Lecture VIII: Fourier seriesmaxim.ece.illinois.edu/teaching/fall08/lec8.pdf · Trigonometric Fourier series Fact: the sequence of T-periodic functions {φk(t)}∞ k=0 deﬁned by

Lecture VIII: Fourier series

Maxim Raginsky

BME 171: Signals and Systems

Duke University

September 19, 2008

Maxim Raginsky Lecture VIII: Fourier series

This lecture

Plan for the lecture:

1 Review of vectors and vector spaces

2 Vector space of continuous-time signals

3 Vector space of T -periodic signals

4 Complete orthonormal systems of functions

5 Trigonometric Fourier series

6 Complex exponential Fourier series


Review: vectors

Recall vectors in n-space Rn. Each such vector u can be uniquely

represented as a linear combination of n unit vectors e1, . . . , en:

u = α1e1 + α2e2 + . . . + αnen,

where α1, . . . , αn are real numbers (coefficients). These can becomputed using the scalar (or dot) product as follows:

αk = u · ek, k = 1, . . . , n

Note that the vectors e1, . . . , en are:

normalized — ek · ek = 1 for all k

orthogonal — ek · em = 0 for k 6= m

e1

e2

u

v

x

y

6.53

3

5

0

Example: u,v in R2

u = 3e1 + 5e2

v = 6.5e1 + 3e2


Vector spaces

By definition, vectors are objects that can be added together andmultiplied by scalars:

if u =∑n

k=1αkek and v =

∑nk=1

βkek, then we can form their sum

u + v =

n∑

k=1

(αk + βk)ek

if u =∑n

k=1αkek and β is a scalar, then we can form the vector

βu =

n∑

k=1

βαkek

More generally, a collection of objects that can be added and/or

multiplied by scalars is called a vector space.


Signal spaces

We have already seen one example of a vector space — the space ofcontinuous-time signals. We can easily verify:

we can form the sum of any two signals x1(t) and x2(t) to getanother signal

x(t) = x1(t) + x2(t)

we can multiply any signal x(t) by a constant c to get another signal

z(t) = cx(t)

Unlike the n-space Rn, the vector space of all continuous-time signals is

huge. In fact, it is infinite-dimensional.


The space of periodic signals

Let us consider all T -periodic signals. Any such signal x(t) satisfies

x(t + T ) = x(t) for all t

for some given T > 0.

It is easy to see that T -periodic signals form a vector space:

if x1(t) and x2(t) are T -periodic, then

x1(t + T ) + x2(t + T ) = x1(t) + x2(t),

so their sum is T -periodic

if x(t) is T -periodic, then

cx(t + T ) = cx(t),

so any scaled version of x(t) is also T -periodic

If we impose even more conditions on our T -periodic signals (theso-called Dirichlet conditions, which hold for all signals encountered inpractice), then we can represent signals as infinite linear combinations oforthogonal and normalized (orthonormal) vectors.


Complete orthonormal sets of functions

First of all, we can define a scalar (or dot) product of two T -periodicsignals x1(t) and x2(t) as

〈x1, x2〉 =

∫ T

0

x1(t)x2(t)dt

(note that we can integrate over any whole period, not necessarily fromt = 0 to t = T ).

Then we can take any sequence of T -periodic functions (signals)φ0(t), φ1(t), φ2(t), . . . that are

1 normalized: 〈φk, φk〉 =

∫ T

0

φ2

k(t)dt = 1

2 orthogonal: 〈φk, φm〉 =

∫ T

0

φk(t)φm(t)dt = 0 if k 6= m

3 complete: if a signal x(t) is such that

〈x, φk〉 =

∫ T

0

x(t)φk(t)dt = 0 for all k, then x(t) ≡ 0


Fourier series

Let {φk(t)}∞k=1be a complete, orthonormal set of functions. Then any

“well-behaved” T -periodic signal x(t) can be uniquely represented as aninfinite series

x(t) =

∞∑

k=0

αkφk(t).

This is called the Fourier series representation of x(t). The scalars(numbers) αk are called the Fourier coefficients of x(t) (with respect to{φk(t)}) and are computed as follows:

αk = 〈x, φk〉 =

∫ T

0

x(t)φk(t)dt

In analogy to vectors in n-space, you can think of αk as the projection(or component) of x(t) in the direction of φk(t).


Fourier coefficients

To derive the formula for αk, write

x(t)φk(t) =

∞∑

m=0

αmφm(t)φk(t),

and integrate over one period:

∫ T

0

x(t)φk(t)dt

︸︷︷︸

〈x,φk〉

=

∫ T

0

(∞∑

m=0

αmφm(t)φk(t)

)

dt

=

∞∑

m=0

αm

∫ T

0

φm(t)φk(t)dt

︸︷︷︸

〈φm,φk〉

= αk,

where in the last line we use the fact that {φk} form an orthonormal

system of functions.


Convergence of Fourier series

It can be proved that, because the functions {φk} form a completeorthonormal system, the partial sums of the Fourier series

x(t) =

∞∑

k=0

αkφk(t)

converge to x(t) in the following sense:

limN→∞

∫ T

0

(

x(t) −N∑

k=1

αkφk(t)

)2

dt = 0

So, we can use the partial sums

xN (t) =

N∑

k=1

αkφk(t)

to approximate x(t).


Trigonometric Fourier series

Fact: the sequence of T -periodic functions {φk(t)}∞k=0defined by

φ0(t) =1√T

and φk(t) =

√2

T sin(kω0t), if k ≥ 1 is odd√

2

T cos(kω0t), if k > 1 is even

is complete and orthonormal. Here,

ω0 =2π

T

is called the fundamental frequency. Orthonormality is quite easy toshow, completeness — not so much.

Thus, any well-behaved T -periodic signal x(t) can be represented as aninfinite sum of sinusoids (plus a constant term α0φ0):

x(t) =∞∑

k=0

αkφk(t)



A more common way of writing down the trigonometric Fourier series ofx(t) is this:

x(t) = a0 +∞∑

k=1

ak cos(kω0t) +∞∑

k=1

bk sin(kω0t)

Then the Fourier coefficients can be computed as follows:

a0 =1

T

∫ T

0

x(t)dt

ak =2

T

∫ T

0

x(t) cos(kω0t)dt

bk =2

T

∫ T

0

x(t) sin(kω0t)dt

Recall that ω0 = 2π/T .



To relate this to the orthonormal representation in terms of the φk, wenote that we can write

a0 =1√T

∫ T

0

x(t)φ0(t)dt =1√T

α0

ak =

√

2

T

∫ T

0

x(t)φ2k(t)dt =

√

2

Tα2k

bk =

√

2

T

∫ T

0

x(t)φ2k−1(t)dt =

√

2

Tα2k−1

Thus, we have

x(t) = a0 +

∞∑

k=1

ak cos(kω0t) +

∞∑

k=1

bk sin(kω0t)

=1√T

αk ·√

Tφ0(t) +

√

2

T

(∞∑

k=1

α2k ·√

T

2φ2k(t) +

∞∑

k=1

α2k−1 ·√

T

2φ2k−1(t)

)

=∞∑

k=1

αkφk(t)


Symmetry properties

Things to watch out for when computing the Fourier coefficients:

if x(t) is an even function, i.e., x(t) = x(−t) for all t, then all itssine Fourier coefficients are zero:

bk =2

T

∫ T/2

−2/T

x(t) sin(kω0t)dt = 0

if x(t) is an odd function, i.e., x(t) = −x(−t), then all its cosineFourier coefficients are zero:

ak =2

T

∫ T/2

−2/T

x(t) cos(kω0t)dt = 0,

and

a0 =1

T

∫ T/2

−2/T

x(t)dt = 0


Trigonometric Fourier series: example

Consider a 2-periodic signal x(t) given by repeating the square wave:

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

a0 = 0 and ak = 0 for all k(the signal has oddsymmetry)

bk =

∫0

−1

sin(kπt)dt −∫

1

0

sin(kπt)dt

= − 1

kπ[cos(kπt)]

0

−1+

1

kπ[cos(kπt)]

1

0

=1

kπ

(

cos(−kπ) − 1 + cos(kπ) − 1)

=1

kπ(2 cos(kπ) − 2)

= −4 sin2(kπ/2)

kπ

x(t) = −∞∑

k=1

4 sin2(kπ/2)

kπsin(kπt) = −

∞∑

k odd

4

kπsin(kπt)


Trigonometric Fourier series: example

Approximating x(t) by partial sums of its Fourier series:

−1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

1.5N = 10

−1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

1.5N = 15

−1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

1.5N = 50

−1 −0.5 0 0.5 1−1.5

−1

−0.5

0

0.5

1

1.5N = 150

Note the Gibbs phenomenon: the Fourier series (over/under)shoots the

actual value of x(t) at points of discontinuity. In signal processing, this

effect is also called ringing.


Complex exponential Fourier series

Another useful complete orthonormal set is furnished by the complexexponentials:

φk(t) =1√T

ejkω0t, k = . . . ,−2,−1, 0, 1, 2, . . .

where ω0 = 2π/T , as before.

Note that these functions are complex-valued, and we need to redefinethe dot product as

〈x1, x2〉 =

∫ T

0

x1(t)x∗2(t)dt,

where x∗1(t) denotes the complex conjugate of x2(t). Then it is

straightforward to show that

〈φk, φm〉 =1

T

∫ T

0

ejkω0te−jmω0tdt =

{1, k = m0, k 6= m


Complex exponential Fourier series

Thus, we can expand any T -periodic x(t) as

x(t) =∞∑

k=−∞

ckejkω0t

The Fourier coefficients are given by

ck =1

T

∫ T

0

x(t)e−jkω0tdt

To derive this, multiply the series representation of x(t) on the right by

e−jkω0t and integrate from 0 to T .


Symmetry properties for real signals

Consider a real-valued T -periodic signal x(t). Then

ck =1

T

∫ T

0

x(t)e−jkω0tdt and c−k =1

T

∫ T

0

x(t)ejkω0tdt = c∗k

Write ck in polar form:

ck = Akejθk , Ak = |ck|, θk = ∠ck

Thenck = c∗−k ⇒ Ak = A−k and ∠ck = −∠c−k

Thus, for real signals the amplitude spectrum Ak has even symmetry,

while the phase spectrum θk has odd symmetry.


Amplitude and phase spectra

We can use the amplitudes Ak and the phases θk to represent thespectrum (or frequency content) of x(t) graphically as follows:

ω02ω03ω0

ω02ω03ω0

−3ω0−2ω

0−ω

0

−3ω0−2ω

0−ω

0

0

0

A(ω)

θ(ω)

ω

ω

Observe that the amplitude spectrum A(ω) has even symmetry, while

the phase spectrum θ(ω) has odd symmetry.


Documents

Lecture VIII: Fourier seriesmaxim.ece.illinois.edu/teaching/fall08/lec8.pdf · Trigonometric Fourier series Fact: the sequence of T-periodic functions {φk(t)}∞ k=0 deﬁned by