# Solution manual fundamentals of electric circuits 3rd edition chapter17

Chapter 17, Problem 1.
Evaluate each of the following functions and see if it is periodic. If periodic, find its
period.
(a) f (t ) = cos π t + 2 cos 3 π t + 3 cos 5 π t
(b) y(t) = sin t + 4 cos 2 π t
(c) g(t) = sin 3t cos 4t
(d) h(t) = cos 2 t
(e) z(t) = 4.2 sin(0.4 π t + 10º)
+ 0.8 sin(0.6 π t + 50º)
(f) p(t) = 10
(g) q(t) = e −πt

Chapter 17, Solution 1.

(a)

This is periodic with ω = π which leads to T = 2π/ω = 2.

(b)

y(t) is not periodic although sin t and 4 cos 2πt are independently

periodic.

(c)

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)],
g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t
which is harmonic or periodic with the fundamental frequency
ω = 1 or T = 2π/ω = 2π.

(d)

h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and
a constant is also periodic, h(t) is periodic. ω = 2 or T = 2π/ω = π.

(e)

The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.
ω = 0.2π or T = 2π/ω = 10.

(f)

p(t) = 10 is not periodic.

(g)

g(t) is not periodic.

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Chapter 17, Problem 2.

Using MATLAB, synthesize the periodic waveform for which the Fourier
trigonometric Fourier series is
1 4 ⎛
1

1

f (t ) = − 2 ⎜ cos t + cos 3t + cos 5t + ⋅ ⋅ ⋅ ⎟
2 π ⎝
9
25

Chapter 17, Solution 2.
The function f(t) has a DC offset and is even. We use the following MATLAB code to
plot f(t). The plot is shown below. If more terms are taken, the curve is clearly indicating
a triangular wave shape which is easily represented with just the DC component and
three, cosinusoidal terms of the expansion.

for n=1:100
tn(n)=n/10;
t=n/10;
y1=cos(t);
y2=(1/9)*cos(3*t);
y3=(1/25)*cos(5*t);
factor=4/(pi*pi);
y(n)=0.5- factor*(y1+y2+y3);
end
plot(tn,y)

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Chapter 17, Problem 3.

Give the Fourier coefficients a0, an, and bn of the waveform in Fig. 17.47. Plot the
amplitude and phase spectra.

Figure 17.47
For Prob. 17.3.
Chapter 17, Solution 3.

T = 4, ωo = 2π/T = π/2
g(t) = 5,
10,
0,

012
T

1

0

0

2

ao = (1/T) ∫ g( t )dt = 0.25[ ∫ 5dt + ∫ 10dt ] = 3.75
an = (2/T)

T

∫ g( t ) cos(nω t )dt
o

0

1

= (2/4)[

1

∫ 5 cos( 2
0

2

t )dt + ∫ 10 cos(
1

t )dt ]
2

2

1

2

2

sin
t + 10
sin
t ] = (–1/(nπ))5 sin(nπ/2)
= 0.5[ 5

2 0

2 1
an =

bn = (2/T)

T

(5/(nπ))(–1)(n+1)/2,
0,
1

∫ g( t ) sin(nωo t )dt = (2/4)[

∫ 5 sin(

1

2

0

0

n = odd
n = even

2

t )dt + ∫ 10 sin( t )dt ]
1
2
2

− 2x5

2 x10

cos
t –
cos
t ] = (5/(nπ))[3 – 2 cos nπ + cos(nπ/2)]
= 0.5[

2 0

2 1
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n
1
2
3
4
5
6
7
8

an
–1.59
0
0.53
0
–0.32
0
0.23
0

bn
7.95
0
2.65
0.80
1.59
0
1.15
0.40

An
8.11
0
2.70
0.80
1.62
0
1.17
0.40

phase
–101.31
0
–78.69
–90
–101.31
0
–78.69
–90

8

An

0

π 2π 3π 4π 5π 6π 7π 8π

ω

0

π 2π 3π 4π 5π 6π 7π 8π

ω

–78.69˚
90˚
–101.31˚

φ

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Chapter 17, Problem 4.

Find the Fourier series expansion of the backward sawtooth waveform of Fig. 17.48.
Obtain the amplitude and phase spectra.

