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
.
t)
ω
= 1200
π
or T = 2
π
/
ω
= 2
π
/(1200
π
) = 1/600.
(d) f
4
(t) = e
j10t
= cos 10t + jsin 10t.
ω
= 10 or T = 2
π
/
ω
= 0.2
π
. Chapter 17, Solution 3.
∫
ω
T
0
o
dt)tncos()t(g
∫
π
1
0
dt)t
2
n
cos(5+
∫
π
2
1
dt)t
2
n
cos(10 ]
= 0.5[
1
0
t
2
n
sin
= (2/T) = (2/4)[
∫
ω
T
0
o
dt)tnsin()t(g
∫
π
1
0
dt)t
2
n
sin(5+
∫
π
2
1
dt)t
2
n
sin(10 ]
= 0.5[
1
0
t
2
n
= 2π/T = π
a
o
= (1/T) = (1/2) =
∫
T
0
dt)t(f
∫
−
2
0
dt)t510(
2
0
2
)]2/t5(t10[5.0 −
= 5
a
n
= (2/T) = (2/2)
∫
ω
T
0
o
dt)tncos()t(f
∫
0
tnsin
n
t5
π
π
= [–5/(n
2
π
2
)](cos 2nπ – 1) = 0
b
n
= (2/2)
∫
π−
2
0
dt)tnsin()t510(
=
–
∫
π
2
0
dt)tnsin()10(
∫
π
π
+
∑
∞
=
5
Chapter 17, Solution 5. 1T/2,2T =π=ωπ=
5.0]x2x1[
2
1
dt)t(z
T
1
a
T
0
o
−=π−π
π
==
∫
1
ntdtcos1
1
dtncos)t(z
T
2
a
∫∫∫
π
π
π
π
π
π
=
=
π
=
π
+
π
−=
π
−
π
n
6
5.0)t(z
oddn
1n
∑
∞
=
=
π
+−=
Chapter 17, Solution 6. .0a,functionoddanisthisSince
3
2
6
)1x21x4(
2
1
dt)t(y
2
1
a
2
2
π
=
π−π
π
−−π
π
−
=π
π
−π
π
−
=
π+π=ω=
oddn
1n
evenn,0
oddn,
n
4
2
1
1
0
2
1
1
0
2
0
Chapter 17, Solution 7. 0a,
6
T/2,12T
0
=
π
=π=ω=
∫∫∫
π−+π=ω=
−
10
4
4
2
T
0
on
]dt6/tncos)10(dt6/tncos10[
6
1
dtncos)t(f
T
2
a
4
2
T
0
on
]dt6/tnsin)10(dt6/tnsin10[
6
1
dtnsin)t(f
T
2
b[]
3/n2sin23/ncos3/n5cos
n
10
6/ntncos
n
10
6/tncos
n
10
10
4
4
2
π−π+π
π
2ttdt)1t(2
2
1
dt)t(f
T
1
a
1
1
1
1
2
T
0
o
=+=+==
−
−
∫∫0tnsin
n
1
tnsin
n
t
tncos
n
π
=π+=ω=
−
−
∫∫
π
π
−=
π
π
−π
π
−π
π
−=π+=ω=
−
−
∫∫
ncos
n
4
π
−=
1n
n
tncos
n
)1(4
2)t(f Chapter 17, Solution 9. f(t) is an even function, b
n
=0.
