# Solution Manual for Principles of Communications, 7th Edition

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Chapter 2
Signal and Linear System Analysis
2.1
Problem Solutions
Problem 2.1
a. For the single-sided spectra, write the signal as
x1 (t) = 10 cos(4 t + =8) + 6 sin(8 t + 3 =4)
= 10 cos(4 t + =8) + 6 cos(8 t + 3 =4
=2)
= 10 cos(4 t + =8) + 6 cos(8 t + =4)
i
h
= Re 10ej(4 t+ =8) + 6ej(8 t+ =4)
For the double-sided spectra, write the signal in terms of complex exponentials using Eulerโs
theorem:
x1 (t) = 5 exp[j(4 t + =8)] + 5 exp[ j(4 t + =8)]
+3 exp[j(8 t + 3 =4)] + 3 exp[ j(8 t + 3 =4)]
The spectra are plotted in Fig. 2.1.
b. Write the given signal as
h
i
x2 (t) = Re 8ej(2 t+ =3) + 4ej(6 t+ =4)
to plot the single-sided spectra. For the double-side spectra, write it as
x2 (t) = 4ej(2 t+ =3) + 4e j(2 t+ =3) + 2ej(6 t+ =4) + 2e j(6 t+ =4)
The spectra are plotted in Fig. 2.2.
1
2
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. Change the sines to cosines by subtracting =2 from their arguments to get
x3 (t) = 2 cos (4 t + =8
=2) + 12 cos (10 t
= 2 cos (4 t 3 =8) + 12 cos (10 t
h
i
= Re 2ej(4 t 3 =8) + 12ej(10 t =2)
=2)
=2)
= ej(4 t 3 =8) + e j(4 t 3 =8) + 6ej(10 t
=2)
+ 6e j(10 t
=2)
Spectral plots are given in Fig. 2.3.
d. Use a trig identity to write
3 sin (18 t + =2) = 3 cos (18 t)
and get
x4 (t) = 2 cos (7 t + =4) + 3 cos (18 t)
i
h
= Re 2ej(7 t+ =4) + 3ej18 t
= ej(7 t+ =4) + e j(7 t+ =4) + 1:5ej18 t + 1:5e j18 t
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
2 and 3 at frequencies of 3.5 and 9 Hz, respectively, and the phase spectrum consists of
a line of height =4 at 3.5 Hz. The double-sided amplitude spectrum consists of lines of
amplitudes 1, 1, 1.5, and 1.5 at frequencies of 3.5, -3.5, 9, and -9 Hz, respectively. The
double-sided phase spectrum consists of lines of heights =4 and
=4 at frequencies 3.5
Hz and 3:5 Hz, respectively.
e. Use sin (2 t) = cos (2 t
=2) to write
x5 (t) = 5 cos (2 t
h
= Re 5ej(2 t
= 2:5ej(2 t
=2) + 4 cos (5 t + =4)
i
=2)
+ 4ej(5 t+ =4)
=2)
+ 2:5e j(2 t
=2)
+ 2ej(5 t+ =4) + 2e j(5 t+ =4)
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
5 and 4 at frequencies of 1 and 2.5 Hz, respectively, and the phase spectrum consists of
lines of heights
=2 and =4 at 1 and 2.5 Hz, respectively. The double-sided amplitude
spectrum consists of lines of amplitudes 2.5, 2.5, 2, and 2 at frequencies of 1, -1, 2.5, and
-2.5 Hz, respectively. The double-sided phase spectrum consists of lines of heights
=2,
=2, =4, and
=4 at frequencies of 1, -1, 2.5, and -2.5 Hz, respectively.
2.1. PROBLEM SOLUTIONS
3
Single sided
Double sided
6
8
Amplitude
Amplitude
10
6
4
4
2
2
0
0
1
2
3
f, Hz
4
0
-5
5
0
f, Hz
5
0
f, Hz
5
0.5
0.6
Phase, rad
Phase, rad
0.8
0.4
0.2
0
-0.5
0
0
1
2
3
f, Hz
4
5
f. Use sin (10 t + =6) = cos (10 t + =6
-5
=2) = cos (10 t
x6 (t) = 3 cos (4 t + =8) + 4 cos (10 t
=3)
i
h
= Re 3ej(4 t+ =8) + 4ej(10 t =3)
= 1:5ej(4 t+ =8) + 1:5e j(4 t+ =8) + 2ej10 t
=3) to write
=3)
+ 2e j(10 t
=3)
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
3 and 4 at frequencies of 2 and 5 Hz, respectively, and the phase spectrum consists of
lines of heights =8 and
=3 at 2 and 5 Hz, respectively. The double-sided amplitude
spectrum consists of lines of amplitudes 1.5, 1.5, 2, and 2 at frequencies of 2, -2, 5, and -5
Hz, respectively. The double-sided phase spectrum consists of lines of heights =8,
=8,
=3, and =3 at frequencies of 2, -2, 5, and -5 Hz, respectively.
4
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Single sided
Double sided
5
8
6
Amplitude
Amplitude
4
4
2
0
2
1
0
1
2
f, Hz
3
0
-4
4
-2
0
f, Hz
2
4
-2
0
f, Hz
2
4
1
Phase, rad
1
Phase, rad
3
0.5
0
0.5
0
-0.5
-1
0
1
2
f, Hz
3
4
-4
2.1. PROBLEM SOLUTIONS
5
Single sided
Double sided
6
Amplitude
Amplitude
10
5
0
0
2
4
4
2
0
6
-5
0
f, Hz
5
-5
0
f, Hz
5
f, Hz
0
Phase, rad
Phase, rad
1
-0.5
-1
0
-1
-1.5
0
2
4
f, Hz
6
6
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.2
By noting the amplitudes and phases of the various frequency components from the plots,
the result is
x(t) = 4ej(8 t+ =2) + 4e j(8 t+ =2) + 2ej(4 t
= 8 cos (8 t + =2) + 4 cos (4 t
=
8 sin (8 t) + 4 cos (4 t
=4)
+ 2e j(4 t
=4)
=4)
=4)
Problem 2.3
a. Not periodic because f1 = 1= Hz and f2 = 3 Hz are not commensurable.
b. Periodic. To โฆnd the period, note that
30
6
= 3 = n1 f0 and
= 15 = n2 f0
2
2
Therefore
15
n2
=
3
n1
Hence, take n1 = 1, n2 = 5; and f0 = 3 Hz (we want the largest possible value for f0 with
n1 and n2 integer-valued).
c. Periodic. Using a similar procedure as used in (b), we โฆnd that n1 = 4, n2 = 21; and
f0 = 0:5 Hz.
d. Periodic. Using a similar procedure as used in (b), we โฆnd that n1 = 4, n2 = 7;
n3 = 11; and f0 = 0:5 Hz.
e. Periodic. We โฆnd that n1 = 17, n2 = 18; and f0 = 0:5 Hz.
f. Periodic. We โฆnd that n1 = 2, n2 = 3; and f0 = 0:5 Hz.
g. Periodic. We โฆnd that n1 = 7, n2 = 11; and f0 = 0:5 Hz.
h. Not periodic. The frequencies of the separate terms are incommensurable.
i. Periodic. We โฆnd that n1 = 19, n2 = 21; and f0 = 0:5 Hz.
j. Periodic. We โฆnd that n1 = 6, n2 = 7; and f0 = 0:5 Hz.
2.1. PROBLEM SOLUTIONS
7
Problem 2.4
a. The single-sided amplitude spectrum consists of a single line of amplitude 5 at 6 Hz
and the phase spectrum consists of a single line of height
=6 rad at 6 Hz. The
double-sided amplitude spectrum consists of lines of amplitude 2.5 at frequencies 6
Hz. The double -sided phase spectrum consists of a line of height =6 at -6 Hz and
a line of height
=6 at 6 Hz.
b. Write the signal as
x2 (t) = 3 cos(12 t
=2) + 4 cos(16 t)
From this it is seen that the single-sided amplitude spectrum consists of lines of heights 3
and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists
of a line of height
=2 radians at frequency 6 Hz (the phase at 8 Hz is 0). The doublesided amplitude spectrum consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz,
respectively, and lines of height 1.5 and 2 at frequencies 6 and 8 Hz, respectively. The
double-sided phase spectrum consists of a line of height
=2 radians at frequency 6 Hz
and a line of height =2 radians at frequency 6 Hz.
c. Use the trig identity cos x cos y = 0:5 cos (x + y) + 0:5 cos (x
y) to write
x3 (t) = 2 cos 20 t + 2 cos 4 t
From this we see that the single-sided amplitude spectrum consists of lines of height 2 at 2
and 10 Hz, and the single-sided phase spectrum is 0 at these frequencies. The double-sided
amplitude spectrum consists of lines of height 1 at frequencies of 10, 2, 2, and 10 Hz.
The double-sided phase spectrum is 0.
d. Use trig identies to get
x4 (t) = 4 sin (2 t) [1 + cos (10 t)]
= 4 sin (2 t)
2 sin (8 t + ) + 2 sin (12 t)
= 4 cos (2 t
h
= Re 4ej(2 t
=2) + 2 cos (8 t + =2) + 2 cos (12 t
i
=2)
+ 2ej(8 t+ =2) + 2ej(12 t =2)
= 2ej(2 t
=2)
+ 2e j(2 t
=2)
=2)
+ ej(8 t+ =2) + e j(8 t+ =2) + ej(12 t
=2)
+ e j(12 t
From this we see that the single-sided amplitude spectrum consists of lines of heights 4,
2, and 2 at frequencies 1, 4, and 6 Hz, respectively and the single-sided phase spectrum is
=2 radians at 1 and 6 Hz and =2 radians at 4 Hz. The double-sided amplitude spectrum
=2)
8
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
consists of lines of height 2 at frequencies of 1 and 1 Hz and of height 1 at frequencies of
4, -4, 6, and -6 Hz. The double-sided phase spectrum is =2 radians at -1, 4, and -6 Hz and
=2 radians at 1, -4, and 6 Hz.
e. Clearly, from the form of the cosine sum, the single-sided amplitude spectrum has
lines of heights 1 and 7 at frequencies of 3 and 15 Hz, respectively. The single-sided
phase spectrum is zero. The double-sided amplitude spectrum has lines of heights 0.5,
0.5, 3.5, and 3.5 at frequencies of 3, -3, 15, and -15 Hz, respectfully. The double-sided
phase spectrum is zero.
f. The single-sided amplitude spectrum has lines of heights 1 and 9 at frequencies of
2 and 10.5 Hz, respectively. The single-sided phase spectrum is
=2 radians at
10.5 Hz and 0 otherwise. The double-sided amplitude spectrum has lines of heights
0.5, 0.5, 4.5, and 4.5 at frequencies of 2, -2, 10.5, and -10.5 Hz, respectfully. The
double-sided phase spectrum is =2 radians at -10.5 Hz and
=2 radians at 10.5 Hz
and 0 otherwise.
g. Convert the sine to a cosine by subtracting =2 from its argument. It then follows
that the single-sided amplitude spectrum is 2, 1, and 6 at frequencies of 2, 3, and
8.5 Hz and 0 otherwise. The single-sided phase spectrum is
=2 radians at 8.5 Hz
and 0 otherwise. The double-sided amplitude spectrum is 1, 1, 0.5, 0.5, 3, and 3
at frequencies of 2, 2, 3, 3 8:5, and 8.5 Hz, respectively, and 0 otherwise. The
double-sided phase spectrum is =2 radians at a frequency of 8:5 Hz and
=2
radians at a frequency of 8.5 Hz. It is 0 otherwise.
Problem 2.5
a. This function has area
Area =
Z1
1
=
Z1
sin( u) 2
du = 1
( u)
sin( t= ) 2
dt
( t= )
1
1
where a tabulated integral has been used for sinc2 u. A sketch shows that no matter how
small is, the area is still 1. With ! 0; the central lobe of the function becomes narrower
and higher. Thus, in the limit, it approximates a delta function.