Figure 17.48
For Probs. 17.4 and 17.66.
Chapter 17, Solution 4.

f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π
T

ao = (1/T)

∫ f ( t )dt

an = (2/T)

∫ f ( t ) cos(nω t )dt

=

2

2

0

0

= (1/2) ∫ (10 − 5t )dt = 0.5[10t − (5t 2 / 2)] = 5

0

T

o

0

2

2

0

0

= (2/2)

2

∫ (10 − 5t ) cos(nπt )dt
0

∫ (10) cos(nπt )dt – ∫ (5t ) cos(nπt )dt
2

2

−5
5t
= 2 2 cos nπt +
sin nπt = [–5/(n2π2)](cos 2nπ – 1) = 0
n π
n
π
0
0
bn = (2/2)
=

2

∫ (10 − 5t ) sin(nπt )dt
0

2

∫ (10) sin(nπt )dt –
0

2

∫ (5t ) sin(nπt )dt

2

0

2

−5
5t
= 2 2 sin nπt +
cos nπt = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)

n π
0
0
Hence

f(t) = 5 +

10 ∞ 1
∑ sin(nπt )
π n =1 n

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Chapter 17, Problem 5.
Obtain the Fourier series expansion for the waveform shown in Fig. 17.49.

Figure 17.49
For Prob. 17.5.
Chapter 17, Solution 5.
T = 2π,

ω = 2π / T = 1
T

1
1
a o = ∫ z( t )dt = [1xπ − 2 xπ] = −0.5
T

0

an =

T

π

0

0

π

T

π

0

0

1
1
2
z( t ) cos nωo dt = ∫ 1 cos ntdt −

T
π
π

2
1
1
b n = ∫ z( t ) cos nωo dt = ∫ 1sin ntdt −
π
π
T

1

∫ 2 cos ntdt = nπ sin ..nt

2
π

sin nt π = 0
0 nπ

⎧ 6
1
2
π
2π ⎪
, n = odd
∫ 2 sin ntdt = − nπ cos nt 0 + nπ cos nt π = ⎨ nπ
⎪⎩0, n = even
π

Thus,
z( t ) = − 0.5 +

6
sin nt
n =1 nπ

n =odd

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Chapter 17, Problem 6.
Find the trigonometric Fourier series for
⎧ 5, 0 < t < π
f (t ) = ⎨
and f (t + 2π ) = f (t ).
⎩10, π < t < 2π

Chapter 17, Solution 6.
T=2π, ωo=2π/T = 1
T
π

1
1 ⎡
1
ao = ∫ f(t)dt =
(5π + 10π ) = 7.5
⎢ ∫ 5dt + ∫ 10dt ⎥ =
T0
2π ⎣ 0
π
⎦ 2π

T
π

2
2 ⎡
(
)cos
5cos
10cos ntdt ⎥ = 0
f
t
n
ω
tdt
ntdt
=
+
⎢∫
o

2π ⎣ 0
T0
π

T
π

⎤ 1⎡ 1
π 1
2π ⎤
2
2 ⎡
bn = ∫ f(t)sin nωotdt =

⎢ ∫ 5 sin ntdt + ∫ 10 s inntdt ⎥ = ⎢ − cos nt − cos nt
T0
2π ⎣ 0
0 n
π ⎥⎦
π
⎦ π ⎢⎣ n
⎧ 10
, n = odd
5
⎪−
=
⎡⎣cos π n − 1⎤⎦ = ⎨ nπ

⎪⎩ 0, n = even

an =

Thus,
f(t) = 7.5 −

10
sin nt
n= odd nπ

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Chapter 17, Problem 7.
* Determine the Fourier series of the periodic function in Fig. 17.50.

Figure 17.50
For Prob. 17.7.
* An asterisk indicates a challenging problem.

