Lecture Notes Fourier Analysis

Prof. Xu Chen Department of Mechanical Engineering University of Washington chx AT uw.edu X. Chen Fourier Analysis September 29, 2019

Contents

1 Background 1 1.1 Periodic functions...... 1 1.2 Orthogonal decomposition...... 1

2 Fourier series 3 2.1 Arbitrary period...... 4 2.2 Even and odd functions...... 5 2.3 Application example: ODE with special inputs...... 6

3 Complex Fourier series7

4 Fourier integral8

5 Fourier transform 10

6 *Discrete Fourier analysis 11 6.1 Discrete Fourier series...... 11 6.2 Discrete-time Fourier transform (DTFT)...... 12 6.3 Discrete Fourier transform (DFT)...... 13 X. Chen Fourier Analysis September 29, 2019

Notations: If x is complex, < (x) and = (x) denote, respectively, the real and imaginary parts of x. hx, yi denotes the inner product of x and y.

1 Background

Fourier analysis is important in modeling and solving partial differential equations related to boundary and initial value problems of mechanics, heat flow, electrostatics, and other fields.

1.1 Periodic functions A function f(x) is called periodic if f(x) is deﬁned for all real x, except possibly at some points, and if there is some positive number p, called a period of f(x), such that

f (x + p) = f (x)

Remark 1. If p is a period of f(x), then clearly 2p, 3p, . . . are also periods of f (x). The smallest positive period is called the fundamental period.

1.2 Orthogonal decomposition In vector spaces equipped with inner products, we have the following essential theorem.

Theorem 2. Suppose {e1, e2, . . . , en} is an orthogonal basis of a vector space V. Then for every v ∈ V, we have hv, e1i hv, eni v = e1 + ··· + en he1, e1i hen, eni or, by using the norm notation,

hv, e1i hv, eni v = 2 e1 + ··· + 2 en ||e1|| ||en||

Proof. Let v ∈ V. Because {e1, e2, . . . , en} is a basis, there exists scalars α1, . . . , αn such that

v = α1e1 + ··· + αnen

Taking the inner product with ej (j = 1, . . . , n) yields

hv, eji = hα1e1 + ··· + αnen, eji = α1 he1, eji + ··· + αn hen, eji

But the basis is orthogonal, hence hei, eji = 0 if i 6= j. This yields

hα1e1 + ··· + αnen, eji = αj hej, eji

1 X. Chen Fourier Analysis September 29, 2019 and thus hv, eji hv, eji = αj hej, eji ⇐⇒ αj = hej, eji

Recall that we have the following fact. Fact (Inner product for functions, function spaces). The set of all real-valued continuous functions f (x), g (x), . . . x ∈ [α, β] is a real vector space under the usual addition of functions and multiplication by scalars. An inner product on this function space is Z β hf, gi = f (x) g (x) dx α and the norm of f is s Z β ||f (x) || = f (x)2 dx α As a very useful special case, the trigonometric system is a function space with orthogonal basis ∞ {sin nx, cos nx, 1}n=1. To check, we ﬁrst notice that the functions are all periodic with period 2π. The inner product in this case is Z π hf, gi = f (x) g (x) dx −π Noting that, by checking the area under the graphs and noting the shape of sine and cosine functions, we have Z π cos nxdx = 0, ∀n = 1, 2,... (1) −π Z π sin nxdx = 0, ∀n = 1, 2,... −π This conﬁrms the orthogonality between 1 and sin nx or cos nx. Furthermore, using (1) we immediately know Z π Z π cos (n + m) x + cos (n − m) x hcos nx, cos mxi = cos nx cos mxdx = dx −π −π 2 1 Z π 1 Z π = cos (n + m) xdx + cos (n − m) xdx 2 −π 2 −π = 0 + 0 as long as n 6= m namely, cos nx and cos mx are orthogonal if m 6= n. Similarly, as long as n 6= m (by assumption, both m and n are positive integers), sin nx and sin mx are orthogonal. This is because Z π Z π cos (n − m) x − cos (n + m) x hsin nx, sin mxi = sin nx sin mxdx = dx = 0 −π −π 2

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Furthermore, regardless of the values of m and n, we have

Z π Z π sin (n + m) x + sin (n − m) x hsin nx, cos mxi = sin nx cos mxdx = dx = 0 −π −π 2 So sin nx and cos mx are always orthogonal. Sketch the plots of sine and cosine functions. You should ﬁnd the above results intuitive. To perform the orthogonal decomposition under the trigonometric system, we need just one more thing–the norms of the basis functions. They are Z π Z π 1 − cos 2nx || sin nx||2 = hsin nx, sin nxi = (sin nx)2 dx = dx = π −π −π 2 Z π Z π cos 2nx + 1 || cos nx||2 = (cos nx)2 dx = dx = π −π −π 2 Z π ||1||2 = 1dx = 2π −π 2 Fourier series

