Fourier Transforms with Applications to FTIR

Jacob Mullins

Samford University

May 7, 2013

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 1 / 26

Outline

1 Goal

2 Inner Product Spaces

3 Discrete Fourier Transform

4 Chemistry

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 2 / 26

Outline

1 Goal

2 Inner Product Spaces

3 Discrete Fourier Transform

4 Chemistry

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 2 / 26

Outline

1 Goal

2 Inner Product Spaces

3 Discrete Fourier Transform

4 Chemistry

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 2 / 26

Outline

1 Goal

2 Inner Product Spaces

3 Discrete Fourier Transform

4 Chemistry

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 2 / 26

Goals

Fourier Transforms are incredibly useful in many different fields, one ofwhich is computational work in chemistry.

The true Fourier Transform takes too long to process for large amounts ofdata.

In these cases, a spinoff of the Fourier Transform is used, known as theDiscrete Fourier Transform.

This is also followed by computational methods to allow for fewercalculations, such as Horner’s Method or the Fast Fourier Transform.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 3 / 26

Goals

Fourier Transforms are incredibly useful in many different fields, one ofwhich is computational work in chemistry.

The true Fourier Transform takes too long to process for large amounts ofdata.

In these cases, a spinoff of the Fourier Transform is used, known as theDiscrete Fourier Transform.

This is also followed by computational methods to allow for fewercalculations, such as Horner’s Method or the Fast Fourier Transform.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 3 / 26

Goals

Fourier Transforms are incredibly useful in many different fields, one ofwhich is computational work in chemistry.

The true Fourier Transform takes too long to process for large amounts ofdata.

In these cases, a spinoff of the Fourier Transform is used, known as theDiscrete Fourier Transform.

This is also followed by computational methods to allow for fewercalculations, such as Horner’s Method or the Fast Fourier Transform.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 3 / 26

Goals

Fourier Transforms are incredibly useful in many different fields, one ofwhich is computational work in chemistry.

The true Fourier Transform takes too long to process for large amounts ofdata.

In these cases, a spinoff of the Fourier Transform is used, known as theDiscrete Fourier Transform.

This is also followed by computational methods to allow for fewercalculations, such as Horner’s Method or the Fast Fourier Transform.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 3 / 26

Inner Product Spaces

Definition

Let V be a complex vector space. An inner product on V is acomplex-valued function < u, v > of u and v ∈ V having the followingproperties:

< u, v >= < v , u > (1)

< αu + βv ,w >= α < u,w > +β < v ,w > (2)

< u, u >≥ 0 (3)

< u, u >= 0 =⇒ u = 0 (4)

< u, αv + βw >= α < u, v > +β < u,w > (5)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 4 / 26

Inner Product Spaces

Definition

Let V be a complex vector space. An inner product on V is acomplex-valued function < u, v > of u and v ∈ V having the followingproperties:

< u, v >= < v , u > (1)

< αu + βv ,w >= α < u,w > +β < v ,w > (2)

< u, u >≥ 0 (3)

< u, u >= 0 =⇒ u = 0 (4)

< u, αv + βw >= α < u, v > +β < u,w > (5)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 4 / 26

Inner Product Spaces

Definition

Let V be a complex vector space. An inner product on V is acomplex-valued function < u, v > of u and v ∈ V having the followingproperties:

< u, v >= < v , u > (1)

< αu + βv ,w >= α < u,w > +β < v ,w > (2)

< u, u >≥ 0 (3)

< u, u >= 0 =⇒ u = 0 (4)

< u, αv + βw >= α < u, v > +β < u,w > (5)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 4 / 26

Inner Product Spaces

Definition

Let V be a complex vector space. An inner product on V is acomplex-valued function < u, v > of u and v ∈ V having the followingproperties:

< u, v >= < v , u > (1)

< αu + βv ,w >= α < u,w > +β < v ,w > (2)

< u, u >≥ 0 (3)

