Chapter 16 Solutions
16
Fourier Series
Assessment Problems
AP 16.1 av =
1
T
ak =
2
T
=
0
Vm dt +
2T /3
0
4Vm
3kω0 T
= bk =
2T /3
2
T
2T /3
0
4Vm
3kω0 T
1
T
Vm
3
T
2T /3
Vm cos kω0 t dt + sin 4kπ
3
=
Vm sin kω0 t dt +
1 − cos
4kπ
3
7 dt = Vm = 7π V
9
Vm cos kω0 t dt
3
T
2T /3
6
4kπ
sin k 3
Vm
sin kω0 t dt
3
T
2T /3
=
6 k 1 − cos
4kπ
3
AP 16.2 [a] av = 7π = 21.99 V
[b] a1 = −5.196 b1 = 9
a2 = 2.598
a3 = 0 a4 = −1.299
a5 = 1.039
b2 = 4.5
b3 = 0
b5 = 1.8
b4 = 2.25
2π
= 50 rad/s
T
[d] f3 = 3f0 = 23.87 Hz
[c] w0 =
[e] v(t) = 21.99 − 5.2 cos 50t + 9 sin 50t + 2.6 sin 100t + 4.5 cos 100t
−1.3 sin 200t + 2.25 cos 200t + 1.04 sin 250t + 1.8 cos 250t + · · · V
AP 16.3 Odd function with both half- and quarter-wave symmetry. vg (t) =
6Vm t, T
0 ≤ t ≤ T /6;
av = 0,
ak = 0 for all k
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
16–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
16–2
CHAPTER 16. Fourier Series bk = 0 for k even bk =
8
T
T /4
8
T
T /6
=
0
f(t) sin kω0 t dt,
0
k odd
6Vm
8
t sin kω0 t dt +
T
T
T /4
T /6
Vm sin kω0 t dt
12Vm kπ sin
2
2 k π
3
=
vg (t) =
12Vm π2 ∞
1 nπ sin sin nω0 t V
2
3 n=1,3,5,... n
A2 = 2.6 − j4.5 = 5.2/ − 60◦
AP 16.4 [a] A1 = −5.2 − j9 = 10.4/ − 120◦ ;
A3 = 0;
A4 = −1.3 − j2.25 = 2.6/ − 120◦
A5 = 1.04 − j1.8 = 2.1/ − 60◦ θ1 = −120◦ ;
θ2 = −60◦ ;
θ4 = −120◦ ;
θ5 = −60◦
θ3 not defined;
[b] v(t) = 21.99 + 10.4 cos(50t − 120◦ ) + 5.2 cos(100t − 60◦ )
+2.6 cos(200t − 120◦ ) + 2.1 cos(250t − 60◦ ) + · · · V
AP 16.5 The Fourier series for the input voltage is
∞
8A
1
nπ sin sin nω0 (t + T /4)
2
2
π
Fourier Series
Assessment Problems
AP 16.1 av =
1
T
ak =
2
T
=
0
Vm dt +
2T /3
0
4Vm
3kω0 T
= bk =
2T /3
2
T
2T /3
0
4Vm
3kω0 T
1
T
Vm
3
T
2T /3
Vm cos kω0 t dt + sin 4kπ
3
=
Vm sin kω0 t dt +
1 − cos
4kπ
3
7 dt = Vm = 7π V
9
Vm cos kω0 t dt
3
T
2T /3
6
4kπ
sin k 3
Vm
sin kω0 t dt
3
T
2T /3
=
6 k 1 − cos
4kπ
3
AP 16.2 [a] av = 7π = 21.99 V
[b] a1 = −5.196 b1 = 9
a2 = 2.598
a3 = 0 a4 = −1.299
a5 = 1.039
b2 = 4.5
b3 = 0
b5 = 1.8
b4 = 2.25
2π
= 50 rad/s
T
[d] f3 = 3f0 = 23.87 Hz
[c] w0 =
[e] v(t) = 21.99 − 5.2 cos 50t + 9 sin 50t + 2.6 sin 100t + 4.5 cos 100t
−1.3 sin 200t + 2.25 cos 200t + 1.04 sin 250t + 1.8 cos 250t + · · · V
AP 16.3 Odd function with both half- and quarter-wave symmetry. vg (t) =
6Vm t, T
0 ≤ t ≤ T /6;
av = 0,
ak = 0 for all k
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
16–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
16–2
CHAPTER 16. Fourier Series bk = 0 for k even bk =
8
T
T /4
8
T
T /6
=
0
f(t) sin kω0 t dt,
0
k odd
6Vm
8
t sin kω0 t dt +
T
T
T /4
T /6
Vm sin kω0 t dt
12Vm kπ sin
2
2 k π
3
=
vg (t) =
12Vm π2 ∞
1 nπ sin sin nω0 t V
2
3 n=1,3,5,... n
A2 = 2.6 − j4.5 = 5.2/ − 60◦
AP 16.4 [a] A1 = −5.2 − j9 = 10.4/ − 120◦ ;
A3 = 0;
A4 = −1.3 − j2.25 = 2.6/ − 120◦
A5 = 1.04 − j1.8 = 2.1/ − 60◦ θ1 = −120◦ ;
θ2 = −60◦ ;
θ4 = −120◦ ;
θ5 = −60◦
θ3 not defined;
[b] v(t) = 21.99 + 10.4 cos(50t − 120◦ ) + 5.2 cos(100t − 60◦ )
+2.6 cos(200t − 120◦ ) + 2.1 cos(250t − 60◦ ) + · · · V
AP 16.5 The Fourier series for the input voltage is
∞
8A
1
nπ sin sin nω0 (t + T /4)
2
2
π