2.1. PROBLEM SOLUTIONS
9
b. The area for the function is
Z1
Area =
1
exp( t= )u (t) dt =
1
Z1
exp( u)du = 1
0
A sketch shows that no matter how small is, the area is still 1. With ! 0; the function
becomes narrower and higher. Thus, in the limit, it approximates a delta function.
R 1
R1
c. Area =
(1 jtj = ) dt =
As ! 0, the function becomes
1 (t) dt = 1.
narrower and higher, so it approximates a delta function in the limit.
Problem 2.6
1
(t) to write (2t 5) =
a. Make use of the formula (at) = jaj
1
5
t 2 and use the sifting property of the -function to get
2
Ia =
1
2
5
2
2
+
1
exp
2
2
5
2
=
[2 (t
5=2)] =
25 1
+ exp [ 5] = 3:1284
8
2
b. Impulses at 10, 5, 0, 5, 10 are included in the integral. Use the sifting property
after writing the expression as the sum of โฆve integrals to get
Ib = ( 10)2 + 1 + ( 5)2 + 1 + 02 + 1 + 52 + 1 + 102 + 1 = 255
c. Matching coeยข cients of like derivatives of -functions on either side of the equation
gives A = 5, B = 10, and C = 3.
d. Use
I = 14
1
(at) = jaj
(t) to write
e
4 ( 3=4) + tan
10
(4t + 3) = 14
3
4
= 14
e3
t + 34 .
The integral then becomes
+ tan ( 7:5 ) =
9:277
1013 .
e. Use property 5 of the unit impulse function to get
d2
d
Ie = ( 1)2 2 cos 5 t + e 3t t=2 =
5 sin 5 t 3e 3t t=2
dt
dt
h
i
=
(5 )2 cos 5 t + 9e 3t
= (5 )2 cos 10 + 9e 6 = 246:73
t=2
10
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.7
(a), (c), and (e) are periodic. Their periods are 2 s (fundamental frequency of 0.5 Hz),
2 s, and 3 s, respectively. The waveform of part (c) is a periodic train of triangles,
each 2 units wide, extending from -1 to 1 spaced by 2 s ((b) is similar except that it
is zero for t < 1 thus making it aperiodic). Waveform (d) is aperiodic because the
frequencies of its two components are incommensurable. The waveform of part (e) is a
doubly-inโฆnite train of square pulses, each of which is one unit high and one unit wide,
centered at
; 6; 3; 0; 3; 6;
. Waveform (f) is identical to (e) for t
1=2 but 0
for t < 1=2 thereby making it aperiodic.
Problem 2.8
a. The result is
x(t) = cos (6 t)+2 cos (10 t
j(10 t
=2)
1
1
+ e j6 t + ej6 t + ej(10 t
2
2
=2)
=2) = Re e
j6 t
+Re 2e
h
= Re e
j6 t
+ 2e
j(10 t
b. The result is
x(t) = e j(10 t
=2)
c. The single-sided amplitude spectrum consists of lines of height 1 and 2 at frequencies
of 3 and 5 Hz, respectively. The single-sided phase spectrum consists of a line of height
=2 at frequency 5 Hz. The double-sided amplitude spectrum consists of lines of
height 1, 1/2, 1/2, and 1 at frequencies of 5; 3; 3; and 5 Hz, respectively. The
double-sided phase spectrum consists of lines of height =2 and
=2 at frequencies
of 5 and 5 Hz, respectively.
Problem 2.9
a. Power. Since it is a periodic signal, we obtain
Z T0
Z T0
1
1
2
P1 =
4 cos (4 t + 2 =3) dt =
2 [1 + cos (8 t + 4 =3)] dt = 2 W
T0 0
T0 0
where T0 = 1=2 s is the period. The cosine in the above integral integrates to zero because
the interval of integratation is two periods.
b. Energy. The energy is
E2 =
Z 1
1
e
2 t 2
u (t)dt =
Z 1
0
e 2 t dt =
1
J
2
=2)
i
2.1. PROBLEM SOLUTIONS
11
c. Energy. The energy is
E3 =
Z 1
e
2 t 2
u ( t)dt =
Z 0
e2 t dt =
1
1
1
J
2
d. Energy. The energy is
E4 =
=
=
Z T
dt
1
= lim 2
2 + t2 )
T
!1
(
T
lim
T !1
1
lim
T !1
1h
2
2
T
t
tan 1
i
= lim
dt
1 + (t= )2
T
1
tan 1 (T = )
T !1
T
=
Z T
tan 1 ( T = )
J
e. Energy. Since it is the sum of x2 (t) and x3 (t), its energy is the sum of the energies
of these two signals, or E5 = 1= J.
f. Energy. The energy is
E6 =
=
=
=
=
=
lim
T !1
lim
T !1
lim
T !1
lim
T !1
lim
Z T h
T
Z T h
t
u (t)
(t 1)
e
e 2 t u2 (t)
e
2 t
dt
e
1)
(t 1)
e
i2
dt
e
2 t
dt
2
+
Z T
e
1
2 (t 1)
dt +
0
e 2 t dt0 +
0
e 2
2
+e
0
1
1
=
2
1
1) + e 2 (t 1) u2 (t
u (t) u (t
1
8 0
< e 2 t T
e
2
u (t
e
Z T
0
Z T
t
e
T
Z T
T !1 :
1
2
e
Z T
0
t0
1
e
2
T 1
e 2
2
0
J
e 2 (t 1) dt
1
Z T 1
t0
T
0
0
e 2 t dt0
9
1=
;
i
1) dt
12
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.10
a. Power. Since the signal is periodic with period 2 =!, we have
!
P =
2
Z 2 =!
0
!
A jsin (!t + )j dt =
2
2
2
Z 2 =!
0
A2
f1
2
cos [2 (!t + )]g dt =
A2
W
2
b. Neither. The energy calculation gives
Z T
Z T
(A )2 dt
(A )2 dt
p
p
p
dt ! 1
E = lim
dt = lim
2 + t2
T !1
T !1
+ jt
jt
T
T
The power calculation gives
1
P = lim
T !1 2T
Z T
(A )2 dt
(A )2
p
ln
dt = lim
2 + t2
T !1 2T
T
c. Energy:
E=
Z 1
0
1
A t exp ( 2t= ) dt = A2
8
2 2
r
2
p
1 + T 2= 2
p
1 + 1 + T 2= 2
1+
!
=0W
W (use a table of integrals)
d. Energy: This is a "top hat" pulse which is height 2 for jtj
=2, height 1 for
=2 < jtj
, and 0 everywhere else. Making use of the even symmetry about t = 0,
the energy is
!
Z =2
Z
E=2
22 dt +
12 dt = 5 J
0
=2
e. Energy. The signal is a "house" two units wide and one unit up to the eves with a
equilateral triangle for a roof. Because of symmetry, the energy calculation need be
carried out for positive t and doubled. The calculation is
Z 1
1
2
2 2 8
14
E=2
(2 t)2 dt =
(2 t)3 =
+
=
J
3
3
3
3
0
0
f. Power. Since the two terms are harmonically related, we may add their respective
powers and get
A2 B 2
P =
+
W
2
2
2.1. PROBLEM SOLUTIONS
13
Problem 2.11
a. Using the fact that the power contained in a sinusoid is its amplitude squared divided
by 2, we get
22
P =
=2W
2
b. This is a periodic train of "box cars" 3 units high, 2 units wide, and occurring every
4 units (period of 4 seconds). The power calculation is
Z
1 1 2
32 2
P =
3 dt =
= 4:5 W
4 1
4
c. This is a train of triangles 1 unit high, 4 s wide, and occuring every 6 s. Using the
waveform period centered at 0, the power calculation is
1
P =
6
Z 2
1
2
t
2
2
dt =
12
63
1
t
2
3 2
=
0
2
W
9
d. This is a train of "houses" each of which is 2 s wide, 1 unit high to the eves, with an
isoceles triangle on top for the roof. They are separated by 4 s (the period). Using
the even symmetry of each house, the power calculation is
2
Pd =
4
Z 1
2
(2
t) dt =
0
1 (2 t)3
2
3
1
=
0
1
2
1
3
23
3
=
7
W
6
Problem 2.12
a. The energy is
E =
Z 1
0
= 36
6e
Z 1
( 3+j4 )t
e 6t dt =
0
2
dt = 36
Z 1
e( 3+j4 )t e( 3 j4 )t dt
0
1
e 6t
36
=6J
6 0
The power is 0 W.
b. This signal is a "top hat" pulse which is 2 for 2 t 4, 1 for 0 t < 2 and 4 < t
and 0 everywhere else. It is clearly an energy signal with energy
E=2
12 + 2
22 + 2
12 = 12 J
6,
14
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Its power is 0 W.
c. This is a power signal with power
Z T
Z
1
49 T
49
j6 t
j6 t
P = lim
49e
e
u (t) dt = lim
dt =
= 24:5 W
T !1 2T
T
!1
2T 0
2
T
Its energy is inโฆnite.
2
d. This is a periodic signal with power P = 22 = 2 W. Its energy is inโฆnite.
e. This is neither an energy nor a power signal. Its energy is inโฆnite and its power is
1
T !1 2T
P = lim
Z T
1 t3
T !1 2T 3
T
t2 dt = lim
T
1 2T 3
!1
T !1 2T 3
= lim
T
f. This is neither an energy nor a power signal. Its energy is
Z 1
t 1 dt = ln (t)j1
E=
1 !1
1
and its power is
1
P = lim
T !1 2T
Problem 2.13
Z T
1
ln (t)j1
1 =0
T !1 2T
t 1 dt = lim
1
a. This is a cosine burst from t = 6 to t = 6 seconds. The energy is E1 =
R6
2 0 12 + 12 cos (12 t) dt = 6 J
R6
6 cos
b. The energy is
E2 =
Z 1h
1
=
2
e
e
jtj=3
2t=3
2=3
i2
dt = 2
Z 1
e 2t=3 dt (by even symmetry)
0
1
=3J
0
Since the result is โฆnite, this is an energy signal.
c. The energy is
E3 =
Z 1
1
f2 [u (t)
u (t
2
8)]g dt =
Z 8
0
4dt = 32 J
2 (6
t) dt =
2.1. PROBLEM SOLUTIONS
15
Since the result is โฆnite, this is an energy signal.
d. Note that
r (t) ,
Z t
u( )d =
1
0; t < 0
t; t 0
which is called the unit ramp. Thus the given signal is a triangle between 0 and 20. The
energy is
Z 1
Z 10
2
2000
10
2
E4 =
[r (t) 2r (t 10) + r (t 20)] dt = 2
t2 dt = t3 0 =
J
3
3
1
0
where the last integral follows because the integrand is a symmetrical triangle about t = 10.
Since the result is โฆnite, this is an energy signal.
