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Principles of Communication 5Ed 习题答案.pdf
Chapter 2
Signal and Linear System Theory
2.1 Problem Solutions
Problem 2.1
For the single-sided spectra, write the signal in terms of cosines:
x(t) = 10 cos(4πt + π/8) + 6 sin(8πt + 3π/4)
= 10 cos(4πt + π/8) + 6 cos(8πt + 3π/4 − π/2)
= 10 cos(4πt + π/8) + 6 cos(8πt + π/4)
For the double-sided spectra, write the signal in terms of complex exponentials using Euler’s
theorem:
x(t) = 5 exp[(4πt + π/8)] + 5 exp[−j(4πt + π/8)]
+3 exp[j(8πt + 3π/4)] + 3 exp[−j(8πt + 3π/4)]
The two sets of spectra are plotted in Figures 2.1 and 2.2.
Problem 2.2
The result is
x(t) = 4ej(8πt+π/2) + 4e−j(8πt+π/2) + 2ej(4πt−π/4) + 2e−j(4πt−π/4)
= 8 cos (8πt + π/2) + 4 cos (4πt − π/4)
= −8 sin (8πt) + 4 cos (4πt − π/4)
1
2
10
5
CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
Single-sided amplitude
π/4
π/8
f, Hz
Single-sided phase, rad.
f, Hz
0 2 4 6
0 2 4 6
Figure 2.1:
Problem 2.3
(a) Not periodic.
(b) Periodic. To find the period, note that
6π
2π
= n1f0 and
20π
2π
= n2f0
Therefore
=
10
3
Hence, take n1 = 3, n2 = 10, and f0 = 1 Hz.
(c) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 7, and
f0 = 1 Hz.
(d) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 3, n3 = 11,
and f0 = 1 Hz.
n2
n1
Problem 2.4
(a) The single-sided amplitude spectrum consists of a single line of height 5 at frequency 6
Hz, and the phase spectrum consists of a single line of height -π/6 radians at frequency 6
Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of
6 and -6 Hz, and the double-sided phase spectrum consists of a line of height -π/6 radians
at frequency 6 Hz and a line of height π/6 at frequency -6 radians Hz.
(b) Write the signal as
xb(t) = 3 cos(12πt − π/2) + 4 cos(16πt)
From this it is seen that the single-sided amplitude spectrum consists of lines of height 3
and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists
2.1. PROBLEM SOLUTIONS
3
Double-sided amplitude
5
-6 -4 -2 0 2 4 6
Double-sided phase, rad.
π/4
π/8
-6 -4 -2
f, Hz
f, Hz
0 2 4 6
-π/8
-π/4
Figure 2.2:
4
CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
of a line of height -π/2 radians at frequency 6 Hz. The double-sided 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 -π/2 radians at frequency 6 Hz and a line of height π/2 radians
at frequency -6 Hz.
Problem 2.5
(a) This function has area
Area =
=
dt
(πt/²) ¸2
²−1· sin(πt/²)
∞Z−∞
(πu) ¸2
∞Z−∞ · sin(πu)
du = 1
A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the central
lobe of the function becomes narrower and higher. Thus, in the limit, it approximates a
delta function.
(b) The area for the function is
Area =
∞Z−∞
1
²
exp(−t/²)u (t) dt =
∞Z0
exp(−u)du = 1
A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the function
becomes narrower and higher. Thus, in the limit, it approximates a delta function.
(c) Area =R ²
−1 Λ (t) dt = 1. As ² → 0, the function becomes narrower
² (1 − |t| /²) dt =R 1
and higher, so it approximates a delta function in the limit.
1
−²
Problem 2.6
(a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9.
Problem 2.7
(a), (c), (e), and (f) are periodic. Their periods are 1 s, 4 s, 3 s, and 2/7 s, respectively.
