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.
Chapter 17, Solution 2.
(a) The frequency ratio is 6/5 = 1.2. The highest common factor is 1.
ω = 1 = 2π/T or T = 2π.
(b) ω = 2 or T = 2π/ω = π.
(c) f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t)
ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600.
(d) f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.
Chapter 17, Solution 3.
T = 4, ωo = 2π/T = π/2
g(t) = 5, 0 2
bn = (2/2) ∫ (10 − 5t ) sin(nπt )dt
0
2 2
= ∫ (10) sin(nπt )dt – ∫ (5t ) sin(nπt )dt
0 0
2 2
−5 5t
= 2 2 sin nπt + cos nπt = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)
n π 0 nπ 0
10 ∞ 1
Hence f(t) = 5 + ∑ sin(nπt )
π n =1 n
Chapter 17, Solution 5.
T = 2π, ω = 2π / T = 1
T
1 1
a o = ∫ z( t )dt = [1xπ − 2 xπ] = −0.5
T 2π
0
T π 2π
2 1 1 1 π 2 2π
an =
T ∫ z(t ) cos nωo dt = π ∫ 1cos ntdt − π ∫ 2 cos ntdt = nπ sin ..nt −
0 nπ
sin nt π = 0
0 0 π
T π 2π 6
2 1 1 1 π 2 2π , n = odd
b n = ∫ z( t ) cos nωo dt = ∫ 1sin ntdt − ∫ 2 sin ntdt = − nπ cos nt 0 + nπ cos nt π = nπ
T π π 0, n = even
0 0 π
Thus,
∞
6
z( t ) = − 0.5 + ∑ sin nt
n =1 nπ
n =odd
Chapter 17, Solution 6.
2π
T = 2, ωo = =π
2
1 2 1 6
ao =
2 ∫0 y(t )dt = 2 (4x1 + 2x1) = 2 = 3
Since this is an odd function, a n = 0.
2 2 1 2
bn =
2 ∫0 y(t ) sin(nωo t )dt = ∫0 4 sin(nπt )dt + ∫1 2 sin(nπt )dt
−4 1 2 2 −4 2
= cos(nπt ) 0 − cos(nπt ) 1 = (cos(nπ) − 1) − (cos(2nπ) − cos(nπ))
nπ nπ nπ nπ
4 2 2 0, n = even
= (1 − cos(nπ)) − (1 − cos(nπ)) = (1 − cos(nπ)) = 4
nπ nπ nπ , n = odd
nπ
4 ∞ 1
y( t ) = 3 + ∑ sin(nπt )
π n =1 n
n = odd
Chapter 17, Solution 7.
π
T = 12, ω = 2π / T = , a0 = 0
6
T 4 10
2 1
a n = ∫ f ( t ) cos nωo dt = [ ∫ 10 cos nπt / 6dt + ∫ (−10) cos nπt / 6dt ]
T 6
0 −2 4
10 10 10 10
=
4
sin nπt / 6 − 2 − sin nπt / 6 4 = [2 sin 2nπ / 3 + sin nπ / 3 − sin 5nπ / 3]
nπ nπ nπ
T 4 10
2 1
b n = ∫ f ( t ) sin nωo dt = [ ∫ 10 sin nπt / 6dt + ∫ (−10) sin nπt / 6dt ]
T 6
0 −2 4
10 10 10 10
=−
4
cos nπt / 6 − 2 + cos nπnt / 6 4 = [cos 5nπ / 3 + cos nπ / 3 − 2 sin 2nπ / 3]
nπ nπ nπ
∞
f (t) = ∑ (a n cos nπt / 6 + b n sin nπt / 6)
n =1
where an and bn are defined above.
Chapter 17, Solution 8.
f ( t ) = 2(1 + t ), - 1 < t < 1, T = 2, ωo = 2π / T = π
T 1 1
1 1
a o = ∫ f ( t )dt = ∫ 2( t + 1)dt = t 2 + t =2
T 2
0 −1 −1
T 1 1
2 2 1 t 1
an =
T ∫ f (t ) cos nωo dt = 2 ∫ 2(t + 1) cos nπtdt = 2 n 2 π 2 cos nπt + nπ sin nπt + nπ sin nπt = 0
−1
0 −1
T 1 1
2 2 1 t 1 4
bn =
T ∫ f (t ) sin nωodt = 2 ∫ 2(t + 1) sin nπtdt = 2 − n 2 π2 sin nπt − nπ cos nπt − nπ cos nπt = − nπ cos nπ
−1
0 −1
4 ∞ (−1) n
f (t) = 2 − ∑
π n =1 n
cos nπt
Chapter 17, Solution 9.
f(t) is an even function, bn=0.
