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Homework 8 Solutions - Signals and Systems I | ECE 633, Assignments of Signals and Systems

Material Type: Assignment; Class: Signals and Systems I; Subject: Electrical&Comp Engineering; University: University of New Hampshire-Main Campus; Term: Fall 2005;

Typology: Assignments

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ECE633 Signals and Systems I, Fall 2009 – Homework 8 Solutions
1
From: Linear Systems and Signals, 2nd 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
1
T
Cftdt
T
=
()
0
00
22
cos
nT
nt
aft dt
TT
π
⎛⎞
=⎜⎟
⎝⎠
()
0
00
22
sin
nT
nt
bft dt
TT
π
⎛⎞
=⎜⎟
⎝⎠
22
nnn
Cab=+
1
tan n
n
n
b
a
θ
⎛⎞
=⎜⎟
⎝⎠
()
010
2
cos
nn
n
nt
ft C C T
π
θ
=
⎛⎞
=+ +
⎜⎟
⎝⎠
pf3
pf4
pf5
pf8
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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

=