The waveform of part (c) is a periodic train of impulses extending from -∞ to ∞ spaced
by 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). The
waveform of part (e) is a doubly-infinite 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 a raised
cosine of minimum and maximum amplitudes 0 and 2, respectively.
2.1. PROBLEM SOLUTIONS
5
Problem 2.8
(a) The result is
x(t) = Re¡ej6πt¢ + 6 Re³ej(12πt−π/2)´ = Rehej6πt + 6ej(12πt−π/2)i
(b) The result is
x(t) =
ej6πt +
1
2
1
2
e−j6πt + 3ej(12πt−π/2) + 3e−j(12πt−π/2)
(c) The single-sided amplitude spectrum consists of lines of height 1 and 6 at frequencies
of 3 and 6 Hz, respectively. The single-sided phase spectrum consists of a line of height
−π/2 at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height
3, 1/2, 1/2, and 3 at frequencies of −6, −3, 3, and 6 Hz, respectively. The double-sided
phase spectrum consists of lines of height π/2 and −π/2 at frequencies of −6 and 6 Hz,
respectively.
Problem 2.9
(a) Power. Since it is a periodic signal, we obtain
P1 =
1
T0Z T0
0
4 sin2 (8πt + π/4) dt =
1
T0Z T0
0
2 [1 − cos (16πt + π/2)] dt = 2 W
where T0 = 1/8 s is the period.
(b) Energy. The energy is
E2 =Z ∞
−∞
e−2αtu2(t)dt =Z ∞
0
e−2αtdt =
1
2α
(c) Energy. The energy is
E3 =Z ∞
−∞
e2αtu2(−t)dt =Z 0
−∞
e2αtdt =
1
2α
(d) Neither energy or power.
E4 = lim
T→∞Z T
−T
dt
(α2 + t2)1/4 = ∞
P4 = 0 since limT→∞
x2(t), its energy is the sum of the energies of these two signals, or E5 = 1/α.
(α2+t2)1/4 = 0.(e) Energy.
−T
Since it is the sum of x1(t) and
dt
1
T R T
6
CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
(f) Power. Since it is an aperiodic signal (the sine starts at t = 0), we use
P6 = lim
T→∞
= lim
T→∞
0
1
2T Z T
2T · T
1
sin2 (5πt) dt = lim
T→∞
1
2
sin (20πt)
20π
2 −
=
W
1
4
¸T
0
1
2T Z T
0
1
2
[1 − cos (10πt)] dt
Problem 2.10
(a) Power. Since the signal is periodic with period π/ω, we have
ω
πZ π/ω
P =
A2 |sin (ωt + θ)|2 dt =
(b) Neither. The energy calculation gives
0
ω
πZ π/ω
0
A2
2 {1 − cos [2 (ωt + θ)]} dt =
A2
2
E = lim
T→∞Z T
−T
(Aτ )2 dt
√τ + jt√τ − jt
The power calculation gives
1
2T Z T
−T
(Aτ )2 dt
√τ 2 + t2
= lim
T→∞
P = lim
T→∞
(c) Energy:
(Aτ )2
2T
= lim
−T
(Aτ )2 dt
√τ 2 + t2 → ∞
T→∞Z T
lnà 1 +p1 + T 2/τ 2
−1 +p1 + T 2/τ 2! = 0
E =Z ∞
0
A2t4 exp (−2t/τ ) dt =
3
4
A2τ 5 (use table of integrals)
(d) Energy:
E = 2ÃZ τ /2
0
22dt +Z τ
τ /2
12dt! = 5τ
Problem 2.11
(a) This is a periodic train of “boxcars”, each 3 units in width and centered at multiples of
6:
Pa =
1
6Z 3
−3
Π2µ t
3¶ dt =
1
6Z 1.5
−1.5
dt =
1
2
W
2.1. PROBLEM SOLUTIONS
7
(b) This is a periodic train of unit-high isoceles triangles, each 4 units wide and centered
at multiples of 5:
Pb =
1
5Z 2.5
−2.5
Λ2µ t
2¶ dt =
2
0 µ1 −
5Z 2
t
2¶2
dt = −
2
5
2
3µ1 −
2
0
=
4
15
W
(c) This is a backward train of sawtooths (right triangles with the right angle on the left),
each 2 units wide and spaced by 3 units:
t
2¶3¯¯¯¯¯
(d) This is a full-wave rectified cosine wave of period 1/5 (the width of each cosine pulse):
Pc =
1
0 µ1 −
3Z 2
t
2¶2
dt = −
1
3
2
3µ1 −
Pd = 5Z 1/10
−1/10 |cos (5πt)|2 dt = 2 × 5Z 1/10
0
· 1
2
+
0
2
t
=
2
9
W
2¶3¯¯¯¯¯
cos (10πt)¸ dt =
1
2
1
2
W
Problem 2.12
(a) E = ∞, P = ∞; (b) E = 5 J, P = 0 W; (c) E = ∞, P = 49 W; (d) E = ∞, P = 2 W.
Problem 2.13
(a) The energy is
(b) The energy is
+
−6
E =Z 6
−∞he−|t|/3 cos (12πt)i2
0 · 1
cos2 (6πt) dt = 2Z 6
dt = 2Z ∞
2
0
1
2
cos (12πt)¸ dt = 6 J
e−2t/3· 1
1
2
+
2
cos (24πt)¸ dt
E =Z ∞
where the last integral follows by the eveness of the integrand of the first one. Use a table
of definte integrals to obtain
E =Z ∞
0
e−2t/3dt +Z ∞
0
e−2t/3 cos (24πt) dt =
3
2
+
2/3
(2/3)2 + (24π)2
Since the result is finite, this is an energy signal.
(c) The energy is
E =Z ∞
−∞ {2 [u (t) − u (t − 7)]}2 dt =Z 7
0
4dt = 28 J
8
CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
Since the result is finite, this is an energy signal.
(d) Note that
which is called the unit ramp. The energy is
−∞
Z t
u (λ) dλ = r (t) =½ 0, t < 0
10¶2
0 µ t
[r (t) − 2r (t − 10) + r (t − 20)]2 dt = 2Z 10
t, t ≥ 0
E =Z ∞
−∞
dt =
20
3
J
where the last integral follows because the integrand is a symmetrical triangle about t = 10.
Since the result is finite, this is an energy signal.
Problem 2.14
(a) Expand the integrand, integrate term by term, and simplify making use of the orthog-
onality property of the orthonormal functions.
(b) Add and subtract the quantity suggested right above (2.34) and simplify.
(c) These are unit-high rectangular pulses of width T /4.
T /8, 3T /8, 5T /8, and 7T /8.
other and fill the interval [0, T ].
(d) Using the expression for the generalized Fourier series coefficients, we find that X1 =
1/8, X2 = 3/8, X3 = 5/8, and X4 = 7/8. Also, cn = T /4. Thus, the ramp signal is
approximated by
They are centered at t =
Since they are spaced by T /4, they are adjacent to each
t
T
=
1
8
φ1 (t) +
3
8
φ2 (t) +
5
8
φ3 (t) +
7
8
φ4 (t) , 0 ≤ t ≤ T
where the φn (t)s are given in part (c).
(e) These are unit-high rectangular pulses of width T /2 and centered at t = T /4 and 3T /4.
We find that X1 = 1/4 and X2 = 3/4.
(f) To compute the ISE, we use
²N =ZT |x (t)|2 dt −
NXn=1
cn¯¯X 2
n¯¯
64¢ = 5.208 × 10−3T .
16¢ = 2.083 × 10−2T .
0 (t/T )2 dt = T /3. Hence, for (d),
64 + 25
64 + 49
16 + 9
Note thatRT |x (t)|2 dt =R T
4¡ 1
64 + 9
3 − T
2¡ 1
3 − T
ISEd = T
For (e), ISEe = T
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