Chapter 17, Solution 7.
T = 3,
ao =

ωo = 2π / T = 2π / 3

T
2
3
⎤ 1
1
1⎡
=
+
f
(
t
)
dt
2
dt
(−1)dt ⎥ = (4 − 1) = 1
⎢∫

T0
3 ⎣0
2
⎦ 3

T
2
3
2
2nπ t
2⎡
2nπ t
2nπ t ⎤
an = ∫ f(t)cos
dt = ⎢ ∫ 2 cos
dt + ∫ (−1)cos
dt ⎥
T0
3
3 ⎣0
3
3
2

2⎡ 3
2nπt
3
2nπt
sin
sin
= ⎢2
−1
3 ⎢ 2 nπ
3 0
2 nπ
3

2

3⎤

3
4 nπ
⎥=
sin
3
2 ⎥⎦ nπ

T
2
3
2
2nπ t
2⎡
2nπ t
2nπ t ⎤
=
+

f
(
t
)sin
dt
2
sin
dt
(
1
)s
in
dt ⎥

∫2
T ∫0
3
3 ⎣ ∫0
3
3

2⎡
3
2nπ t 2
3
2nπ t 3 ⎤ 3
4nπ
cos
cos
(1− 2 cos
)
= ⎢ −2 x
+
⎥=
3 ⎢⎣
2nπ
3 0 2nπ
3 2 ⎥⎦ nπ
3
1 ⎛
4nπ ⎞ 3 ⎛
4nπ ⎞
+ 1⎟ =
= ⎜ 2 − 3 cos
⎜1 − cos

nπ ⎝
3
3 ⎠
⎠ nπ ⎝

bn =

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Hence,
f(t) = 1 +

⎡3

∑ ⎢⎣ nπ sin

n =0

4nπ
2nπt 3 ⎛
4nπ ⎞
2nπt ⎤
cos
+
⎜1 − cos
⎟ sin
3
3
nπ ⎝
3 ⎠
3 ⎥⎦

We can now use MATLAB to check our answer,

>> t=0:.01:3;
>> f=1*ones(size(t));
>> for n=1:1:99,
f=f+(3/(n*pi))*sin(4*n*pi/3)*cos(2*n*pi*t/3)+(3/(n*pi))*(1cos(4*n*pi/3))*sin(2*n*pi*t/3);
end
>> plot(t,f)

2 .5
2
1 .5
1
0 .5
0
-0 . 5
-1
-1 . 5

0

0 .5

1

1 .5

2

2 .5

3

Clearly we have the same figure we started with!!

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Chapter 17, Problem 8.
Obtain the exponential Fourier series of the function in Fig. 17.51.

Figure 17.51
For Prob. 17.8.
Chapter 17, Solution 8.
T = 2,

ωo = 2π / T = π
⎧5(1− t), 0 < t < 1
f(t) = ⎨
⎩ 0, 1 < t < 2
1

cn =

1 T
1
f(t)e− jnωot dt = ∫ 5(1− t)e− jnπ t dt

0
20
T

1
1
5 1 − jnπ t
5
5 e− jnπ t 1 5 e− jnπ t
(− jnπ t − 1)
= ∫ e dt − ∫ te− jnπ t dt =

2
2 0
20
2 − jnπ 0 2 (− jnπ )
0
− jnπ
− 1⎦⎤ 5 e− jnπ
5 ⎣⎡e
5 (−1)
=

(− jnπ − 1) +
2 2
2 − jnπ
2 −n π
2 −n2π 2

But e− jnπ = cos π n − j sin nπ = cos nπ + 0 = (−1)n
cn =

2.5[1− (−1)n ] 2.5(−1)n[1+ jnπ ] 2.5

+ 2 2
jnπ
n2π 2

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Chapter 17, Problem 9.
Determine the Fourier coefficients an and bn of the first three harmonic terms of the
rectified cosine wave in Fig. 17.52.

Figure 17.52
For Prob. 17.9.
Chapter 17, Solution 9.
f(t) is an even function, bn=0.
T = 8, ω = 2π / T = π / 4
ao =

T
2
⎤ 10 4
1
2⎡
f
(
t
)
dt
=
⎢ ∫ 10 cos πt / 4dt + 0⎥ = ( ) sin πt / 4

T 0
8 ⎣0
⎦ 4 π
T /2

2

2
0

=

10

π

= 3.183

2

4
40
f (t ) cos nω o dt = [ ∫ 10 cos πt / 4 cos nπt / 4dt +0] = 5∫ [cos πt (n + 1) / 4 + cos πt (n − 1) / 4]dt

