FOURIER TRANSFORMS WITH APPLICATIONS TOFTIR

JACOB MULLINS

1. Introduction

Fourier Transforms can be used in many different fields, such aschemistry, physics, and of course, mathematics. In this paper, chem-istry will be the background and give way to the necessity of eachdefinition, theorem, and proof. We will concentrate on Fourier Trans-form Infrared Radiation, a method of understanding the atoms thatmake up a given material by observing the absorption of light at cer-tain wavelengths. This method is accomplished by splitting a beam oflight, and causing each half beam to go in and out of phase to obtainan interferogram of different wavelenghts of light at different intensi-ties. This interferogram is then transformed by the Fourier Transforminto what is known as an Infrared Radiation Spectrum, which is easilyanalyzed by chemists to detect certain atoms.

First, we will lay the basic foundation necessary for the rest of thepaper by defining a complex vector space and stating the propertiesthat follow from that space. Next, we will show that we can estimatea function by a trigonometric polynomial using interpolation. Then,we will show the ease we can calculate coefficients by looking at anequation as a polynomial. Afterwards, we will show the usefulness ofthe Fourier Transform in chemisty. Finally, we will estimate the FourierTransform using Riemann Sum.

2. Inner Product Spaces

Although we will go deeper in that idea later, first, we must gothrough some definitions and theorems to build up to that point.

Definition 2.1. Let V be a complex vector space. An inner producton V is a complex-valued function < u, v > of u and v ∈ V having thefollowing properties:

(1) 〈u, v〉 = 〈v, u〉

Date: October 21, 2013.1

2 JACOB MULLINS

(2) 〈αu+ βv, w〉 = α〈u,w〉+ β〈v, w〉

(3) 〈u, u〉 ≥ 0

(4) 〈u, u〉 = 0 =⇒ u = 0

(5) 〈u, αv + βw〉 = α〈u, v〉+ β〈u,w〉

We will now define what is meant by the orthogonality of two vectorsin the space in Definition 2.1

Definition 2.2. Two vectors u and v are called orthogonal if 〈u, v〉 = 0and orthonormal if each vector has norm one. The norm of u is definedas ‖u‖ =

√〈u, u〉

Theorem 2.1. [1] If ρ1, ρ2, ..., ρN is an ON basis in an N dimensionalinner product space V , then every u ∈ V can be written asu =

∑Nj=1〈u, ρj〉ρj, and furthermore one has

‖u‖2 =N∑j=1

|〈u, ρj〉|2.

For the inner product of two vectors one also has the following formula:

〈u, v〉 =N∑j=1

〈u, ρj〉〈v, ρj〉.

3. Interpolation

In the book by Vretblad [1], he considers the interval P = [0, 2π], butwe will consider the interval P = [−π, π]. Subdivide P into N = 2mintervals, m ∈ Z Define xk = −π + πk

mLet lm be the set of complex-valued functions with domain xk : k = 0, 1, ..., 2m− 1We define, on lm,

〈f, g〉 =2m−1∑k=0

f(xk)g(xk).

〈f, g〉 satisfies Definition 2.1, and therefore contains all of the propertiesstated. For n = −m,−m+ 1, ...,m− 1 let

ϕn(x) = einx

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 3

Now suppose l 6= n Then

〈ϕl, ϕn〉 =2m−1∑k=0

ϕl(xk)ϕn(xk)

=2m−1∑k=0

eilxke−inxk

=2m−1∑k=0

ei(l−n)xk

=2m−1∑k=0

ei(l−n)(−π+( πm

)k)

=2m−1∑k=0

eiπ(l−n)ei(l−nm

)πk

Remembering that eiπx = (−1)x,

〈ϕl, ϕn〉 = (−1)l−nei(l−nm

)πk.

Letting rk = ei(l−nm

)πk, we get

〈ϕl, ϕn〉 =2m−1∑k=0

(−1)l−nrk

By the geometric sum, we obtain

〈ϕl, ϕn〉 =2m−1∑k=0

(−1)l−n(1− r2m

1− r)

But notice that r2m = ei(l−nm

)π·2m = ei(l−n)2π = 1.So 〈ϕl, ϕn〉 =

∑2m−1k=0 (−1)l−n(1−1

1−r ) = 0 Now let l = n. Then 〈ϕl, ϕn〉 =〈ϕn, ϕn〉.So 〈ϕn, ϕn〉 =

∑2m−1k=0 (1) = 2m

〈ϕn, ϕn〉 has 2m elements, which can be shown to be the dimensionof lm, we know it is a basis of lm. Therefore, ϕ−m, . . . , ϕm−1 is anorthogonal basis for lm but not orthonormal.Since we have an orthogonal basis, we can write an arbitrary f ∈ lm asf =

∑m−1n=−m cnϕn.