< u, u >= 0 =⇒ u = 0 (4)

< u, αv + βw >= α < u, v > +β < u,w > (5)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 4 / 26

Inner Product Spaces

Definition

Let V be a complex vector space. An inner product on V is acomplex-valued function < u, v > of u and v ∈ V having the followingproperties:

< u, v >= < v , u > (1)

< αu + βv ,w >= α < u,w > +β < v ,w > (2)

< u, u >≥ 0 (3)

< u, u >= 0 =⇒ u = 0 (4)

< u, αv + βw >= α < u, v > +β < u,w > (5)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 4 / 26

Inner Product Spaces

Definition

Let V be a complex vector space. An inner product on V is acomplex-valued function < u, v > of u and v ∈ V having the followingproperties:

< u, v >= < v , u > (1)

< αu + βv ,w >= α < u,w > +β < v ,w > (2)

< u, u >≥ 0 (3)

< u, u >= 0 =⇒ u = 0 (4)

< u, αv + βw >= α < u, v > +β < u,w > (5)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 4 / 26

Orthonormal

Note that two vectors u and v are called orthogonal if < u, v >= 0 andorthonormal if each vector has norm one. The norm of u is defined as‖u‖ =

√< u, u >.

Theorem

If ρ1, ρ2, ..., ρN is an ON basis in an N dimensional inner product space V ,then every u ∈ V can be written as u =

∑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 >.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 5 / 26

Orthonormal

Note that two vectors u and v are called orthogonal if < u, v >= 0 andorthonormal if each vector has norm one. The norm of u is defined as‖u‖ =

√< u, u >.

Theorem

If ρ1, ρ2, ..., ρN is an ON basis in an N dimensional inner product space V ,then every u ∈ V can be written as u =

∑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 >.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 5 / 26

Discrete Fourier Transform

Consider the interval [−π, π].

Define xk = −π + πkm

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

〈f , g〉 =2m−1∑k=0

f (xk)g(xk)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 6 / 26

Discrete Fourier Transform

Consider the interval [−π, π].Define xk = −π + πk

mLet lm be the set of complex-valued functions with domainxk : k = 0, 1, ..., 2m − 1

We define, on lm,

〈f , g〉 =2m−1∑k=0

f (xk)g(xk)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 6 / 26

Discrete Fourier Transform

Consider the interval [−π, π].Define xk = −π + πk

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

〈f , g〉 =2m−1∑k=0

f (xk)g(xk)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 6 / 26

Discrete Fourier Transform

For n = −m,−m + 1, ...,m − 1 let

ϕn(x) = e inx

It can be shown that for l 6= n,

〈ϕl , ϕn〉 =2m−1∑k=0

ϕl(xk)ϕn(xk) = 0

but for l = n,

〈ϕl , ϕl〉 =2m−1∑k=0

ϕl(xk)ϕ(xk) = 2m.

Therefore, ϕ−m, . . . , ϕm−1 is an orthogonal basis for lm but notorthonormal.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 7 / 26

Discrete Fourier Transform

For n = −m,−m + 1, ...,m − 1 let

ϕn(x) = e inx

It can be shown that for l 6= n,

〈ϕl , ϕn〉 =2m−1∑k=0

ϕl(xk)ϕn(xk) = 0

but for l = n,

〈ϕl , ϕl〉 =2m−1∑k=0

ϕl(xk)ϕ(xk) = 2m.

Therefore, ϕ−m, . . . , ϕm−1 is an orthogonal basis for lm but notorthonormal.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 7 / 26

Discrete Fourier Transform

For n = −m,−m + 1, ...,m − 1 let

ϕn(x) = e inx

It can be shown that for l 6= n,

〈ϕl , ϕn〉 =2m−1∑k=0

ϕl(xk)ϕn(xk) = 0

but for l = n,

〈ϕl , ϕl〉 =2m−1∑k=0

ϕl(xk)ϕ(xk) = 2m.

