Signals and Systems with MATLAB
Computing and Simulink
Modeling, Fourth Edition
7
−
5
Copyright
©
Orchard Publications
Evaluation of the Coefficients
Figure 7.5.
Graphical proof of
Figure 7.6.
Graphical proof of
To evaluate any coefficient in (7.10), say
, we multiply both sides of (7.10) by
. Then,
Next, we multiply both sides of the above expression by
, and we integrate over the period
to
. Then,
(7.11)
x
sin
x
2
sin
mt
sin
(
)
2
t
d
0
2
π
∫
π
=
x
cos
x
2
cos
m
cos
t
(
)
2
t
d
0
2
π
∫
π
=
b
2
2t
sin
f t
( )
2t
sin
1
2
--a
0
2t
sin
a
1
t
2t
sin
cos
a
2
2t
2t
sin
cos
a
3
3t
2t
sin
a
4
4t
2t
sin
cos
+
cos
…
+
+
+
+
=
b
1
t
2t
sin
sin
b
2
2t
sin
(
)
2
b
3
3t
2t
sin
b
4
4t
2t
sin
sin
+
sin
…
+
+
+
dt
0
2
π
f t
( )
2t
sin
t
d
0
2
π
∫
1
2
--a
0
2t
sin
t
d
0
2
π
∫
a
1
t
2t
sin
cos
t
d
0
2
π
∫
a
2
2t
2t
sin
cos
t
d
0
2
π
∫
+
+
=
+ a
3
3t
2t
sin
cos
t
d
0
2
π
∫
…
+
+ b
1
t
sin
2t
sin
t
d
0
2
π
∫
b
2
2t
sin
(
)
2
t
d
0
2
π
∫
b
3
3t
sin
2t
sin
t
…
+
d
0
2
π
∫
+
+

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Chapter 7
Fourier Series
7
−
6
Signals and Systems with MATLAB
Computing and Simulink
Modeling, Fourth Edition
Copyright
©
Orchard Publications
We observe that every term on the right side of (7.11) except the term
is zero as we found in (7.6) and (7.7). Therefore, (7.11) reduces to
or
and thus we can evaluate this integral for any given function
. The remaining coefficients
can be evaluated similarly.
The coefficients
,
,
and
are found from the following relations.
(7.12)
(7.13)
(7.14)
The integral of (7.12) yields the average (
) value of
.
7.3
Symmetry in Trigonometric Fourier Series
With a few exceptions such as the waveform of the half
−
rectified waveform, Page 7
−
17, the most
common waveforms that are used in science and engineering, do not have the average, cosine,
and sine terms all present. Some waveforms have cosine terms only, while others have sine terms
only. Still other waveforms have or have not
components. Fortunately, it is possible to pre-
dict which terms will be present in the trigonometric Fourier series, by observing whether or not
the given waveform possesses some kind of symmetry.
b
2
2t
sin
(
)
2
t
d
0
2
π
∫
f t
( )
2t
sin
t
d
0
2
π
∫
b
2
2t
sin
(
)
2
t
d
0
2
π
∫
b
2
π
=
=
b
2
1
π
--
f t
( )
2t
sin
t
d
0
2
π
∫
=
f t
( )
a
0
a
n
b
n
1
2
--a
0
1
2
π
------
f t
( )
t
d
0
2
π
∫
=
a
n
1
π
--
f t
( )
nt t
d
cos
0
2
π
∫
=
b
n
1
π
--
f t
( )
nt t
d
sin
0
2
π
∫
=
DC
f t
( )
DC