4//2,8
ππω
=== TT183.3
10
4/sin)
4
(
4
10
04/cos10
T
a
T
on
∫∫∫
−++=+==
2
0
2
0
2/
0
4/)1(cos4/)1(cos5]04/cos4/cos10[
8
40
cos)(
4
ππππω
For n = 1,
102/sin
2
5]12/[cos5
2
0
2
0
1
=
)1(
sin
)1(
20
2
0
−
−
+
+
+
=
−
−
+
+
+
=
n
n
n
n
n
n
tn
n
a
n
π
π
3213210
0,0,362.6,10,183.3 bbbaaaa =======
Chapter 17, Solution 10. π=π=ω= T/2,2T
o
π−
−
π−
=
2
1
dte)t(h
T
1
c
[]
,
even n ,0
oddn ,
n
j6
]6ncos6[
n2
j
e2e24e4
n2
j
c
jnn2jnj
n
=
=
π
−
=
Chapter 17, Solution 11. 2/T/2,4T
o
π=π=ω=
∫∫ ∫
++==
−
π−π−
ω−
T
0
0
1
1
0
2/tjn2/tjn
π−
1
0
2/tjn
0
1
2/tjn
22
2/tjn
n
e
jn
2
e
jn
2
)12/tjn(
4/n
e
4
1
c
2222
=But
2/nsinj2/nsinj2/ncose,2/nsinj2/nsinj2/ncose
2/jn2/jn
π−=π−π=π=π+π=
π−π
[]
2/nsinn2/nsin)12/jn(j1
n
1
c
22
n
ππ+π−π+
π
=
[]
2/tjn
22
n
e2/nsinn2/nsin)12/jn(j1
n
1
)t(y
v
(
t
) = 2
π
t
–
t
2
, 0 <
t
< 2
π
,
T
= 2
π
,
ω
o
= 2
π
/
T
= 1
a
o
π
π
=−π
π
=
πa
n
=
π
π
+
π
π
=−π
∫
2
0
2
T
0
2
1
)11(
n
2
−
=ππ
π
−−=
b
n
=
∫∫
−
π
=− dt)ntsin()tnt2(
1
dt)ntsin()tnt2(
T
2
2
T
0
2π
π
−+
π
=
−
π
1n
2
2
)ntcos(
n
4
3
2
Chapter 17, Solution 13. T = 2π, ω
o
= 1
a
o
= (1/T) dttsin10[
2
1
dt)t(h
0
T
0
= (2/T)
∫
ω
T
0
o
dt)tncos()t(h
= [2/(2π)]
π−+
∫∫
ππ
π0
2
dt)ntcos()tsin(20dt)ntcos(tsin10
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)
a
n
=
−
−
+
+
+
+
−
−
−
+
+
−
π
π
π
π 2
0
n1
)t]n1cos([2
n1
)t]n1cos([2
n1
But, [1/(1+n)] + [1/(1-n)] = 1/(1–n
2
)
cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ
a
n
= (5/π)[(6/(1–n
2
)) + (6 cos(nπ)/(1–n
2
))]
= [30/(π(1–n
2
))](1 + cos nπ) = [–60/(π(n–1))], n = even
= 0, n = odd
b
n
= (2/T)
∫
ω
T
0
o
+
2
)]n1(}
=
+
π+
+
−
π−
−
π n1
)]n1sin([
n1
)]n1sin([5
= 0
Thus, h(t) =
∑
∞
=
−
π
−
33
)nt2sin()4/nsin(
1n
10
)nt2cos()4/ncos(
1n
10
2
Chapter 17, Solution 15. (a) Dcos ωt + Esin ωt = A cos(ωt - θ)
where A =
22
ED + , θ = tan
-1
(E/D)
A =
622
n
1
)1n(
+
+
16
622
n4
1n
tannt10cos
n
1
)1n(
16(b) Dcos ωt + Esin ωt = A sin(ωt + θ)
where A =
22
ED + , θ = tan
-1
(D/E)
f(t) =
∑
+
10
∞
=
−
v
2
*
(t) = v
2
(t) + 1.
1
v
2
*
(t)
2
t
-2 -1 0 1 2 3 4 5 Comparing v
2
*
(t) with v
1
(t) shows that
2
(t) + 1 = 2v
1
((t+1)/2)
v
2
(t) = -1 + 2v
1
((t+1)/2)
= -1 + 1
+
+
π+
2v
2
(t) =
+
π
+
π
+
π
t
cos
8
2
v
2
(t) =
+
π
+
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) t
2
– 1,
even(c) cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt,
odd(d) sin
2
n(-t) = (-sin πt)
2
= sin
2
πt,
even
f
2
(t) = f
2
(-t), showing that f
2
(t) is
even
.
(c) T = 4 leads to ω
o
= π/2 f
3
(t) is
even and half-wave symmetric
. Chapter 17, Solution 19. This is a half-wave even symmetric function.
a
o
= 0 = b
](1 − cos nπ) = 8/(n
2
π
2
), n = odd
= 0, n = even
f (t) =
∑
∞
=
π
π
oddn
22
2
tn
cos
n
18
dt)t(f
T
2 =
−+− )23(4)t4t2(
3
1
2
1
2
= 2
a
n
=
∫
π
4/T
0
dt)3/tncos()t(f
T
4
n
3
3
tn
sin
n
t3
3
tn
cos
n
9
6
16
π
π
+
π = [24/(n
2
π
2
)][cos(2nπ/3) − cos(nπ/3)]
Thus f(t) =
∑
∞
=
π
f(2) = 2 + (24/π
2
)[(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, Solution 21. This is an even function.
