Periodicity, Real Fourier Series, and Fourier Transforms
Samantha R Summerson
5 October, 2009
1
Periodicity and Fourier Series
The period of the a function is the smallest value T ∈ R such that ∀t ∈ R and any k ∈ Z,
s(t) = s(t + kT ).
The fundamental frequency of a signal is
1
.
T
The harmonics are the terms are integer multiples of the the fundamental frequency, i.e. the k th harmonic
is at frequency Tk .
f0 =
Example 1. Find the period of a basic sine wave. Suppose s(t) = sin(2πf t). We want to find T such that
s(t)
⇒ sin(2πf t)
= s(t − T ),
= sin(2πf (t − T )),
= sin(2πf t − 2πf T ).
In order for the last inequality to be true, we require f T ∈ Z. The smallest value of T that meets this
requirement is
1
T = ,
f
as we should expect.
Example 2. Find the period and Fourier series of
s(t) = sin(2π440t) + sin(2π550t) + sin(2π660t).
This signal represents the “major triad” consisting of notes A, E, and C#. To find the period, we used the
same trick as before.
s(t − T )
=
sin(2π440(t − T )) + sin(2π550(t − T )) + sin(2π660(t − T )),
=
sin(2π440t − 2π440T ) + sin(2π550t − 2π550T ) + sin(2π660t − 2π660T ).
We want the smallest T such that 440T, 550T , and 660T , are all integers. Equivalently, we want the largest
f0 = T1 such that
440 550 660
,
,
∈ Z.
T
T
T
This f0 is the greatest common divisor (GCD) of the three numbers. (Note: there is a GCD function in
Matlab.) It should be clear here that f0 = 110Hz, so
T =
1
.
110
1
What is the real Fourier series for this signal? First of all, we know it is a periodic odd signal, so ak = 0 for
all k. We re-write the signal:
2π5t
2π6t
2π4t
+ sin
+ sin
.
s(t) = sin
1
1
1
110
From the above we know
110
bk =
1
0
110
k = 4, 5, 6
.
o/w
Example 3. Find the period and Fourier series of
s(t) = 10sin(2πt) + (10 + 2j)cos(2πt).
First, we find the period:
s(t − T )
=
10sin(2π(t − T )) + (10 + 2j)cos(2π(t − T )),
=
10sin(2πt − 2πT ) + (10 + 2j)cos(2πt − 2πT ).
Since we want 2πT to be an integer multiple of 2π, we have that T = 1. Note that since the signal is
complex-valued, we cannot find the real Fourier series. To find the values of ck , we use Euler’s formula:
s(t)
=
=
=
=
10 j2πt 10 −j2πt 10 + 2j j2πt 10 + 2j −j2πt
e
− e
+
e
+
e
,
2j
2j
2
2
8 + 10j j2πt 10j − 12 −j2πt
e
+
e
,
2j
2j
4 + 5j j2πt 5j − 6 −j2πt
e
+
e
,
j
j
(5 − 4j)ej2πt + (5 + 6j)e−j2πt .
Thus,

 5 + 6j
5 − 4j
ck =

0
k = −1
k=1 .
o/w
Example 4. Find the period and Fourier series of
√
s(t) = sin(2πt) + sin( 2πt).
To find the period, we examine s(t − T ).
s(t − T )
√
= sin(2π(t − T )) + sin( 2π(t − T )),
√
√
= sin(2πt − 2πT ) + sin( 2πt − 2πT ).
We need
2πT = 2πk1 , k1 ∈ Z,
√
2πT = 2πk2 ∈ Z.
This implies that T = √12 k2 and T = k1 , which is not possible since T cannot be an integer in both cases.
Thus, T = 0. The signal is not periodic and therefore does not have a Fourier series.
2
s(t)
1
2
4
−1
Example 5. Consider the triangle wave. Find the Fourier series.
