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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 3 t cos 4 t
(d) h ( t ) = t
2 cos
(e) z ( t ) = 4.2 sin(0.4^ π^ t + 10º)
(f) p ( t ) = 10
(g) q ( t ) =
t e
− π
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 = 2 π.
(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.
Using MATLAB, synthesize the periodic waveform for which the Fourier
trigonometric Fourier series is
f t = − t + t + cos 5 t +⋅⋅⋅ 25
cos 3 9
cos
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: tn(n)=n/10;
t=n/10;
y1=cos(t);
y2=(1/9)cos(3t);
y3=(1/25)cos(5t); factor=4/(pi*pi);
y(n)=0.5- factor*(y1+y2+y3);
end
plot(tn,y)
n a (^) n bn An phase
1 –1.59 7.95 8.11 –101.
2 0 0 0 0
3 0.53 2.65 2.70 –78.
4 0 0.80 0.80 –
5 –0.32 1.59 1.62 –101.
6 0 0 0 0
7 0.23 1.15 1.17 –78.
8 0 0.40 0.40 –
0 π 2 π 3 π 4 π 5 π 6 π 7 π 8 π
An
ω
0 π 2 π 3 π 4 π 5 π 6 π 7 π 8 π
90 ˚
φ
ω
–101.31˚
–78.69˚
Find the Fourier series expansion of the backward sawtooth waveform of Fig. 17.48.
Obtain the amplitude and phase spectra.
Figure 17.
For Probs. 17.4 and 17.66.
Chapter 17, Solution 4.
f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π
a o = (1/T) ∫
T
0
f (t)dt = (1/2) ∫ −
2
0
( 10 5 t)dt=
2
0
2
- 5 [ 10 t− ( 5 t / 2 )] = 5
a n = (2/T) ∫ ω
T
0 o^
f (t)cos(n t)dt = (2/2) ∫ − π
2
0
( 10 5 t)cos(n t)dt
2
0
( 10 )cos(n t)dt– ∫ π
2
0
( 5 t)cos(n t)dt
2
0
2 2 cosn t n
π π
2
0
sinn t n
5 t π π
= [–5/(n
2 π
2 )](cos 2nπ – 1) = 0
bn = (2/2)
− π
2
0
( 10 5 t)sin(n t)dt
2
0
( 10 )sin(n t)dt– ∫ π
2
0
( 5 t)sin(n t)dt
2
0
2 2 sinn t n
π π
2
0
cosn t n
5 t π π
= 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)
Hence f(t) = sin(n t) n
n 1
π π
∞
=
Find the trigonometric Fourier series for
= and ( 2 ) (). 10 , 2
( ) f t f t t
t
f t π
Chapter 17, Solution 6.
T=2π, ωo =2π/T = 1
2
0 0
T
ao f t dt dt dt T
π π
π
2
0 0
( )cos 5cos 10cos 0 2
T
an f t n (^) otdt ntdt ntdt T
π π
π
2
0 0
( )sin 5 sin 10 s cos cos 2 0
T
bn f t n (^) otdt ntdt inntdt nt nt T n n
π π
π
⎣ ⎦ ⎢⎣^ ⎥⎦
cos 1
0,
n odd n (^) n n n even
⎪−^ =
Thus,
10 ( ) 7.5 sin n odd
f t nt
n π
∞
=
- Determine the Fourier series of the periodic function in Fig. 17.50.
Figure 17.
For Prob. 17.7.
- An asterisk indicates a challenging problem.
Chapter 17, Solution 7.
T = 3, ω o = 2 π / T = 2 π / 3
2 3
0 0 2
T
ao f t dt dt dt T
2 3
0 0 2
( )cos 2cos ( 1)cos 3 3 3 3
T
n
n t n t n t a f t dt dt dt T
π ⎡^ π π ⎤
4 n sin n
2 n t sin 2 n
2 n t sin 2 n
3
2
2
0
π
π
π
π
π
π
2 3
0 0 2
( )sin 2 sin ( 1)s 3 3 3 3
T
n
n t n t n t b f t dt dt in dt T
π ⎡^ π π ⎤
2 cos cos (1 2cos ) 3 2 3 0 2 3 2 3
n t n t n x n n n
⎛ π − π
π − π 3
4 n 1 cos n
4 n 2 3 cos n
