Fourier Series, Fourier Integral and Discrete Fourier Transform.Author: Arkadi Kagan. arkadi_kagan@hotmail.com

Introduction.Fourier Transform is learned centuries and first strict prove is found around year 1829 by Leguen Derehle. For this reason I can be sure that in my text there is exactly nothing that was not proven by somebody else.The main source of information for me was grate textbook “Base of Math Analysis” by G.M.Fihtengoltz. Else I was using some other sources.Although the sources I was using give grate explanation, prove for Fourier Transform is given only partial. Math textbook assume that you seek for parts of prove in other topic. There is a lot of material in textbook, therefore search for particular topic can blur attention from the actual problem that must be solved.

1. Orthonormal and Orthogonal sets of functions.• f nx and nx are sets of functions defined for x in range [a, b].• n – index of functions defined in range [c, d].• for each n in [c, d]

• f nx and nx are almost continuous in range [a, b]• exist n0f nx and nx are called Orthogonal in range [a, b] if and only if

∫a

b

f n xn*x dx=0 where both m and n∈[c , d ] .

n* x is Complex Conjugate of nx .

if a=ar jai then Complex Conjugate of a is a*=ar− jai .

Notice that if b=e jx , then b*=e− jx . This can be proven by Euler's Formula.

Set of functions f n x is called Pairwise Orthogonal if each pair of functions f n x

and f m x are orthogonal unless n=m : ∫a

b

f nxn* xdx={ 0 n≠m

n n=m}

If n=1 for each n, then set of functions f nx is called Orthonormal.

2. Fourier Series.Let f x to be a function defined in range [a, b]Let nx to be a set of Pairwise Orthogonal functions in range [a, b],while n∈[c , d ] .• c, d and n are integers.• a and b are real numbers.Function f(x) for each x in range [c, d] can be approximated by Fourier Series.Fourier Series:

f x=∑n=c

d

F nnx

where

F k = 1k∫a

b

f xk* xdx and k=∫

a

b

k x k* dx

Lets prove equation:

f x =∑n=c

d

F nnx

Multiply left and right parts by k* x with k∈[c ,d ] .

f x k* x =∑

n=c

d

F nnx k* x

Take integral by variable x in range [a, b] from the left and right parts of equation.

∫a

b

f x k*x dx=∫

a

b

∑n=c

d

F nn xk* xdx =

( by known properties of integral )

= ∑n=c

d

F n∫a

b

n xk* xdx =

∫a

b

n xk* xdx={ 0 k≠n

k k=n}because of orthogonality

= k⋅F k .Therefore

F k = 1k∫

a

b

f xk* x dx .

That is the searched equation.

3. Fourier Integral.Let f x to be a function defined in range [a, b]Let nx to be continuous set of Pairwise Orthogonal functions in range [a, b],while n∈[c , d ] .• c, d and n are real numbers since no need to perform summation.• a and b are real numbers too.Real function f(x) for each x in range [c, d] can be represented as Fourier Integral.Fourier Integral:

f x=∫c

d

F y y xdy

where

F z= 1z∫a

b

f x z*x dx and z=∫

a

b

z x z* x dx

Lets prove equation:

f x=∫c

d

F y y xdy

Multiply left and right part by z* x , where z

* x is Complex Conjugate of z x .

f xz* x =∫

c

d

F y y x dy z* x

Take integral by x from left and right parts in range [a, b]

∫a

b

f x z* xdx=∫

a

b

∫c

d

F y y xdy z* xdx =

( by known properties of integral )

= ∫c

d

F y ∫a

b

y x z* xdxdy =

∫a

b

y xz* x dx={ 0 y≠z

z y=zbecause of orthogonality

= ∫c

d

F yX z ydy where X z y ={ 0 y≠z z y=z

= F z z

The last equality can and will be proven by definition of integral.

∫a

b

y x z* xdx=X z y , X z y={ z z=y

0 z≠y

Lets suppose the equality is right.

In this case F z = 1z∫a

b

f xz*xdx .

That is the searched equation.

Lets prove the equality.

If X z y={ 0 y≠zz y=z

then ∫c

d

F yX z ydy=F zz

By definition of integral and Riemann Sums

∫c

d

t xdx= limN ∞

d−cN ∑

n=0

N

t cn d−cN

Therefore

∫c

d

F y X z ydy= limN ∞

d−cN ∑

n=0

N

F cn d−cN X zcn d−c

N

Now lets find n, z=cn d−cN

.