b
n
= 0, T = 4, ω
o
= 2π/T = π/2.
f(t) = 2 − 2t, 0 < t < 1
= 0, 1 < t < 2
a
π
−=ω
1
0
2/T
0
o
dt
2
tn
cos)t1(2
4
4
dt)tncos()t(f
T
4 = [8/(π
2
n
2
)][1 − cos(nπ/2)]
f(t) =
∑
∞
=
cos1
n
8
2
1
Chapter 17, Solution 22. Calculate the Fourier coefficients for the function in Fig. 16.54.
f(t)
4
t
-5 -4 -3 -2 -1 0 1 2 3 4 5 Figure 17.61 For Prob. 17.22
This is an even function, therefore b
n
= 0. In addition, T=4 and ω
2T
0
o
dt)2/tncos(t4
4
4
dt)ntcos()t(f
T
4
1
0
22
)2/tnsin(
n
t2
)2/tncos(
n
4
4
π
π
o
= 0 = a
n
, T = 2, ω
o
= 2π/T = π
b
n
=
∫∫
π=ω
1
0
2/T
0
o
dt)tnsin(t
2
4
dt)tnsin()t(f
T
4 =
[]
1
0
22
(a) This is an odd function.
a
o
= 0 = a
n
, T = 2π, ω
o
= 2π/T = 1
b
n
=
∫
ω
2/T
0
o
dt)ntsin()t(f
T
4 f(t) = 1 + t/π, 0 < t < π
b
n
=
∫
)ntcos(
n
12 = [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)
n+1
] a
2
=
0
, b
2
= [2/(2π)][1 + 2(−1)] = −1/π = −
0.3183(b) ωn = nω
o
= 10 or n = 10
a
10
= 0, b
10
= [2/(10π)][1 − cos(10π)] = −1/(5π)
π−
π
1n
)ntsin()]ncos(21[
n
2
f(π/2) =
∑
∞
=
ππ−
π
1n
)2/nsin()]ncos(21[
n
2
π
For n = 1, f
1
= (2/π)(1 + 2) = 6/π
For n = 2, f
2
= 0
For n = 3, f
3
= [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π)
1/7 + - - - Chapter 17, Solution 25. This is an odd function since f(−t) = −f(t).
a
o
= 0 = a
n
, T = 3, ω
o
= 2π/3.
b
n
=
∫∫
π=ω
1
0
2/T
0
o
dt)3/nt2sin(t
3
4
dt)tnsin()t(f
π
π
−
π
π =
π
π
π
π
π
−
π
π
1n
22
3
++=
∫∫∫∫
4
3
3
1
1
0
T
0
dt1dt2dt1
4
1
dt)t(f
T
1
= 1
a
n
= dt)tncos()t(f
T
T
0
o
∫
ω
2
π
π
+
π
π
+
π
π
4
3
3
2
2
1
2
tn
sin
n
2
2
tn
sin
n
4
2
tn
sin
n
o
dt)tnsin()t(f
T
2 =
π
+
π
+
π
∫∫∫
4
3
3
2
2
1
dt
2
tn
sin1dt
2
2
2
1
2
tn
cos
n
2
2
tn
cos
n
4
2
tn
cos
n
2
2
=
[]
1)ncos(
n
4
−π
πHence
f(t) = t, 0 < t < 1
= 0, 1 < t < 2
b
n
=
1
0
22
1
0
2
tn
cos
n
t2
2
tn
sin
n
4
dt
2
tn
sint
4
−
π
π = 4(−1)
(n−1)/2
/(n
2
π
2
), n = odd −2(−1)
n/2
/(nπ), n = even
a
3
=
0
, b
3
= 4(−1)/(9π
2
) =
0.045
ba
2
1
a F
rms
2
= 0.5Σb
n
2
= [1/(2π
2
)][(16/π
2
) + 1 + (16/(8π
2
)) + (1/4) + (16/(625π
2
))]
= (1/19.729)(2.6211 + 0.27 + 0.00259)
F
rms
= 14659.0 =
0.3829
2/T
0
o
dt)tncos()t22(
2
4
dt)tncos()t(f
T
4 =
1
0
22
)tnsin(
n
t
)tncos(
n
1
)tnsin(
n
1
4
dt)tnsin()t1(
=
1
0
22
)tncos(
n
t
)tnsin(
n
1
)tncos(
n
1
4
π
π
+π
π
−π
π
−
Chapter 17, Solution 29. This function is half-wave symmetric.