Therefore, ϕ−m, . . . , ϕm−1 is an orthogonal basis for lm but notorthonormal.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 7 / 26

Discrete Fourier Transform

For n = −m,−m + 1, ...,m − 1 let

ϕn(x) = e inx

It can be shown that for l 6= n,

〈ϕl , ϕn〉 =2m−1∑k=0

ϕl(xk)ϕn(xk) = 0

but for l = n,

〈ϕl , ϕl〉 =2m−1∑k=0

ϕl(xk)ϕ(xk) = 2m.

Therefore, ϕ−m, . . . , ϕm−1 is an orthogonal basis for lm but notorthonormal.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 7 / 26

Discrete Fourier Transform

We can now write any f ∈ lm in terms of the orthonormal basis, ϕn√2m

, forn = −m, . . . ,m − 1.

The result is

f =m−1∑n=−m

f̂ (n)ϕn

where

f̂ (n) =〈f , ϕn〉

2m

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 8 / 26

Discrete Fourier Transform

We can now write any f ∈ lm in terms of the orthonormal basis, ϕn√2m

, forn = −m, . . . ,m − 1.The result is

f =m−1∑n=−m

f̂ (n)ϕn

where

f̂ (n) =〈f , ϕn〉

2m

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 8 / 26

Interpolation

It follows that for k = 0, 1, ..., 2m − 1,

f (xk) = S(xk)

where

S(x) =m−1∑n=−m

f̂ (n)ϕn(x)

=m−1∑n=−m

f̂ (n)e inx

= f̂ (−m)e−imx +−1∑

n=−(m−1)

f̂ (n)e inx + f̂ (0) +m−1∑n=1

f̂ (n)e inx

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 9 / 26

Interpolation

Define

an =1

m

2m−1∑k=0

f (xk) cos nxk

and

bn =1

m

2m−1∑k=0

f (xk) sin nxk

Writing S(xk) in terms of an and bn, it can be shown that

S(xk) =a0 + am cosmx

2+

m−1∑k=1

(ak cos kx + bk sin kx)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 10 / 26

Interpolation

Define

an =1

m

2m−1∑k=0

f (xk) cos nxk

and

bn =1

m

2m−1∑k=0

f (xk) sin nxk

Writing S(xk) in terms of an and bn, it can be shown that

S(xk) =a0 + am cosmx

2+

m−1∑k=1

(ak cos kx + bk sin kx)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 10 / 26

Assumption

If f is even, bn = 0.

Therefore, it can be shown that

S(x) = f̂ (0) +m−1∑k=1

2f̂ (k) cos kx + f̂ (m) cosmx

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 11 / 26

Assumption

If f is even, bn = 0.Therefore, it can be shown that

S(x) = f̂ (0) +m−1∑k=1

2f̂ (k) cos kx + f̂ (m) cosmx

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 11 / 26

Example

Letf (x) = x6 − 3x4 + 7x2 − 10.

Letting m = 2,

S(x) = f̂ (0) + 2f̂ (1) cos x + f̂ (2) cos 2x .

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 12 / 26

Example

Letf (x) = x6 − 3x4 + 7x2 − 10.

Letting m = 2,

S(x) = f̂ (0) + 2f̂ (1) cos x + f̂ (2) cos 2x .

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 12 / 26

m=2

-3 -2 -1 1 2 3

200

400

600

Figure : Interpolation of f(x) with m=2

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 13 / 26

m = 25

-3 -2 -1 1 2 3

100

200

300

400

500

600

700

Figure : Interpolation of f(x) with m = 25

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 14 / 26

Waves

So now that we have the tool of the Fourier Transform, let us move on toapplying it to chemistry.

A Fourier Transform Infrared Radiation Spectrum (FTIR) shines a lightonto a beam splitter, which causes the light to follow two separate paths.One path is reflected off a fixed mirror and the other off a moving mirror.The combined light then shines through a material and the amount oflight absorbed by the material is detected.