Fourier series is an extension of the orthogonal decomposition reviewed in the last section. Under the trigonometric system, the orthogonal decomposition of a function f (x)—provided that the decomposition exists (we will talk about the existence very soon)—is

∞ X f (x) = a0 + (an cos nx + bn sin nx) (2) n=1 where hf (x) , 1i 1 Z π a0 = = f (x) dx (3) h1, 1i 2π −π and hf (x) , cos nxi 1 Z π hf (x) , sin nxi 1 Z π an = = f (x) cos nxdx, bn = = f (x) sin nxdx (4) hcos nx, cos nxi π −π hsin nx, sin nxi π −π (2) is called the Fourier series of f (x).(3) and (4) are called the corresponding Fourier coeﬃcients.

Example 3. Find the Fourier coeﬃcients of ( −k if − π < x < 0 f (x) = , and f (x + 2π) = f (x) k if 0 < x < π

Such functions occur as external forces acting on mechanical systems, electromotive forces in electric circuits, etc.

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(Solution: 4k 1 1 f (x) = sin x + sin 3x + sin 5x + ... π 3 5 )

Theorem 4 (Existance of Fourier Series). Let f (x) be periodic with period 2π and piecewise continuous in the interval −π ≤ x ≤ π. Furthermore, let f (x) have a left-hand derivative and a right-hand derivative at each point of that interval. Then the Fourier series of f (x) converges. Its sum is f (x), except at points xo where f (x) is discontinuous. There the sum of the series is the average of the left- and right-hand limits of f (x) at xo. Hence a periodic function f (x) = f (x + 2π) with

f (x) = 1/x, x ∈ [−π, π] does not have a Fourier series extension, as it does not have a left- or right-hand derivative at x = 0. A discontinuous periodic function, with period 2π and ( 1, −π < x < 0 f (x) = −1, π > x ≥ 0 is piecewise continuous. At x = 0, it has a left-hand derivative of 0 and a right-hand derivative of 0. Hence the function has a Fourier series expansion.

2.1 Arbitrary period The transition from period 2π to period p = 2L can be done by a suitable change of scale. If f (x) has a period 2L, we let x v = π L You can see, that x = ±L corresponds to v = ±π. So

L f (x) = f v π is periodic in v with period 2π. L Now forget about f (x). Focus just on the new f π v with period 2π. The Fourier series is

∞ L X f v = a + (a cos nv + b sin nv) π 0 n n n=1 where 1 Z π L 1 Z π L 1 Z π L a0 = f v dv, an = f v cos nvdv, bn = f v sin nvdv (5) 2π −π π π −π π π −π π

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Changing back to the x notation, by using L x π f (x) = f v , v = π, dv = dx π L L we have ∞ X nπ nπ f (x) = a + a cos x + b sin x (6) 0 n L n L n=1 with 1 Z L 1 Z L nπ 1 Z L nπ a0 = f (x) dx, an = f (x) cos xdx, bn = f (x) sin xdx (7) 2L −L L −L L L −L L Example 5. Find the Fourier series of  0 if − 2 < x < −1  f (x) = k if − 1 < x < 1 , p = 2L = 4,L = 2 0 if 1 < x < 2 Answer: k 2k π 1 3π 1 5π f (x) = + cos x − cos x + cos x − + ... 2 π 2 3 2 5 2

2.2 Even and odd functions It turns out that, if f(x) is even or odd, the Fourier series can be significantly simplified. This is because, if f (x) is an even function, then Z L f (x) sin nxdx = 0 −L as the integral of any odd function is zero. Hence in (7) bn = 0 so that (6) simplifies to ∞ X nπ f (x) = a + a cos x 0 n L n=1 Furthermore, since f (x) is even, we have 1 Z L 1 Z L 1 Z L nπ 2 Z L nπ a0 = f (x) dx = f (x) dx, an = f (x) cos xdx = f (x) cos xdx 2L −L L 0 L −L L L 0 L Similarly, if f (x) is an odd function, then Z L f (x) cos nxdx = 0 −L

5 X. Chen Fourier Analysis September 29, 2019 and ∞ X nπ 2 Z L nπ f (x) = b sin x, b = f (x) sin xdx n L n L L n=1 0 The following Theorem is intuitive and very useful.

Theorem 6. The Fourier series of a sum f1 + f2 are the sums of the corresponding Fourier coefficients of f1 and f2. The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f. Here is one example of using the results in this subsection.

Example 7 (Sawtooth wave). Find the Fourier series of the function

f (x) = x + π if − π < x < π and f (x + 2π) = f (x)

The basic idea is to decompose the function as

f (x) = f1 (x) + f2 (x) , f1 (x) = x, f2 (x) = π f1 (x) is an odd function; f2 (x) is an even function. Their Fourier coeﬃcients are simpler to ﬁnd.