4 JACOB MULLINS

Fix l ∈ (−m,−m+ 1, . . . ,m− 1). Then

〈f, ϕl〉 =m−1∑−m

cn〈ϕn, ϕl〉

= cl〈ϕl, ϕl〉

by theory above.This implies

cl =〈f, ϕl〉〈ϕl, ϕl〉

=〈f, ϕl〉

2m.

Then

(6) f =m−1∑n=−m

cnϕn =m−1∑n=−m

〈f, ϕn〉2m

ϕn =m−1∑n=−m

f(n)ϕn,

letting

(7) f(n) =〈f, ϕn〉

2m.

It follows that for k = 0, 1, . . . , 2m− 1, f(xk) = S(xk), where

S(x) =m−1∑n=−m

f(n)ϕn(x)

=m−1∑n=−m

f(n)einx

= f(−m)e−imx +m−1∑n=1

(f(n)einx + f(−n)e−inx) + f(0).

Since einx = cosx+ i sinx, we call S(x) a trigonometric polynomial.

Theorem 3.1. Assume f(t) is even on [−π, π] and real. Then f(t)can be interpolated by the equation

S(x) = f(0) +m−1∑k=1

2f(k) cos kx+ f(m) cosmx

I.e. f(xk) = S(xk) for k = 0, . . . , 2m− 1.

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 5

Proof. We will look at each piece of S(x). First we will start with

f(−m)

f(−m) =1

2m〈f, ϕ−m〉

=1

2m

2m−1∑k=0

f(xk)ϕ−m(xk)

Notice

ϕ−m(xk) = e−imxk

= e−im(−π+πkm

)

= eimπe−ikπ

= eiπ(m−k) = (−1)m−k

Therefore

f(−m) =1

2m

2m−1∑k=0

f(xk)(−1)m−k

=(−1)m

2m

2m−1∑k=0

f(xk)(−1)k

Now look at f(n).

f(n) = 12m

∑2m−1k=0 f(xk)ϕn(xk).

ϕn(xk) = einxk . By Euler’s formula, einxk = cosnxk − i sinnxk. So

f(n) = 12m

∑2m−1k=0 f(xk) cosnxk − i

2m

∑2m−1k=0 f(xk) sinnxk

We will now define

an =1

m

2m−1∑k=0

f(xk) cosnxk

and

bn =1

m

2m−1∑k=0

f(xk) sinnxk.

6 JACOB MULLINS

Then for 1 ≤ n ≤ m − 1, f(n) = 12(an − ibn). Similarly, f(−n) =

12(an + ibn). Since all real data, f(n) and f(−n) are both real. Now

f(n)einx + f(−n)e−inx =1

2(an − ibn)einx +

1

2(an + ibn)e−inx

= Re[(an − ibn)einx]

= Re[(an − ibn)(cosnx+ i sinnx)]

= an cosnx+ bn sinnx

But notice that if we let l = k −m.

bn =1

m

m−1∑l=−m

f(xm+l) sinnxm+l

Now

xm+l = −π +π(m+ l)

m

= −π +πm+ πl

m

=πl

m.

Therefore

bn =1

m

m−1∑l=−m

f(πl

m) sin (

nπl

m)

=1

m(f(0) sin 0 + f(−π) sin (−nπ) +

m−1∑l=1

[f(πl

m) sin (

nπl

m) + f(

−πlm

) sin−nπlm

]

= 0

since f was assumed to be even.Now remembering back to the fact above that f(n) = 1

2(an − ibn) and

bn = 0, we get

(8) f(n) =1

2an and f(−n) =

1

2an.

So f(n)einx + f(−n)e−inx = an cosnx

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 7

Now looking at f(m), by equation above, we get

f(m) =1

2am

=1

2m

2m−1∑k=0

f(xk) cosmxk

=1

2m

2m−1∑k=0

f(xk) cosm(−π +πk

m)

Notice that

cosm(−π +πk

m) = Re[ei(m(−π+πk

m))]

= Re[ei(−mπ+πk)]

= Re[e−imπeiπk]

= Re[(cos−mπ + i sin−mπ)(cosπk + i sin πk)]

= Re[(cosmπ)(cos πk)]

= (−1)m(−1)k.