A single laser results in the detection of the transmittance that can begraphed with the equation f (δ) = cos (2πδν) multiplied by some constant.Here δ is twice the distance the mirror is moved, and ν is the inverse ofthe wavelength, also known as the wavenumber.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 15 / 26

Waves

So now that we have the tool of the Fourier Transform, let us move on toapplying it to chemistry.

A Fourier Transform Infrared Radiation Spectrum (FTIR) shines a lightonto a beam splitter, which causes the light to follow two separate paths.One path is reflected off a fixed mirror and the other off a moving mirror.The combined light then shines through a material and the amount oflight absorbed by the material is detected.

A single laser results in the detection of the transmittance that can begraphed with the equation f (δ) = cos (2πδν) multiplied by some constant.Here δ is twice the distance the mirror is moved, and ν is the inverse ofthe wavelength, also known as the wavenumber.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 15 / 26

Waves

So now that we have the tool of the Fourier Transform, let us move on toapplying it to chemistry.

A Fourier Transform Infrared Radiation Spectrum (FTIR) shines a lightonto a beam splitter, which causes the light to follow two separate paths.One path is reflected off a fixed mirror and the other off a moving mirror.The combined light then shines through a material and the amount oflight absorbed by the material is detected.

A single laser results in the detection of the transmittance that can begraphed with the equation f (δ) = cos (2πδν) multiplied by some constant.Here δ is twice the distance the mirror is moved, and ν is the inverse ofthe wavelength, also known as the wavenumber.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 15 / 26

Interferogram

Now, assuming a continuous spectrum of light, and adding in a function tocorrect the fact that each wavelength of light may be at a differentintensity. This results in the following function of delta being received bythe detector.

I (δ) =

∫ ∞0

B(ν) cos 2πδνdν

This is known as an interferogram, which can be interpretted as a linearcombination of single lasers.By the Fourier Inversion Formula, it can be shown that

B(ν) = 2

∫ ∞−∞

I (δ) cos 2πδνdδ

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 16 / 26

Interferogram

Now, assuming a continuous spectrum of light, and adding in a function tocorrect the fact that each wavelength of light may be at a differentintensity. This results in the following function of delta being received bythe detector.

I (δ) =

∫ ∞0

B(ν) cos 2πδνdν

This is known as an interferogram, which can be interpretted as a linearcombination of single lasers.

By the Fourier Inversion Formula, it can be shown that

B(ν) = 2

∫ ∞−∞

I (δ) cos 2πδνdδ

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 16 / 26

Interferogram

Now, assuming a continuous spectrum of light, and adding in a function tocorrect the fact that each wavelength of light may be at a differentintensity. This results in the following function of delta being received bythe detector.

I (δ) =

∫ ∞0

B(ν) cos 2πδνdν

This is known as an interferogram, which can be interpretted as a linearcombination of single lasers.By the Fourier Inversion Formula, it can be shown that

B(ν) = 2

∫ ∞−∞

I (δ) cos 2πδνdδ

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 16 / 26

Interferogram

Figure : Interferogram of eicosanoic acid

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 17 / 26

Truncation

A function often used in Chemistry is called the boxcar function in order totruncate the infinite integral. The boxcar function sets everything inside ofa certain interval to 1 and everything outside of the interval to 0.

Definition

The Fourier Transform of a function f (t) is defined as

f̂ (ω) = F [f (t)](ω) =

∫ ∞−∞

f (t)e−iωtdt

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 18 / 26

Truncation

A function often used in Chemistry is called the boxcar function in order totruncate the infinite integral. The boxcar function sets everything inside ofa certain interval to 1 and everything outside of the interval to 0.

Definition

The Fourier Transform of a function f (t) is defined as

f̂ (ω) = F [f (t)](ω) =

∫ ∞−∞

f (t)e−iωtdt

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 18 / 26

If we consider the boxcar function on the interval (−∆,∆), then it can beshown that its Fourier Transform is given by

2∆sinc(ω∆)

where sinc (t) = sin tt .

