E1C02 09/14/2010
13:35:18 Page 59
The reader can verify that all A n ¼ 0 for this function. For even values of n, B n is identically zero
and for odd values of n,
B n ¼
4
np
The resulting Fourier series is then
y t
ð Þ ¼
4
p
sin
2p
10
t þ
4
3p
sin
6p
10
t þ
4
5p
sin
10p
10
t þ Á Á Á
Note that the fundamental frequency is v ¼
2p
10
rad=s and the subsequent terms are the oddnumbered harmonics of v.
COMMENT Consider the function given by
y t
ð Þ ¼ 1 0 < t < 5
We may represent y(t) by a Fourier series if we extend the function beyond the specified range
(0 – 5) either as an even periodic extension or as an odd periodic extension. (Because we are
interested only in the Fourier series representing the function over the range 0 < t < 5, we can
impose any behavior outside of that domain that helps to generate the Fourier coefficients!) Let’s
choose an odd periodic extension of y(t); the resulting function remains identical to the function
shown in Figure 2.14.
Example 2.4
Find the Fourier coefficients of the periodic function
y t
ð Þ ¼ À5 when À p < t < 0
y t
ð Þ ¼ þ5 when 0 < t < p
and y(t þ 2p) ¼ y(t). Plot the resulting first four partial sums for the Fourier series.
KNOWN Function y(t) over the range Àp to p
FIND Coefficients A n and B n
SOLUTION The function as stated is periodic, having a period of T ¼ 2p (i.e., v ¼ 1 rad/s), and
is identical in form to the odd function examined in Example 2.3. Since this function is also odd, the
Fourier series contains only sine terms, and
B n ¼
1
p
ð p
Àp
y t
ð Þsin nvtdt ¼
1
p
ð 0
Àp
À5
ð Þsin ntdt þ
ð p
0
þ5
ð Þsin ntdt
!
which yields upon integration
1
p
5
ð Þ
cos nt
n
! 0
Àp
À 5
ð Þ
cos nt
n
! p
0
(
)
Thus,
B n ¼
10
np
1 À cos np
ð
Þ
2.4 Signal Amplitude And Frequency 59
13:35:18 Page 59
The reader can verify that all A n ¼ 0 for this function. For even values of n, B n is identically zero
and for odd values of n,
B n ¼
4
np
The resulting Fourier series is then
y t
ð Þ ¼
4
p
sin
2p
10
t þ
4
3p
sin
6p
10
t þ
4
5p
sin
10p
10
t þ Á Á Á
Note that the fundamental frequency is v ¼
2p
10
rad=s and the subsequent terms are the oddnumbered harmonics of v.
COMMENT Consider the function given by
y t
ð Þ ¼ 1 0 < t < 5
We may represent y(t) by a Fourier series if we extend the function beyond the specified range
(0 – 5) either as an even periodic extension or as an odd periodic extension. (Because we are
interested only in the Fourier series representing the function over the range 0 < t < 5, we can
impose any behavior outside of that domain that helps to generate the Fourier coefficients!) Let’s
choose an odd periodic extension of y(t); the resulting function remains identical to the function
shown in Figure 2.14.
Example 2.4
Find the Fourier coefficients of the periodic function
y t
ð Þ ¼ À5 when À p < t < 0
y t
ð Þ ¼ þ5 when 0 < t < p
and y(t þ 2p) ¼ y(t). Plot the resulting first four partial sums for the Fourier series.
KNOWN Function y(t) over the range Àp to p
FIND Coefficients A n and B n
SOLUTION The function as stated is periodic, having a period of T ¼ 2p (i.e., v ¼ 1 rad/s), and
is identical in form to the odd function examined in Example 2.3. Since this function is also odd, the
Fourier series contains only sine terms, and
B n ¼
1
p
ð p
Àp
y t
ð Þsin nvtdt ¼
1
p
ð 0
Àp
À5
ð Þsin ntdt þ
ð p
0
þ5
ð Þsin ntdt
!
which yields upon integration
1
p
5
ð Þ
cos nt
n
! 0
Àp
À 5
ð Þ
cos nt
n
! p
0
(
)
Thus,
B n ¼
10
np
1 À cos np
ð
Þ
2.4 Signal Amplitude And Frequency 59