Problem 2.14
a. This is a cosine burst nonzero between 0 and 2 seconds. Its power is 0. Its energy is
Z
Z 2
1 2
2
[1 + cos (20 t)] dt = 1 J
E1 =
cos (10 t) dt =
2 0
0
b. This is a periodic sequence of triangles of period 3 s. Its energy is inโฆnite. Its power
is
Z
2 2
4
P2 =
(1 t=2)2 dt = J
3 0
9
c. This is an energy signal. Its power is 0. Using evenness of the integrand, its energy
is
Z 1
Z 1
Z 1
2t
2
2t
E3 = 2
e cos (2 t) dt =
e dt +
e 2t cos (4 t) dt
0
=
0
0
1
2
+
J
2 4 + 16 2
.
d. This is an energy signal. Its energy is
E4 = 2
Z 1
0
(2
t)2 dt =
2
(2
3
1
t)3
=
0
14
J
3
16
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.15
a. Use the exponential representation of the sine to get the Fourier coeยข cients as (note
that the period = 12 f10 ).
x1 (t) =
ej2 f0 t
e j2 f0 t
2j
from which we โฆnd that
X 1 = X1 =
2
1
e j4 f0 t
4
=
2 + ej4 f0 t
1
1
; X0 =
4
2
All other coeยข cients are zero.
b. Use the exponential representations of the sine and cosine to get
1
1
1
x2 (t) = ej2 f0 t + e j2 f0 t + ej4 f0 t
2
2
2j
1 j4 f0 t
e
2j
Therefore, the Fourier coeยข cients for this case are
X 1 = X1 =
1
1
and X2 = X 2 =
2
2j
All other coeยข cients are zero.
c. Use a trig identity to write this signal as
x3 (t) =
1
1
sin 8 f0 t = ej8 f0 t
2
4j
1 j8 f0 t
e
4j
The fundamental frequency is 4f0 Hz. From this it follows that the Fourier coeยข cients are
X1 = X 1 =
1
4j
All other coeยข cients are zero.
d. Use trig identities and the exponential forms of cosine to write this signal as
x4 (t) =
=
1
3
cos 2 f0 t + cos 6 f0 t
4
4
3 j2 f0 t 3 j2 f0 t 1 j6 f0 t 1 j6 f0 t
+ e
+ e
+ e
e
8
8
8
8
2.1. PROBLEM SOLUTIONS
17
The fundamental frequency is f0 Hz. It follows that the Fourier coeยข cients for this case
are
3
1
X 1 = X1 = ; X 3 = X3 =
8
8
All other Fourier coeยข cients are zero.
e. Use trig identies to write
x5 (t) =
=
1
1
1
sin (2 f0 t)
sin (6 f0 t) + sin (10 f0 t)
2
4
4
1 j2 f0 t
1 j2 f0 t
1 j6 f0 t
1
1
e
e
e
+ e j6 f0 t + ej10 f0 t
4j
4j
8j
8j
8j
1 j10 f0 t
e
8j
The fundamental frequency is f0 Hz. It follows that the Fourier coeยข cients for this case
are
j
j
j
X 1 = X1 =
; X 3 = X3 = ; X 5 = X5 =
4
8
8
All other Fourier coeยข cients are zero.
f. Use trig identities to write
x6 (t) =
=
1
1
1
cos (6 f0 t)
cos ( f0 t)
cos (11 f0 t)
2
4
4
1 j6 f0 t 1 j6 f0 t 1 j f0 t 1 j f0 t 1 j11 f0 t
e
+ e
e
e
e
4
4
8
8
8
1 j11 f0 t
e
8
The fundamental frequency is f0 =2 Hz. It follows that the Fourier coeยข cients for this case
are
1
1
1
X 1 = X1 =
; X 12 = X12 = ; X 22 = X22 =
8
4
8
All other Fourier coeยข cients are zero.
Problem 2.16
The expansion interval is T0 = 4 so that f0 = 1=4 Hz. The Fourier coeยข cients are
Z
Z
2 2 2
1 2 2 jn( =2)t
Xn =
2t e
dt =
t (cos n t=2 j sin n t=2) dt =
4 2
4 2
Z
2 2 2
n t
dt
=
2t cos
4 0
2
which follows by the oddness of the second integrand and the eveness of the โฆrst integrand.
Let u = n t=2 to obtain the form
Z
2 3 n 2
16
Xn =
u cos u du =
( 1)n n 6= 0 (use a table of integrals)
2
n
(n )
0
18
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
If n = 0, the integral for the coeยข cients is
1
X0 =
4
Z 2
2t2 dt =
2
8
3
The Fourier series is therefore
8
x (t) = +
3
1
X
( 1)n
n= 1; n6=0
16 jn( =2)t
e
(n )2
Problem 2.17
Parts (a) through (c) were discussed in the text. For (d), break the integral for Xn up into
a part for t 0. Then use the odd half-wave symmetry condition.
The development follows:
Xn =
=
=
1
T0
1
T0
1
T0
1
=
T0
(
=
“Z
T0 =2
x (t) e j2 nf0 t dt +
T0 =2
x (t) e j2 nf0 t dt +
0
“Z
Z T0 =2
#
x t0 + T0 =2 e j2
nf0 (t0 +T0 =2)
0
T0 =2
x (t) e
j2 nf0 t
dt
0
“Z
x (t) e j2 nf0 t dt
T0 =2
0
“Z
Z 0
Z T0 =2
0
x t e
j2 nf0 t0 jn
x (t) e
j2 nf0 t
dt
( 1)
n
0
0; n even
R T0 =2
2
x (t) e j2 nf0 t dt; n odd
T0 0
Z T0 =2
0
0
x t e
dt ; t0 = t + T0 =2
dt ; f0 = 1=T0
0
T0 =2
#
#
j2 nf0 t0
dt
#
2.1. PROBLEM SOLUTIONS
19
Problem 2.18
This is a matter of integration. Only the solution for part (b) will be given here.
integral for the Fourier coeยข cients is
Xn =
=
=
=
=
=
Z T0 =2
A
sin (! 0 t) e
dt =
ej!0 t e j!0 t e jn!0 t dt
2jT0 0
0
#
“Z
Z T0 =2
T0 =2
A
e j(1+n)!0 t dt
ej(1 n)!0 t dt
2jT0 0
0
2
3
T0 =2
T0 =2
j(1
n)!
t
j(1+n)!
t
0
0
A 4 e
e
5
2jT0 j (1 n) ! 0
j (1 + n) ! 0
0
0
”
#
A
ej(1 n)
1 e j(1+n)
1
+
; n 6= 1 (! 0 T0 =2 = )
4
1 n
1+n
A
T0
A
4
(
Z T0 =2
The
jn! 0 t
( 1)n + 1 ( 1)n + 1
+
; n 6=
1 n
1+n
0; n odd and n 6=
A
; n even
(1 n2 )
1
1
For n = 1, the integral is
X1 =
A
2jT0
=
A
2jT0
Z T0 =2
ej!0 t
e j!0 t e j!0 t dt
0
Z T0 =2
1
e j2!0 t dt =
0
jA
=X 1
4
This is the same result as given in Table 2.1.
Problem 2.19
a. Use Parsevalโs theorem to get
Pjnf0 j
1=
=
N
X
n= N
jXn j2 =
N
X
n= N
A
T0
2
sinc2 (nf0 )
where N is an appropriately chosen limit on the sum. We are given that only frequences for
which jnf0 j 1= are to be included. This is the same as requiring that jnj 1= ( f0 ) =
20
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
T0 = = 2: Also, for a pulse train, Ptotal = A2 =T0 and, in this case, Ptotal = A2 =2: Thus
2
Pjnf0 j 1=
Ptotal
=
2 X
A2
n= 2
A
2
2
sinc2 (nf0 )
2
=
1 X
sinc2 (nf0 )
2
n= 2
=
=
b. In this case, jnj
Pjnf0 j 1=
Ptotal
c. In this case, jnj
Pjnf0 j 1=
Ptotal
=
1
1 + 2 sinc2 (1=2) + sinc2 (1)
2″
#
1
2 2
1+2
= 0:9053
2
5, Ptotal = A2 =5, and
5
1 X
=
sinc2 (n=5)
5
n= 5
h
io
1n
=
1 + 2 (0:9355)2 + (0:7568)2 + (0:5046)2 + (0:2339)2
5
= 0:9029
10, Ptotal = A2 =10, and
10
1 X
sinc2 (n=10)
10
n= 10
1
1+2
10
= 0:9028
=
d. In this case, jnj
(0:9836)2 + (0:9355)2 + (0:8584)2 + (0:7568)2 + (0:6366)2
+ (0:5046)2 + (0:3679)2 + (0:2339)2 + (0:1093)2
20, Ptotal = A2 =20, and
Pjnf0 j 1=
Ptotal
=
=
20
1 X
sinc2 (n=20)
20
n= 20
(
)
20
X
1
1+2
sinc2 (n=20)
20
n=1
= 0:9028
2.1. PROBLEM SOLUTIONS
21
Problem 2.20
a. The integral for Yn is
Z
Z T0
1
1
jn! 0 t
y (t) e
dt =
x (t
Yn =
T0 T0
T0 0
Let t0 = t
t0 ) e jn!0 t dt; ! 0 = 2 f0
t0 , which results in
Z T0 t0
1
0
Yn =
x t0 e jn!0 t dt0 e jn!0 t0 = Xn e j2 nf0 t0
T 0 t0
b. Note that
y (t) = A cos ! 0 t = A sin (! 0 t + =2) = A sin [! 0 (t + =2! 0 )]
Thus, t0 in the theorem proved in part (a) here is
can be expressed as
1
sin (! 0 t) = ej!0 t
2j
=2! 0 . By Eulerโs theorem, a sine wave
1 j!0 t
e
2j
1
Its Fourier coeยข cients are thereforeX1 = 2j
and X 1 =
proved in part (a), we multiply these by the factor
e jn!0 t0 = e jn!0 (
=2! 0 )
1
2j .
According to the theorem
= ejn =2
For n = 1, we obtain
Y1 =
For n =
1 j =2 1
e
=
2j
2
1, we obtain
1 j =2 1
e
=
2j
2
which gives the Fourier series representation of a cosine wave as
Y 1=
1
1
y (t) = ej!0 t + e j!0 t = cos ! 0 t
2
2
We could have written down this Fourier representation directly by using Eulerโs theorem.
Problem 2.21
a. Use the Fourier series of a square wave (specialize the Fourier series of a pulse train)
with A = 1 and t = 0 to obtain the series
1=
4
1
1 1
+
3 5
1
+
7
22
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Multiply both sides by 4 to get the series in the problem statement. Hence, the sum is 4 .
b. Use the Fourier series of a triangular wave as given in Table 2.1 with A = 1 and t = 0
to obtain the series
1=
+
4
4
4
4
4
4
+
+ 2+ 2+ 2+
+
25 2 9 2
9
25 2
2
Multiply both sides by 8 to get the series in given in the problem. Hence, its sum is
2
8 .
Problem 2.22
a. In the expression for the Fourier series of a pulse train (Table 2.1), let t0 =
and = T0 =4 to get
n
nf0
A
exp j
Xn = sinc
4
4
4
T0 =8
The spectra are shown in Fig. 2.4.
b. The amplitude spectrum is the same as for part (a) except that X0 = 3A
4 . Note that
this can be viewed as having a sinc-function envelope with zeros at multiples of 3T4 0 .
The phase spectrum can be obtained from that of part (a) by subtracting a phase shift
of for negative frequencies and adding for postitive frequencies (or vice versa).
The Fourier coeยข cients are given by
Xn =
3A
sinc
4
3n
4
exp
j
3 nf0
4
See Fig. 2.4 for amplitude and phase plots.
Problem 2.23
a. Use the rectangular pulse waveform of Table 2.1 specialized to
xa (t) = 2A
t
T0 =4
T0 =2
A; jtj < T0 =2
and periodically extended. Hence, from Table 2.1, we have
2AT0 =2
nT0 =2
2 nT0 =4
sinc
exp
j
T0
T0
T0
= Asinc (n=2) exp ( j n=2) ; n 6= 0
sin (n =2)
= A
exp ( j n=2) ; n 6= 0
n =2
XnA =
2.1. PROBLEM SOLUTIONS
23
Part (a)
Part (b)
0.6
Amplitude
Amplitude
0.6
0.4
0.2
0
-5
0
f, Hz
0.2
0
5
-5
0
f, Hz
5
-5
0
f, Hz
5
2
Phase, rad
2
Phase, rad
0.4
0
-2
0
-2
-5
0
f, Hz
5
24
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
where the superscript A refers to xA (t). The dc component is 0 so X0A = 0. The Fourier
coeยข cients are therefore
X0A = 0
X1A =
j2A= ; X A1 = j2A=
X2A = 0 = X A2
X3A = j2A=3 ; X A3 =
j2A=3
b. Note that
xA (t) =
dxB (t)
dt
where A = T2B
= 4B
T0 obtained from matching the amplitude of xA (t) with the slope of
0 =2
xB (t) or B = T40 A. The relationship between spectral components is therefore
XnA = (jn! 0 ) XnB = j2 nf0 XnB
or
XnB =
XnA
T0 XnA
=
j2 nf0
j2 n
where the superscript A refers to xA (t) and B refers to xB (t). For example,
X1B =
j2A=
=
j2 f0
2B
2
= X B1
Problem 2.24
a. This is a decaying exponential starting at t = 0 and its Fourier transform is
Z 1
Z 1
X1 (f ) = A
e t= e j2 f t dt = A
e (1= +j2 f )t dt
0
0
(1= +j2 f )t
=
Ae
1= + j2 f
=
A
1 + j2 f
1
=
0
A
1= + j2 f
2.1. PROBLEM SOLUTIONS
25
b. Since x2 (t) = x1 ( t) we have, by the time reversal theorem, that
X2 (f ) = X1 (f ) = X1 ( f )
A
=
1 j2 f
c. Since x3 (t) = x1 (t)
x2 (t) we have, after some simpliโฆcation, that
X3 (f ) = X1 (f ) X2 (f )
A
A
=
1 + j2 f
1 j2 f
j4A f
=
1 + (2 f )2
d. Since x4 (t) = x1 (t) + x2 (t) we have, after some simpliโฆcation, that
X4 (f ) = X1 (f ) + X2 (f )
A
A
=
+
1 + j2 f
1 j2 f
2A
=
1 + (2 f )2
This is the expected result since x4 (t) is really a double-sided decaying exponential.
e. By part a and the delay theorem
X5 (f ) = X1 (f ) e j10 f =
A e j10 f
1 + j2 f
f. By parts a and e and superposition
h
X6 (f ) = X1 (f ) 1
i A
e j10 f =
1 e j10 f
1 + j2 f
Problem 2.25
a. Using a table of Fourier transforms and the time reversal theorem, the Fourier transform of the given signal is
X (f ) =
1
+ j2 f
1
j2 f
26
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Note that x (t) ! sgn(t) in the limit as
transform as ! 0, we deduce that
F [sgn (t)] =
! 0.
