1.1Fourier transform and Fourier Series
We have already seen that the Fourier transform is important. For an LTI system, 𝑥(𝑡) = 𝑒 𝑖𝜔𝑡 ,
then the complex number determining the output 𝑦(𝑡) = 𝐻(𝑓)𝑒 𝑖2𝜋𝑓𝑡 is given by the Fourier
transform of the impulse response:
∞
𝐻(𝑓) = ∫ ℎ(𝑡)𝑒 −𝑖2𝜋𝑓𝑡 𝑑𝑡
−∞
Well what if we could write arbitrary inputs as superpositions of complex exponentials, i.e. via
sums or integrals of the following kind:
𝑥(𝑡) = ∑ 𝑋𝑘 𝑒 𝑖2𝜋𝑓𝑘 𝑡
𝑘
Then notice, outputs of LTI systems y(t) will always take the form
𝑦(𝑡) = ∑ 𝑋𝑘 𝐻(𝑓𝑘 )𝑒 𝑖2𝜋𝑓𝑘 𝑡
𝑘
This is the root of the Fourier series.
Proposition 1.1. Let x(t) be period with period T, so that the frequencies 𝑓𝑘 =
𝑘
𝑥(𝑡) = ∑𝑘 𝑋𝑘 𝑒 𝑖2𝜋𝑇𝑡 - SYNTHESIS EQUATION
= ∑ 𝑋𝑘 𝑒 𝑖2𝜋𝑘𝑓0 𝑡
𝑘
Then, 𝑥(𝑡) = 𝑥(𝑡 ± 𝑚𝑇), and
1
𝑘
𝑇
𝑋𝑘 = 𝑇 ∫0 𝑥(𝑡)𝑒 −𝑖2𝜋𝑇𝑡 𝑑𝑡 - ANALYSIS EQUATION
𝑇
1 ⁄2
= ∫ 𝑥(𝑡)𝑒 −𝑖2𝜋𝑘𝑓0 𝑡 𝑑𝑡
𝑇 −𝑇⁄
2
Proof: Use the property that
𝑇
(𝑚−𝑛)
𝑡
𝑇
𝑑𝑡
∫ 𝑒 𝑖2𝜋
= 𝑇𝛿[𝑚 − 𝑛]
0
Then we have
𝑇
𝑚
𝑇
𝑘
𝑚
∫ 𝑥(𝑡)𝑒 −𝑖2𝜋 𝑇 𝑡 𝑑𝑡 = ∫ ∑ 𝑋𝑘 𝑒 𝑖2𝜋𝑇𝑡 𝑒 −𝑖2𝜋 𝑇 𝑡 𝑑𝑡
0
0
𝑘
𝑘
𝑇
= 𝑘𝑓0 , and
𝑇
= ∑ 𝑋𝑘 ∫ 𝑒 𝑖2𝜋
(𝑘−𝑚)
𝑡
𝑇
𝑑𝑡
0
𝑘
= ∑ 𝑋𝑘 𝑇𝛿[𝑘 − 𝑚]
𝑘
OK, so how do we use this. Well, for periodic signals with period T, then we just have to evaluate
the Fourier series coefficients 𝑋𝑘 .
Example 1.1.
1.
x(t)=constant, then 𝑋0 =constant and 𝑋𝑘 = 0, 𝑘 ≠ 0 for any period T.
2.
𝑥(𝑡) = 𝑒 𝑖2𝜋𝑓0 𝑡 , then 𝑇 = 𝑓 , 𝑋1 = 1, 𝑋𝑘 = 0, 𝑘 ≠ 1.
3.
𝑥(𝑡) = cos(2𝜋𝑓0 𝑡), then 𝑇 = , 𝑋1 = 𝑋−1 = , 𝑋𝑘 = 0, 𝑘 ≠ ±1.
4.
𝑥(𝑡) = sin(2𝜋𝑓0 𝑡), then 𝑇 = 𝑓 , 𝑋1 = 2𝑗, 𝑋−1 = − 2𝑗, 𝑋𝑘 = 0, 𝑘 ≠ ±1.
1
0
1
𝑓0
1
1
2
1
1
0
1.2 Relationship of Fourier Series and Fourier Transform
So, Fourier series is for periodic signals. Fourier transform is for non-periodic signals. Let’s
examine and construct the Fourier transform by allowing the period of the periodic signals go to
∞, see what we get.