Therefore

f(m) =1

2m

2m−1∑k=0

f(xk)(−1)m(−1)k

=(−1)m

2m

2m−1∑k=0

f(xk)(−1)k

= f(−m)

Therefore, remembering f(x) is assumed to be even,

S(x) = f(−m)e−imx +m−1∑n=1

(f(n)einx + f(−n)e−inx) + f(0)

= f(−m) cosmx+m−1∑n=1

(f(n)einx + f(−n)e−inx) + f(0)

= f(m) cosmx+m−1∑k=1

an cosnx+ f(0)

= f(m) cosmx+ 2m−1∑k=1

f(k) cos kx+ f(0)

�

8 JACOB MULLINS

For a quick example, we will now look at

f(x) = x6 − 3x4 + 7x2 − 10.

Letting m = 2,

S(x) = f(0) + 2f(1) cosx+ f(2) cos 2x.

The figure below shows the interpolation of f(x) with m = 2. We cantell this interpolation is equal to f at every partition point, but it isnot a very good estimation in between the points.

-3 -2 -1 1 2 3

200

400

600

Figure 1. Interpolation of f(x) with m=2

Now letting m = 25, we can tell it is still equal at every partitionpoint, but the estimation is much better.

-3 -2 -1 1 2 3

100

200

300

400

500

600

700

Figure 2. Interpolation of f(x) with m = 25

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 9

4. Fast Methods For Finding Coefficients

Now, for 0 ≤ n ≤ m,

f(n) =1

2m〈f, ϕn〉

=1

2m

2m−1∑k=0

f(xk)ϕn(xk)

=1

2m

2m−1∑k=0

f(xk)e−inxk

=1

2m

2m−1∑k=0

f(xk)e−in(−π+πk

m)

=1

2m

2m−1∑k=0

f(xk)(einπ)(e−

inπm )k

=(−1)n

2m

2m−1∑k=0

f(xk)(ωn)k where ωn = e−inπm

We will look at Horner’s method, where we look at f(n) as a k degreepolynomial in the ωn variable.

Let

f(x) = x5 − 2x4 + 5x3 − 3x2 + x− 2

First, we will calculate f(3) using a direct method.

f(3) = 35 − (2 ∗ 34) + (5 ∗ 33)− (3 ∗ 32) + 3− 2

= 243− (2 ∗ 81) + (5 ∗ 27)− (3 ∗ 9) + 3− 2

= 243− 162 + 135− 27 + 3− 2

= 190

which takes up to 14 steps. In Horner’s Method i.e. synthetic division,we will divide out x − 3. First, we write out the coefficients of thepolynomial. Then we drop down the first coefficient, multiply it by 3and subtract it from the second coefficient. Drop that number downand continue. In the last step, the number dropped down is the valueat f(3), shown below.

1 − 2 5 − 3 1 − 2

3 3 3 24 63 192

1 1 8 21 64 190

10 JACOB MULLINS

which takes only 10 steps.

The Fast Fourier Transform, by Cooley and Tukey, is another methodthat takes advantage of the fact that f(n) can be written as a polyno-mial. This method is the one actually used in chemistry on an infraredradiation spectrum, but will not be explained in this paper. Now wewill lead into a discussion of IR spectra and the instrument itself, aswell as the Fourier Transform.

5. Fourier Transform on IR Spectra

Infrared Radiation Spectroscopy is a very important tool in chem-istry because it allows one to see inside the molecule and obtain whatatoms or groups are contained. In order to do this, the instrumentsends light into a beam splitter, which allows fifty percent of the lightto continue straight, and the other fifty percent to be reflected by thefixed mirror and sent back to the beam splitter, where some of thelight is sent back to the source, and some is sent through the sampleto the detector. The light that went straight is reflected off a movablemirror that allows for the wavelengths of light to travel different dis-tances than the fixed mirror path. The light is then reflected back tothe beam splitter, where some of the light is sent to the source andsome is sent through the sample and into the detector. The light fromthe fixed mirror path and the movable mirror path interact with eachother to show a phenomena known as constructive and destructive in-terference. If two wavelengths of light are in phase, meaning if theyreach their maxima and the minima together, then it is known as con-structive interference, and the intensity of the light at that wavelengthis intensified. If the two wavelengths of light are out of phase, meaningone reaches its maxima while the other is reaching its minima, then itis known as destructive interference, and the intensity of the light isdampened. The detector reads these intensities and measures it as afunction of time and volts. Then the data is ran through the FourierTransform, and ratioed with data that was obtained without a sample,to obtain what is known as the Infrared Radiation Spectrum.

Now that we have shown how the instrument works, we will showthe math behind the method.A wavenumber is defined as the inverse wavelength of light.Let f(δ) = cos 2πδν. So f(δ) is a sinusoid that can describe a laserwith wavenumber ν. Now, assuming a continuous spectrum of light,and adding in B(ν),a function to correct the fact that each wavelengthof light may be at a different intensity,

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 11

Figure 3. Infrared Radiation Spectroscopy Instrument [2]

(9) I(δ) =

∫ ∞0

B(ν) cos 2πδνdν

where ν = 1λ

Remembering Euler’s formula, eix = cosx + ı sin x, and assumingB(ν) is even,

12 JACOB MULLINS

(10) I(δ) =

∫ ∞0

B(ν)e−2πıδνdν.

Definition 5.1. [1] The function f defined by

f(ω) =

∫R

f(t)e−iωtdt,

where ω ∈ R is called the fourier transform of f .

Now we can see that Equation 10 is a Fourier Transform. Thereforewe can use the inversion formula to find B(ν) [1], which is the InfraredRadiation Spectrum.

(11) f(t) =1

2π

∫ ∞−∞

f(ω)eiωtdω

Now, with equation above, let t = 2πν and g(t) = B( t2π

). So

I(δ) =1

2π

∫ ∞0

g(t)e−ıδtdt.

Now we let g(t) be even, which is a valid assumption because of the in-strumentation. Also notice the right side of the equation is the FourierTransform of g(t). So

g(δ) =

∫ ∞−∞

g(t) cos δtdt

= 2

∫ ∞0

g(t) cos δtdt

Now we substitute I(δ) in using the equation above, and we obtain

g(δ) = 2(2πI(δ)) = 4πI(δ).

Notice that g(δ) is even. Now using the inverse transform on g(δ), weget

g(t) =1

2π

∫ ∞−∞

g(δ)eiδtdδ.

Remembering that g(t) is even, substituting in I(δ), and slightly sim-plifying,

g(t) = 2

∫ ∞−∞

I(δ) cos δtdδ.

Substitution in for g(t) and t,

(12) B(ν) = 2

∫ ∞−∞

I(δ) cos 2πνδdδ.

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 13

which is the formula for the spectrum of an IR instrument.For use in the next proof, we will now define a notation and state a

theorem.

Definition 5.2. F [f(t)](ω) =∫∞−∞ f(t)eintdt

Theorem 5.1. [1] F [eiλtf(t)](ω) = f(ω − λ)

A function often used in Chemistry is called the boxcar functionin order to truncate the infinite set above. The boxcar function setseverything inside of a certain interval to 1 and everything outside ofthe interval to 0. Chemists often use truncation methods in order tospeed up data collection and to warrant use of the Discrete FourierTransform.

Theorem 5.2. After truncating by the boxcar function,

B(ν) ≈ 2∆[sinc(2π∆(ν − ν0)) + sinc(2π∆(ν + ν0)).

where we assume I(δ) = cos (2πδν0).

Proof. Let B(ν) = 2∫∞−∞ I(δ) cos 2πνδdδ.

Now we will estimateB(ν) by the equationB∆(ν) = 2∫ ∆

−∆I(δ) cos 2πνδdδ,

since there is no instrument that can measure from −∞ to ∞.Substituting in for I(δ), we get

B∆(ν) = 2

∫ ∆

−∆

cos 2πν0δ cos 2πνδdδ

= 2

∫ ∆

−∆

cos 2πν0δe−2πνδidδ

since we know B(ν) is an even function.Multiplying by the boxcar function we get

B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ.

where

χA(δ) =

{1 δ ∈ A0 δ 6∈ A.