1
j2 f
Taking the limit of the above Fourier
1
1
=
j2 f
j f
b. Using the given relationship between the unit step and the signum function and the
linearity property of the Fourier transform, we obtain
F [u (t)] =
=
1
1
F [sgn (t)] + F [1]
2
2
1
1
+
(f )
j2 f
2
c. The same result as obtained in part (b) is obtained.
Problem 2.26
a. One diยคerentiation gives
dxa (t)
=
dt
(t
0:5)
(t
2:5)
Two diยคerentiations give
d2 xa (t)
= (t)
dt2
(t
1)
(t
2) + (t + 3)
Application of the diยคerentiation theorem of Fourier transforms gives
(j2 f )2 Xa (f ) = 1
=
1 e j2 f
ej3 f
ej f
= 2 (cos 3 f
1 e j4 f + 1 e j6 f
e j f + e j3 f e j3 f
cos f ) e j3 f
where the time delay theorem and the Fourier transform of a unit impulse have been used.
Dividing both sides by (j2 f )2 , we obtain
Xa (f ) =
cos f cos 3 f j3 f
2 (cos 3 f cos f ) e j3 f
=
e
2
2 2f 2
(j2 f )
2.1. PROBLEM SOLUTIONS
27
Use the trig identity sin2 x = 12
1
2 cos 2x or cos 2x = 1
2 sin2 x to rewrite this result as
2 sin2 (0:5 f ) 1 + 2 sin2 (1:5 f ) j3 f
e
2 2f 2
sin2 (0:5 f ) + sin2 (1:5 f ) j3 f
=
e
2f 2
sin2 (1:5 f ) j3 f sin2 (0:5 f ) j3 f
=
e
e
2f 2
2f 2
"
#
2
2
sin
1:5
f
sin
0:5
f
= 1:52
0:52
e j3 f
1:5 f
0:5 f
Xa (f ) =
=
1
0:52 sinc2 (0:5f ) e j3 f
1:52 sinc2 (1:5f )
This is the same result as would have been obtained by writing
t
xa (t) = 1:5
1:5
1:5
t
0:5
1:5
0:5
and using the Fourier transform of the triangular pulse along with the superposition and
time delay theorems.
b. Two diยคerentiations give (sketch dxb (t) =dt to see this)
d2 xb (t)
= (t)
dt2
2 (t
1) + 2 (t
3)
(t
4)
Application of the diยคerentiation theorem gives
(j2 f )2 Xb (f ) = 1
2e j2 f + 2e j6 f
e j8 f
Dividing both sides by (j2 f )2 , we obtain
Xb (f ) =
1
2e j2 f + 2e j6 f
4 2f 2
e j8 f
Further manipulation may be applied to this result to convert it to
h
i
Xb (f ) = sinc2 (f ) e j2 f e j6 f
h
i
= sinc2 (f ) ej2 f e j2 f e j4 f
= 2j sin (2 f ) sinc2 (f ) e j4 f
which would have resulted from Fourier transforming the waveform written as
xb (t) =
(t
1)
(t
3)
28
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. Two diยคerentiations give (sketch dxc (t) =dt to see this)
d2 xc (t)
= (t)
dt2
2 (t
1) + 2 (t
2)
2 (t
3) + (t
4)
Application of the diยคerentiation theorem gives
(j2 f )2 Xc (f ) = 1
2e j2 f + 2e j4 f
2e j6 f + e j8 f
Dividing both sides by (j2 f )2 , we obtain
Xc (f ) =
2e j2 f + 2e j4 f 2e j6 f + e j8 f
(j2 f )2
1
This result may be further arranged to give
2
h
Xc (f ) = sinc (f ) e
j2 f
j6 f
+e
i
= 2 cos (2 f ) sinc2 (f ) e j4 f
which would have resulted from Fourier transforming the waveform written as
xc (t) =
(t
1) +
(t
3)
d. Two diยคerentiations give (sketch dxd (t) =dt to see this)
d2 xd (t)
= (t)
dt2
2 (t
1) + (t
1:5) + (t
2:5)
2 (t
3) + (t
Application of the diยคerentiation theorem gives
(j2 f )2 Xd (f ) = 1
2e j2 f + e j3 f + e j5 f
2e j6 f + e j8 f
Dividing both sides by (j2 f )2 , we obtain
Xd (f ) =
1
2e j2 f + e j3 f + e j5 f
(j2 f )2
2e j6 f + e j8 f
This result may be further arranged to give
h
i
Xd (f ) = sinc2 (f ) e j2 f + 0:5e j4 f + e j6 f
which would have resulted from Fourier transforming the waveform written as
xd (t) =
(t
1) + 0:5 (t
2) +
(t
3)
4)
2.1. PROBLEM SOLUTIONS
29
Problem 2.27
See the solutions to Problem 2.26.
Problem 2.28
a. The steps in โฆnding the Fourier transform for (i) are as follows:
(t)
! sinc (f )
(t) exp [j4 t]
(t
1) exp [j4 (t
! sinc (f
1)]
! sinc (f
2)
2) exp ( j2 f )
The steps in โฆnding the Fourier transform for (ii) are as follows:
(t)
! sinc (f )
(t) exp [j4 t]
! sinc (f
(t + 1) exp [j4 (t + 1)]
! sinc (f
2)
2) exp (j2 f )
b. The steps in โฆnding the Fourier transform for (i) are as follows:
(t)
(t
(t
1)
1) exp [j4 (t
1)]
! sinc (f )
! sinc (f ) exp ( j2 f )
=
(t
1) exp (j4 t)
! sinc (f
2) exp [ j2 (f
2)] = sinc (f
2) exp ( j2 f )
which follows because exp( jn2 ) = 1 where n is an integer. The steps in โฆnding the
Fourier transform for (ii) are as follows:
(t)
(t + 1)
(t + 1) exp [j4 (t + 1)]
! sinc (f )
! sinc (f ) exp (j2 f )
=
(t + 1) exp (j4 t)
! sinc (f
2) exp [j2 (f
2)] = sinc (f
2) exp (j2 f )
30
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.29
a. The time reversal theorem states that x ( t)
so
xa (t)
!
=
! X ( f ) 6= X (f ) if x (t) is complex,
1
1
1
1
X1 (f ) + X1 ( f ) = sinc (f 2) exp ( j2 f ) + sinc ( f
2
2
2
2
1
1
sinc (f 2) exp ( j2 f ) + sinc (f + 2) exp (j2 f )
2
2
2) exp (j2 f )
Note that
xa (t) =
=
1
2
1
2
(t
1) exp [j4 (t
1)] +
(t
1) exp (j4 t) +
1
2
1
2
( t
1) exp [j4 ( t
1)]
(t + 1) exp ( j4 t) (by the eveness of
(u) )
b. Similarly to (a), we obtain
xb (t)
1
1
1
! X2 (f )+ X2 ( f ) = sinc (f
2
2
2
1
2) exp (j2 f )+ sinc (f + 2) exp ( j2 f )
2
2.1. PROBLEM SOLUTIONS
31
Problem 2.30
a. The result is
X1 (f ) = 2sinc (2f ) exp ( j2 f )
b. The result is
X2 (f ) = 2
1
2
f
2
exp ( j2 f )
c. The result is
X3 (f ) = 8sinc2 (8f ) exp ( j4 f )
d. The result is
X4 (f ) = 4 (4f ) exp ( j6 f )
e. The result is
X5 (f ) = 5
= 5
1
2
f
2
f
2
exp ( j2 f ) + 5
1
2
f
2
exp (j2 f )
cos (2 f )
f. The result is
X6 (f ) = 16sinc2 (8f ) exp ( j4 f ) + 16sinc2 (8f ) exp (j4 f )
= 32sinc2 (8f ) cos (4 f )
Problem 2.31
a. This is an odd signal, so its Fourier transform is odd and purely imaginary.
b. This is an even signal, so its Fourier transform is even and purely real.
c. This is an odd signal, so its Fourier transform is odd and purely imaginary.
d. This signal is neither even nor odd, so its Fourier transform is complex.
e. This is an even signal, so its Fourier transform is even and purely real.
f. This signal is even, so its Fourier transform is real and even.
32
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.32
In the Poisson sum formula, we identify p (t) = (t=2) which has Fourier transform P (f ) =
2sinc(2f ). Thus, for this case, the Poisson sum formula becomes
1
X
p (t
mTs ) =
m= 1
or
1
X
t
4m
2
m= 1
1
X
m= 1
t
1
1 X
2sinc
4 n= 1
=
2n
4
ej2 (n=4)t = fs
1
X
P (nfs ) ej2 nfs t
n= 1
1
X
1
n j (n=2)t
=
sinc
e
2
2
n= 1
4m
2
The fundamental frequency is 0.25 Hz and the Fourier coeยข cients areX0 = 1=2; X1 =
X 1 = 12 sinc 12 = 1 , X2 = X 2 = 0, X3 = X 3 = 21 sinc 32 = 31 , etc.
Problem 2.33
a. The Fourier transform of this signal is
X1 (f ) =
10
2
=
5 + j2 f
1 + j2 f =5
Thus, the energy spectral density is
G1 (f ) =
4
1 + (2 f =5)2
b. The Fourier transform of this signal is
f
2
X2 (f ) = 5
Thus, the energy spectral density is
X2 (f ) = 25
2
f
2
f
2
= 25
c. The Fourier transform of this signal is
3
X3 (f ) = sinc
2
so the energy spectral density is G3 (f ) = 94 sinc2
f
2
f
2
2.1. PROBLEM SOLUTIONS
33
d. The Fourier transform of this signal is
X4 (f ) =
3
sinc
4
f
5
f +5
2
+ sinc
2
so the energy spectral density is
G4 (f ) =
9
sinc
16
f
5
2
+ sinc
f +5
2
!
1
+ j2 f
2
Problem 2.34
a. Use the transform pair
t
x1 (t) = e
u (t)
Using Rayleighโs energy theorem, we obtain the integral relationship
Z 1
Z 1
Z 1
Z 1
df
1
2
2
jX1 (f )j df =
df =
jx1 (t)j dt =
e 2 t dt =
2 + (2 f )2
2
1
1
1
0
b. Use the transform pair
1
t
=
Z 1
x2 (t) =
Rayleighโs energy theorem gives
Z 1
jX2 (f )j2 df
1
1
=
Z 1
1
! sinc ( f ) = X2 (f )
2
sinc ( f ) df =
1
2
2
t
Z 1
dt =
Z
1
jx2 (t)j2 dt
=2
=2
dt
2
=
1
c. Use the transform pair
e
jtj
!
