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ME261  Solved Examples for Homework #5
Problem 1: (a)
x
(
t
) = 5
2 sin (10
) + 3 sin (15
)
; (b)
x
(
t
) = 3 sin (20
4) + 3 sin (30
+
6)
;
(i) Represent the above functions in the forms that explicitly show magnitudes and phases for their onesided and
twosided magnitude and phase spectra;
(ii)
For each waveform indicate fundamental frequency and harmonic number
of each component, and draw the corresponding magnitude and phase spectra; (iii) Now, assume the functions above
are passed through a sampler. Choose a sampling frequency in such a manner that the lowest discrete time harmonic
in each corresponding sampled function has eight samples per cycle. Rewrite the sampled functions to explicitly show
magnitudes and phases for their discrete time onesided and twosided magnitude and phase spectra, indicate the
harmonic number of each component and draw the corresponding discrete time magnitude and phase spectra.
Solution:
(a)
x
(
t
) = 5
2 sin (10
) + 3 sin (15
) = 5 + 2 cos
2
(5) +
2
±
+ 3 cos
2
15
2
±
2
±
;
cos
±
+
2
±
=
sin (
±
)
,
cos
±
2
±
= sin (
±
)
x
(
t
) = 5 + 2 cos
2
o
k
1
t
+
2
±
+ 3 cos
2
o
k
2
t
2
±
=
= 5 +
e
j
(
2
)
e
j
2
(
k
1
f
o
)
t
+
e
j
2
e
j
2
(
k
1
f
o
)
t
+ 1
:
5
e
j
2
e
j
2
(
k
2
f
o
)
t
+ 1
:
5
e
j
(
2
)
e
j
2
(
k
2
f
o
)
t
)
k
1
f
o
= 5;
k
2
f
o
=
15
2
)
fundamental frequency:
f
o
=
greatest common divisor
=
5
2
= 2
:
5
Hz
;
harmonic numbers:
k
1
= 2;
k
2
= 3;
Figure 1.
Discrete Time:
f
l
= 5
Hz
(lowest frequency in x(t) is 5 Hz),
x
(
n
) = 5 + 2 cos(2
1
F
o

{z
}
F
l
n
+
2
) + 3 cos
2
2
F
o
n
2
±
F
l
=
f
l
S
f
=
1
8
cycle/sample
S
f
=
f
l
(8) = (5) (8) = 40
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x
[
n
] = 5 + 2 cos
2
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 Spring '08
 Staff

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