Particularly it will be right for n= z−cNd−c

.

This value of n could be not necessary right integer. To ensure that n is indeedinteger we can choose N to be multiply of (d-c), i.e. exist integer k, N = k(d-c).Since N is growing to infinity, this assumption is not reducing from generality.By definition of X z y , X z z =z and X z y =0 for each n,

cn d −cN

= y≠z .

Finally, ∫c

d

F yX z ydy=F zz .

Equality proven.

Discrete Fourier Transform.Select set of M Pairwise Orthogonal Complex vectors, N elements each.By definition of orthogonality two vectors m and k of set are

orthogonal iff ∑n=0

N−1m[n ]k

* [n ]=0 , where x* is Complex Conjugate of complex number

x .

Sign m= ∑n=0

N−1m[n ]m

* [n ] .

Define source vector f with M complex elements.Exist vector F of N complex elements.

For each 0≤mM , f [m]∈ f , f [m]= 1m

∑n=0

N−1F [n ]m [n]

and for each 0≤nN , F [n ]∈F , F [n]=∑m=0

M −1

f [m]m* [n ]

f [k ]= 1k

∑n=0

N−1F [n ]k [n]=

1k

∑n=0

N−1∑

m=0

M −1f [m]m

* [n ]k [n]=

= 1k

∑m=0

M −1f [m ] ∑

n=0

N−1m

* [n]k [n ]=1k

f [k ]k= f [ k]

By this simple computation we prove that:For any set of M complex pairwise orthogonal vectors dimension N , if exist m≠0 for each 0≤mM , m∈ and

m=∑n=0

N−1

m[n]m* [n] , then any complex vector f of dimension

M can be recalculated like this:

f [m]= 1m∑n=0

N−1

F [n]m[n] , 0≤mM

where F [n]=∑m=0

M −1

f [m]m* [n ] , 0≤nN .

This two formulas are called Discrete Fourier Transform and corresponding Inverse Discrete Fourier Transform.

4. Known sets of Pairwise Orthogonal functions.There are endless possible sets of pairwise orthogonal functions.For Discrete Fourier Transform it is not enough to define set of pairwise orthogonal continuous functions. We need rather a set of pairwise orthogonal vectors of given dimension.Define m and n integers, c and r real, j=−1 .

x∈[−c ,c ]n , m and c∈−∞ ,∞

This formulas define three most popular sets of pairwise orthogonal vectors of length N:

1. n[m]=ej 2

N mn

This set of pairwise orthogonal vectors consists of no more than N vectors.This set of vectors have interesting property:

n [N−m]=n* [m] and N−n [m]=n

* [m] .

2. n[m]=cos 2N mn

This set of pairwise orthogonal vectors consists of no more than N/2 vectors.Assume you have more.Than

cos 2N

m N−n=

= cos 2N

mN−2N

mn=

= cos2mcos 2N

mnsin2m sin 2N

mn=

= cos 2N

mn

Therefore n and N−n are not orthogonal and here from is the limit of N/2 vectors.

3. n[m]=sin 2N mn

This set of pairwise orthogonal vectors consists of no more than N/2 vectors.Assume you have more.Than

sin 2N

m N−n=

= sin 2N

mN −2N

mn=

= cos 2msin 2N

mn−sin 2mcos 2N

mn=

= sin 2N

mn

Therefore n and N−n are not orthogonal and here from is the limit of N/2 vectors.

Proving known trigonometric equalities.This equalities are well known and could be proven somewhere else,however it could be better to have the entire prove chain, at least for less popular formulas.

cos n−m x−cos nm x =cos ab=cos acos b −sin a sin b

= cos nxcosmxsin nx sin mx −cos nx cos mx −sin nxsin mx == 2sin nx sin mx

sin nx sin mx = 12 cos n−m x−cosnm x

cos n−m xcos nm x =cos ab=cos acos b −sin a sin b

= cos nxcosmxsin nx sin mx cos nx cos mx −sin nxsin mx == 2cos nxcosmx

cos nxcos mx=12 cos n−mxcos nm x

sin nmx−sin n−mx =sin ab=sin acos bcosasin b

= sin nxcos mxcos nx sin mx −sin nxcos mx −cos nxsin mx == 2cos nxsin mx

cos nx sin mx = 12 sin nm x−sin n−m x

sin nmxsin n−mx =sin ab=sin acos bcosasin b

= sin nxcos mxcos nx sin mx sin nxcos mx −cos nxsin mx == 2sin nx cos mx

sin nx cos mx = 12 sin nm xsin n−m x

5. Euler Identities.Lets take known Taylor-Maclaurin sequences for e jx , e− jx , cos x and sin x :

e jx=1 jx− x 2

2 !− j x3

3! x4

4 !...

e− jx=1− jx− x 2

2 ! j x3

3 !x4

4 !−...

cos x =1− x2

2 !x4

4!−...

sin x=x− j x3

3!x5

5! −...

