-
Problem 19.1
Solution: Known quantities:
The square wave signal: ,
-
b. Frequency spectrum of the signal for %50= is shown in Figure
19.xx and for %30= is shown in Figure 19.xx
5000 4000 3000 2000 1000 0 1000 2000 3000 4000 5000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mag
nitu
de
Freq (Hz)
5000 4000 3000 2000 1000 0 1000 2000 3000 4000 5000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mag
nitu
de
Freq (Hz)
19.2
-
Problem 19.2
Solution: Known quantities: Functional form of a full-wave
rectified sinusoidal signal of time period T |)sin(|)( sec, ttx o=
, and natural
frequency srad 2000 = .
Find: a. The Fourier series coefficients. b. Frequency spectrum
of the signal. Analysis: The rectified sine wave signal is an even
function. Hence, we need to compute only the coefficients of the
Fourier series.
na
[ ]
++=
=
+=
=
=
=
=
==
=
===
=
==
)6cos(652)4cos(15
2)2cos(3214)(
0odd :n 0
even :n )1)(1(
241524
0324
)()sin()3sin(214
)()2cos()sin(4)()2cos(|)sin(|2
0
)()2sin(214)()cos()sin(4)()cos(|)sin(|2
4
)()sin(2)(|)sin(|1
000
4
3
2
0
0
2
02
1
00
2
01
0
0
2
00
ttttx
b
nna
a
a
a
tdtt
tdtttdtta
a
tdttdtttdtta
a
tdttdta
n
n
ooo
oooooo
oooooooo
oooo
b. The Frequency spectrum for the full wave rectified sinusoid
of frequency srad 2000 = is shown in
Figure 19.xx.
19.2
-
800 600 400 200 0 200 400 600 800
2
4
6
8
10
12
Mag
nitu
de
Freq (Hz)
Problem 19.3
Solution: Known quantities: Functional form of a full-wave
rectified cosine wave of time period T |)sin(|)( sec, ttx o= , and
natural
frequency srad 1500 = .
Find: a. The Fourier series coefficients. b. Frequency spectrum
of the signal. Analysis:
The rectified cosine wave signal is same as the rectified sine
wave with a phase shift of rad 2
. The Fourier series
coefficients are to be found over a period of rad. If we
consider the period from rad 2
3 to2
, the analysis is
same as a full wave rectified sinusoid. Hence, the Fourier
series coefficients are the same as computed in Problem 19.2.
4
)()sin(2)(|)cos(|1
0
23
2
2
00
=
==
a
tdttdta oooo
19.2
-
[ ]
++=
=
+=
=
=
==
=
===
)6cos(652)4cos(15
2)2cos(3214)(
0odd :n 0
even :n )1)(1(
24324
)()sin()3sin(214
)()2cos()sin(4)()2cos(|)cos(|2
0
)()sin(214)()cos()sin(4)()cos(|)cos(|2
000
2
23
2
23
2
2
02
1
0
23
2
2
01
ttttx
b
nna
a
tdtt
tdtttdtta
a
tdttdtttdtta
n
n
ooo
oooooo
oooooooo
b. The Frequency spectrum for the full wave rectified cosine
wave of frequency srad 1500 = is shown in
Figure 19.xx.
800 600 400 200 0 200 400 600 800
2
4
6
8
10
12
Mag
nitu
de
Freq (Hz)
19.2
-
Problem 19.4
Solution: Known quantities: Functional form of a cosine burst as
shown in Fig. 19.xx and mathematically defined as:
)cos()( tT
tx =
Find: Fourier series coefficients for the cosine burst.