T 0
8 0
0
For n = 1,
an =

2

2

⎡2

a1 = 5∫ [cos πt / 2 + 1]dt = 5⎢ sin πt / 2dt + t ⎥ = 10
⎣π
⎦0
0
For n>1,
2

an =

20
20
20
20
π (n + 1)t
π (n − 1)
π (n + 1)
π (n − 1)
+
=
+
sin
sin
sin
sin
4
4
2
2
π (n + 1)
π (n − 1)
π (n + 1)
π (n − 1)
0

a2 =

20
20
sin 1.5π + sin π / 2 = 4.244,

π

a3 =

20
10
sin 2π + sin π = 0

π

Thus,
a 0 = 3.183,

a 1 = 10,

a 2 = 4.244,

a 3 = 0,

b1 = 0 = b 2 = b 3

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Chapter 17, Problem 10.
Find the exponential Fourier series for the waveform in Fig. 17.53.

Figure 17.53
For Prob. 17.10.

Chapter 17, Solution 10.
T = 2π ,

ωo = 2π / T = 1

T
π
Vo
Vo e= jnt π
1
− jnωot
− jnt
cn = ∫ f(t)e
dt =
(1)e dt =
T0
2π ∫0
2π − jn 0

=

Vo
jV
⎡⎣ je− jnπ − j ⎤⎦ = o (cos nπ − 1)
2nπ
2nπ

f(t) =

jVo

∑ 2nπ (cos nπ − 1)e

jnt

n=−∞

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Chapter 17, Problem 11.
Obtain the exponential Fourier series for the signal in Fig. 17.54.

Figure 17.54
For Prob. 17.11.
Chapter 17, Solution 11.
T = 4,

ω o = 2π / T = π / 2
T

1
1
1 0
c n = ∫ y( t )e − jnωo t dt = ⎡⎢ ∫ ( t + 1)e − jnπt / 2 dt + ∫ (1)e − jnπt / 2 dt ⎤⎥

1
0
T
4⎣

0

cn =
=

1 ⎡ e − jnπt / 2
2 − jnπt / 2 0
2 − jnπt / 2 1 ⎤

e
e
⎢ 2 2 (− jnπt / 2 − 1) −
−1 jnπ
0⎥
4 ⎣⎢ − n π / 4
jnπ
⎦⎥
1⎡ 4
2
4
2 jnπ / 2
2 − jnπ / 2
2 ⎤
e jnπ / 2 ( jnπ / 2 − 1) +
e
e

+

+

2
2
2
2
4 ⎣n π
jnπ n π
jnπ
jnπ
jnπ ⎥⎦

But
e jnπ / 2 = cos nπ / 2 + j sin nπ / 2 = j sin nπ / 2,
cn =

1
2 2

n π

e − jnπ / 2 = cos nπ / 2 − j sin nπ / 2 = − j sin nπ / 2

[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]

y( t ) =

1
2 2

n = −∞ n π

[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]e jnπt / 2

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Chapter 17, Problem 12.
* A voltage source has a periodic waveform defined over its period as
v(t) = t(2 π − t) V,

0 < t < 2π

Find the Fourier series for this voltage.
* An asterisk indicates a challenging problem.

Chapter 17, Solution 12.
A voltage source has a periodic waveform defined over its period as
for all 0 < t < 2π
v(t) = t(2π - t) V,
Find the Fourier series for this voltage.
v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1
T

ao = (1/T) ∫ f ( t )dt =
0

1 2π
1
(πt 2 − t 3 / 3)
(2πt − t 2 )dt =

0

0

=

4π 3
2π 2
(1 − 2 / 3) =

3

2 T
1 ⎡ 2π
2πt

sin(nt )⎥
an = ∫ (2πt − t 2 ) cos(nt )dt = ⎢ 2 cos(nt ) +
T 0
n
π ⎣n
⎦0

bn =
=

[

1
2nt cos(nt ) − 2 sin(nt ) + n 2 t 2 sin( nt )
3
πn

=

−4
2
1
(1 − 1) − 3 4nπ cos(2πn ) = 2
2
πn
n
n

]

0

2 T
1
(2nt − t 2 ) sin( nt )dt = ∫ (2nt − t 2 ) sin(nt )dt

0
T
π

1
2n 1
π
(sin(nt ) − nt cos(nt )) 0 − 3 (2nt sin(nt ) + 2 cos(nt ) − n 2 t 2 cos(nt ))
2
0
π n
πn

− 4 π 4π
+
=0
n
n
2π 2 ∞ 4
− ∑ 2 cos(nt )
f(t) =
3
n =1 n
=

Hence,

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Chapter 17, Problem 13.