From this sequences we directly derive Euler Identities.Euler Identities:

e jx=cos x j sin x

cos x=e jxe− jx

2

sin x=e jx−e− jx

2j

6. Known Integral equalities.To prove orthogonality of chosen pairs of functions we will need several simpleintegral equalities.

∫−c

c

dx=c−−c =2

or simple ∫−c

c

dx=2

For real a, a≠0 :

∫−c

c

cos ax dx=sin axa

∣ c−c

=

= 1a sin ac−sin ac− =

=1a sin acos acsin ac cos a−sin accosa−sin acos ac =

= 0⋅1a=0

or simple ∫−c

c

cos ax dx=0

∫−c

c

sin axdx=−cos ax a

∣ c−c

=

= −1a cos a c−cos a −c =

=1a cosac−−cos a c =

=1a cos accos asin acsin a−cos accos a−sin ac sin a =

=1a 2sin acsin a=0

therefore ∫−c

c

sin axdx=0 for any a, including 0.

7. Prove of continuous orthogonality.Lets start to prove orthogonality for continuous trigonometric functions:

∫−c

c

sin nxsin mxdx = sin nx sin mx= 12cosn−m x−cosnm x

=12 ∫−c

c

cos n−m x dx− ∫−c

c

cos nm x dx =

∫−c

c

cosax dx={ 0 a≠02 a=0

= { 0 n≠m n=m

∫−c

c

sin nxsin mxdx={ 0 n≠m n=m

∫−c

c

cos nx cos mx dx = cosnx cosmx =12cosn−m xcosnm x

= 12 ∫−c

c

cos n−m x dx ∫−c

c

cos nm x dx =

∫−c

c

cosax dx={ 0 a≠02 a=0

} = { 0 n≠m n=m

∫−c

c

cos nx cos mx dx={ 0 n≠m n=m

This two equalities prove that:• sin nx is a set of Pairwise Orthogonal functions.• cos nx is a set of Pairwise Orthogonal functions.The last set of Pairwise Orthogonal functions to prove is for z-Transform.

∫−c

c

r e jx nr e jx−m dx = rn−m ∫−c

c

e j n−m x dx = e jax=cos ax j sin ax

= r n−m ∫−c

c

cos n−m x dx j ∫−c

c

sin n−m x dx = { 0 n≠m2 n=m

∫−c

c

r e jx nr e jx−m dx={ 0 n≠m2 n=m

Prove discrete orthogonality of vectors.I will start from inductive prove of three useful theorems:Theorem 1: For any integer number N and 0≤mN

∑n=0

N−1

ej2 mn

N ={ 0 for m % N ≠0N for m % N =0

Theorem 2: For any even number N and 0≤m N2

∑n=0

N−1

cos2 mnN ={ 0 for m % N≠0

N for m % N=0Theorem 3: For any even number N and 0≤m N

2

∑n=0

N−1

sin 2 mnN

=0

Prove Theorem 1: Lets start induction on N .N =2 :

∑n=0

N−1

ej2 mn

N =ej2 m⋅0

2 ej2m⋅1

2 =1e jm =

e jm=cosm jsin m=cos m=−1m

= 1−1m={ 0 for m% 2≠02 for m% 2=0

N =4 :

S4=ej2 m⋅0

4 ej2m⋅1

4 ej2 m⋅2

4 ej2m⋅3

4

ej2m⋅0

4 =1

ej2m⋅1

4 =ej

2m=cos

2 jsin

2

m

=0 j m= jm

ej2m⋅2

4 =e jm=cos jsin m=−1m

ej2m⋅3

4 =ej

2m=cos

2 jsin

2

m

=− j m

S4=1 jm−1m− j m

Split calculation to the four cases.Case 1: m % 4=0−1m=1jm= j⋅j ⋅ j⋅j =−1⋅−1=1− j m=− j⋅− j ⋅− j⋅− j =1s4=1111=4

Case 2: m % 4=1−1m=−1jm= j− j m=− jS4=1−1 j− j

Case 3: m % 4=2−1m=−1⋅−1=1jm= j⋅ j=−1− j m=− j⋅− j=−1S4=11−1−1=0

Case 4: m % 4=3−1m=−1⋅−1⋅−1=−1jm= j⋅ j⋅j=− j− j m=− j⋅− j⋅− j= jS 4=1−1− j j=0

Assume by induction ∑n=0

N2 −1

ej2 mn

N / 2=0 and m %N /2≠0 .

