Download Homework 8 Solutions - Signals and Systems I | ECE 633 and more Assignments Signals and Systems in PDF only on Docsity!
From: Linear Systems and Signals, 2
nd
ed. B. P. Lathi, Oxford University Press, 2005.
6.1-1 For each of the periodic signals shown in Fig. P6.1-1, find the compact trigonometric
Fourier series and sketch the amplitude and phase spectra. If either the sine of cosine terms
are absent in the Fourier series, explain why.
0
0
0
T
C f t dt
T
0
0 0
cos
n
T
nt
a f t dt
T T
0
0 0
sin
n
T
nt
b f t dt
T T
2 2
n n n
C = a + b
1
tan
n
n
n
b
a
−
0
1 0
cos
n n
n
nt
f t C C
T
∞
=
(a)
0
C = mean = 0
1 3
1 3
1 1
1 1
cos cos sin sin
n
nt nt nt nt
a dt dt
n n
π π π π
π π
−
−
sin sin sin sin 3sin sin
n
n n n n n n
a
n n
3 3
3sin 3sin 4sin sin sin integers
n
n n n n n
a
n n n
π π π π π
π π π
n
b = (even symmetry)
sin
n n
n
C a
n
1
tan
n
n
n n
a
a a
θ
−
1
sin cos cos cos cos cos
n
n nt t t t t
f t
n
∞
=
(c)
2
2
2
0 2
0
0
t t
C dt
π
π
2
2 2
2 2 2
0 0
0
cos cos cos sin
n
t t
a nt dt t nt dt nt nt
n n
π
π π
2 2 2 2 2 2
cos 2 sin 2 cos 0 sin 0 0
n
a n n
n n n n n n
2
2 2
2 2 2
0 0
0
sin sin sin cos
n
t t
b nt dt t nt dt nt nt
n n
π
π π
2 2 2
sin 2 cos 2 sin 0 cos 0
n
b n n
n n n n n
n n
C b
n π
1
tan
n
n
n b
n
−
1
cos 90
cos 90 cos 2 90 cos 3 90 cos 4 90
sin sin 2 sin 3 sin 4 sin 5 sin 6
n
f t nt
n
nt nt nt nt
nt nt nt nt nt nt
∞
=
(d)
0
C = mean = 0
n
a = (odd symmetry)
/
/4 2
2 2 2
/4 0
/
sin 2 sin 2 sin 2 cos 2
n
t t
b nt dt t nt dt nt nt
n n
π
π π
π
π
−
−
[ ] [ ]
2 2 2
sin 2 cos 2 sin 2 cos 2
n
b n n n n
n n n n
π π π π π π
π
2 2
sin cos sin cos
n
n n n n
b
n n n n
sin cos
n n
n n
C b
n n
1
tan
n
n
n
n
b
b
b
θ
−
1
2 2
2 2
2
sin cos cos 2
cos 2 90 cos 4 90 cos 6 90 cos 8 90
cos 10 90 cos 12 90 cos 14 90 cos 16 90
sin 2 s
n
n n
b
n n
f t nt
b n n
t t t t
t t t t
t
∞
=
in 6 sin 10 sin 4 sin 8 sin 12
t t t t t
2 2
2 2
1
1
2
2 2 cos 2 sin
cos sin
cos tan
cos sin 1
n
n n n n
n
n n n
f t
n t
n n n
∞
=
−
2 1
2 1
2
2 1
2
2
2
6 3 cos tan
27 16 12 3 cos tan cos 2 90
27 64 24 3 cos tan
27 100 30 3 cos
t
t
t
t
−
−
−
1
tan cos 4 90
0.1667 0.2947 cos 80.69 0.2003cos 166.53 0.1061cos 2 90
0.07276 cos 41.29 0.06953cos 157.54 0.0531c
t
t
t t
t
t t
−
os 4 π t + 90 ° +"
(f)
0
C = mean = 0
n
b = (even symmetry)
1 1 2
2 1 1
2 cos cos 2 cos
n
nt nt nt
a t dt dt t dt
−
− −
1 1
2 2
2 1
2
2 2
1
cos sin sin
cos sin
n
nt t nt nt
a
n n n
nt t nt
n n
−
− −
2 2
cos cos
n
n n
a
n
2 2
cos cos
n n
n n
C a
n
1
tan
n
n
n n
a
a a
−
2 2
1
2
cos cos cos
cos 0 cos 180 0 cos 0 cos
0 cos 3 180 0 cos 0 cos
n
n n
a n n nt
f t
n a
t t t
t
t t
t
∞
=