Analysis: The Fourier series coefficients can be calculated as
follows:
)12)(12(2)1(
2)12(sin
)12(1
2)12(sin
)12(11
)12(sin)12(
)12(sin)12(
1
)12(cos)12(cos212
)2cos()cos(2
2
)sin(1)cos(1)(1
1
2
2
2
2
2
2
0
2
2
2
2
0
+=
+
+
+=
+
+
+=
+
+=
=
=
===
+
nna
nn
nn
a
tTn
nTt
Tn
nT
T
dttTnt
Tn
T
dttTnt
TTa
a
tT
TT
dttTT
dttxT
a
nn
n
T
T
T
T
T
Tn
T
TT
T
T
Since the cosine burst is an even signal the coefficients are
all zero. Hence nb 0bn = . The cosine burst signal can be written
as:
+
+= t
Tt
Ttx
4cos
1522cos
322)(
Problem 19.5
Solution: Known quantities: Functional form of a triangular
pulse signal as shown in Fig. 19.xx and mathematically defined
as:
19.2
-
( ))()(||1)( TtuTtuTtAtx +
=
Find: a. Fourier transform of the function. b. Plot the
frequency spectrum of the triangular pulse of period, T 0.01 = sec
and amplitude, . 0.5 A =Analysis: The mathematical equation for the
triangular pulse can be split into a function defined over
different periods as follows:
-
800 600 400 200 0 200 400 600 800
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5x 10
3
Mag
nitu
de
Freq (Hz)
Problem 19.6
Solution: Known quantities: Functional form of a exponential
pulse signal as shown in Fig. 19.xx and mathematically defined
as:
=
0for,(exp)-0for ,00for ,(exp)
)(ttt
txat
at
Find: Fourier transform for the exponential pulse. Analysis: We
can formulate a compact notation for the pulse signal by using
signum function which equals +1 for positive time and 1 for
negative time. This function is defined as:
=0for,1-0for ,00for ,1
)sgn(ttt
t
The signal can be written as: )(tx)sgn(|)|exp()( ttatx =
The Fourier transform is now calculated as follows:
= dtftjttafX )2exp()sgn(|)|exp()(
19.2
-
222
0
0
44)(
21
21
))2(exp())2exp((
fafjfX
fjafja
dttfjadttfja
+
=
++
=
++=
b. The Frequency spectrum for the signal is shown in Figure
19.xx.
80 60 40 20 0 20 40 60 80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Mag
nitu
de
Freq (Hz)
Problem 19.7
Solution: Known quantities:
)()2cos(exp(-at))( tutftx c=Functional form of a damped sinusoid
signal as shown in Fig. 19.xx and mathematically defined as: Find:
Fourier transform for the damped sinusoid.
19.2
-
Analysis:
=
=
dtftjtutfat
dtftjtxfX
c )2exp()()2cos()exp(
)2exp()()(
[ ]
{ }
++
++
=
++++=
+=
)(21
)(21
21)(
))](2[exp())](2[exp(21
)2exp()2exp()2exp(21)exp(
0
0
cc
cc
cc
ffjaffjafX
dttffjatffja
dtftjtfjtfjat
b. The Frequency spectrum from Matlab is shown in Figure
19.xx.
200 150 100 50 0 50 100 150 200
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Mag
nitu
de
Freq (Hz)
Problem 19.8
Solution: Known quantities:
Functional form of an ideal sampling function of frequency
Hz0T
1 as shown in Figure 19.xx and having
mathematical equation:
=
=m
T mTt )( 00
19.2
-
Find: a. The Fourier transform for the periodic signal. b.
Frequency spectrum of the signal for T sec 0.01 o = . Analysis: In
a limiting sense, Fourier transforms can be defined for periodic
signals. Therefore, it is reasonable to represent that a periodic
signal can be represented in terms of a Fourier transform, provided
that this transform is permitted to include delta functions. An
ideal sampling function consists of an infinite sequence of
uniformly spaced delta functions. We observe that the generating
function for the ideal sampling function is simply a delta function
)(t . The periodic signal can be represented in terms of the
complex exponential Fourier series:
=
=m
n tnfjcT )2exp(00
where is the complex Fourier series coefficients defined as:
nc
)(
)2exp()(1
00
00
nfGf
dttnfjtT
cn
=
=
whereG is the Fourier transform of )( 0nf )(t evaluated at the
frequency nf . For the delta function: 0n allfor 1)( 0 =nfG
Therefore, using the relation for Fourier transform pair for a
periodic signal with a generating function
and period T : )(
0tgT )(tg
0
=
=
nm
nffnfGfmTtg )()()( 0000
We get the Fourier transform pair for the ideal sampling
function as:
=
=
nm
nfffmTt )()( 000
We can see that the Fourier transform of a periodic train of
delta functions, spaced T seconds apart, consists of
another set of delta functions weighted by a factor
0
00
1T
f = and regularly spaced Hz apart along the frequency
axis.
0f
Problem 19.9
Solution: Known quantities: Functional form of the modulating
signal m , the carrier signal , and the modulation index )(t )(tc
.