Notice that, substituting in Fourier notation,

B∆(ν) = F [2 cos 2πν0δχ[−∆,∆](δ)](2πν)

= F [2ei2πν0δχ[−∆,∆](δ)](2πν)

+ F [2e−i2πν0δχ[−∆,∆](δ)](2πν)

14 JACOB MULLINS

By Theorem 1.3, we know

F [2ei2πν0δχ[−∆,∆](δ)](2πν)

= F [χ[−∆,∆](δ)](2πν − 2πν0)

Therefore, we will first look at F [χ[−∆,∆](δ)](2πν). Let ω = 2πν. Lett = δ. So

F [χ[−∆,∆](δ)](2πν)

=

∫ ∆

−∆

e−iωtdt Let s =t

∆

= ∆

∫ 1

−1

e−i(ω∆)sds

= 2∆sinc(ω∆)

Now

F [2ei2πν0δχ[−∆,∆](δ)](2πν)

= F [χ[−∆,∆](δ)](2πν − 2πν0)

= 2∆sinc((2πν − 2πν0)∆)

= 2∆sinc(2π(ν − ν0)∆)

and

F [2e−i2πν0δχ[−∆,∆](δ)](2πν)

= F [χ[−∆,∆](δ)](2πν + 2πν0)

= 2∆sinc(2π(ν + ν0)∆)

Therefore

B(ν) ≈ B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ

= 2∆[sinc(2π∆(ν − ν0)) + sinc(2π∆(ν + ν0))]

�

There are many other truncation equations that have been tested.One of which is the triangle function f(x) = 1 − |x|. The Fouriertransform of the triangle function can also be written as a sinc function.

Theorem 5.3. After truncating by the triangle function,

B(ν) ≈ ∆(sinc2(π∆(ν − ν0)) + sinc2(π∆(ν + ν0))).

Although these approximations are extremely useful, we must takeone more approximation in order to make the Fourier Transform prac-tical for these types of calculations.

FOURIER TRANSFORMS WITH APPLICATIONS TO FTIR 15

6. Application of Fast Methods to the Spectra

We will use Riemann Sum approximation to take advantage of thetheory earlier in the paper. This will speed up the calculations andmake Fourier Transform theory a possible tool in chemistry. First, wewill once again approximate Equation 12 by

B∆(ν) ≈ 2

∫ ∆

∆

I(δ) cos (2πνδ)dδ

= 2

∫ ∆

−∆

I(δ)e−i2πνδdδ.

Now we will let t = πδ∆

and dδ = ∆πdt. So

B∆(ν) =2∆

π

∫ π

−πI(

∆

πt)e−i(2δν)tdt.

With Riemann Sum approximation in mind, we partition the interval[−π, π] again with xk. Therefore using the approximation,

B∆(ν) =2∆

m

2m−1∑0

I(∆

π(xk))e

−i(2δν)(xk).

Let f(xk) = I(∆πxk) and notice e−i2∆νxk = ϕ2∆ν(xk). By Theorem 2.1,

B∆(ν) =2∆

m

2m−1∑0

f(xk)ϕ2∆ν(xk)

=4∆

2m〈f, ϕ2∆ν〉.

Remembering Equation 7,

(13) B∆(ν) = 4∆f(2∆ν)

If B∆(ν) can be written as a polynomial, all of the theory earlier canbe useful to speed up calculations. Therefore, we state the followingtheorem.

Theorem 6.1. B∆ can be written as a polynomial.

Proof. Note that B∆(−ν) = B∆(ν). Now choose ν so that 2∆ν = n,

for −m ≤ n ≤ m− 1. So B∆( n2∆

) = 4∆f(n). Notice

f(xk) = I(∆

π(−π +

πk

m))

= I(−∆ +∆k

m)

16 JACOB MULLINS

Now, if n = 1, · · · ,m− 1, and δ = (∆π

(−π + πkm

)), then

B∆(n

2∆) =

4∆

2m

2m−1∑k=0

I(δk)e−in(−π+πk

m)

=2∆

m

2m−1∑k=0

I(δk)(−1)ne−iπnkm

=2∆(−1)n

m

2m−1∑k=0

I(δk)e−iπnkm .

Therefore B∆( n2∆

) is written in a polynomial of degree 2m− 1 �

Since it is written as a polynomial, Horner’s method or the FastFourier Transform [3] can be used to speed up calculations and allowfor the Fourier Transform to be a viable method for chemical instru-mentation.

References

[1] A. Vretblad, Fourier Analysis and Its Application, Springer, New York 2003[2] W. D. Perkins ”Fourier Transform-Infrared Spectroscopy”, J. Chem. Educ.

1986, 63, A5- A10.[3] J.W. Cooley; J.W. Tukey. ”An Algorithm for the machine Calculation of com-

plex Fourier Series”, Math. Comp 1965, it 19, 297-301.