Solutions to End of Chapter Problems
1.
(a) Calculate the wavelength of signals with frequencies of 1.5 kHz, 18 MHz, and 22 GHz.
(b) Since an antenna that is needed to transmit these frequencies must be at least a tenth of the wavelength, which signal
frequency would NOT be practical for direct (i.e., baseband) transmission?
The 1500 Hz signal would require an antenna that is too long to be practical.
(c) Name and define a technique that could be used to transmit the frequency in part (b).
Modulation could be used to upshift the frequency, thus requiring a shorter antenna. In modulation, the
information signal is used to manipulate the amplitude, frequency or phase of a high frequency carrier. This
action increases the frequency to be transmitted.
2. An AM signal is comprised of the following two signals:
vm(t) = 80 cos (2π5000t) volts
vc(t) = 100 cos (2π800,000t) volts
where vm(t) is the message and vc(t) is the “unmodulated carrier” (i.e., the output of the modulator when
information signal is present).
no
(a) Find the carrier frequency, the upper-sideband and lower-sideband frequencies, and the percent modulation (m).
fc = 800 kHz (given). fm = 5 kHz (given). fUSB = fc + fm = 805 kHz, fLSB = fc - fm = 795 kHz.
m(%) = m x 100% = 80%.
m=Vm/ Vc = 80/100 = 0.8.
(b) Suppose vm(t) changes to 120 cos(2π5000t). Find the new percent modulation (m). Give the technical term for this
condition and explain the effects of this condition occurring.
m=Vm/ Vc = 120/100 = 1.2. This is overmodulation, which causes a distortion in the transmitted AM signal’s
envelope. Since the information is carried in the envelope, the information that the receiver recovers is
distorted.
3.
A radio station, 1280AM, is conducting a monthly test of the Emergency Alert System. The test begins with an
annoying sound comprised of two pure tones at 853 Hz and 960 Hz. The signal being broadcast has exactly five
frequency components, i.e., the signal could be written as follows:
vAM(t) = V1 sin(2π f1 t) + V2 cos(2π f2 t) - V3 cos(2π f3 t) + V4 cos(2π f4 t) - V5 cos(2π f5 t) volts
(a) Find the five frequencies that comprise the AM signal being broadcast. Recall that the carrier frequency and the two
sideband frequencies for each of the emergency alert tones will be involved.
The carrier is always one of the frequencies transmitted (1280 kHz). In addition, for each alert tone (fA = 853
Hz, fB = 960 Hz), there are two frequencies transmitted: fc + fA, fc – fA, fc + fB, and fc – fB. So the five
frequencies transmitted are:
1280 kHz, 1280 kHz + 853 Hz = 1280.853 kHz, 1280 kHz – 853 Hz = 1279.147 kHz
1280 kHz + 960 Hz = 1280.960 kHz, and 1280 kHz – 960 Hz = 1279.040 kHz.
(b) Find the bandwidth for this particular broadcast.
BW = 2 fmax = 2 (960 Hz) = 1920 Hz. Or, BW equals the highest frequency transmitted minus the lowest
frequency transmitted, so BW = 1280.960 kHz – 1279.040 kHz = 1920 Hz.
(c) Determine which of these two emergency alert tones (853 Hz or 960 Hz) determines the bandwidth.
fmax = 960 Hz (the maximum frequency in the information signal).
(d) What is the bandwidth assigned to a commercial AM radio station in the United States?
For commercial AM, BW = 10 kHz.
4.
Musical notes can be viewed as pure tones (if we ignore the “warmth” added by any particular instrument). Pure tones
are signals that contain only one frequency. Chords are combinations of notes, such as the C-Major chord on the piano,
comprised of notes C, E, and G.