I want to prove that ∑n=0

N−1

ej2 mn

N =0 .

Split the sum to odd and even summators.

∑n=0

N−1

ej2 mn

N =∑n=0

N2 −1

ej2 m2n

N ∑n=0

N2 −1

ej2 m2n1

N =

=∑n=0

N2−1

ej2 mn

N /2ej2 m

N⋅∑n=0

N2−1

ej2 mn

N /2 =

∑n=0

N2−1

ej2 mn

N / 2=0 as assumed

= 0ej2 m

N⋅0=0

Assume by induction ∑n=0

N2 −1

ej2 mn

N / 2=N2

and m %N /2=0 .

∑n=0

N−1

ej2 mn

N =∑n=0

N2 −1

ej2 mn

N /2ej2 m

N⋅∑n=0

N2 −1

ej2 mn

N /2 =

= N2e

j2 mN⋅N

2= { 0 m% N ≠0

N m% N =0

ej2 m

N=cos 2 mN jsin 2 m

N ;

assume m % N=0 : cos 2 mN=1 , sin 2 m

N=0 ;

ej2 m

N=1 N2N

2⋅e

j2 mN= N

2 N

2=N

assume m % N≠0 :

known m % N /2=0m % N =N /2 ; cos2 mN=−1 , sin 2 m

N=0 ;

ej2 m

N=−1 N2− N

2⋅e

j2 mN= N

2−N

2=0

From the above calculations we get an exact prove for “Theorem 1” that is

∑n=0

N−1

ej2 mn

N ={ 0 for m % N ≠0N for m % N =0

.

Prove Theorem 2 and 3:

From the Theorem 1 ∑n=0

N−1

ej2 mn

N ={ 0 for m % N ≠0N for m % N =0

.

Remind Euler formula: e jx=cos x jsin x .

∑n=0

N−1

ej2 mn

N =∑n=0

N −1

cos 2 mnN

j⋅∑n=0

N −1

sin 2 mnN

Therefore

∑n=0

N−1

cos2 mnN

={ 0 m % N ≠0N m % N =0

∑n=0

N−1

sin 2 mnN =0 for any 0≤mN

2 .

By this Theorems 2 and 3 are proven.

By proven theorems it is easy to get orthogonality of vector sets.Select two values q and p : 0≤qN , 0≤ pN .That mean −Nq− pN .In this condition q− p% N=0q− p=0 .Write the Theorem 1 equation while m=q− p .

∑n=0

N−1

ej2 q−p n

N ={ 0 for q− p≠0N for q− p=0

∑n=0

N−1

ej2 qn

N e− j2 pn

N ={ 0 for q≠ pN for q= p

The last equation is exactly the definition of orthogonality.

Conclusion: set of vectors defined by ej2 nm

N is a set of pairwise orthogonal vectors

for any integer N and 0≤nN , 0≤mN .Similar for Theorem 2 while m=q− p .

∑n=0

N−1

cos2 q− pnN

={ 0 for q− p≠0N for q− p=0

.

Remind equation cos nxcos mx=12 cos n−mxcos nm x .

By this equation the above sum can be rewritten:

∑n=0

N−1

cos 2 pnN

cos 2 qnN

=

= 12∑

n=0

N−1

cos 2 p−qnN

∑n=0

N −1

cos 2 pqnN

0≤ pN , 0≤qN therefore pq% N=0 p=q=0 .If p−q% N=0 and p≠0 then:

∑n=0

N−1

cos2 pqnN

=0 ; ∑n=0

N−1

cos2 p−qnN

=∑n=0

N−1

cos 0=N

∑n=0

N−1

cos2 pnN

cos2 qnN

= N2

.

else, if p=q=0 then:

∑n=0

N−1

cos2 pqnN

=∑n=0

N−1

cos 2 p−qnN

=∑n=0

N−1

cos 0=N

∑n=0

N−1

cos2 pnN

cos2 qnN

=N

else, if p≠q then:

∑n=0

N−1

cos2 pqnN

=∑n=0

N−1

cos 2 p−qnN

=0

∑n=0

N−1

cos2 pnN

cos2 qnN

=0 .