Theorem (THE SHIFTING THEOREM)

F [e iλt f (t)] = f̂ (ω − λ)

Theorem

After truncating by the boxcar function,

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

where we assume I (δ) = cos (2πδν0).

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 19 / 26

If we consider the boxcar function on the interval (−∆,∆), then it can beshown that its Fourier Transform is given by

2∆sinc(ω∆)

where sinc (t) = sin tt .

Theorem (THE SHIFTING THEOREM)

F [e iλt f (t)] = f̂ (ω − λ)

Theorem

After truncating by the boxcar function,

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

where we assume I (δ) = cos (2πδν0).

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 19 / 26

If we consider the boxcar function on the interval (−∆,∆), then it can beshown that its Fourier Transform is given by

2∆sinc(ω∆)

where sinc (t) = sin tt .

Theorem (THE SHIFTING THEOREM)

F [e iλt f (t)] = f̂ (ω − λ)

Theorem

After truncating by the boxcar function,

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

where we assume I (δ) = cos (2πδν0).

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 19 / 26

Proof.

Let B(ν) = 2∫∞−∞ I (δ) cos 2πνδdδ.

Now we will estimate B(ν) by the equation

B∆(ν) = 2∫ ∆−∆ I (δ) cos 2πνδdδ, since there is no instrument that can

measure from −∞ to ∞.So B∆(ν) = 2

∫ ∆−∆ cos 2πν0δ cos 2πνδdδ

Multiplying by the boxcar function, and using the fact that I (δ) is even,we get

B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ.

where

χA(δ) =

{1 δ ∈ A0 δ 6∈ A.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 20 / 26

Proof.

Let B(ν) = 2∫∞−∞ I (δ) cos 2πνδdδ.

Now we will estimate B(ν) by the equation

B∆(ν) = 2∫ ∆−∆ I (δ) cos 2πνδdδ, since there is no instrument that can

measure from −∞ to ∞.

So B∆(ν) = 2∫ ∆−∆ cos 2πν0δ cos 2πνδdδ

Multiplying by the boxcar function, and using the fact that I (δ) is even,we get

B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ.

where

χA(δ) =

{1 δ ∈ A0 δ 6∈ A.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 20 / 26

Proof.

Let B(ν) = 2∫∞−∞ I (δ) cos 2πνδdδ.

Now we will estimate B(ν) by the equation

B∆(ν) = 2∫ ∆−∆ I (δ) cos 2πνδdδ, since there is no instrument that can

measure from −∞ to ∞.So B∆(ν) = 2

∫ ∆−∆ cos 2πν0δ cos 2πνδdδ

Multiplying by the boxcar function, and using the fact that I (δ) is even,we get

B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ.

where

χA(δ) =

{1 δ ∈ A0 δ 6∈ A.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 20 / 26

Proof.

Let B(ν) = 2∫∞−∞ I (δ) cos 2πνδdδ.

Now we will estimate B(ν) by the equation

B∆(ν) = 2∫ ∆−∆ I (δ) cos 2πνδdδ, since there is no instrument that can

measure from −∞ to ∞.So B∆(ν) = 2

∫ ∆−∆ cos 2πν0δ cos 2πνδdδ

Multiplying by the boxcar function, and using the fact that I (δ) is even,we get

B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ.

where

χA(δ) =

{1 δ ∈ A0 δ 6∈ A.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 20 / 26

Continue Proof

Notice that, substituting in Fourier notation,

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

= F [2e i2πν0δχ[−∆,∆](δ)](2πν)

+ F [2e−i2πν0δχ[−∆,∆](δ)](2πν)

By the earlier theorem, we know

F [2e i2πν0δχ[−∆,∆](δ)](2πν)

= F [χ[−∆,∆](δ)](2πν − 2πν0)

Therefore, we will first look at F [χ[−∆,∆](δ)](2πν).