T = 8, ω = 2π / T = π / 4
2 10 4
T 2
1 10
∫ f (t )dt = ∫ 10 cos πt / 4dt + 0 = ( ) sin πt / 4
2
ao = = = 3.183
4 π π
0
T 0 8 0
T /2 2 2
4 40
∫ [ 10 cos πt / 4 cos nπt / 4dt +0] = 5∫ [cos πt (n + 1) / 4 + cos πt (n − 1) / 4]dt
8 ∫
an = f (t ) cos nω o dt =
T 0 0 0
For n = 1,
2 2
2
a1 = 5∫ [cos πt / 2 + 1]dt = 5 sin πt / 2dt + t = 10
0 π 0
For n>1,
2
20 π (n + 1)t 20 π (n − 1) 20 π (n + 1) 20 π (n − 1)
an = sin + sin = sin + sin
π (n + 1) 4 π (n − 1) 4 0 π (n + 1) 2 π (n − 1) 2
10 20 20 10
a2 = sin π + sin π / 2 = 6.3662, a3 = sin 2π + sin π = 0
π π 4π π
Thus,
a 0 = 3.183, a1 = 10, a 2 = 6.362, a3 = 0, b1 = 0 = b2 = b3
Chapter 17, Solution 10.
T = 2, ωo = 2π / T = π
T
1 1 1 1 4e − jnπt 1 2e − jnπt 2
h ( t )e − jnωo t dt = ∫ 4e − jnπt dt + ∫ (−2)e − jnπt dt =
2
T∫
cn = 2 − jnπ 0 − − jnπ 1
2 0
1
0
[ ]
6j
j j − , n = odd
cn = 4e − jπn − 4 − 2e − j2nπ + 2e − jnπ = [6 cos nπ − 6] = nπ ,
2nπ 2nπ 0, n = even
Thus,
∞
− j6 jnπt
f (t ) = ∑ e
n =−∞ nπ
n =odd
Chapter 17, Solution 11.
T = 4, ω o = 2π / T = π / 2
T
1 1 0
c n = ∫ y( t )e − jnωo t dt = ∫ ( t + 1)e − jnπt / 2 dt + ∫ (1)e − jnπt / 2 dt
1
T −1
4 0
0
1 e − jnπt / 2 2 − jnπt / 2 0 2 − jnπt / 2 1
cn = 2 2 (− jnπt / 2 − 1) − e −
−1 jnπ
e 0
4 − n π / 4
jnπ
1 4 2 4 jnπ / 2 2 jnπ / 2 2 − jnπ / 2 2
= 2 2 − jnπ + 2 2 e ( jnπ / 2 − 1) + e − e +
4 n π n π jnπ jnπ jnπ
But
e jnπ / 2 = cos nπ / 2 + j sin nπ / 2 = j sin nπ / 2, e − jnπ / 2 = cos nπ / 2 − j sin nπ / 2 = − j sin nπ / 2
1
cn = [1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]
n 2π2
∞
1
y( t ) = ∑ 2 2
[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]e jnπt / 2
n = −∞ n π
Chapter 17, Solution 12.
A voltage source has a periodic waveform defined over its period as
v(t) = t(2π - t) V, for all 0 < t < 2π
Find the Fourier series for this voltage.