Conclusion: Set of vectors, defined by function f p[n ]=cos 2 pnN

is Pairwise Orthogonal in range [ 0, N2

) :

∑n=0

N−1

cos2 pnN

cos2 qnN

={

0 p≠qN2

p=q≠0

N p=q=0Similar computation can be done for set of vectors defined by f p[n ]=sin 2 pn

N

in range [ 0, N2 ) . This time I will use equation sin nx sin mx =1

2cos n−m x−cos nm x .

∑n=0

N−1

sin 2 pnN

sin 2 qnN

=

= 12∑

n=0

N−1

cos 2 p−qnN

−∑n=0

N −1

cos 2 pqnN

The calculations are exactly the same as it was for case of cosine.Conclusion: Set of vectors, defined by function f p[n ]=sin 2 pn

N

is Pairwise Orthogonal in range [ 0, N2

) :

∑n=0

N−1

sin 2 pnN

sin 2 qnN

={

0 p≠qN2

p=q≠0

0 p=q=0

Known Discrete Fourier Transforms.Given the proven above sets of pairwise orthogonal vectors, we can define Discrete Fourier Transform with different basis functions.

1.

∑n=0

N−1

ej2 qn

N e− j2 pn

N ={ 0 for q≠ pN for q= p

By general definition of Discrete Fourier Transform

f [m]= 1m∑n=0

N−1

F [n]m[n] , 0≤mM

where F [n]=∑m=0

M −1

f [m]m* [n ] , 0≤nN .

In this case N=M , therefore

f [m]= 1N ∑

n=0

N−1

F [n ]ej2mn

N , where 0≤mN , is called

Complex Sinusoid Discrete Fourier Transform.

F [n]=∑m=0

M −1

f [m]e− j2 mn

N , where 0≤nN , is called

Inverse Complex Sinusoid Discrete Fourier Transform.

2.

∑n=0

N−1

cos2 pnN

cos2 qnN

={

0 p≠qN2

p=q≠0

N p=q=0where 0≤ p−q N

2.

By general definition of Discrete Fourier Transform

f [m]= 1m∑n=0

N−1

F [n]m[n] , 0≤mM

where F [n]=∑m=0

M −1

f [m]m* [n ] , 0≤nN .

In this case M =N2 .

f [m]= 1N ∑

n=0

N−1

F [n ]cos 2⋅0= 1N ∑

n=0

N−1

F [n ] for m=0 .

f [m]= 2N ∑

n=0

N−1

F [n ]cos 2 mnN

for m0 .

F [n]=∑m=0

M −1

f [m]cos2 mnN

.

However, we know that

cos2 mN−nN

=cos 2 mNN

−2 mnN=

= cos2mcos2 mnNsin 2m sin 2 mn

N=

= cos2 mnN

Therefore F [N−n]= ∑m=0

M−1f [m ]cos 2 mN −n

N =F [n ] .

Here from, for any m0 ,

f [m]= 2N ∑

n=0

N−1

F [n ]cos2 mnN

=

= 2N∑

n=0

N2−1

F [n ]cos 2 mnN

∑n=0

N2−1

F [N−n]cos2 mN−n N

=

= 2N /2 ∑n=0

N2−1

F [n]cos mnN /2

Remind M =N2 , therefore f [m]= 2

M ∑n=0

M−1

F [n]cos mnM

for m0 .

for m=0 :

f [m]= 1N ∑

n=0

N−1

F [n ]= 1N∑

n=0

N2−1

F [n ]∑n=0

N2−1

F [N−n]= 1N /2 ∑n=0

N2−1

F [n ]

or f [m]= 1M ∑

n=0

M−1

F [n] .

Writing another way: f [m]=m ∑n=0

M−1

F [n]cos mnM

, where m={

1M

m=0

2M

m≠0

This formula is called Cosine Discrete Fourier Transform.Corresponding inverse transform is recalculated like this:

F [n]=∑m=0

M −1

f [m]cos2 mnN

=∑m=0

M−1

f [m]cos mnM

.

Finally, formula F [n]=∑m=0

M −1

f [m]cos mnM

is called Inverse Cosine Discrete

Fourier Transform.