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 21 / 26

Continue Proof

Notice that, substituting in Fourier notation,

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

= F [2e i2πν0δχ[−∆,∆](δ)](2πν)

+ F [2e−i2πν0δχ[−∆,∆](δ)](2πν)

By the earlier theorem, we know

F [2e i2πν0δχ[−∆,∆](δ)](2πν)

= F [χ[−∆,∆](δ)](2πν − 2πν0)

Therefore, we will first look at F [χ[−∆,∆](δ)](2πν).

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 21 / 26

Continue Proof

Notice that, substituting in Fourier notation,

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

= F [2e i2πν0δχ[−∆,∆](δ)](2πν)

+ F [2e−i2πν0δχ[−∆,∆](δ)](2πν)

By the earlier theorem, we know

F [2e i2πν0δχ[−∆,∆](δ)](2πν)

= F [χ[−∆,∆](δ)](2πν − 2πν0)

Therefore, we will first look at F [χ[−∆,∆](δ)](2πν).

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 21 / 26

Continue Proof

Let ω = 2πν. Let t = δ. So

F [χ[−∆,∆](δ)](2πν)

=

∫ ∆

−∆e−iωtdt Let s =

t

∆

= ∆

∫ 1

−1e−i(ω∆)sds

= 2∆sinc(ω∆)

Now

F [2e i2πν0δχ[−∆,∆](δ)](2πν)

= 2∆sinc(2π(ν − ν0)∆)

andF [2e−i2πν0δχ[−∆,∆](δ)](2πν) = 2∆sinc(2π(ν + ν0)∆)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 22 / 26

Continue Proof

Let ω = 2πν. Let t = δ. So

F [χ[−∆,∆](δ)](2πν)

=

∫ ∆

−∆e−iωtdt Let s =

t

∆

= ∆

∫ 1

−1e−i(ω∆)sds

= 2∆sinc(ω∆)

Now

F [2e i2πν0δχ[−∆,∆](δ)](2πν)

= 2∆sinc(2π(ν − ν0)∆)

andF [2e−i2πν0δχ[−∆,∆](δ)](2πν) = 2∆sinc(2π(ν + ν0)∆)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 22 / 26

Continue Proof

Let ω = 2πν. Let t = δ. So

F [χ[−∆,∆](δ)](2πν)

=

∫ ∆

−∆e−iωtdt Let s =

t

∆

= ∆

∫ 1

−1e−i(ω∆)sds

= 2∆sinc(ω∆)

Now

F [2e i2πν0δχ[−∆,∆](δ)](2πν)

= 2∆sinc(2π(ν − ν0)∆)

andF [2e−i2πν0δχ[−∆,∆](δ)](2πν) = 2∆sinc(2π(ν + ν0)∆)

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 22 / 26

Proof.

Therefore

B(ν) ≈ B∆(ν) =

∫ ∞−∞

[2 cos 2πν0δχ[−∆,∆](δ)]e−2πνδidδ

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

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 23 / 26

Sinc Graph

-4 -2 2 4

-1

1

2

3

4

5

Figure : Truncated spectrum with ∆ = 5 and ν0 = 2

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 24 / 26

Spectrum

Figure : Spectrum after ratioing

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 25 / 26

Computational Work

B(ν) = 2

∫ ∞−∞

I (δ) cos 2πδνdδ

These calculations can take too long to be practical for computationalwork.

Therefore the fact that it can be written as a polynomial, methods such asHorner’s method or the FFT can be used to speed up the calculations.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 26 / 26

Computational Work

B(ν) = 2

∫ ∞−∞

I (δ) cos 2πδνdδ

These calculations can take too long to be practical for computationalwork.

Therefore the fact that it can be written as a polynomial, methods such asHorner’s method or the FFT can be used to speed up the calculations.

Jacob Mullins (Samford University) Fourier Transforms with Applications to FTIR May 7, 2013 26 / 26