v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1
ao =
T 1 2π 1 2π 4π 3 2π 2
(1/T) ∫ f ( t )dt = ∫0 (2πt − t )dt = 2π (πt − t / 3)
2 2 3
= (1 − 2 / 3) =
0 2π 0
2π 3
2π
2 T 1 2π 2πt
an = ∫ (2πt − t 2 ) cos(nt )dt = 2 cos(nt ) + sin(nt )
T 0 π n n 0
−
1
πn 3
[
2nt cos(nt ) − 2 sin(nt ) + n 2 t 2 sin( nt ) ] 2π
0
2 1 −4
= (1 − 1) − 3 4nπ cos(2πn ) = 2
n 2
πn n
2 T 1
∫0 (2nt − t ) sin(nt )dt = π ∫ (2nt − t ) sin(nt )dt
2 2
bn =
T
2n 1 π 1 2π
= (sin(nt ) − nt cos(nt )) 0 − 3 (2nt sin(nt ) + 2 cos(nt ) − n 2 t 2 cos(nt ))
π n 2
πn 0
− 4 π 4π
= + =0
n n
2π 2 ∞ 4
Hence, f(t) = − ∑ 2 cos(nt )
3 n =1 n
Chapter 17, Solution 13.
T = 2π, ωo = 1
T 1 π 2π
ao = (1/T) ∫ h( t )dt = [ ∫ 10 sin t dt + ∫ 20 sin( t − π) dt ]
0 2π 0 π
=
1
2π
[ π 2π
− 10 cos t 0 − 20 cos( t − π) π =
30
π
]
T
an = (2/T) ∫ h( t ) cos(nω t )dt
0
o
= [2/(2π)] ∫ 10 sin t cos( nt )dt + 20 sin( t − π) cos( nt )dt
π 2π
0
∫π
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)
1 π
10∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt − 20∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt
2π
an =
2π 0
π
5 cos([1 + n ]t ) cos([1 − n ]t ) π 2 cos([1 + n ]t ) 2 cos([1 − n ]t ) 2 π
= − − + +
π
1+ n 1− n 0 1+ n 1− n π
5 3 3 3 cos([1 + n ]π) 3 cos([1 − n ]π)
an =
π 1 + n + 1 − n −
1+ n
−
1− n
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 2π
= [2/(2π)][ ∫ 10 sin t sin nt dt + ∫ 20( − sin t ) sin nt dt
0 π
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
2π
+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1 + n )] π }
5 sin([1 − n ]π) sin([1 + n ]π)
=
π −
1− n
+
1+ n = 0
30 60 ∞ cos( 2kt )
Thus, h(t) =
π
− ∑
π k = 1 ( 4k 2 − 1)
Chapter 17, Solution 14.
Since cos(A + B) = cos A cos B – sin A sin B.
∞
10 10
f(t) = 2 + ∑ 3 cos(nπ / 4) cos( 2nt ) − 3 sin(nπ / 4) sin( 2nt )
n =1 n + 1 n +1
Chapter 17, Solution 15.
(a) Dcos ωt + Esin ωt = A cos(ωt - θ)
where A = D 2 + E 2 , θ = tan-1(E/D)
16 1
A = + 6 , θ = tan-1((n2+1)/(4n3))
( n + 1)
2 2
n
∞
16 1 −1 n + 1
2
f(t) = 10 + ∑ +
( n 2 + 1) 2 n 6
cos 10nt − tan
4n 3
n =1
(b) Dcos ωt + Esin ωt = A sin(ωt + θ)
where A = D 2 + E 2 , θ = tan-1(D/E)
∞
16 1 4n 3
f(t) = 10 + ∑ +
( n 2 + 1) 2 n 6
sin 10nt + tan
−1
n2 + 1
n =1
Chapter 17, Solution 16.
If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e.
v2*(t) = v2(t) + 1.
v2*(t)
2
1
t
-2 -1 0 1 2 3 4 5
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)
8 t + 1 1 t + 1 1 t + 1
= -1 + 1 − cos π 2 + 9 cos 3π 2 + 25 cos 5π 2 +
π2
8 πt π 1 3πt 3π 1 5πt 5π
v2(t) = − cos 2 + 2 + 9 cos 2 + 2 + 25 cos 2 + 2 +
π2
8 π t 1 3 πt 1 5 πt
v2(t) = − sin 2 + 9 sin 2 + 25 sin 2 +
π2
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, 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, Solution 19.
This is a half-wave even symmetric function.
ao = 0 = bn, ωo = 2π/T π/2
4 T/2 4t
an =
T ∫0 1 − T cos(nωo t )dt
= [4/(nπ)2](1 − cos nπ) = 8/(n2π2), n = odd
= 0, n = even
8 ∞
1 nπt
f (t) =
π2
∑
n = odd n 2
cos
2
Chapter 17, Solution 20.