A periodic function is defined over its period as
0⎧ 10 sin t ,
h(t ) = ⎨
⎩20 sin(t − π ), π < t < 2π
Find the Fourier series of h(t).
Chapter 17, Solution 13.

T = 2π, ωo = 1
T

ao = (1/T) ∫ h( t )dt =
0

=
an = (2/T)

1 π
[ ∫ 10 sin t dt + ∫ 20 sin( t − π) dt ]
π
2π 0

[

]

1
30
π

− 10 cos t 0 − 20 cos( t − π) π =
π

T

∫ h( t ) cos(nω t )dt
o

0

= [2/(2π)] ⎡ ∫ 10 sin t cos( nt )dt +
⎢⎣ 0
π

π

20 sin( t − π) cos( nt )dt ⎤
⎥⎦

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)]
sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t]
sin(t – π) = sin t cos π – cost sin π = –sin t
sin(t – π)cos(nt) = –sin(t)cos(nt)
an =

1 ⎡ π
10∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt − 20∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt ⎤
⎥⎦
π
2π ⎢⎣ 0

5
=
π

⎡⎛ cos([1 + n ]t ) cos([1 − n ]t ) ⎞ π ⎛ 2 cos([1 + n ]t ) 2 cos([1 − n ]t ) ⎞ 2 π ⎤

+
⎢⎜ −
⎟ +⎜
⎟ ⎥
1+ n
1− n
1+ n
1− n
⎠0 ⎝
⎠ π ⎥⎦
⎢⎣⎝

an =

5⎡ 3
3
3 cos([1 + n ]π) 3 cos([1 − n ]π) ⎤
+

⎥⎦
π ⎢⎣1 + n 1 − n
1+ n
1− n

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But,

[1/(1+n)] + [1/(1-n)] = 1/(1–n2)

cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ
an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))]
= [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even
= 0,
n = odd
T

bn = (2/T) ∫ h ( t ) sin nωo t dt
0

π

0

π

= [2/(2π)][ ∫ 10 sin t sin nt dt + ∫ 20( − sin t ) sin nt dt
But,

sin A sin B = 0.5[cos(A–B) – cos(A+B)]
sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)]
π

bn = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ (1 + n )] 0

+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1 + n )] π }
=

Thus,

5
π

⎡ sin([1 − n ]π) sin([1 + n ]π) ⎤
+
⎢⎣ −
⎥⎦ = 0
1− n
1+ n
h(t) =

30 60 ∞ cos( 2kt )

π
π k = 1 ( 4k 2 − 1)

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Chapter 17, Problem 14.
Find the quadrature (cosine and sine) form of the Fourier series
nπ ⎞
10

cos⎜ 2nt +

4 ⎠

n =1 n + 1

f (t ) = 2 + ∑

3

Chapter 17, Solution 14.
Since cos(A + B) = cos A cos B – sin A sin B.

10
⎛ 10

cos(nπ / 4) cos( 2nt ) − 3
sin(nπ / 4) sin( 2nt ) ⎟
f(t) = 2 + ∑ ⎜ 3
n +1

n =1 ⎝ n + 1

Chapter 17, Problem 15.
Express the Fourier series

4
1
cos 10nt + 3 sin 10nt
n
n =1 n + 1

f (t ) = 10 + ∑

2

(a) in a cosine and angle form.
(b) in a sine and angle form.