Since the function is odd, we know that ak = 0 for all k. Thus, we only need to find the bk coefficients.
bk
=
2
4
=
1
2
=
1
2
=
1
2
=
=
1
2
⎧
⎨
⎩
2
4
∫
2πkt
)dt,
4
(∫ 1
)
∫ 3
∫ 4
πkt
πkt
πkt
)dt +
(2 − t)sin(
)dt +
(t − 4)sin(
)dt ,
tsin(
2
2
2
0
1
3
(
)
∫ 1
∫ 3
∫ 4
−2(2 − t)
−2(t − 4) 4
−2t
πkt 1
πkt
πkt 3
πkt
πkt
2
2
2
cos(
)0+
cos(
)dt +
cos(
)1−
cos(
)dt +
cos(
)dt
+
,
3
πk
2
2
πk
2
2
πk
2
0 πk
1 πk
3 πk
(
)
−2
πk
4
πkt 1
2
3πk
2
πk
4
πkt 3 −2
3πk
4
πkt 4
cos( ) +
sin(
) +
cos(
)+
cos( ) −
sin(
) +
cos(
)+
sin(
)
,
πk
2
(πk)2
2 0 πk
2
πk
2
(πk)2
2 1
πk
2
(πk)2
2 3
(
)
8
πk
8
3πk
sin(
)
−
sin(
)
,
(πk)2
2
(πk)2
2
s(t)sin(
0
8
(πk)2
−8
(πk)2
0
k = 1, 5, 9, ...
k = 3, 7, 11, ... .
o/w
Fourier Transforms
For a any continuous signal, s(t), we define its Fourier transform to be
Z ∞
S(f ) =
s(t)e−j2πf t dt.
−∞
The inverse Fourier transform returns the signal from its transform:
Z ∞
s(t) =
S(f )ej2πf t dt.
−∞
Note that for the Fourier transform we integrate over t (time), so that our expression becomes a function of
f (frequency).
Example 6. Find the Fourier transform of
s(t) = e−at u(t),
3
for some real number a.
Z
S(f )
∞
=
−∞
Z ∞
=
e−at u(t)e−j2πf t dt,
e−(a+j2πf )t dt,
0
=
=
∞
−1
e−(a+j2πf )t ,
a + j2πf
0
1
.
a + j2πf
Example 7. Find the Fourier transform of the pulse of length ∆ centered about zero.
Z ∆2
P∆ (f ) =
e−j2πf t dt,
−∆
2
=
=
=
=
−1 −j2πf t ∆2
e
∆,
j2πf
−2
∆
∆
−1
e−j2πf 2 − ej2πf 2 ,
j2πf
1
sin(πf ∆),
πf
∆sinc(πf ∆).
It turns out that if the time-domain signal is a sinc function, the Fourier transform is a pulse.
1 |f | < W
s(t) = 2W sinc(2πW t) ↔ S(f ) =
0 |f | > W
There are actually many nice properties of Fourier transforms. Below is a list of some of them:
• Linearity:
a1 s1 (t) + a2 s2 (t) ↔ a1 S1 (f ) + a2 S2 (f ).
• Conjugate symmetry: if s(t) ∈ R, then S(f ) = S(−f )∗ . This implies that |S(f )| is an even function
and ∠S(f ) is an odd function (similar to the property for Fourier series of a real periodic signal).
• Even symmetry: if s(t) is an even function, then S(f ) is an even function.
• Odd symmetry: if s(t) is an odd function, then S(f ) is an odd function.
• Scaling: for a ∈ R,
1
s(at) ↔
S
|a|
f
.
a
• Time delay: a delay of τ in the time domain corresponds to multiplication by an exponential in the
frequency domain.
s(t − τ ) ↔ e−j2πf τ S(f )
• Complex modulation: multiplication by an exponential in the time domain corresponds to a frequency
shift in the frequency domain.
ej2πf0 t s(t) ↔ S(f − f0 )
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