)2cos()()2cos()(tfAtmtfAtc
mm
cc
==
Find: The average power delivered to a 1-ohm resistor. Analysis:
The AM signal is given by:
[ ] )2cos()2cos(1)( tftfAts cmc +=
19.2
-
Expressing the product of two cosines as the sum of sinusoidal
waves, we get:
[ ] [ ]tffAtffAtfAts mccmcccc )(2cos21)(2cos
21)2cos()( +++=
The Fourier transform of is therefore: )(ts
[ ]
[ ]
[ ])()(41
)()(41
)()(21)(
mcmcc
mcmcc
ccc
ffffffA
ffffffA
ffffAfS
++++
++++
++=
Thus the spectrum of an AM wave, for sinusoidal modulation,
consists of delta functions at as seen from its Fourier transform.
mcmcc fffff ,,
In practice, the AM wave is a voltage or current wave. In either
case, the average power delivered to a 1-ohm resistor by is
comprised of three components:
)(ts)(ts
dttxT T= )(
1 power Average 2
Using Parsevals energy relation we can find average power in
frequency domain as: 0 fat power Average 2 == |X(f)|
Hence the carrier frequency, upper side-frequency and lower
side-frequency power is:
[ ]
[ ]
[ ]
22
2
22
2
2
2
81
)()(41 power frequency -sideLower
81
)()(41 power frequency -sideUpper
21
)()(21 wer Carrier Po
c
mcmcc
c
mcmcc
c
ccc
A
ffffA
A
ffffA
A
ffA
=
++=
=
++=
=
+=
For a load resistor R different from 1-ohm, which is usually the
case in practice, the expression for carrier power, upper
side-frequency power, and lower side-frequency power are merely
scaled by the factor R
1 or R , depending
on whether the modulated wave is a voltage or current,
respectively. )(ts
19.2
-
Problem 19.10
Solution: Known quantities: Carrier signal frequency, , upper
side-band frequency components at frequencies MHz82.0=cf
MHz,83.0 MHz84.0 MHz,825.0 321 == sf= ss ff , their amplitudes
and the modulation index 1= . Find: a. Modulating signal equation.
b. Plot spectrum of the modulating signal. c. Plot the spectrum of
the AM signal including the lower side-band. Analysis: a. We know
from the theory for AM that the upper side-band frequency has
frequency components at frequencies:
signal modulating thein componentsfrequency ofnumber theis n
wheremncsn fff +=
and, their amplitudes in the AM signal are 21 times the original
amplitude of the modulating signal for a
modulation index 1= . Hence, we can find the modulating signal
components to be:
)200002sin(5.0)()100002sin(4.0)(
)50002sin(8.0)(
3
2
1
ttmttmttm
===
Hence the modulating signal is:
)200002sin(5.0)100002sin(4.0)50002sin(8.0)( ttttm ++=
b. The spectrum for the modulating signal is shown in Figure
19.xx c. The spectrum for the AM modulated signal with the lower
side-band is shown in Figure 19.xx
Problem 19.11 Known quantities: AM frequency spectrum from 525 ,
bandwidth for each channel is 10 MHz1.7 tokHz kHz Find: a. Number
of channels that can be transmitted in the given frequency range b.
The maximum modulating frequency that can be transmitted without
overlap. Analysis: Assume: No guardband between channels. a. The
frequency range allocated for AM broadcast is
kHz 11755251700 ==Rf This range is partitioned to allow 10 of
separation between each channel; therefore, the total number of
channels,
kHz N is
channels 11810
1175=N
b. The carriers of two separate channels are separated by 10 .
If we let the maximum frequency of the message signal increase, the
outer edges of both sidebands move away from the carrier frequency
and into each other, thereby increasing the bandwidth of each AM
channel. The maximum allowable message frequency will occur at the
midpoint of the spacing between the carriers. Hence, the maximum
message frequency is half the frequency spacing between the
carriers.
kHz
kHz 5(max) =mf
19.2
-
19.2
Problem 19.1Solution:Known quantities:Find:Analysis:
Problem 19.2Solution:Known quantities:Find:Analysis:
Problem 19.3Solution:Known quantities:Find:Analysis:
Problem 19.4Solution:Known quantities:Analysis:
Problem 19.5Solution:Known quantities:Analysis:
Problem 19.6Solution:Known quantities:Analysis:
Problem 19.7Solution:Known quantities:Analysis:
Problem 19.8Solution:Known quantities:Find:Analysis:
Problem 19.9Solution:Known quantities:Functional form of the
modulating signal , the carrier signal , and the modulation index
.Find:Analysis:
Problem 19.10Solution:Known quantities:Find:Analysis:
Problem 19.11Known quantities:Find:Analysis:
