lectures on fourier series

ECE 3640

Lecture 4 – Fourier series: expansions of periodic functions.

Objective: To build upon the ideas from the previous lecture to learn aboutFourier series, which are series representations of periodic functions.

Periodic signals and representations

From the last lecture we learned how functions can be represented as a series ofother functions:

f(t) =

n∑

k=1

ckik(t).

We discussed how certain classes of things can be built using certain kinds of basisfunctions. In this lecture we will consider specifically functions that are periodic,and basic functions which are trigonometric. Then the series is said to be a Fourierseries.

A signal f(t) is said to be periodic with period T0 if

f(t) = f(t+ T0)

for all t. Diagram on board. Note that this must be an everlasting signal. Alsonote that, if we know one period of the signal we can find the rest of it by periodic

extension. The integral over a single period of the function is denoted by∫

T0

f(t)dt.

When integrating over one period of a periodic function, it does not matter whenwe start. Usually it is convenient to start at the beginning of a period.

The building block functions that can be used to build up periodic functionsare themselves periodic: we will use the set of sinusoids. If the period of f(t) isT0, let ω0 = 2π/T0. The frequency ω0 is said to be the fundamental frequency; thefundamental frequency is related to the period of the function. Furthermore, letF0 = 1/T0. We will represent the function f(t) using the set of sinusoids

i0(t) = cos(0t) = 1

i1(t) = cos(ω0t+ θ1)

i2(t) = cos(2ω0t+ θ2)

...

Then,

f(t) = C0 +

∞∑

n=1

Cn cos(nω0t+ θn)

The frequency nω0 is said to be the nth harmonic of ω0.Note that for each basis function associated with f(t) there are actually two

parameters: the amplitude Cn and the phase θn. It will often turn out to be moreuseful to represent the function using both sines and cosines. Note that we canwrite

Cn cos(nω0t+ θn) = Cn cos(θn) cos(nω0t) − Cn sin(θn) sin(nω0t).

ECE 3640: Lecture 4 – Fourier series: expansions of periodic functions. 2

Now letan = Cn cos θn bn = −Cn sin θn

ThenCn cos(nω0t+ θn) = an cos(nω0t) + bn sin(nω0t)

Then the series representation can be

f(t) = C0 +∞∑

n=1

Cn cos(nω0t+ θn)

= a0 +

∞∑

n=1

an cos(nω0t) + bn sin(nω0t)

The first of these is the compact trigonometric Fourier series. The second isthe trigonometric Fourier series.. To go from one to the other use

C0 = a0

Cn =√

a2n + b2n

θn = tan−1(−bn/an).

To complete the representation we must be able to compute the coefficients.But this is the same sort of thing we did before. If we can show that the set offunctions {cos(nω0t), sin(nω0t)} is in fact an orthogonal set, then we can use thesame formulas we did before. Check:

∫

T0

cos(nω0t) cos(mω0t) dt =

{

0 n 6= mT0

2 n = m 6= 0

So pairs of cosine functions are orthogonal. Similarly,

∫

T0

sin(nω0t) sin(mω0t) dt =

{

0 n 6= mT0

2 n = m 6= 0

So pairs of sin functions are orthogonal. Also,

∫

T0

sin(nω0t) cos(mω0t) dt = 0 for all n and m.

So the sines and cosines are orthogonal to each other. This means that we can useour formulas:

an =〈f(t), cos(nω0t)〉

〈 cos(nω0t), cos(nω0t)〉=

2

T0

∫

T0

f(t) cos(nω0t) dt

bn =〈f(t), sin(nω0t)〉

〈 sin(nω0t), sin(nω0t)〉=

2

T0

∫

T0

f(t) sin(nω0t) dt

The only exception is the coefficient a0:

a0 =〈f(t), 1〉〈1, 1〉 =

1

T0

∫

T0

f(t) dt

Example 1

Let us start with a square wave:


f(t) =

{

1 −π/2 < t < π/20 t ∈ [−π,−π/2), (π, 2, π]

and its periodic extension. The period is T0 = 2π, so ω0 = 2π/T0 = 1. The series is

f(t) = a0 +

∞∑

n=1

an cosnt+ bn sinnt

The coefficients can be found as follows:

a0 =1

2π

∫ π/2

−π/2

dt =1

2

an =1

π

∫ π/2

−π/2

cosnt dt =2

nπsin(nπ/2) =

0 n even2

πn n = 1, 5, 9, 13, . . .− 2

πn n = 3, 7, 11, 15, . . .

bn =1

π

∫ π/2

−π/2

sinnt dt = 0.

We can write

f(t) =1

2+

2

π

(

cos t− 1

3cos 3t+

1

5cos 5t− 1

7cos 7t+ · · ·

)

.

Here is an interesting tidbit: Evaluate this at t = 0:

1 =1

2+

2

π

(

1 − 1

3+

1

5− 1

7+ · · ·

)

.

Solving, we get

π = 4

(

1− 1

3+

1

5− 1

7+ · · ·

)

.

In the compact form,

C0 =1

2

Cn =

{

0 n even2

πn n odd

θn =

{

−π n = 3, 7, 11, 15, . . . ,0 otherwise

2

It is interesting to see how the function gets built up at the pieces are added


together:

The Fourier Spectrum

For a function f(t) having a compact trigonometric Fourier series, the set of am-plitudes {Cn} and the set of phases {θn} provide all the information necessary torepresent the function. (This is interesting, if you think about it: a function whichis defined at every point in a continuum can be represented with only a countablenumber of points.) A plot of the amplitudes {Cn} vs. the ω is the amplitude

spectrum of the signal. A plot of the phase {θn} vs. ω is the phase spectrum

of the signal.

Example 2 Show the magnitude and phase spectrum of the square wave from theprevious example.

Actually, if we allow the amplitude to show a shift of π by allowing a signed


quantity, then the task is somewhat easier:

In this case the spectrum is said to be an amplitude spectrum, rather than amagnitude spectrum. The amplitude spectrum is convenient to work with wheneverall the sine terms are zero, so all the information is conveyed in the {an}. 2

Example 3

Find the Fourier series of the following signal, and plot its magnitude and phasespectrum.

The signal is periodic with T0 = 2. The fundamental frequency is

ω0 =2π

2= π.

The function can be written analytically (for the purposes of integration) as

f(t) =

{

2At − 12 ≤ t ≤ 1

22A(1 − t) 1

2 ≤ t ≤ 32

This gives is only one period of the function, but that is sufficient for our purposes.The Fourier series coefficients can be written as

an =2

T0

∫ 3/2

−1/2

f(t) cosnπtdt

=

∫ 1/2

−1/2

2At cosnπtdt+

∫ 3/2

1/2

2A(1 − t) cosnπtdt

= 0

(We will take this at face value for now. Soon we will learn some tricks that willhelp evaluate this sort of thing sometimes. But notice the utility of being able

to use a symbolic integration package. You may want to go back and review


what we did early last quarter with Matlab: the symbolic integration exerciseswere written with this sort of integration mind!)

bn =2

T0

∫ 3/2

−1/2

f(t) sinnπtdt

=

∫ 1/2

−1/2

2At sinnπtdt+

∫ 3/2

1/2

2A(1 − t) sinnπtdt

=8A

n2π2sin(nπ/2)

Actually, I checked this using my current favorite symbolic tool, Mathematica. Thisis the result:

In[6] := bn = Integrate[2 A t Sin[n Pi t],{t,-1/2,1/2}] +

Integrate[2 A (1-t) Sin[n Pi t],{t,1/2,3/2}]

n Pi n Pi

A (n Pi Cos[----] - 2 Sin[----])

2 2

Out[6]= -(--------------------------------) +

2 2

n Pi

n Pi n Pi

A (n Pi Cos[----] + 2 Sin[----])

2 2

> -------------------------------- +

2 2

n Pi

n Pi

-(A n Pi Cos[----])

2 n Pi

2 (------------------- + A Sin[----])

2 2

> ------------------------------------- +

2 2

n Pi

3 n Pi 3 n Pi

A (n Pi Cos[------] - 2 Sin[------])

2 2

> ------------------------------------

2 2

n Pi

In[7]:= Simplify[bn]

n Pi n Pi n Pi 2

-4 A (n Pi Cos[----] - 2 Sin[----]) Sin[----]

2 2 2

Out[7]= ----------------------------------------------

2 2


n Pi

In problems of this sort, it is considered bad form to leave it in terms of all the sinsand cosines (go thy way and sin no more!), since they can occlude the underlyingvalues of the coefficients. So we will work through the detail to get the answershown in the book. In doing this, however, do not lose sight of the forest for thetrees. We already have the coefficients; we are simply manipulating them a bit. Inpractice, these types of manipulations could be skipped or be done by a computer(say, if we were plotting the magnitude spectrum).

In the simplification, note that the term sin2(nπ/2) is zero when n is even. Sowe only need to consider odd n. For odd n, sin2(nπ/2) = 1, so we can focus on theother part. First, consider odd n of the form n = 4k + 1. Then

cos(nπ/2) = cos((4k + 1)π/2) = cos(π/2) = 0

andsin(nπ/2) = sin((4k + 1)π/2) = sin(π/2) = 1

sin(3nπ/2) = sin(3(4k + 1)π/2) = sin(3π/2) = −1

When n = 4k + 3,

cos(nπ/2) = cos((4k + 3)π/2) = cos(3π/2) = 0

sin(nπ/2) = sin((4k + 3)π/2) = sin(3π/2) = −1

sin(3nπ/2) = sin(3(4k + 3)π/2) = sin(9π/2) = 1

Combining all this together gives the desired answer.Using the symbolic toolbox in Matlab, I obtained

syms A n t

int(’2*A*t*sin(n*pi*t)’,t,-1/2,1/2)

ans =

-2*A*(-2*sin(1/2*n*pi)+n*pi*cos(1/2*n*pi))/n^2/pi^2

The Fourier series is

f(t) =∑

n

bn sin(nπt)

=8A

π2

∑

n=1,3,5,...

(−1)(n−1)/2

n2sin(nπt)

=8A

π2

∑

n=1,3,5,...

sin(nπ/2)

n2sin(nπt)

=8A

π2

[

sin(πt) − 1

9sin(3πt) +

1

25sin(5πt) − 1

49sin(7πt) + · · ·

]

To plot the Fourier spectrum the compact trigonometric Fourier series is needed.Recall that this expresses the function in terms of cosines. We only have it in termsof sines. We we have to do a little phase-shift trick:

sin(kt) = cos(kt− 90◦)

− sin(kt) = cos(kt+ 90◦)


Then the compact Fourier series is

f(t) =8A

π2

[

cos(πt− 90◦) +1

9cos(3πt+ 90◦) +

1

25cos(5πt− 90◦) +

1

49cos(7πt+ 90◦) + · · ·

]

.

The magnitude and phase spectrum are:

2

Symmetry and its effects

An even function f(t) is a function such that

f(t) = f(−t)

Examples of even functions are f(t) = cos(t) and f(t) = t2. An odd function is afunction such that

f(t) = −f(−t)Examples of odd functions are f(t) = sin(t) and f(t) = t3. There are severalfacts about even and odd functions that can help us simplify and interpret somecomputations.

1. The product rules:

• even × even = even

• even × odd = odd × even = odd

• odd × odd = even.

For example, t2 cos(t) is an even function. t cos(t) is an odd function. (Therules are the same as the rules for adding even and odd numbers.)

2. Integration. When integrating over a symmetric interval about the origin,

∫ T0/2

−T0/2

even(t)dt = 2

∫ T0/2

0

even(t)dt

∫ T0/2

−T0/2

odd(t)dt = 0


Let us use these facts in relation to Fourier series. Suppose we want to computethe F.S. of an even function (such as the square wave signal example). Then

a0 =1

T0

∫ T0/2

−T0/2

f(t)dt =2

T0

∫ T0/2

0

f(t)dt.

an =2

T0

∫ T0/2

−T0/2

f(t) cos(nω0t)dt =4

T0

∫ T0/2

0

f(t) cos(nω0t)dt.

bn =2

T0

∫ T0/2

−T0/2

f(t) sin(nω0t)dt = 0

To compute the F.S. of an odd signal,

a0 =1

T0

∫ T0/2

−T0/2

f(t)dt = 0

an =2

T0

∫ T0/2

−T0/2

f(t) cos(nω0t)dt = 0

bn =2

T0

∫ T0/2

−T0/2

f(t) sin(nω0t)dt =4

T0

∫ T0/2

0

f(t) sin(nω0t)dt.

Review the signals transformed so far in light of these symmetries.

Determining the fundamental frequency

The trigonometric Fourier series can be used to represent any periodic function.In periodic functions, every frequency in the Fourier series representation is anintegral multiple of some fundamental frequency. Such frequencies are said to beharmonically related. The ratio of any two harmonically related frequencies isa rational number (i.e., a number which can be represented as the ratio of twointegers). (Interesting mathematical fact: there are more irrational numbers thanthere are rational numbers. ) Any number which involves a transcendental numbersuch as π or e, or which involves square roots which cannot be simplified down toratios of integers (such as

√2) is an irrational number. For functions which are

harmonically related, the fundamental frequency is the greatest common divisor ofthe frequencies.

Example 4 Is the function f1(t) = 2 + 7 cos( 12 t+ θ1) +

√3 cos( 2

3 t+ θ2) + 5 cos( 76 t)

a periodic function? If it is, what is the fundamental frequency?We need to determine if the frequencies are harmonically related. We neglect

the DC term. Taking ratios:1223

=3

4

which is rational.1276

=3

7

which is rational.2376

=4

7

which is rational. So f(t) is a periodic function. The fundamental frequency is thegreatest common divisor

GCD(1/2, 2/3, 7/6) = 1/6.


(The GCD(num)/LCM(den)). 2

Example 5 Is the function f(t) = 3 sin(3√

2t+ θ) + 7 cos(6√

3t) periodic?The ratio of frequencies is

3√

2

6√

3=

√2

2√

3

which is irrational. 2

Example 6 Is f(t) = 3 sin(3√

2t+ θ) + 7 cos(6√

2t) periodic?The ratio of frequencies is

3√

2

6√

2

which is rational. The fundamental frequency is

GCD(3√

2, 6√

2) = 3√

2.

2

Interpretation of the smoothness of the function

Functions which are smooth (e.g. continuous) have most of their variations at lowerfrequencies. Functions which are not smooth have variations at higher frequencies.We can look at the rate of decay of the amplitude spectrum to determine somethingabout the smoothness of the function.

For example, the square wave function has abrupt jumps and is not even contin-uous. The coefficients of the F.S. decay as 1/n. By contrast, the sawtooth functionwe examined is smoother, since it is continuous. Its coefficients decay more quickly,decaying down as 1/n2.

The Gibbs phenomenon

According to our theory, with a complete set of basis functions we can representany function exactly. We furthermore know how to obtain the best approximationif we use only a finite set of functions. Interestingly, even the best approximationcan still have some substantial errors. Consider the error in the square-wave series.Observe that there is a jump just before the point of discontinuity. As it turns out,no matter how large n is, this error remains, and it has an amplitude of about9% of the discontinuity. As n gets larger and larger, this wiggle becomes narrowerand moves closer to the point of the discontinuity, but it never goes away. Thisovershoot phenomenon is known as the Gibbs phenomenon.

One of the important ramifications of this is in how we define functions to beequal. It is true that

∫

T0

(f(t) −∞∑

n=0

Cn cos(nω0t+ θn))2dt = 0

that is, there is zero error between the function and the Fourier series, as definedby this squared integral criterion. But it does not mean that the functions arepoint-for-point equal. In this case, the error region simply becomes so small thatthe integral is zero. But this does not mean that

f(t) =

∞∑

n=0

Cn cos(nω0t+ θn)


at every point. The mathematicians who like to leave no such stones unturned havemade this an object of tremendous study, and consider such qualifications as equal“almost everywhere” (a.e.) or equal “with probability one”. These are both indistinction to equal “everywhere” or equal “always.” We have to be careful whatwe mean when we say two things are equal! (Ayn Rand would probably have troublewith this: perhaps it is not the case that A = A is always true!)

Using Fourier Series for Signal Analysis

Suppose we have a system with transfer function H(s) and a periodic input signalf(t). What is the output signal? One way to do this, of course, is to convolvethe input signal with the impulse response. But we all know how much we loveconvolution, and there is not a lot of insight to be gained from such a brute forcecomputation. Another approach is to represent f(t) in terms of sinusoids, then usethe properties of L.T.I. systems. Specifically, if

f(t) = C0 +

∞∑

n=1

Cn cos(nω0t+ θn).

then the output will be the sum of the responses due to each input:

y(t) = C0H(0) +∞∑

n=1

Cn|H(jnω0)| cos(nω0t+ θn + ∠H(jnω0))

Discuss what happens in terms of filtering. In the homework you will work anexample of this.

Exponential Fourier Series

We have seen how sin and cosine functions can be represented in terms of complexexponentials. It turns out that we can use complex exponentials to represent Fourierseries. In many respects, this makes for a simpler representation.

Let’s go back to the compact Fourier series representation function, and expressit in terms of complex exponentials:

Cn cos(nω0t+ θn) = Dnejnω0t +D−ne

−jnω0t

where

Dn =1

2ejθnCn

D−n =1

2e−jθnCn

We can write the Fourier series:

f(t) = C0 +∞∑

n=1

Cn cos(nω0t+ θn)

= D0 +

∞∑

n=1

Dnejnω0t +D−ne

−jnω0t

= D0 +∑

n6=0

Dnejnω0t

=∞∑

n=−∞Dne

jnω0t


Now, how do we find the coefficients? First, note that if we define the inner productcorrectly, the exponentials are orthogonal. Up to now, inner products have beendefined for real functions. We will extend this now to inner products over complexfunctions. If f(t) and g(t) are a periodic complex functions with period T0, definethe inner product as

〈f(t), g(t)〉 =

∫

T0

f(t)g(t)dt

where the over-line means complex conjugate. The rules for this inner product arethe same as before, except that

〈f, g〉 = 〈g, f〉

With this inner product, note that

〈ejnω0t, ejmω0t〉 =

∫

T0

ejnω0te−jmω0tdt =1

j(n−m)ω0[ej(n−m)2π − 1]

which is 0 if n −m 6= 0 and T0 is n = m. So under this inner product, we have awhole set of orthogonal functions. The geometry of orthogonal functions we talkedabout before applies, including the orthogonality theorem. We can therefore write

Dn = proj(f(t), ejnω0t) =〈f, ejnω0t〉

〈ejnω0t, ejnω0t〉 =1

T0

∫

T0

f(t)e−jnω0tdt

(Derive this another way also.)Note that this is true for all values of n (there are no special cases when n = 0)

and there is only one formula (not two, as for sins and cosines). This is my preferredform! In fact, due to its similarity with the Fourier transform to be discussed soon,it is the most common form of the Fourier series. It is, of course, possible to convertfrom one form to another. For example,

Dn =1

2(an − jbn)

Suppose (as is most often the case) that f(t) is a real function. Then

Dn =1

T0

∫

T0

f(t)e−jnω0tdt =1

T0

∫

T0

f(t)ejnω0tdt = D−n.

Example 7 Find the exponential F.S. of the square wave function with period 2πof p. 428.

Dn =1

2π

∫ π/2

−π/2

(1)e−jntdt =1

2

sin(nπ/2)

nπ/2

This is a good time to introduce a function that is near and dear to the heart ofengineers:

sinc(x) =sinx

x

So we can write

Dn =1

2sinc(nπ/2)

Note: sinc(0) = 1. (How do we know?) 2

Important observation: To compute the F.S. coefficients, multiply the function byan exponential with negative exponent. There are many transforms that electricalengineers use — Laplace transforms, Z-transforms, Fourier series, Fourier trans-forms, etc. In all of these, the exponent is negative. Don’t forget it!


Exponential Fourier Spectra

As for the trigonometric F.S. we can make a plot of the F.S. by plotting the mag-nitude and phase of the complex numbers. For the example above,

D0 = 1/2 |D0| = 1/2 ∠D0 = 0

D1 = 1/π |D1| = 1/π ∠D1 = 0

D−1 = 1/π |D−1| = 1/π ∠D−1 = 0

D2 = 0 D−2 = 0

When plotting the spectrum, both positive and negative values of n need, ingeneral, to be plotted.

Bandwidth of a signal

The bandwidth of a signal is the amount of “frequency” required to sustain thesignal unmodified. Actually, there are about a bajillion different definitions forbandwidth, depending on the application of the problem.

Example 8 Spectra on p. 445. The bandwidth is the highest nonzero frequency −the lowest nonzero frequency = 9− 0 = 9.

2

Example 9 The bandwidth of the square wave function is infinite! What is oftendone from a practical power of view is to go out “far enough” – till the terms notincluded are small enough to worry about. Just what is “far enough” depends onthe particular application. 2

Example 10 In this example, we will find the F.S. of an important function, theperiodic set of pulses. This is used (as we will see in chapter 8) as a representationof the sampling process.

δT0(t) =

∑

n

δ(t− nT0)


Plot and explain. This is clearly periodic, and hence has a F.S. (in some sense).We can write

δT0(t) =

∞∑

n=−∞Dne

jnω0t

where we can compute the coefficients from

Dn =1

T0

∫

T0

δT0(t)e−jnω0tdt

Taking the integral from −T0/2 to T0/2 we get

Dn =1

T0

∫ T0/2

−T0/2

δ(t)e−jnω0tdt =1

T0

This gives the important formula

δT0(t) =

1

T0

∞∑

n=−∞ejnω0t

The spectrum does not decay! |Dn| = 1T0

and ∠Dn = 0. 2

Energy of signals and Parseval’s relationships

It is possible and often theoretically useful to examine the energy of signals in boththe time domain and the frequency (Fourier series) domain. We will develop animportant relationship. Suppose f(t) is a periodic function with F.S. representation

f(t) =∑

n

Dnejnω0t

and g(t) is a periodic function with the same period and a F.S. representation

g(t) =∑

n

Enejnω0t

Now consider an average energy kind of term

1

T0

∫

T0

f(t)g∗(t)dt

Substituting in for each of the F.S. gives (taking advantage of the orthogonality ofthe exponential function)

1

T0

∫

T0

f(t)g∗(t)dt =∑

n

DnE∗n

We can write this in a convenient inner product notation. We can define the innerproduct between two series {Dn} and {En} as

〈Dn, En〉 =∑

n

DnE∗n

Then we can write (using our complex inner product for functions)

1

T0〈f(t), g(t)〉 = 〈Dn, En〉


A relationship such as this is known as a Parseval’s relationship, named after someguy.

As a special case, take g(t) = f(t). Then 1T0

〈f(t), f(t)〉 is the average energy off(t). By the Parseval’s relationship,

1

T0〈f(t), f(t)〉 = 〈Dn, Dn〉

Example 11 Find the sum of the series

∑

n

(π

2sinc(nπ/2)

)2

By recognizing the terms from the F.S. for a square wave with period 2π, we canuse Parseval’s relationship:

∑

n

(π

2sinc(nπ/2)

)2

=1

2π

∫ π/2

−π/2

(1)2dt =1

2

This would have been hard to do any other way! 2

A return to the geometric viewpoint

We have seen that functions can be represented as series of orthogonal functions,and have seen examples of orthogonal functions, the trigonometric and the complexexponentials. Historically, these were first examined by Jean Baptiste Fourier, whoused these to solve the partial differential equations related to heat flow. At firsthis methods were considered unconventional by mathematicians. Now the gener-alization of Fourier’s methods form one of the largest and most fruitful areas ofmathematics.

Are there any other useful orthogonal functions? Given a set of functions thatare not orthogonal, is it possible to make it orthogonal somehow? Both answers areyes.

We begin looking at a set of orthogonal polynomials, defined over the inter-val [−1, 1]. It is easy to check that the set of polynomials {1, t, t2, t3, . . .} is notorthogonal. For example,

〈t, t3〉 =

∫ 1

−1

(t)(t3)dt =2

5

Consider, however, the polynomials

P0(t) = 1

P1(t) = t

P2(t) =3

2t2 − 1

2

P3(t) =5

2t3 − 3

2t

In general,

Pn =1

2nn!

dn

dtn(t2 − 1)n.


These polynomials are known as the Legendre polynomials. It can be shown that

∫ 1

−1

Pm(t)Pn(t)dt =

{

0 m 6= n2

2m+1 m = n

For functions that are defined over any finite interval, Legendre polynomials can beused as a functional representation.

Another way that orthogonal functions arise is by means of another inner prod-uct. While we have not mentioned it in the past, it is possible to introduce a positiveweighting factor into the inner product. Every one of these produces a new innerproduct, each with properties that may be useful for particular applications. If w(t)is a non-negative function, then we can define an inner product

〈f, g〉w =

∫

I

w(t)f(t)g(t)dt

where the subscript w indicates the weighting and I is some interval of integration.For example, for w(t) = 1√

1−t2and I = [−1, 1] we might define an inner product as

〈f(t), g(t)〉w =

∫ 1

−1

1√1 − t2

f(t)g(t)dt

A set of polynomials that is orthogonal with respect to this norm is the set ofChebyshev polynomials:

T0(t) = 1


Tn(t) = cosn cos−1 t

For example,T1(t) = t

T2(t) = cos 2 cos−1 t = 2 cos2(cos−1 t) − 1 = 2t2 − 1

T3(t) = cos 3 cos−1 t = 4 cos3(cos−1 t) − 3 cos(cos−1 t) = 4t3 − 3t

Notice that these have the “equal ripple” property. This makes them useful infunction approximation (and in fact, Chebyshev functions are used at the heart ofthe Remez algorithm).

Another set of interesting orthogonal functions that have been the topic of anincredible amount of research lately are wavelet functions. Unlike the familiar andcommon trigonometric functions, there are actually several families of wavelets (thisis part of what makes it confusing). One type of wavelets is described by two setsof functions: φ(t) (known as a scaling function), and ψ(t) (known as a waveletfunction). In dealing with these functions, instead of looking at different frequencies,we scale and shift these functions. We define

φj,k(t) = 2j/2φ(2jt− k)

ψj,k(t) = 2j/2ψ(2jt− k)

Make plots and explain. Then (somewhat miraculously), the following orthogonalityproperties exist:

〈φj,k(t), φj,m(t)〉 = δk,m

〈ψj,k(t), ψl,m(t)〉 = δj,lδk,m

〈φj,k(t), ψl,m(t)〉 = δj,lδk,m

These are pretty remarkable! This gives us a whole bunch of functions that we canuse to signal representations:

f(t) =∑

k

aJ,kφJ,k(t) +∑

j,k

bj,kψj,k(t)


This is useful for a variety of things which will become more clear after we talkabout Fourier transforms (next chapter). But notice that we can represent any

function, not just periodic functions, localizing both frequency information andtime information. There are also very efficient algorithms that are faster than theFast Fourier Transform (FFT) to compute a discrete wavelet transform.

lectures on fourier series

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