Chapter 17, Solution 15.
(a)
Dcos ωt + Esin ωt = A cos(ωt - θ)
where

f(t) = 10 +

A =

D 2 + E 2 , θ = tan-1(E/D)

A =

16
1
+ 6 , θ = tan-1((n2+1)/(4n3))
2
( n + 1)
n

n =1

2

2

16
1
−1 n + 1 ⎞

cos⎜⎜ 10nt − tan
+
( n 2 + 1) 2 n 6
4n 3 ⎟⎠

Dcos ωt + Esin ωt = A sin(ωt + θ)

(b)
where

f(t) = 10 +

A =

n =1

D 2 + E 2 , θ = tan-1(D/E)

16
1
4n 3 ⎞
−1

+
+
sin
10
nt
tan

( n 2 + 1) 2 n 6
n 2 + 1 ⎟⎠

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Chapter 17, Problem 16.
The waveform in Fig. 17.55(a) has the following Fourier series:
v1 (t ) =

1 4

2 π2

1
1

⎜ cos πt + cos 3πt + cos 5πt + ⋅ ⋅ ⋅⎟ V
9
25

Obtain the Fourier series of v 2 (t ) in Fig. 17.55(b).

Figure 17.55
For Probs. 17.16 and 17.69.

Chapter 17, Solution 16.
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If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e.
v2*(t) = v2(t) + 1.
*

Comparing v2*(t) with v1(t) shows that
v2*(t) = 2v1((t + to)/2)
where (t + to)/2 = 0 at t = -1 or to = 1
Hence

v2*(t) = 2v1((t + 1)/2)

But

v2*(t) = v2(t) + 1
v2(t) + 1 = 2v1((t+1)/2)
v2(t) = -1 + 2v1((t+1)/2)

= -1 + 1 −

v2(t) = −

8
π2

8
π2

⎛ t + 1⎞
⎛ t + 1⎞ 1
⎛ t + 1⎞ 1
⎢cos π⎜ 2 ⎟ + 9 cos 3π⎜ 2 ⎟ + 25 cos 5π⎜ 2 ⎟ + L⎥

⎡ ⎛ πt π ⎞ 1
⎛ 5πt 5π ⎞
⎛ 3πt 3π ⎞ 1
⎢cos⎜ 2 + 2 ⎟ + 9 cos⎜ 2 + 2 ⎟ + 25 cos⎜ 2 + 2 ⎟ + L⎥

⎣ ⎝

v2(t) = −

8
π2

⎡ ⎛ π t ⎞ 1 ⎛ 3πt ⎞

1
⎛ 5 πt ⎞
⎢sin⎜ 2 ⎟ + 9 sin⎜ 2 ⎟ + 25 sin⎜ 2 ⎟ + L⎥

⎣ ⎝ ⎠

Chapter 17, Problem 17.
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Determine if these functions are even, odd, or neither.
(b) t 2 − 1
(e) e −t

(a) 1 + t
(d) sin 2 πt

(c) cos nπt sin nπt

Chapter 17, Solution 17.
We replace t by –t in each case and see if the function remains unchanged.
(a)

1 – t,

neither odd nor even.

(b)

t2 – 1,

even

(c)

cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt,

odd

(d)

sin2 n(-t) = (-sin πt)2 = sin2 πt,

even

(e)

e t,

neither odd nor even.

Chapter 17, Problem 18.
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Determine the fundamental frequency and specify the type of symmetry present in the
functions in Fig. 17.56.

Figure 17.56
For Probs. 17.18 and 17.63.
Chapter 17, Solution 18.
(a)
T = 2 leads to ωo = 2π/T = π
f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.
(b)

T = 3 leads to ωo = 2π/3
f2(t) = f2(-t), showing that f2(t) is even.

(c)

T = 4 leads to ωo = π/2
f3(t) is even and half-wave symmetric.

Chapter 17, Problem 19.
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Obtain the Fourier series for the periodic waveform in Fig. 17.57.

Figure 17.57
For Prob. 17.19.
Chapter 17, Solution 19.
T = 4, ωo = 2π / T = π / 2
⎧ 10t, 0 < t < 1
f(t) = ⎨
⎩10(2 − t), 1 < t < 2
T
1
2
1
1
1
1 2 1 10
t2 2
a0 = ∫ f(t)dt = ∫ 10tdt + ∫ 10(2 − t)dt = 5t + (2t − ) = 2.5
40
41
4
2 1
T0
0 4