2.3 Application example: ODE with special inputs Consider a mass spring damper system

d2 d m y + b y + ky = r (t) dt2 dt We have learned how to solve the nonhomogeneous ODE if r (t) is a standard function such as sine, cosine, power functions. Diﬃculty arises if r (t) is not a smooth function such as

( π t + 2 if − π < t < 0 r (t) = π , r (t + 2π) = r (t) −t + 2 if 0 < t < π

We can solve the problem by decomposing r (t) as a Fourier series and then use linearity of the ODE to obtain the solution. The solution is actually very interesting. For the case of

d2 d y + 0.05 y + 25y = r (t) dt2 dt the solution looks like that in Fig.1.

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X. Chen Fourier Analysis September 29, 2019 494 CHAP. 11 Fourier Analysis

0.3

Output 0.2

0.1

–3 –2 –1 012 3 t

–0.1 Input

–0.2

Fig. 277. Input and steady-state output in Example 1 Figure 1: Fig. 277 from [EK]

3 Complex Fourier series PROBLEM SET 11.3 jnx ∞ Instead of sin and cos functions, we can use {e }j=−∞ as the (complex) basis for Fourier series. For 1. Coefficients C . Derive the formula for C from A the program to Probs. 7 and 11 with initial values of your better physicaln intuitions, we usuallyn writen n = ωsl. The formula for Fourier series (when it exists) is and Bn. choice. 2. Change of spring and damping. In Example 1, what ∞ jω lx X f (x) , e s happens to the amplitudes Cn if we take a stiffer fspring,(x) = ejωslx (8) say, of k ϭ 49 ? If we increase the damping? 13–16hejωSTEADY-STATEslx, ejωslxi DAMPED OSCILLATIONS l=−∞ 3. Phase shift. Explain the role of the Bn ’s. What happens Find the steady-state oscillations of ys ϩ cyr ϩ y ϭ r (t) if we let c 0 ? Fact (Inner: product for complex functions)with. The c Ͼ set0 and of r ( allt) as complex-valued given. Note that the continuousspring constant functions f (x), 4. Differentiation of input. In Example 1, what happens is k ϭ 1 . Show the details. In Probs. 14–16 sketch r (t) .

ifg we( xreplace), . . r . (t) x with∈ its[α, derivative, β] is a the complex rectangular vector wave? space underN the usual addition of functions and multiplication Whatby scalars.is the ratio An of the inner new productCn to the old on ones? this function13. spacer (t) ϭ isa (an cos nt ϩ bn sin nt) 5. Sign of coefficients. Some of the A in Example 1 are nϭ1 n Z β positive, some negative. All Bn are positive. Is this Ϫ1ifϪp Ͻ t Ͻ 0 physically understandable? hg, fi14.=r (t) ϭg (x)f (x) dx and r(t ϩ 2p) ϭ r(t) α 1if0Ͻ t Ͻ p 2 2 6–11 GENERAL SOLUTION 15. r (t) ϭ t (p Ϫ t ) if Ϫp Ͻ t Ͻ p and and the norm of f is 2 ϩ ϭ Find a general solution of the ODE ys ϩ v y ϭ r (t) with r(t 2p) r(t) s β r (t) as given. Show the details of your work. 16. r (t) ϭ Z p b 2 2 2 2 6. r (t) ϭ sin at ϩ sin bt, v $ a , b ||f (x) || = hf, fi =t if Ϫp 2 |Ͻf t(Ͻx)p| 2dx α and r(t ϩ 2p) ϭ r(t) 7. r (t) ϭ sin t, v ϭ 0.5, 0.9, 1.1, 1.5, 10 √ p Ϫ t if p>2 Ͻ t Ͻ 3p> 2 8. Rectifier. r (t) jωϭsplx/4 cos t if Ϫp Ͻ t Ͻ p and p Fact 8. e hasƒ aƒ norm of Ts = 2π/ωs e r (t ϩ 2p) ϭ r (t), ƒ v ƒ $ 0, 2, 4, Á > > 9. WhatProof. kind By of deﬁnitionsolution is excluded in Prob. 8 by 17–19 RLC-CIRCUIT v $ 0, 2, 4, Á ? ƒ ƒ Find the steady-state current I (t) in the RLC-circuit in s s ؊1 10. Rectifier. r (t) ϭ p/4 ƒ sin t ƒ if 0 Ͻ t Ͻ 2p and Fig. 275,Z Twheres/2 R ϭ 10 ⍀, L ϭ 1 H, ZC ϭTs10/2 Fand with 2 jωslx p jωslx jωslx jωslx p r (t ϩ 2p) ϭ r (t), ƒ v||ƒe$ 0, 2,|| 4,=Á he , e i =E (t) V as follows|e and |periodicdx = with period 2dpx . Graph= orTs Ϫ1 if Ϫp Ͻ t Ͻ 0 sketch the−T sfirst/2 four partial sums. Note −thatTs /the2 coefficients 11. r (t) ϭ ƒ v ƒ $ 1, 3, 5, Á of the solution decrease rapidly. Hint. Remember that the 1 if 0 Ͻ t Ͻ p, ODE contains Er(t) , not E (t) , cf. Sec. 2.9. 12. CAS Program. Write a program for solving the ODE 2 Ϫ50t if Ϫp Ͻ t Ͻ 0 just considered and for jointly graphing input and output The Fourier series expansion (8) hence simpliﬁes17. E (t) ϭ to of an initial value problem involving that ODE. Apply 50t 2 if 0 Ͻ t Ͻ p b ∞ 1 X f (x) = f (x) , ejωslx ejωslx Ts l=−∞ b