2
e jtj
or
2
2 + (2 f )
2
!
1
2 + (2 f )2
34
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
The desired integral, by Rayleighโs energy theorem, is
Z 1
2
e 2 jtj
1
df
=
dt
2
2 + (2 f )2
1 4
1
Z 1
1
1 e 2 t
2
1
2 t
e
dt
=
=
4 2 0
2 2 2 0
4 3
Z 1
I3 =
=
d. Use the transform pair
1
t
! sinc2 ( f )
The desired integral, by Rayleighโs energy theorem, is
Z 1
Z 1
2
jX4 (f )j df =
sinc4 ( f ) df
I4 =
1
1
Z 1
Z
1
2
2
=
(t= ) dt = 2
[1 (t= )]2 dt
2
=
2
Z 1
1
(1
0
2
u)2 du =
1
1
3
0
1
u2
=
0
2
3
Problem 2.35
a. The convolution operation gives
8
< 0;
1
1 e
y1 (t) =
: 1
e (t
(t
+1=2)
;
1=2)
e
(t
t
1=2
1=2 + 1=2
b. The convolution of these two signals gives
y2 (t) =
(t) + trap (t)
where trap(t) is a trapezoidal function given by
8
0;
t 3=2
>
>
<
1;
1=2 t 1=2
trap (t) =
3=2 + t; 3=2 t
>
:
3=2 t;
1=2 t
t
1=2
>
< tR 1=2 e d ; R
0
t+1=2
y3 (t) =
e
d ; 1=2
>
: R t+1=2 e
d ;
t > 1=2
t 1=2
Integration of these three cases gives
8 1
e (t+1=2) e (t 1=2) ;
<
1
y3 (t) =
e (t 1=2) e (t+1=2) ;
: 1
e (t 1=2) e (t+1=2) ;
d. The convolution gives
y4 (t) =
Z t
t
1=2 1=2
x( )d
1
Problem 2.36
a. The inverse FT of (f ) is sinc(t). By the time delay theorem, the inverse Fourier
transform of (f ) exp ( j4 f ) is sinc(t 2). The product of this and 2 cos (2 f ) in
the frequency domain has an inverse Fourier transform which is the convolution of
their respective Fourier transforms. Thus
x1 (t) = sinc (t
= sinc (t
b. The inverse Fourier transform of
2) [ (t
3) + sinc (t
1) + (t + 1)]
1)
(f =2) is 2sinc2 (2t). By the time delay theorem
x2 (t) = 2sinc2 [2 (t
2:5)]
c. The inverse Fourier transform of (f =2) is 2sinc(2t). By the modulation theorem,
f +4
f 4
the inverse Fourier transform of
+
is sinc(2t) cos (8 t). By the time
2
2
delay theorem
x3 (t) = sinc [2 (t 4)] cos [2 (t 4)]
36
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.37
a. From before, the total energy is E1; total = 21 . The Fourier transform of the given
signal is
1
X1 (f ) =
+ j2 f
so that the energy spectral density is
G1 (f ) = jX1 (f )j2 =
1
2 + (2
f )2
By Rayleighโs energy theorem, the normalized inband energy is
Z W
E1 (jf j W )
=2
E1; total
df
2
= tan 1
2
2 + (2 f )
W
2 W
b. The total energy is E2; total = . The Fourier transform of the given signal and its
energy spectral density are, respectively,
X2 (f ) = sinc (f ) and G2 (f ) = jX2 (f )j2 =
2
sinc2 (f )
By Rayleighโs energy theorem, the normalized inband energy is
E2 (jf j W )
1
=
E2; total
Z W
2
sinc2 (f ) df = 2
W
Z
W
sinc2 (u) du
0
The integration must be carried out numerically.
c. The total energy is
Z 1h
E3; total =
e
t
t
e
0
=
1
2
2
+
+
i2
dt =
1
=
2
Z 1h
e 2 t
2e ( + )t + e 2 t
0
( + )
2
i2
dt
4 + ( + )
(
)2
=
( + )
2 ( + )
The Fourier transform of the given signal and its energy spectral density are, respectively,
X3 (f ) =
1
+ j2 f
1
+ j2 f
2.1. PROBLEM SOLUTIONS
37
and
1
+ j2 f
2
G3 (f ) = jX3 (f )j =
1
2 + (2 f )2
1
2 + (2 f )2
=
=
1
+ (2 f )2
j2 f
+ j2 f
1
2 Re
+ 2
2 + (2 f )2 2 + (2 f )2
+ (2 f )2
2
3
(
) j2 f + (2 f )2 5
1
2 Re 4
+ 2
2
2
2
2 + (2 f )
+ (2 f )2
+ (2 f )
2 Re
1
=
2 + (2
f )2
1
2 + (2 f )2
=
2
1
+ j2 f
1
1
+
+ j2 f
j2 f
2
+ (2 f )2
2
2
2 + (2
2
f)
+
2
2
+ (2 f )
1
+ (2 f )2
The normalized inband energy is
E3 (jf j W )
1
=
E3; total
E3; total
Z W
G3 (f ) df
W
The โฆrst and third terms may be integrated easily as inverse tangents. The second term
may be integrated after partial fraction expansion:
+ (2 f )2
2
2 + (2
2
2
f)
+ (2 f )
where
A
2 + (2
2 +
f)
Therefore
=
=
=
=
=
1
E3; total
Z W
1
E3; total
1
E3; total
1
E3; total
2
(
)
1
2 and B = 2
2
1
2 + (2
A
2
f)
1
tan
1
2 W
A
B
tan 1
2 W
(1
B
+ (2 f )2
A) tan 1
1
+
tan 1
2
2
2 + (2
W
”
2
2
2
A=2
E3 (jf j W )
E3; total
=
2
tan 1
2 W
tan
2
f)
1
2 W
+ 1 tan 1
2 W
+
2
1
2 W
B
+
+ (2 f )2
#
B) tan 1
(1
2 W
1
tan 1
2 W
tan 1
2 W
2 W
2
1
df
+ (2 f )2
38
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Pulse
Fraction of total energy
1
E1/E1,tot
x 1(t)
1
0.5
0
-2
0
2
0.5
0
4
0
2
ฮฑt
E2/E2,tot
x 2(t)
0.5
0
2
0
4
0
2
6
4
6
ฯW
1
1
ฮฒ = 2ฮฑ
E3/E3,tot
x 3(t)
4
0.5
t/ฯ
0.5
0
-2
6
1
1
0
-2
4
W/ ฮฑ
0
2
ฮฑt
4
0.5
0
0
2
W/ ฮฑ
Plots of the signals and inband energy for all three cases are shown in Fig. 2.5
2.1. PROBLEM SOLUTIONS
39
Problem 2.38
a. By the modulation theorem
X (f ) =
=
AT0
4
AT0
4
sinc (f
sinc
1
2
f0 )
f
f0
T0
T0
+ sinc (f + f0 )
2
2
1 f
1 + sinc
+1
2 f0
b. Use the superposition and modulation theorems to get
X (f ) =
AT0
4
sinc
f
2f0
+
1
1
sinc
2
2
f
f0
2
+ sinc
1
2
f
+2
f0
c. In this case, p(t) = x(t) and P (f ) = X(f ) of part (a) and Ts = T0 : From part (a),
we have
AT0
n 1
n+1
P (nf0 ) =
sinc
+ sinc
4
2
2
Using this in (2.149), we have the Fourier transform of the half-wave rectiโฆed cosine waveform as
1
X
n 1
n+1
A
X(f ) =
sinc
+ sinc
(f nf0 )
4
2
2
n= 1
Note that sinc(x) = 0 for integer values of its argument and it is 1 for its argument 0. Also,
use sinc(1=2) = 2= ; sinc(3=2) = 2=3 , etc. to get
X (f ) =
A
(f ) +
A
[ (f
4
A
[ (f
15
f0 ) + (f + f0 )] +
A
[ (f
3
2f0 ) + (f + 2f0 )]
4f0 ) + (f + 4f0 )] +
Problem 2.39
Signals x1 (t), x2 (t), and x6 (t) are real and even. Therefore their Fourier transforms are
real and even. Signals x3 (t), x4 (t), and x5 (t) are real and odd. Therefore their Fourier
transforms are imaginary and odd. Using the Fourier transforms for a square pulse and a
triangle along with superposition, time delay, and scaling, the Fourier transforms of these
signals are the following.
a. X1 (f ) = 2sinc2 (2f ) + 2sinc(2f )
b. X2 (f ) = 2sinc(2f ) sinc2 (f )
40
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. X3 (f ) = sinc(f ) ej f sinc(f ) e j f = 2j sin ( f )sinc(f )
d. X4 (f ) = sinc2 (f ) e j2 f sinc2 (f ) ej2 f =
2j sin (2 f )sinc2 (f )
e. By duality sgn(t) ! j= ( f ) so, by the convolution theorem of Fourier transforms,
X5 (f ) =sinc2 (f ) j= ( f ). The convolution cannot be carried out in closed form,
but it is clear that the result is imaginary and odd.
f. By the modulation theorem X6 (f ) = 12 sinc2 (f
even.
1) + 21 sinc2 (f + 1) which is real and
Problem 2.40
a. Write
6 cos (20 t) + 3 sin (20 t) = R cos (20 t
)
= R cos cos (20 t) + R sin sin (20 t)
Thus
R cos
= 6 cos (20 t)
R sin
= 3 sin (20 t)
Square both equations, add, and take the square root to obtain
p
p
R = 62 + 32 = 45 = 6:7082
Divide the second equation by the โฆrst to obtain
tan
= 0:5
= 0:4636 rad
So
x (t) = 3 +
p
45 cos (20 t
0:4636)
Following Example 2.19, we obtain
Rx ( ) = 3 2 +
45
cos (20
2
)
b. Taking the Fourier transform of Rx ( ) we obtain
Sx (f ) = 9 (f ) +
45
[ (f
4
10) + (f + 10)]
2.1. PROBLEM SOLUTIONS
41
Problem 2.41
Use the facts that the power spectral density integrates to give total power, it must be even,
and contains no phase information.
a. The total power of this signal is 22 =2 = 2 watts which is distributed equally at the
frequencies 10 hertz. Therefore, by inspection we write
S1 (f ) = (f
10) + (f + 10) W/Hz
b. The total power of this signal is 32 =2 = 4:5 watts which is distributed equally at the
frequencies 15 hertz. Therefore, by inspection we write
S2 (f ) = 2:25 (f
15) + 2:25 (f + 15) W/Hz
c. The total power of this signal is 52 =2 = 12:5 watts which is distributed equally at the
frequencies 5 hertz. Therefore, by inspection we write
S3 (f ) = 6:25 (f
5) + 6:25 (f + 5) W/Hz
d. The power of the โฆrst component of this signal is 32 =2 = 4:5 watts which is distributed
equally at the frequencies 15 hertz. The power of the second component of this
signal is 52 =2 = 12:5 watts which is distributed equally at the frequencies 5 hertz.
Therefore, by inspection we write
S4 (f ) = 2:25 (f
15) + 2:25 (f + 15) + 6:25 (f
5) + 6:25 (f + 5) W/Hz
Problem 2.42
Since the autocorrelation function and power spectral density of a signal are Fourier transform pairs, we may write down the answers by inspection using the Fourier transform pair
A cos (2 f0 t) ! A2 (f f0 ) + A2 (f + f0 ). The answers are the following.
a. R1 ( ) = 8 cos (30
); Average power = R1 (0) = 8 W.
b. R2 ( ) = 18 cos (40
); Average power = R2 (0) = 18 W.
c. R3 ( ) = 32 cos (10
); Average power = R3 (0) = 32 W.
d. R4 ( ) = 18 cos (40
) + 32 cos (10
); Average power = R4 (0) = 50 W.
42
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Problem 2.43
The autocorrelation function must be (1) even, (2) have an absolute maximum at
and (3) have a Fourier transform that is real and nonnegative.