1

T

an =

2

2
2
2
f(t)cos nωotdt = ∫ 10t cos nωotdt + ∫ 10(2 − t)cos nωotdt

T0
40
41

1 10
2
2
20
5
5t
t
cos nωot +
sin nωot +
sin nωot + 2 2 cos nωot +
sin nωot
nωo
nωo
nωo
0 nωo
1 n ωo
1
20
1
10
5
=
(cos nπ / 2 − 1) +
sin nπ / 2 +
(sin nπ − sin nπ / 2) + 2 2
cos nπ
nωo
nωo
nωo
n π /4

=

5
n π /4
2

2

cos nπ / 2 +

10
5
sin nπ −
sin nπ / 2
nωo
nπ / 2

T

1

2

2
2
2
bn = ∫ f(t)s innωotdt = ∫ 10t s innωotdt + ∫ 10(2 − t)sin nωotdt
T0
40
41

=

=

1 10
1
2
2
5
5
t
sin nωot −
cos nωot − 2 2 sin nωot +
cos nωot
nωo
0 nωo
0 n ωo
1 nωo
1
5

2

2
o

sin nπ / 2 −

10
5
(cos π n − cos nπ / 2) − 2 2 (sin π n − sin nπ / 2)
nωo
n ωo

2
cos π n / 2
cos nπ −
nωo
nωo

Chapter 17, Problem 20.
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Find the Fourier series for the signal in Fig. 17.58. Evaluate f(t) at t = 2 using the
first three nonzero harmonics.

Figure 17.58
For Probs. 17.20 and 17.67.

Chapter 17, Solution 20.
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This is an even function.
bn = 0, T = 6, ω = 2π/6 = π/3
2
T

ao =

T/2

0

f ( t )dt =

3
2⎡ 2
( 4 t − 4)dt ∫ 4 dt ⎤

⎥⎦
2
6 ⎢⎣ 1

2
1⎡ 2
( 2 t − 4 t ) + 4(3 − 2)⎤ = 2
⎥⎦
1
3 ⎢⎣

=

4
T

an =

T/4

0

f ( t ) cos( nπt / 3)dt
2

= (4/6)[ ∫ ( 4 t − 4) cos( nπt / 3)dt +
1

3

2

4 cos( nπt / 3)dt ]
2

3

16 ⎡ 9
3
16 ⎡ 3
⎛ nπt ⎞⎤
⎛ nπt ⎞⎤
⎛ nπt ⎞
⎛ nπt ⎞ 3t
=
sin⎜
cos⎜
sin⎜
sin⎜
⎟⎥ +

⎟−
⎟+
2 2

6 ⎣n π
6 ⎣ nπ ⎝ 3 ⎠⎥⎦ 2
⎝ 3 ⎠ nπ ⎝ 3 ⎠ nπ ⎝ 3 ⎠⎦1
= [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)]
f(t) = 2 +

Thus

24 ∞ 1

π2 n =1 n2

⎡ ⎛ 2πn ⎞
⎛ nπt ⎞
⎛ πn ⎞ ⎤
⎢cos⎜ 3 ⎟ − cos⎜ 3 ⎟ ⎥ cos⎜ 3 ⎟

⎠⎦
⎣ ⎝

At t = 2,
f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3)
+ (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3)
+ (1/9)(cos(2π) − cos(π))cos(2π) + -----]
= 2 + 2.432(0.5 + 0 + 0.2222 + -----)
f(2) = 3.756

Chapter 17, Problem 21.
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Determine the trigonometric Fourier series of the signal in Fig. 17.59.

Figure 17.59
For Prob. 17.21.

Chapter 17, Solution 21.
This is an even function.
bn = 0, T = 4, ωo = 2π/T = π/2.
f(t) = 2 − 2t,
= 0,

011

t2 ⎤
2 1
ao =

=

2
(
1
t
)
dt
t
= 0.5

2 ⎥⎦ 0
4 ∫0

an =

4
T

T/2

0

f ( t ) cos( nωo t )dt =

4 1
⎛ nπt ⎞
2(1 − t ) cos⎜
⎟dt

4 0
⎝ 2 ⎠

= [8/(π2n2)][1 − cos(nπ/2)]
f(t) =

1
+
2

8

∑n π
n=1

2

2

⎛ nπt ⎞
⎛ nπ ⎞ ⎤
⎢1 − cos⎜ 2 ⎟ ⎥ cos⎜ 2 ⎟

⎝ ⎠⎦

Chapter 17, Problem 22.