If the radio station 1280AM broadcasts the C-Major chord, it would broadcast the following seven frequencies, listed in
ascending order and annotated by note and sideband:
f LSB-G
f LSB-E
f LSB-C
f carrier
f USB-C
f USB-E
f USB-G
= 1,279,608 Hz
= 1,279,670 Hz
= 1,279,738 Hz
= 1,280,000 Hz
= 1,280,262 Hz
= 1,280,330 Hz
= 1,280,392 Hz
Notice that in the lower sideband, the notes are in reverse order. G, the highest pitch in the chord, is always the
farthest away from the carrier frequency. The carrier frequency is exactly in the middle. Assume that the carrier
amplitude is 100V, and the voltages for the three musical notes are all 20V.
(a) Sketch this broadcast in the frequency domain (label frequencies and amplitudes).
Note: since each musical note has an amplitude of 20V, when modulated, each sideband tone will have an
amplitude of 10V
(b) After demodulation, what frequencies would be heard coming out of a your AM radio’s speaker?
For each musical note (frequencies fC, fE and fG), there are two frequencies transmitted, one in the upper
sideband and one in the lower sideband. So if you just look at the upper sideband, we have fcarrier + fC =
1,280,262 Hz, fcarrier + fE = 1,280,330 Hz and fcarrier + fG = 1,280,392 Hz. Since fcarrier is given as 1.280 MHz, this
means that the three musical notes have frequencies of fC = 262 Hz, fE = 330 Hz and fG = 392 Hz.
(c) Find the bandwidth of the broadcast and determine which note (C, E, or G) sets the bandwidth.
The max frequency in the information signal (musical notes) is 392 Hz, so BW = 2 fmax = 784 Hz.
5. For the following plots of AM signals, determine Vmax, Vmin, Vc, Vm and m. Show your work!
(a)
Vmax =13 V (from plot), Vmin =7 V (from plot), Vc = (Vmax + Vmin)/2 = (13+7)/2 = 10V, Vm = (Vmax - Vmin)/2 = (137)/2 = 3 V, and m = Vm/Vc = 3/10 = 0.3.
(b)
Vmax =18 V (from plot), Vmin =2 V (from plot), Vc = (Vmax + Vmin)/2 = (18+2)/2 = 10V, Vm = (Vmax - Vmin)/2 = (182)/2 = 8 V, and m = Vm/Vc = 8/10 = 0.8.
(c)
Vmax =11 V (from plot), Vmin =9 V (from plot), Vc = (Vmax + Vmin)/2 = (11+9)/2 = 10V, Vm = (Vmax - Vmin)/2 = (119)/2 = 1 V, and m = Vm/Vc = 1/10 = 0.1.
(d)
Vmax =16 V (from plot), Vmin =4 V (from plot), Vc = (Vmax + Vmin)/2 = (16+4)/2 = 10V, Vm = (Vmax - Vmin)/2 = (164)/2 = 6 V, and m = Vm/Vc = 6/10 = 0.6.
6.
Determine fc and fm for any of the AM signals in problem 5 (parts (a)-(d)). (Hint: fc and fm are the same for each case).
Using these values of fc and fm, along with your answers to problem 5, sketch the frequency content for each of these AM
signals (parts (a)-(d)).
Using the plot from part (b) of problem 5. Depending on how closely you’re able to read the axes on the plot,
your answer will probably vary somewhat. From the plot,
fc = 1/Tc = 1/0.25 msec = 4 kHz
fm = 1/Tm = 1/2.2 msec = 455 Hz
7.
For any of the plots shown in problem 5, determine the bandwidth of the transmission. Note that each of these AM
signals has a single sinusoid as the information signal. (Hint: the bandwidth is the same for each case).
Using the plot from part (b) of problem 5, find the period of the information signal.
From the plot, Tm = 2.2 msec, so fm = 1/Tm = 1/2.2 msec = 455 Hz. For AM, BW = 2 fmax. In this problem, fmax is
the same as fm since the information signal is only a single sinusoid. Therefore,
BW = 2 fmax = 2(455 Hz) = 910 Hz.