= 0,
a. Acceptable – all properties satisโฆed;
b. Acceptable – all properties satisโฆed;
c. Not acceptable – none of the properties satisโฆed;
d. Acceptable – all properties satisโฆed;
e. Not acceptable – property (3) not satisโฆed;
f. Not acceptable – none of the properties satisโฆed.
Problem 2.44
2
Given that the autocorrelation function of x (t) = A cos (2 f0 t + ) is Rx ( ) = A2 cos (2 f0 )
(special case of Ex. 2.19), the results are as follows.
a. R1 ( ) = 2 cos (10
);
b. R2 ( ) = 2 cos (10
);
c. R3 ( ) = 5 cos (10
) (write the signal as x3 (t) = Re 5 exp j tan 1 (4=3) exp (j10 t) );
2
2
d. R4 ( ) = 2 +2
cos (10
2
) = 4 cos (10
).
Problem 2.45
This is a matter of applying (2.151) by making the appropriate identiโฆcations with the
parameters given in Example 2.20
Problem 2.46
Fourier transform both sides of the diยคerential equation using the diยคerentiation theorem
of Fourier transforms to get
[j2 f + a] Y (f ) = [j2 bf + c] X (f )
Therefore, the frequency response function is
H (f ) =
Y (f )
c + j2 bf
=
X (f )
a + j2 f
2.1. PROBLEM SOLUTIONS
43
a = 1; b = 2; c = 0
a = 1; b = 0; c = 3
2
3
mag(H)
mag(H)
1.5
1
2
1
0.5
0
0
-5
5
2
2
1
1
angle(H), rad
angle(H), rad
0
-5
0
-1
-2
-5
0
f, Hz
5
0
f, Hz
5
0
-1
-2
-5
5
0
The amplitude response function is
q
and the phase response is
c2 + (2 bf )2
jH (f )j = q
a2 + (2 f )2
arg [H (f )] = tan 1
2 bf
c
tan 1
2 f
a
Amplitude and phase responses for various values of the constants are plotted in Figure 2.6.
Problem 2.47
a. Use the transform pair
Ae
t
u (t)
!
to โฆnd the unit impulse response as
h1 (t) = e 7t u (t)
A
+ j2 f
44
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
b. Long division gives
7
7 + j2 f
H2 (f ) = 1
Use the transform pair (t)
(a) to get
! 1 along with superposition and the transform pair in part
h2 (t) = (t)
7e 7t u (t)
c. Use the time delay theorem along with the result of part (a) to get
h3 (t) = e 7(t 3) u (t
3)
d. Use superposition and the results of parts (a) and (c) to get
h4 (t) = e 7t u (t)
e 7(t 3) u (t
3)
Problem 2.48
Use the transform pair for a sinc function to โฆnd that
Y (f ) =
f
2B
f
2W
a. If W

**B, it follows that Y (f ) = because f 2W = 1 throughout the region where f 2B f 2B is nonzero. c. Part (b) gives distortion because the output spectrum diยคers from the input spectrum. In fact, the output is y (t) = 2Bsinc (2Bt) which clearly diยคers from the input for W > B. 2.1. PROBLEM SOLUTIONS 45 Problem 2.49 a. Replace the capacitors with 1=j!C which is their ac-equivalent impedance. Call the junction of the input resistor, feedback resistor, and capacitors 1. Call the junction at the positive input of the operational ampliโฆer 2. Call the junction at the negative input of the operational ampliโฆer 3. Write down the KCL equations at these three junctions. Use the constraint equation for the operational ampliโฆer, which is V2 = V3 , and the deโฆnitions for ! 0 , Q, and K to get the given transfer function as H (j!) = Vo (j!) =Vi (j!). The node equations are V1 Vi R + j!CV1 + V1 Vo R + j!C(V1 j!C(V2 V2 ) = 0 V1 ) + V2 R Vo = 0 V3 V3 + = 0 Rb Ra (constraint on op amp input) V2 = V3 b. See plot given in Figure 2.7. c. In terms of f , the transfer function magnitude is K jH(f )j = p s 2 (f =f0 ) 1 f f0 2 2 + Q12 f f0 2 It can be shown that p the maximim of jH(f )j is at f = f0 . By substitution, this maximum is jH(f0 )j = KQ= p2. To โฆnd the 3-dB bandwidth, we must โฆnd the frequencies for which jH(f )j = jH(f0 )j= 2. This results in Q p =s 2 (f3 =f0 ) 1 f3 f0 2 2 + Q12 f3 f0 2 which reduces to the quadratic equation f3 f0 4 2+ 1 Q2 f3 f0 2 +1=0 46 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Using the quadratic formula, the solutions to this equation are f 3 f0 f 3 f0 2 = 1 2 s 1 2+ 2 Q 1 1 2+ 2 = 2 Q r 1 1 1 Q 1 Q 2+ r 1+ 1 Q2 4 1 4Q2 1 1 Q f0 = f0 Q 1 ; Q >> 1 2Q Therefore, the 3-dB bandwidth is, for large Q, approximately B = f3 f 3 1+ 1 2Q f0 1 1 2Q d. Combinations of components giving RC = 2:2508 10 4 seconds and Ra = 2:5757 Rb will work. Problem 2.42 a. By voltage division, with the inductor replaced by j2 f L, the frequency response function is R2 + j2 f L R2 =L + j2 f H1 (f ) = = R1 + R2 + j2 f L (R1 + R2 ) =L + j2 f By long division H1 (f ) = 1 R1 =L R1 +R2 + j2 L f Using the transforms of a delta function and a one-sided exponential, we obtain h1 (t) = (t) R1 exp L R1 + R2 t u (t) L 2.1. PROBLEM SOLUTIONS 47 20 |H(f)|, dB 10 0 -10 f0 = 999.9959 Hz; B 3 dB = 300.0242 Hz -20 2 10 3 4 10 10 angle(H(f)), radians 2 1 0 -1 -2 2 10 3 4 10 f, Hz 10 b. Substituting the ac-equivalent impedance for the inductor and using voltage division, the frequency response function is H2 (f ) = = R2 jj (j2 f L) j2 f LR2 where R2 jj (j2 f L) = R1 + R2 jj (j2 f L) R2 + j2 f L j2 f L (R1 k R2 ) =L R2 R2 1 = R2 R1 + R2 RR1+R R + R (R + j2 f L 1 2 1 k R2 ) =L + j2 f 1 2 R2 where R1 k R2 = RR11+R . Therefore, the impulse response is 2 h2 (t) = R2 R1 + R2 (t) R1 R2 exp (R1 + R2 ) L R1 R2 t u (t) (R1 + R2 ) L 2 Both have a high pass amplitude response, with the dc gain of the โฆrst circuit being R1R+R 2 and the second being 0; the high frequency gain of the โฆrst is 1 and that of the second is R2 R1 +R2 . Problem 2.51 Application of the Payley-Wiener criterion gives the integral Z 1 f2 I= df 2 1 1+f 48 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS which does not converge. Hence, the given function is not suitable as the frequency response function of a causal LTI system. Problem 2.52 a. The condition for stability is Z 1 Z 1 jexp ( jh1 (t)j dt = 1 Z11 exp ( = 2 0 which follows because jcos (2 f0 t)j jtj) cos (2 f0 t)j dt t) jcos (2 f0 t)j dt < 2 Z 1 exp ( t) dt = 0 2 <1 1. Hence this system is BIBO stable. b. The condition for stability is Z 1 Z 1 jh2 (t)j dt = jcos (2 f0 t) u (t)j dt 1 1 Z 1 = jcos (2 f0 t)j dt ! 1 0 which follows by integrating one period of jcos (2 f0 t)j and noting that the total integral is the limit of one period of area times N as N ! 1: This system is not BIBO stable. c. The condition for stability is Z 1 Z 1 1 jh3 (t)j dt = u (t 1) dt 1 1 jtj Z 1 dt = ln (t)j1 = 1 !1 t 1 This system is not BIBO stable. d. The condition for stability is Z 1 Z 1 jh4 (t)j dt = je t u (t) e ((t 1)) u (t 1) jdt 1 1 Z 1 e t e (t 1) dt = 1 + e < 1 0 This system is BIBO stable. 2.1. PROBLEM SOLUTIONS 49 e. The condition for stability is Z 1 1 jh5 (t)j dt = Z 1 1 t 2 jdt = 1 < 1 This system is BIBO stable. f. The condition for stability is Z 1 1 jh6 (t)j dt = Z 1 sinc (2t) dt = 1 1 0 2sinc (2f ) exp (j =2) ; f 1 : 8 > 1 = 15 + (2 f ) i2 > 15 (2 f )2 + (16 f )2 ; h i 8 15 + (2 f )2 19 (6 ) 2 2 > : 361 + (6 f ) h 57 + 361 + (6 f )2 225 + 34 (2 f )2 + (2 f )4 The phase delay is Tp2 (f ) = 2 (f ) 2 f tan 1 = 16 f 15 (2 f )2 tan 1 6 f 19 2 f The group and phase delays for (a) and (b) are shown in Fig. 2.9. c. The frequency response is H (f ) = f 2B exp ( j2 t0 f ) The group delay is 1 d [ 2 t0 f ] ; B f B 2 df = t0 ; B f B and 0 otherwise Tg (f ) = The phase delay is Tp (f ) = ( 2 t0 f ) = t0 ; 2 f B f B and 0 otherwise 9 > = > ; 56 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Tg1(f), Tp1(f), s 0.2 Group delay Phase delay 0.15 0.1 0.05 0 -2 10 -1 0 10 10 f, Hz 1 10 2 10 Tg2(f), Tp2(f), s 0.5 0 -0.5 -1 -2 10 -1 0 10 10 f, Hz 1 10 d. The frequency response function is H (f ) = = 5 2 exp ( j2 t0 f ) 3 + j2 f 3 + j2 f 2 [2:5 exp ( j2 t0 f )] 3 + j2 f The phase shift function is (f ) = sin 2 t0 f 2:5 cos 2 t0 f tan 1 2 f 3 The phase delay is Tp (f ) = 2 f sin 2 t0 f 1 tan 1 = 2 f 3 2:5 cos 2 t0 f The group delay is Tg (f ) = = Problem 2.59 1 d 2 f sin 2 t0 f tan 1 = 2 df 3 2:5 cos 2 t0 f 3 1 2:5 cos 2 t0 f 2 + 9 + (2 f ) (2:5 cos 2 t0 f )2 2 10 2.1. PROBLEM SOLUTIONS 57 a. The amplitude response is jH (f ) j = r 2 jf j q 64 + (2 f )2 9 + (2 f )2 b. The phase response is (f ) = 2 sgn (f ) f 4 tan 1 + tan 1 2 f 3 tan 1 2 f 3 c. The phase delay is Tp (f ) = 1 tan 1 2 f f 4 2 sgn (f ) d. The group delay is Tg (f ) = 1=8 1=3 2 + 1 + ( f =4) 1 + ( f =3)2 1 (f ) 4 Problem 2.60 In terms of the input spectrum, the output spectrum is Y (f ) = X (f ) + 0:1X (f ) X (f ) f 10 f + 10 = 2 + 4 4 f + 10 f 10 + +0:4 4 4 f 10 f + 10 = 2 + 4 4 f 20 f +0:4 4 +8 +4 4 4 f 10 4 + f + 10 4 f + 20 4 where (f ) is an isoceles triangle of unit height going from -1 to 1. The student should sketch the output spectrum given the above analytical result. Problem 2.61 a. The amplitude response is jH(f )j = q 2 jf j (9 4 2 f 2 )2 + (0:3 f )2 58 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS 8 |H(f)| 6 4 2 0 -3 -2 -1 0 1 2 3 -2 -1 0 f, Hz 1 2 3 ฮธ(f), radians 2 1 0 -1 -2 -3 b. The phase response is (f ) = 2 sgn (f ) tan 1 0:3 f 9 4 2f 2 These are shown in Fig. 2.10. c. The phase delay is Tp (f ) = 1 2 0:3 f 9 4 2f 2 sgn (f ) tan 1 tan 1 0:3 f 9 4 2f 2 2 d. The group delay is Tg (f ) = = = 1 d 2 df 2 1 6 4 2 81 2 sgn (f ) 0:3 1 (f ) 1+ 0:3 f 9 4 2f 2 1:35 + 0:6 2 f 2 71:91 2 f 2 + 16 4 f 4 2 1 (f ) 2 9 4 2f 2 (9 + (0:3 f ) 8 4 2 f 2 )2 2f 3 7 5 2.1. PROBLEM SOLUTIONS 59 Problem 2.62 Let u = 2 t. We then have y (t) = [cos (u) + cos (3u)]3 = cos3 (u) + 3 cos2 (u) cos (3u) + 3 cos (u) cos2 (3u) + cos3 (3u) Use the trig identities cos2 (z) = cos3 (z) = cos (w) cos (z) = 1 [1 + cos (2z)] 2 3 1 cos (z) + cos (3z) 4 4 1 1 cos (w z) + cos (w + z) 2 2 to get 1 3 3 cos (u) + cos (3u) + [1 + cos (2u)] cos (3u) 4 4 2 3 3 1 + cos (u) [1 + cos (6u)] + cos (3u) + cos (9u) 2 4 4 5 3 3 1 = 3 cos (u) + cos (3u) + cos (5u) + cos (7u) + cos (9u) 2 2 4 4 3 3 1 5 = 3 cos (2 t) + cos (6 t) + cos (10 t) + cos (14 t) + cos (18 t) 2 2 4 4 y (t) = Problem 2.63 Write the transfer function as H (f ) = H0 e j2 f t0 H0 f 2B e j2 f t0 Use the inverse Fourier transform of a constant, the delay theorem, and the inverse Fourier transform of a rectangular pulse function to get h (t) = H0 (t t0 ) 2BH0 sinc [2B (t Problem 2.64 a. The Fourier transform of this signal is p X (f ) = A 2 b2 exp 2 2 2f 2 t0 )] 60 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS By deโฆnition, using a table of integrals, Z 1 p 1 T = jx (t)j dt = 2 x (0) 1 Similarly, 1 W = 2X (0) Therefore, Z 1 1 jX (f )j df = p 2 2 1 p 2 2W T = p 2 2 2 =1 b. The Fourier transform of this signal is X (f ) = The pulse duration is 1 T = x (0) The bandwidth is W = Thus, Z 1 1 2X (0) 2A= 1 + (2 f = )2 1 Z 1 2W T = 2 jx (t)j dt = 1 2 jX (f )j df = 2 4 4 =1 Problem 2.65 a. The poles for a second order Butterworth โฆlter are given by !3 s1 = s2 = p (1 j) 2 where ! 3 is the 3-dB cutoยค frequency of the Butterworth โฆlter. function is H (s) = h !3 s+ p (1 2 Its s-domain transfer !2 !2 i h3 i= p 3 !3 s2 + 2! 3 s + ! 23 j) s + p (1 + j) 2 Letting ! 3 = 2 f3 and s = j! = j2 f , we obtain H (j2 f ) = 4 2f 2 + p 4 2 f32 = 2 (2 f3 ) (j2 f ) + 4 2 f32 f2 p3 f 2 + j 2f3 f + f32 2.1. PROBLEM SOLUTIONS 61 b. If the phase response function of the โฆlter is Tg (f ) = (f ), the group delay is 1 d [ (f )] 2 df For the second-order Butterworth โฆlter considered here, ! p 2f f 3 (f ) = tan 1 f32 f 2 Therefore, the group delay is Tg (f ) = = ” 1 d tan 1 2 df p 2f3 f 2 f3 f 2 !# f f2 + f2 1 1 + (f =f3 )2 p 3 34 p = 2 f3 + f 4 2 f3 1 + (f =f3 )4 This is plotted in Fig. 2.11. c. Use partial fraction expansion of H (s) =s and then inverse Laplace transform it to get the given step response. The expansion is !2 A Bs + C p 3 p = + 2 2 2 s s s + 2! 3 s + ! 3 s + 2! 3 s + ! 23 p where A = 1; B = 1; C = 2! 3 H (s) s = This allows H (s) =s to be written as H (s) s = 1 s = 1 s = 1 s = 1 s = 1 s p s + 2! 3 p s2 + 2! 3 s + ! 23 p s + 2! 3 p s2 + 2! 3 s + ! 23 =2 ! 23 =2 + ! 23 p s + 2! 3 p 2 s + ! 3 = 2 + ! 23 =2 p p p s + ! 3 = 2 ! 3 = 2 + 2! 3 p 2 s + ! 3 = 2 + ! 23 =2 p p s + !3= 2 !3= 2 p 2 p 2 s + ! 3 = 2 + ! 23 =2 s + ! 3 = 2 + ! 23 =2 62 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Step response for a 2nd-order BW filter 1.4 0.3 1.2 0.25 1 0.2 0.8 y s (t) f3Tg(f) Group delay for a 2nd-order BW filter 0.35 0.15 0.6 0.1 0.4 0.05 0.2 0 -1 10 0 10 f/f3 1 10 0 0 0.5 f3t 1 Using the s-shift theorem of Laplace transforms, this inverse transforms to the given expression for the step response (with ! 3 = 2 f3 ). Plot it and estimate the 10% and 90% times from the plot. From the MATLAB plot of Fig. 2.10, f3 t10% 0:08 and f3 t90% 0:42 so that the 10-90 % rise time is about 0:34=f3 seconds. Problem 2.66 a. Slightly less than 0.5 seconds; b & c. Use sketches to show. Problem 2.67 a. The leading edges of the โกat-top samples follow the waveform at the sampling instants. b. The spectrum is Y (f ) = X (f ) H (f ) where X (f ) = fs 1 X n= 1 X (f nfs ) 2.1. PROBLEM SOLUTIONS 63 and H (f ) = sinc (f ) exp ( j f ) The latter represents the frequency response of a โฆlter whose impulse response is a square pulse of width and implements โกat top sampling. If W is the bandwidth of X (f ), very 1 >> W . little distortion will result if Problem 2.68 a. The sampling frequency should be large compared with the bandwidth of the signal. b. The output spectrum of the zero-order hold circuit is Y (f ) = sinc (Ts f ) 1 X X (f nfs ) exp ( j f Ts ) n= 1 where fs = Ts 1 . For small distortion, we want Ts << W 1. Problem 2.69 Use trig identities to rewrite the signal as a sum of sinusoids: x (t) = 10 cos2 (600 t) cos (2400 t) = 5 [1 + cos (600 t)] cos (2400 t) = 5 cos (2400 t) + 2:5 cos (1800 t) + 2:5 cos (3000 t) The lowpass recovery โฆlter can cut oยค in the range 1.5 + kHz to 3 kHz where the superscript + means just above and the superscript means just below. The lower of these is the highest frequency of x (t) and the larger is equal to the sampling frequency minus the highest frequency of x (t). The minimum allowable sampling frequency is just above 3 kHz. Problem 2.70 For bandpass sampling and recovery, all but (b) and (e) will work theoretically, although an ideal โฆlter with bandwidth exactly equal to the unsampled signal bandwidth is necessary. For lowpass sampling and recovery, only (f) will work. 64 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Problem 2.71 The Fourier transform is 1 X (f 2 1 f0 ) + X (f + f0 ) 2 1 1 + [ jsgn (f ) X (f )] (f f0 ) e j =2 + (f + f0 ) ej =2 2 2 1 1 = X (f f0 ) [1 sgn (f f0 )] + X (f + f0 ) [1 + sgn (f + f0 )] 2 2 Y (f ) = Noting that and 1 [1 sgn (f f0 )] = u (f0 f ) 2 1 [1 + sgn (f + f0 )] = u (f + f0 ) 2 this may be rewritten as Y (f ) = X (f f0 ) u (f0 f ) + X (f + f0 ) u (f + f0 ) f Thus, if X (f ) = 2 (a unit-height rectangle 2 units wide centered at f = 0) and f0 = 10 Hz, Y (f ) would consist of unit-height rectangles going from 10 to 9 Hz and from 9 to 10 Hz. Problem 2.72 a. x ba (t) = cos (! 0 t 1 T !1 2T lim =2) = sin (! 0 t), so Z T T x (t) x b (t) dt = = = 1 T !1 2T lim 1 T !1 2T lim Z T sin (! 0 t) cos (! 0 t) dt T Z T 1 sin (2! 0 t) dt T 2 1 cos (2! 0 t) T =0 T !1 2T 4! 0 T lim b. Use trigonometric identities to express x (t) in terms of sines and cosines. Then โฆnd the Hilbert transform of x (t) by phase shifting by =2. Multiply x (t) and x b (t) together term by term, use trigonometric identities for the product of sines and cosines, then integrate. The integrand will be a sum of terms similar to that of part (a). The limit as T ! 1 will be zero term-by-term. 2.1. PROBLEM SOLUTIONS 65 c. For this case x b (t) = A exp (j! 0 t j =2) = jA exp (j! 0 t), take the product, integrate over time to get Z T Z T 1 1 x (t) x b (t) dt = lim [A exp (j! 0 t)] [jA exp (j! 0 t)] dt lim T !1 2T T !1 2T T T Z jA2 T = lim exp (j2! 0 t) dt = 0 T !1 2T T by periodicity of the integrand Problem 2.73 a. Note that F [jb x(t)] = j [ jsgn (f )] X (f ). Hence x1 (t) = = = 2 1 2 1 x (t) + jb x (t) ! X1 (f ) = X (f ) + j [ jsgn (f )] X (f ) 3 3 3 3 2 1 + sgn (f ) X (f ) 3 3 1 3 X (f ) ; f 0 b. It follows that x2 (t) 3 3 x (t) + jb x (t) exp (j2 f0 t) 4 4 3 ) X2 (f ) = [1 + sgn (f f0 )] X (f 4 0; f f0 2 X (f = f0 ) c. This case has the same spectrum as part (a), except that it is shifted right by W Hz. That is, 2 1 x (t) + jb x (t) exp (j2 W t) 3 3 2 1 ! X3 (f ) = + sgn (f W ) X (f 3 3 x3 (t) = W) d. For this signal x4 (t) 1 jb x (t) exp (j W t) 3 2 1 ! X4 (f ) = sgn (f W=2) X (f 3 3 = 2 x (t) 3 W=2) 66 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Problem 2.74 The Hilbert transform of the given signal is x b(t) = 2 sin (52 t) The signal xp (t) is xp (t) = x (t) + jb x(t) = 2 cos (52 t) + j2 sin (52 t) = 2ej52 t a. We have, for f0 = 25 Hz, x ~ (t) = xp (t) e j2 f0 t = 2ej52 t e j50 t = 2ej2 t = 2 cos (2 t) + j2 sin (2 t) xR (t) = 2 cos (2 t) xI (t) = 2 sin (2 t) b. For f0 = 27 Hz, x ~ (t) = 2ej52 t e j54 t = 2e j2 t = 2 cos (2 t) j2 sin (2 t) xR (t) = 2 cos (2 t) xI (t) = 2 sin (2 t) c. For f0 = 10 Hz, x ~ (t) = 2ej52 t e j20 t = 2ej32 t = 2 cos (32 t) + j2 sin (32 t) xR (t) = 2 cos (32 t) xI (t) = 2 sin (32 t) d. For f0 = 15 Hz, x ~ (t) = 2ej52 t e j30 t = 2ej22 t = 2 cos (22 t) + j2 sin (22 t) 2.1. PROBLEM SOLUTIONS 67 xR (t) = 2 cos (22 t) xI (t) = 2 sin (22 t) e. For f0 = 30 Hz, x ~ (t) = 2ej52 t e j60 t = 2e j8 t = 2 cos (8 t) j2 sin (8 t) xR (t) = 2 cos (8 t) xI (t) = 2 sin (8 t) f. For f0 = 20 Hz, x ~ (t) = 2ej52 t e j40 t = 2ej12 t = 2 cos (12 t) + j2 sin (12 t) xR (t) = 2 cos (12 t) xI (t) = 2 sin (12 t) Problem 2.75 For t =2, the result is y (t) = q ( =2) e t 2 + (2 f )2 n e In the above equations, =2 is given by = =2 cos [2 (f0 + f) t + ] o f) t + ] o 68 2.2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Computer Exercises Computer Exercise 2.1 % ce2_1.m: Amplitude spectra and Fourier series synthesized % for various periodic waveforms % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf N = input(โNumber of harmonics in Fourier sum => โ); T = input(โPeriod of periodic waveform => โ); A = input(โAmplitude of waveform => โ); n = -N:1:N; I_type = input(โ1 = rect pulse train; 2 = HR sinewave; 3 = FR sinewave; 4 = triang pulse train; 5 = triangle wave; 6 = sawtooth wave => โ); X = zeros(size(n)); if I_type == 1 tau = input(โWidth of rectangular pulse => โ); t0 = input(โDelay of rectangular pulse center => โ); d = tau/T; X = A*d*sinc(n*d).*exp(-j*2*pi*n*t0/T); elseif I_type == 2 I = โฆnd(n == 1 j n == -1); II = โฆnd(rem(n, 2) == 0); III = โฆnd(rem(abs(n), 2) == 1 & (n ~= 1 & n ~= -1)); X(I) = -0.25*j*n(I)*A; X(II) = A./(pi*(1-n(II).*n(II))); X(III) = 0; elseif I_type == 3 X = 2*A./(pi*(1-4*n.*n)); elseif I_type == 4 tau = input(โHalf width of triangle pulse => โ); t0 = input(โDelay of triangle pulse => โ); d = tau/T; X = A*d*sinc(n*d).*sinc(n*d).*exp(-j*2*pi*n*t0/T); elseif I_type == 5 I = โฆnd(rem(abs(n), 2) == 1); X(I) = 4*A./(pi^2*n(I).^2); elseif I_type == 6 2.2. COMPUTER EXERCISES 69 I = โฆnd(n~=0); X(I) = 2*A*(-exp(-j*2*pi*n(I))+(1 – exp(-j*2*pi*n(I)))./(j*2*pi*n(I)))./(j*2*pi*n(I)); end subplot(2,1,1), stem(n, abs(X)),xlabel(โnโ), ylabel(โjX_njโ), … if I_type == 1 title([โRectangular pulse train; period = โ, num2str(T), โ; delay = โ, num2str(t0), โ; ampli = โ, num2str(A)]) elseif I_type == 2 title([โHalf-rectiโฆed sinewave; period = โ, num2str(T),โ; ampli = โ, num2str(A)]) elseif I_type == 3 title([โFull-rectiโฆed sinewave; period = โ, num2str(T),โ; ampli = โ, num2str(A)]) elseif I_type == 4 title([โTriangle pulse train; โ, num2str(2*N+1), โterms. A = โ, num2str(A), โ, ntau = โ, num2str(tau), โ, T = โ, num2str(T) โs; d = โ, num2str(d), โ; t_0 = โ, num2str(t0), โsโ]) elseif I_type == 5 title([โTriangle waveform; period = โ, num2str(T),โ; ampli = โ, num2str(A)]) elseif I_type == 6 title([โSawtooth waveform; period = โ, num2str(T),โ; ampli = โ, num2str(A)]) end fn = n./T; t = -T:T/500:T; x = real(X*exp(j*2*pi*fnโ*t)); subplot(2,1,2), plot(t, x), xlabel(โtโ), ylabel(โx(t)โ), … >> ce2_1 Number of harmonics in Fourier sum => 25 Period of periodic waveform => 2 Amplitude of waveform => 1 1 = rect pulse train; 2 = HR sinewave; 3 = FR sinewave; 4 = triang pulse train; 5 = triangle wave; 6 = sawtooth wave => 6 >> Computer Exercise 2.2 % ce2_2.m: Plot of line spectra for half-rectiโฆed, % full-rectiโฆed sinewave, square wave, and triangle wave % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf 70 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Sawtooth waveform; period = 2 ; ampli = 1 0.4 |Xn| 0.3 0.2 0.1 0 -25 -20 -15 -10 -5 0 n 5 -0.5 0 t 0.5 10 15 20 25 1.5 2 2 x(t) 1 0 -1 -2 -2 -1.5 -1 1 waveform = input(โEnter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: โ); A = 1; n_max = 13; % maximum harmonic plotted; odd n = -n_max:1:n_max; if waveform == 1 X = A./(pi*(1+eps – n.^2)); % Oยคset 1 slightly to avoid divide by zero for m = 1:2:2*n_max+1 X(m) = 0; % Set odd harmonic lines to zero X(n_max+2) = -j*A/4; % Compute lines for n = 1 and n = -1 X(n_max) = j*A/4; end elseif waveform == 2 X = 2*A./(pi*(1+eps – 4*n.^2)); elseif waveform == 3 X = abs(4*A./(pi*n+eps)); for m = 2:2:2*n_max+1 X(m) = 0; end elseif waveform == 4 X = 4*A./(pi*n+eps).^2; for m = 2:2:2*n_max+1 2.2. COMPUTER EXERCISES 71 X(m) = 0; end end [arg_X, mag_X] = cart2pol(real(X),imag(X)); % Convert to magnitude and phase if waveform == 1 for m = n_max+3:2:2*n_max+1 arg_X(m) = arg_X(m) – 2*pi; % Force phase to be odd end elseif waveform == 2 m = โฆnd(n > 0); arg_X(m) = arg_X(m) – 2*pi; % Force phase to be odd elseif waveform == 4 arg_X = mod(arg_X, 2*pi); end subplot(2,1,1),stem(n, mag_X),ylabel(โjX_njโ) if waveform == 1 title(โHalf-rectiโฆed sine wave spectraโ) elseif waveform == 2 title(โFull-rectiโฆed sine wave spectraโ) elseif waveform == 3 title(โSpectra for square wave with even symmetry โ) elseif waveform == 4 title(โSpectra for triangle wave with even symmetryโ) end subplot(2,1,2),stem(n, arg_X),xlabel(โnf_0โ),ylabel(โangle(X_n)โ) >> ce2_2 Enter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: 1 >> Computer Exercise 2.3 % ce2_3.m: FFT plotting of line spectra for half-rectiโฆed, full-rectiโฆed sinewave, square wave, % and triangular waveforms % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % % User deโฆned functions used: pls_fn( ); trgl_fn( ) % clf 72 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Half-rectified sine wave spectra 0.4 |Xn| 0.3 0.2 0.1 0 -15 -10 -5 0 5 10 15 -10 -5 0 nf0 5 10 15 4 angle(Xn) 2 0 -2 -4 -15 I_wave = input(โEnter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect. sine; 4 = full-wave sine: โ); T = 2; del_t = 0.001; t = 0:del_t:T; L = length(t); fs = (L-1)/T; del_f = 1/T; n = 0:9; if I_wave == 1 x = pls_fn(2*(t-T/4)/T); X_th = abs(0.5*sinc(n/2)); disp(โโ) disp(โ0 – 1 level squarewaveโ) elseif I_wave == 2 x = 2*trgl_fn(2*(t-T/2)/T)-1; X_th = 4./(pi^2*n.^2); X_th(1) = 0; % Set n = even coeยข cients to zero (odd indexed because of MATLAB) X_th(3) = 0; X_th(5) = 0; X_th(7) = 0; X_th(9) = 0; disp(โโ) 2.2. COMPUTER EXERCISES 73 disp(โ0-dc level triangular waveโ) elseif I_wave == 3 x = sin(2*pi*t/T).*pls_fn(2*(t-T/4)/T); X_th = abs(1./(pi*(1-n.^2))); X_th(2) = 0.25; X_th(4) = 0; % Set n = odd coeยข cients to zero (even indexed because of MATLAB) X_th(6) = 0; X_th(8) = 0; X_th(10) = 0; disp(โโ) disp(โHalf-rectiโฆed sinewaveโ) elseif I_wave == 4 x = abs(sin(pi*t/T)); % Period of full-rectiโฆed sinewave is T/2 X_th = abs(2./(pi*(1-4*n.^2))); disp(โโ) disp(โFull-rectiโฆed sinewaveโ) end X = 0.5*ยคt(x)*del_t; % Multiply by 0.5 because of 1/T_0 with T_0 = 2 f = 0:del_f:fs; Y = abs(X(1:10)); Z = [nโYโX_thโ]; disp(โMagnitude of the Fourier coeยข cientsโ); disp(โโ) disp(โn FFT Theoryโ); disp(โ__________________________โ) disp(โโ) disp(Z); subplot(2,1,1),plot(t, x), xlabel(โtโ), ylabel(โx(t)โ) subplot(2,1,2),plot(f, abs(X),โoโ),axis([0 10 0 1]),… xlabel(โnโ), ylabel(โjX_njโ) % Unit-width pulse function % function y = pls_fn(t) y = stp_fn(t+0.5)-stp_fn(t-0.5); function y = stp_fn(t) % Function for generating the unit step % y = zeros(size(t)); 74 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS I = โฆnd(t >= 0); y(I) = ones(size(I)); function y = trgl_fn(t) % This function generates a unit-high triangle centered % at zero and extending from -1 to 1 % y = (1 – abs(t)).*pls_fn(t/2); % % End of script โฆle A typical run follows with a plot given in Fig. 2.13. >> ce2_3 Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect. sine; 4 = full-wave sine: 3 Half-rectiโฆed sinewave Magnitude of the Fourier coeยข cients n FFT Theory __________________________ 0 0.3183 0.3183 1.0000 0.2501 0.2500 2.0000 0.1062 0.1061 3.0000 0.0001 0 4.0000 0.0212 0.0212 5.0000 0.0001 0 6.0000 0.0091 0.0091 7.0000 0.0000 0 8.0000 0.0051 0.0051 9.0000 0.0000 0 Computer Exercise 2.4 Make the time window long compared with the pulse width. Computer Exercise 2.5 % ce2_5.m: Finding the energy ratio in a preset bandwidth % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % % User deโฆned functions used: pls_fn( ); trgl_fn( ) 2.2. COMPUTER EXERCISES 75 x(t) 1 0.5 0 0 0.2 0.4 0.6 0.8 1 t 1.2 1.4 1.6 1.8 2 0 1 2 3 4 5 n 6 7 8 9 10 |Xn| 1 0.5 0 76 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS % I_wave = input(โEnter type of waveform: 1 = rectangular; 2 = triangular; 3 = half-rect. sine; 4 = raised cosine: โ); tau = input(โEnter pulse width: โ); per_cent = input(โEnter percent total desired energy: โ); clf T = 20; f = []; G = []; del_t = 0.001; t = 0:del_t:T; L = length(t); fs = (L-1)/T; del_f = 1/T; % n = [0 1 2 3 4 5 6 7 8 9]; if I_wave == 1 x = pls_fn((t-tau/2)/tau); disp(โโ) disp(โRectangular pulseโ) elseif I_wave == 2 x = trgl_fn(2*(t-tau/2)/tau); disp(โโ) disp(โTriangular pulseโ) elseif I_wave == 3 x = sin(pi*t/tau).*pls_fn((t-tau/2)/tau); disp(โโ) disp(โHalf sinewaveโ) elseif I_wave == 4 x = abs(sin(pi*t/tau).^2.*pls_fn((t-tau/2)/tau)); disp(โโ) disp(โRaised sinewaveโ) end X = ยคt(x)*del_t; f1 = 0:del_f*tau:fs*tau; G1 = X.*conj(X); NN = โกoor(length(G1)/2); G = G1(1:NN); ยค = f1(1:NN); f = f1(1:NN+1); E_tot = sum(G); 2.2. COMPUTER EXERCISES 77 E_f = cumsum(G); E_W = [0 E_f]/E_tot; test = E_W – per_cent/100; L_test = length(test); k = 1; while test(k) > ce2_5 Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 = half-rect. sine; 4 = raised cosine: 3 Enter pulse width: 2 Enter percent total desired energy: 95 Computer Exercise 2.6 The program for this exercise is similar to that for Computer Exercise 2.5, except that the waveform is used in the energy calculation. Computer Exercise 2.7 Use Computer Example 2.2 as a pattern for the solution .**

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