T = 2π, ω
o
= 2π/T = 1, f(t) = −t, 0 < t < π
For odd n, a
n
=
∫
π
−
0
dt)ntcos()t(
T
2
=
[]
π
+
π
−
0
2
)ntsin(nt)ntcos(
n
∑
∞
=
−
π
1k
2
)ntsin(
n
1
)ntcos(
n
2
2
,
n = 2k
−
1
Chapter 17, Solution 30.
(1)
(a) The second term on the right hand side vanishes if f(t) is even. Hence
∫
ω=
2/T
0
on
tdtncos)t(f
T
2
c(b) The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,
∫
ω−=
2/T
0
on
tdtnsin)t(f
T
2j
c
Chapter 17, Solution 31.
'aLet
T'T,/dtd,,t =ααλ=λ=αn
T
0
on
a/dncos)(f
T
2
'a =αλλωλ
α
=
∫Similarly,
nn
b'b =
Chapter 17, Solution 32.
When i
s
2
−∠
+
=
∠+
°∠
−
Thus,
i(t) =
∑
∞
=
−
−
+
+
1n
1
22
))n2(tann3cos(
n41n3
1
3
1
Chapter 17, Solution 33.
π
−π+
=
π
+
π
+
−)5n5.2(jn
4
n
5
n5.2j1
1
n
4
A
n
5
n5.2j1
V
V
22
nn
π
−π
−=Θ
−π+π
=
−
n
5n5.2
tan;
)5n5.2(n
4
A
22
1
n
22222
nV)tnsin(A)t(v
1n
nno
∑
∞
=
Θ+π=
V
V
o
= {−j(2/n)/[2 + jn − j(2/n)]}V = {−j2/[2n + j(n
2
− 2)]}[(10/n
2
)∠(nπ/4)] )]n2/)2n((tan)2/()4/n[(
4nn
20
)n2/)2n((tan)2n(n4n
)2/)4/n((20
21
22
212222
−−π−π∠
+
=
−∠−+
π−π∠
=
−
−
24
n
ntcos
4nn
20Chapter 17, Solution 35.
If v
s
in the circuit of Fig. 17.72 is the same as function f
2
(t) in Fig. 17.57(b),
determine the dc component and the first three nonzero harmonics of v
o
(t). 1
Ω
1 H
1 F
1 Ω
+
−
Figure 17.57(b) For Prob. 17.35
The signal is even, hence, b
n
= 0. In addition, T = 3, ω
o
= 2π/3.
v
s
(t) = 1 for all 0 < t < 1
= 2 for all 1 < t < 1.5
a
o
=
3
4
dt2dt1
3
2
5.1
1
1
0
=
2
)3/tn2sin(
n2
6
)3/tn2sin(
n2
3
3
4
5.1
1
1
0
π
π
−=
π
π
+π
π
v
s
(t) =
+
v
o
−v
S Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3)
Therefore, v
o
= Zv
s
/(Z + 1 + j2nπ/3). Simplifying, we get
v
o
=
)18n4(jn12
v9j
22
s
Θ+
π
+
∑
∞
=where
π
−
π
−=Θ
22
nwhere we can further simplify A
n
to this,
81n4n
)3/n2sin(9
44
n
+ππ
A
π
=
Chapter 17, Solution 36. v
s
(t) =
∑
∞
=
=
θ−
oddn
π
+
π
=
π
π
+
πω
n
= n and 2 H becomes jω
n
L = j2n
Let Z = 1||j2n = j2n/(1 + j2n)
If V
o
is the voltage at the non-reference node or across the 2-H inductor.
V
o
= ZV
s
/(1 + Z) = [j2n/(1 + j2n)]V
s
/{1 + [j2n/(1 + j2n)]}
n161 +
∠tan
−1
4n
=
2
2
n161
n2
2
n
sin9
n
1
+
π
+
π
∠−100.5° − tan
−1
4n
2
∑
∞
=
=
−
−°−
+
π
Chapter 17, Solution 37. From Example 15.1,
v
s
(t) =
∑
∞
=
π
π
+
1k
),tnsin(
n
120
5 n = 2k − 1
100j
20
n
100
j
V
n
100
j
π+π
π
+°−∠
=
π−π
−
=
+
π
−
π
−
−v
o
(t) =
∑
π
+°−π
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