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 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
Single-sided amplitude Single-sided phase, rad.
10 π/4
5 π/8
f, Hz 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 20π
= n1 f0 and = n2 f0
2π 2π
Therefore
10 n2
=
3 n1
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.
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
f, Hz
-6 -4 -2 0 2 4 6
Double-sided phase, rad.
π/4
π/8
-6 -4 -2
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
Z∞ · ¸2
−1 sin(πt/²)
Area = ² dt
(πt/²)
−∞
Z∞ · ¸2
sin(πu)
= du = 1
(πu)
−∞
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
Z∞ Z∞
1
Area = exp(−t/²)u (t) dt = exp(−u)du = 1
²
−∞ 0
A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the function
becomes narrower
R² and higher. Thus,
R 1 in the limit, it approximates a delta function.
(c) Area = −² 1² (1 − |t| /²) dt = −1 Λ (t) dt = 1. As ² → 0, the function becomes narrower
and higher, so it approximates a delta function in the limit.
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.
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 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
Single-sided amplitude Single-sided phase, rad.
10 π/4
5 π/8
f, Hz 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 20π
= n1 f0 and = n2 f0
2π 2π
Therefore
10 n2
=
3 n1
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.
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
f, Hz
-6 -4 -2 0 2 4 6
Double-sided phase, rad.
π/4
π/8
-6 -4 -2
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
Z∞ · ¸2
−1 sin(πt/²)
Area = ² dt
(πt/²)
−∞
Z∞ · ¸2
sin(πu)
= du = 1
(πu)
−∞
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
Z∞ Z∞
1
Area = exp(−t/²)u (t) dt = exp(−u)du = 1
²
−∞ 0
A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the function
becomes narrower
R² and higher. Thus,
R 1 in the limit, it approximates a delta function.
(c) Area = −² 1² (1 − |t| /²) dt = −1 Λ (t) dt = 1. As ² → 0, the function becomes narrower
and higher, so it approximates a delta function in the limit.
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.