This is an even function.
bn = 0, T = 6, ω = 2π/6 = π/3
2 2 2
( 4 t − 4)dt ∫ 4 dt
T/2 3
ao =
T ∫
0
f ( t )dt =
6 ∫1
2
1 2
( 2 t − 4 t ) + 4(3 − 2) = 2
2
=
3
1
4 T/4
an =
T ∫
0
f ( t ) cos( nπt / 3)dt
2 3
= (4/6)[ ∫ ( 4 t − 4) cos( nπt / 3)dt + ∫ 4 cos( nπt / 3)dt ]
1 2
2 3
16 9 nπt 3t nπt 3 nπt 16 3 nπt
= n 2 π 2 cos 3 + nπ sin 3 − nπ sin 3 + 6 nπ sin 3
6 1 2
= [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)]
24 ∞ 1 2πn πn nπt
Thus f(t) = 2 + ∑
π 2 n =1 n2 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, Solution 21.
This is an even function.
bn = 0, T = 4, ωo = 2π/T = π/2.
f(t) = 2 − 2t, 0 = [8/(π2n2)][1 − cos(nπ/2)]
1 ∞
8 nπ nπt
f(t) =
2
+ ∑n π 2 2 1 − cos 2 cos 2
n=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 bn = 0. In addition, T=4 and ωo = π/2.
2 T2 2 1 2 1
ao =
T ∫ 0
f ( t )dt =
4 ∫0 4tdt = t 0 = 1
4 T2 4 1
an =
T ∫ 0
f ( t ) cos(ωo nt )dt =
4 ∫0
4 t cos( nπt / 2)dt
1
4 2t
= 4 2 2 cos( nπt / 2) + sin( nπt / 2)
n π nπ 0
16 8
an = (cos( nπ / 2) − 1) + sin( nπ / 2)
n π
2 2
nπ
Chapter 17, Solution 23.
f(t) is an odd function.
f(t) = t, −1< t < 1
ao = 0 = an, T = 2, ωo = 2π/T = π
4 T/2 4 1
bn =
T ∫ 0
f ( t ) sin( nωo t )dt =
2 ∫0
t sin( nπt )dt
2
= [sin(nπt ) − nπt cos(nπt )] 10
n π 2
2
= −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ)
2 ∞
( −1) n + 1
f(t) =
π
∑
n =1 n
sin( nπt )
Chapter 17, Solution 24.
(a) This is an odd function.
ao = 0 = an, T = 2π, ωo = 2π/T = 1
4 T/2
bn =
T ∫0
f ( t ) sin(ωo nt )dt
f(t) = 1 + t/π, 0 ∞
2
(c) f(t) = ∑ nπ [1 − 2 cos(nπ)] sin(nt ) π
n =1
∞
2
f(π/2) = ∑ nπ [1 − 2 cos(nπ)] sin(nπ / 2) π
n =1
For n = 1, f1 = (2/π)(1 + 2) = 6/π
For n = 2, f2 = 0
For n = 3, f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π)
For n = 4, f4 = 0
For n = 5, f5 = 6/(5π), ----
Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) ---------
= (6/π)[1 − 1/3 + 1/5 − 1/7 + --------]
f(π/2) ≅ 1.3824
which is within 8% of the exact value of 1.5.
(d) From part (c)
f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -]
(3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -]
or π/4 = 1 − 1/3 + 1/5 − 1/7 + - - -
Chapter 17, Solution 25.
This is an odd function since f(−t) = −f(t).
ao = 0 = an, T = 3, ωo = 2π/3.
4 T/2 4 1
bn =
T ∫
0
f ( t ) sin( nωo t )dt =
3 ∫0
t sin(2πnt / 3)dt
1
4 9 2πnt 3t 2πnt
= 4π 2 n 2 sin 3 − 2nπ cos 3
3 0
4 9 2 πn 3 2 πn
= 4π 2 n 2 sin 3 − 2nπ cos 3
3
∞
3 2 πn 2 2π n 2π t
f(t) = ∑ π n 2 2
sin −
3 nπ
cos sin
3 3
n =1
Chapter 17, Solution 26.
