency modulation usually produces more than just two sidebands, but
the sidebands still have a spacing between them equal to the modulating
frequency (5 kHz in this case), so a greater total bandwidth is required.
• The number and strength of sidebands depends on the modulation index
(m), so you would need to know m, or the peak carrier frequency deviation ∆f,
so that you could calculate m.
Problem 8.
Sketch the frequency spectrum (near 1.5 MHz) of a 1.5 MHz carrier which is
amplitude modulated by a music signal (i.e. not just a sine wave) with about 10
kHz bandwidth.
• What are the maximum and minimum frequencies that would be present in
the modulated carrier?
• Would an FM signal require more or less total bandwidth for transmission?
• To calculate the total bandwidth required in the FM case, what additional
information would you need?
Answer: Refer to the answers for problems 5 and 7, except that:
• The spectrum is centred at 1.5 MHz, so that the minimum and maximum
frequencies present are 1500 – 10 = 1490 kHz and 1500 + 10 = 1510 kHz.
• The total bandwidth required is 2 × 10 kHz = 20 kHz.
Problem 9.
Briefly discuss what determines the bandwidth required to transmit a carrier
which has been (i) amplitude modulated, or (ii) frequency modulated by a lowfrequency signal. As part of your answer, you should draw two carefully labelled
graphs showing the respective frequency spectra, where the modulating signal in
each case is a 10 kHz sine wave.
(Note: you do not need to accurately reproduce all the details of the FM spectrum
which depend on the modulation index, but the important features should be
correct.)
Answer:
(See answers for problems 5, 6, 7 etc..)
Problem 10.
Sketch:
• The waveform in time of about 20 cycles of a 100 kHz carrier which has been
amplitude modulated by a 10 kHz square wave, with a modulation index of 1.
• The frequency spectrum from 40 kHz to 160 kHz of the same modulated
signal.
(Note: You will probably need to refer to chapter 4 of your course notes to remind
you of the properties of square waves).
Answer:
There are 10 cycles of carrier for each cycle of modulating square
wave, so that the modulated carrier will look like this (note that since m = 1, the
carrier drops is zero for every second half-cycle).:
The modulating square wave contains not just one frequency, but a fundamental
at 10 kHz and all odd harmonics (see notes in chapter IV), so that its spectrum
should look like this:
40
50
60
70
80
90 100 110 120 130 140 150 160
frequency (kHz)
Tut_Week9_soln.doc rev 27/01/2004 2:12 PM