Let’s define 𝑥̃(𝑡) to be the periodic version of x(t), where x(t) has finite support 𝑥(𝑡) = 0, |𝑡| ≥
𝑇⁄ . Thus, 𝑥̃(𝑡 ± 𝑚𝑇) = 𝑥(𝑡), 𝑡 ∈ [− 𝑇⁄ , 𝑇⁄ ]
2
2
2
Definition 1.1. Define the Fourier transform of x(t) to be
∞
𝑋(𝑓) = ∫ 𝑥(𝑡)𝑒 −𝑖2𝜋𝑓𝑡 𝑑𝑡
−∞
Then we have the relationship between FT and FS.
Proposition 1.2.
1
1
𝑋̃𝑘 = 𝑇 𝑋(𝑘𝑓0 ) where 𝑓0 = 𝑇
where
1
𝑥̃(𝑡) = ∑𝑥 𝑋̃𝑘 𝑒 𝑖2𝜋𝑘𝑓0 𝑡 , where 𝑓0 = 𝑇
Example 1.2. Let 𝑥(𝑡) = 1, 𝑡 ∈ [− 𝐴⁄2 , 𝐴⁄2], and 0 otherwise. Then
∞
𝑋(𝑓) = ∫ 𝑥(𝑡)𝑒 −𝑖2𝜋𝑓𝑡 𝑑𝑡
−∞
𝐴⁄
2
=∫
−𝐴⁄2
𝑒 −𝑖2𝜋𝑓𝑡 𝑑𝑡
𝑒 −𝑖2𝜋𝑓𝑡 |
=
=
𝐴⁄
2
−𝐴⁄2
−𝑖2𝜋𝑓
𝑒 −𝑖𝜋𝑓𝐴 − 𝑒 𝑖𝜋𝑓𝐴
−𝑖2𝜋𝑓
sin(𝜋𝑓𝐴)
𝜋𝑓
=
Let 𝑥̃(𝑡 ± 𝑚𝑇) = 𝑥(𝑡), 𝑡 ∈ [− 𝑇⁄2 , 𝑇⁄2]. Then,
1
𝑥̃(𝑡) = ∑𝑘 𝑋̃𝑘 𝑒 𝑖2𝜋𝑘𝑓0 𝑡 where 𝑓0 =
𝑇
𝑋̃𝑘 =
𝑘
1
sin(𝜋𝑘𝑓0 𝐴) sin(𝜋 𝑇 𝐴)
𝑋(𝑘𝑓0 ) =
=
𝑇
𝑇𝜋𝑘𝑓0
𝜋𝑘
OK, so we see that the Fourier transform can be used to define the Fourier series. Now what we
would like to do is understand how to represent the periodic signals when the period goes to
infinity 𝑇 → ∞, so that we can have a synthesis pair. Let’s remind ourselves that 𝑥̃(𝑡) is the
𝑇
periodic version of x(t), where x(t) has finite support 𝑥(𝑡) = 0, |𝑡| ≥ 2.
Proposition 1.3. Let 𝑥̃(𝑡) be periodic with period T, and 𝑥(𝑡) = 𝑙𝑖𝑚 𝑥̃(𝑡). Then
𝑇→∞
∞
𝑥(𝑡) = ∫ 𝑋(𝑓)𝑒 𝑖2𝜋𝑓𝑡 𝑑𝑓
−∞
To see this,
𝑥(𝑡) = lim 𝑥̃(𝑡) = lim ∑ 𝑋̃𝑘 𝑒 𝑖2𝜋𝑘𝑓0 𝑡
𝑇→∞
𝑇→∞
𝑘
1
= lim ∑ 𝑋(𝑘𝑓0 )𝑒 𝑖2𝜋𝑘𝑓0 𝑡
𝑇→∞
𝑇
𝑘
= lim ∑ 𝑋(𝑘𝑓0 )𝑒 𝑖2𝜋𝑘𝑓0 𝑡
𝑇→∞
𝑘
= lim ∑ 𝑋(𝑘𝑓0 )𝑒 𝑖2𝜋𝑘𝑓0 𝑡 𝑓0
𝑓0 →∞
𝑘
∞
= ∫ 𝑋(𝑓)𝑒 𝑖2𝜋𝑓𝑡 𝑑𝑓
−∞