Notes On The Fourier Transform

17
The Fourier Transform
Assessment Problems
0

AP 17.1 [a] F (ω) =

−τ /2

(−Ae−jωt ) dt +

τ /2
0

Ae−jωt dt

A
[2 − ejωτ /2 − e−jωτ /2] jω ejωτ /2 + e−jωτ /2
2A
=
1−
jω
2
−j2A
=
[1 − cos(ωτ /2)] ω =

[b] F (ω) =
AP 17.2 f(t) =
=

1
2π

∞
0

∞

te−ate−jωt dt =

−2
−3

0

2

4ejtω dω +

−2

te−(a+jω)t dt =
3

ejtω dω +

2

1
(a + jω)2

4ejtω dω

1
{4e−j2t − 4e−j3t + ej2t − e−j2t + 4ej3t − 4ej2t } j2πt 1 3e−j2t − 3ej2t 4ej3t − 4e−j3t
=
+ πt j2 j2 =

1
(4 sin 3t − 3 sin 2t) πt AP 17.3 [a] F (ω) = F (s) |s=jω = L{e−at sinω0 t}s=jω
=

ω0
(s + a)2 + ω02

= s=jω [b] F (ω) = L{f − (t)}s=−jω =

ω0
(a + jω)2 + ω02

1
(s + a)2

= s=−jω 1
(a − jω)2

© 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
17–1 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.

17–2

CHAPTER 17. The Fourier Transform
[c] f + (t) = te−at ,

f − (t) = −te−at
1
,
(s + a)2

L{f + (t)} =

Therefore F (ω) =
AP 17.4 [a] f (t) =
. ·.

2A
,
τ

=

f (t) =

−2A
,
τ

0<t<

τ
2

2A
2A
[u(t + τ /2) − u(t)] −
[u(t) − u(t − τ /2)] τ τ

4A
2A
2A u(t + τ /2) − u(t) + u(t − τ /2) τ τ τ f (t) =

2A τ 4A 2A τ δ t+
−
+ δ t− τ 2 τ τ
2

2A jωτ /2 4A 2A −jωτ /2 e −
+
e τ τ τ [b] F {f (t)} =
=

−1
(s + a)2

1
−j4aω
1
−
= 2
2
2
(a + jω)
(a − jω)
(a + ω 2 )2

−τ
< t < 0;
2

f (t) =

. ·.

L{f − (t)} =

4A ejωτ /2 + e−jωτ /2
4A
ωτ
−1 = cos τ
2
τ
2

[c] F {f (t)} = (jω)2F (ω) = −ω 2F (ω);
Thus we have F (ω) = −
AP 17.5 v(t) = Vm u t +

1 ω2 −1

therefore F (ω) = −

4A ωτ cos τ 2

1
F {f (t)} ω2 −1

τ τ −u t−
2
2

F u t+

τ
2

= πδ(ω) +

1 ejωτ /2 jω F u t−

τ
2

= πδ(ω) +

1 e−jωτ /2 jω Therefore V (ω) = Vm πδ(ω) +

1 jω = j2Vm πδ(ω) sin
=

ejωτ /2 − e−jωτ /2 ωτ 2

+

2Vm ωτ sin ω 2

(Vm τ ) sin(ωτ /2) ωτ /2

© 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 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.

Problems

AP 17.6 [a] Ig (ω) = F {10sgn t} =

17–3

20 jω Vo
Ig
Using current division and Ohm’s law,
4s
4
(−Ig )s =
Ig
Vo = −I2s = −
4+1+s
5+s

[b] H(s) =

H(s) =

4s
,
s+5

H(jω) =

[c] Vo (ω) = H(jω) · Ig (ω) =

j4ω
5 + jω

j4ω
5 + jω

20 jω =

80
5 + jω

[d] vo(t) = 80e−5t u(t) V
[e] Using current division,
1
1 i1 (0− ) = ig = (−10) = −2 A
5
5
+
[f] i1(0 ) = ig + i2 (0+ ) = 10 + i2(0− ) = 10 + 8 = 18 A
[g] Using current division,
4
i2 (0− ) = (10) = 8 A
5
[h] Since the current in an inductor must be continuous, i2 (0+ ) = i2(0− ) = 8 A
[i] Since the inductor behaves as a short circuit for t < 0, vo (0− ) = 0 V
[j] vo(0+ ) = 1i2(0+ ) + 4i1 (0+ ) = 80 V
AP 17.7 [a] Vg (ω) =
H(s) =

1
1
+ πδ(ω) +
1 − jω jω Va
0.5 (1/s)
1
=
=
,
Vg
1 + 0.5 (1/s) s+3 H(jω) =

1
3 + jω

Va (ω) = H(jω)Vg (jω)
1
1 πδ(ω) =
+
+
(1 − jω)(3 + jω) jω(3 + jω) 3 + jω
1/4
1/4
1/3
1/3 πδ(ω) =
+
+
−
+
1 − jω 3 + jω jω 3 + jω 3 + jω
1/4
1/3
1/12
πδ(ω)
=
+
−
+
1 − jω jω 3 + jω 3 + jω
Therefore va(t) =

1 t
1
1
1
e u(−t) + sgn t − e−3tu(t) +
V
4
6
12
6

© 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 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information

Show More

Questions On Non-Monetary Prices

Essay on Lecture 3 Leases 2015 Student Version 2

ACC220 Timetable 2013 Essay

Essay Corporate Finance and Review Practice Problems

Study Guide Essay

Essay on An Introduction to Management Science 12th Edition Anderson Manual

Notes On Micro-Computer Accounting

Syllabus: Foreign Exchange Market Essay

Essay on Course: Chemistry and Chemical Equilibrium

Essay on Days of the Year and Mfc General Information