3.

∑n=0

N−1

sin2 pnN sin2 qn

N ={

0 p≠qN2 p=q≠0

0 p=q=0By general definition of Discrete Fourier Transform

f [m]= 1m∑n=0

N−1

F [n]m[n] , 0≤mM

where F [n]=∑m=0

M −1

f [m]m* [n ] , 0≤nN .

For m=0 , obviously f [m ]=0 .

For m0 , f [m]= 1N /2 ∑n=0

N−1

F [n ]sin 2 mnN

.

Corresponding F [n]=∑m=0

N2−1

f [m]sin 2 mnN

.

Remind that

sin 2 mN−nN

=sin 2 mNN

−2 mnN=

= sin 2m cos2 mnN

−cos2m sin 2 mnN=

= sin 2 mnN

Therefore F [N−n]=F [n] .

Here from

f [m]= 1N /2 ∑n=0

N−1

F [n ]sin 2 mnN

=

= 1N /2

∑n=0

N2−1

F [n]sin 2 mnN

∑n=0

N2−1

F [n ]sin 2 m N−nN

= 2N /2 ∑n=0

N2−1

F [n]sin2 mnN

Finally, formula f [m]= 2M ∑

n=0

M−1

F [n]sin mnM

is called Sine Discrete Fourier

Transform.

Similarly, inverse calculation:

F [n]=∑m=0

N2−1

f [m]sin 2 mnN

=∑m=0

M−1

f [m]sin mnM

.

Formula F [n]= ∑m=0

M −1f [m ]sin mn

M is called Inverse Sine Discrete Fourier

Transform.

8. Known formulas, proven in previous topics.Given the proven sets of pairwise orthogonal functions, we can state this known cases of Continuous Fourier Series:Sine Series:

f x=∑n=−∞

∞

F nsin nx dx

F k = 1∫

−c

c

f xsin kxdx

Cosine Series:

f x=∑n=−∞

∞

F ncos nx dx

F k = 1∫

−c

c

f xcos kx dx

z-Transform:

f x=∑n=−∞

∞

F nr e jxn dx

F k = 1∫

−c

c

f xr e jx−k dx

And the last transform is z-Transform with r = 1, else called Complex Sinusoid Transform, since y=e jx is a formula of Complex Sinusoid.

Complex Sinusoid Series:

f x=∑n=−∞

∞

F ne jnx dx

F k = 1∫

−c

c

f xe− jnx dx

As a natural extension, we can state this known cases of Continuous Fourier Integrals:Sine Fourier Integral:

f x= ∫−c

c

F nsin nxdx

F k = 1∫

−c

c

f xsin kxdx

Cosine Fourier Integral:

f x= ∫−c

c

F ncos nx dx

F k = 1∫

−c

c

f xcos kx dx

z-Transform:

f x= ∫−c

c

F nr e jx n dx

F k = 1∫

−c

c

f x r e jx−k dx

Complex Sinusoid Fourier Integral:

f x= ∫−c

c

F ne jnx dx

F k = 1∫

−c

c

f xe− jnx dx

And finally, we can state this known cases of Discrete Fourier Transforms:Sine Discrete Fourier Transform:

f [m]= 2N ∑

n=0

N−1

F [n ]sin mnN

F [n ]=∑m=0

N−1

f [m ]sin mnN

Cosine Discrete Fourier Transform:

f [m]=m∑n=0

N−1

F [n]cos mnN

, where m={

1N

m=0

2N

m≠0

F [n]=∑m=0

N−1

f [m]cos mnN

Complex Sinusoid Discrete Fourier Transform:

f [m]= 1N ∑

n=0

N−1

F [n ]ej2mn

N

F [n]=∑m=0

M −1

f [m]e− j2 mn

N

References:“Base of Math Analysis” by G.M.Fihtengoltz (Russian edition)

“Base of Math Analysis” by V.A.Ilyin and E.G.Posnyak (Russian edition)

``Mathematics of the Discrete Fourier Transform (DFT), with Music and Audio Applications'', by Julius O. Smith III, W3K Publishing, 2003, ISBN 0-9745607-0-7.

Discrete-Time Speech Signal Processing: Principles and Practice, 1/eThomas F. QuatieriISBN: 013242942X | Prentice Hall PTR

“The FFT Demystified.” by Adrian HeyEngineering Productivity Tools Ltd. http://www.eptools.com/tn/T0001/INDEX.HTM