T = 4, ωo = 2π/T = π/2
1 T 1 1
f ( t )dt = ∫ 1 dt + 2 dt + ∫ 1 dt = 1
3 4
T ∫0 ∫
ao =
40
1 3
2 T
T ∫0
an = f ( t ) cos( nωo t )dt
2 2
2 cos( nπt / 2)dt + ∫ 1 cos( nπt / 2)dt
3 4
4 ∫1 ∫
an = 1 cos( nπt / 2)dt +
2 3
2 nπt
2
4 nπt
3
2 nπt
4
= 2 sin + sin + sin
nπ
2 1 nπ 2 2 nπ 2 3
4 3nπ nπ
=
nπ sin 2 − sin 2
2 T
T ∫0
bn = f ( t ) sin( nωo t )dt
2 2 nπt 3 nπt 4 nπt
= ∫1 1 sin 2 dt +
4 ∫
2
2 sin
2
dt + ∫
3
1 sin
2
dt
2 nπt
2
4 nπt
3
2 nπt
4
= 2− cos − cos − cos
nπ
2 1 nπ 2 2 nπ 2 3
4
= [cos(nπ) − 1]
nπ
Hence
f(t) =
∞
4
1+ ∑ nπ [(sin( 3nπ / 2) − sin(nπ / 2)) cos( nπt / 2) + (cos( nπ) − 1) sin(nπt / 2)]
n =1
Chapter 17, Solution 27.
(a) odd symmetry.
(b) ao = 0 = an, T = 4, ωo = 2π/T = π/2
f(t) = t, 0 < t < 1
= 0, 1Chapter 17, Solution 28.
This is half-wave symmetric since f(t − T/2) = −f(t).
ao = 0, T = 2, ωo = 2π/2 = π
4 T/2 4 1
an =
T ∫ 0
f ( t ) cos( nωo t )dt =
2 ∫0
( 2 − 2 t ) cos( nπt )dt
1
1 1 t
= 4 sin( nπt ) − 2 2 cos( nπt ) − sin( nπt )
nπ n π nπ 0
= [4/(n2π2)][1 − cos(nπ)] = 8/(n2π2), n = odd
0, n = even
1
bn = 4 ∫ (1 − t ) sin( nπt )dt
0
1
1 1 t
= 4 − cos( nπt ) − 2 2 sin( nπt ) + cos( nπt )
nπ n π nπ 0
= 4/(nπ), n = odd
∞
8 4
f(t) = ∑n
k =1 π
2 2
cos( nπt ) +
nπ
sin(nπt ) , n = 2k − 1
Chapter 17, Solution 29.
This function is half-wave symmetric.
T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π
2 2
[cos(nt ) + nt sin(nt )] 0 = 4/(n2π)
π π
For odd n, an =
T ∫ 0
( − t ) cos( nt )dt = −
n π
2
2 2
[sin(nt ) − nt cos(nt )] 0 = −2/n
π π
bn =
π ∫ 0
( − t ) sin( nt )dt = −
n π
2
Thus,
∞
2 1
f(t) = 2∑ 2 cos( nt ) − sin(nt ) , n = 2k − 1
k =1 n π n
Chapter 17, Solution 30.
T/2
1 1 T/2 T/2
cn = ∫ f ( t )e − jnωo t dt = ∫−T / 2 f ( t ) cos nω o tdt − j∫−T / 2 f ( t ) sin nω o tdt (1)
T T
−T / 2
(a) The second term on the right hand side vanishes if f(t) is even. Hence
T/2
2
cn =
T ∫ f (t ) cos nωo tdt
0
(b) The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,
T/2
j2
cn = −
T ∫ f (t ) sin nωo tdt
0
Chapter 17, Solution 31.
2π 2π
If h ( t ) = f (αt ), T' = T / α
→ ωo ' = = = αωo
T' T / α
T' T'
2 2
an '=
T' ∫ h (t ) cos nωo ' tdt = T' ∫ f (αt ) cos nωo ' tdt
0 0
Let αt = λ, , d t = dλ / α , αT ' = T
T
2α
T ∫
an '= f (λ) cos nωo λdλ / α = a n
0
Similarly, bn ' = bn
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