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Example 9 (Fourier series of an impulse train). Show that

∞ ∞ X 1 X jωslt 2π δ (t − lTs) = e , ωs = Ts Ts l=−∞ l=∞ where δ (x) is the Kronecker-delta function satisfying δ (x) = 1 if x = 0 and δ (x) = 0 otherwise. P∞ jωst l=−∞ δ (t − lTs) is periodic with period Ts. You can check that e also has a period of Ts if ωs = 2π/Ts. For the Fourier coeﬃcients, we have

Z Ts/2 f (t) , ejωslt = δ (t) ejωsltdt = 1 −Ts/2 Hence ∞ 1 ∞ 1 ∞ 2π X X jωslt jωslt X jωslt δ (t − lTs) = f (x) , e e = e , ωs = Ts Ts Ts l=−∞ l=−∞ l=∞

4 Fourier integral

What can be done to extend the method of Fourier series to nonperiodic functions? This is the idea of “Fourier integrals.” Let us consider an example of ( 1 if −1 < x < −1 f(x) = (9) 0 otherwise

This is not a periodic function. Construct a rectangular wave  0 if −L < x < −1  fL(x) = 1 if −1 < x < −1 , fL (x + 2L) = fL(x) 0 if 1 < x < L

We are going to make L increase from some small numbers to inﬁnity, which recovers (9). The function is even. The Fourier coeﬃcients are 1 Z 1 1 1 Z 1 nπx 2 Z 1 nπx 2 sin (nπ/L) a0 = dx = , an = cos dx = cos dx = 2L −1 L L −1 L L 0 L L nπ/L As L increases, the frequency of sin (nπ/L) increases. More generally, consider any periodic function fL (x) of period 2L that can be represented by

∞ X nπ f (x) = a + [a cos ω x + b sin ω x] , ω = L 0 n n n n n L n=1

8 X. Chen Fourier Analysis September 29, 2019 where 1 Z L 1 Z L nπ 1 Z L nπ a0 = f (x) dx, an = f (x) cos xdx, bn = f (x) sin xdx 2L −L L −L L L −L L or, in the notation of the newly introduced ωn = nπ/L: ∞ 1 Z L 1 X Z L Z L f (x) = f (v) dv + cos ω x f (v) cos ω vdv + sin ω x f (v) cos ω vdv L 2L L L n L n n L n −L n=1 −L −L (10) Notice that if we deﬁne π 1 ∆ω ∆ω = ω − ω = ⇒ = n+1 n L L π then (10) is actually

1 Z L fL (x) = fL (v) dv 2L −L ∞ 1 X Z L Z L + ∆ω cos (ω x) f (v) cos ω vdv + ∆ω sin (ω x) f (v) cos ω vdv (11) π n L n n L n n=1 −L −L

We want to have an understanding of the case when L → ∞. Assume that f (x) = limL→∞ fL (x) is absolutely integrable, i.e. the following ﬁnite limit exists Z 0 Z b lim |f (x)| dx + lim |f (x)| dx a→−∞ a b→∞ 0 Then 1 Z L lim fL (v) dv = 0 L→∞ 2L −L R L Let g (ω) = cos (ωx) −L fL (v) cos ωvdv. The ﬁrst summation in (11) is ∞ X ∆ωg (ωn) = g (ω1) ∆ω + g (ω2) ∆ω + ... n=1 As ∆ω = π/L → 0 if L → ∞, the last term in (11) thus looks like an integration 1 Z ∞ Z ∞ Z ∞ f (x) = cos ωx f (v) cos ωvdv + sin ωx f (v) cos ωvdv dω π 0 −∞ −∞ Letting 1 Z ∞ 1 Z ∞ A (ω) = f (v) cos ωvdv, B (w) = f (v) sin ωvdv π −∞ π −∞ yields Z ∞ f (x) = [A (ω) cos ωx + B (ω) sin ωx] dω 0 This is called a representation of f (x) (no longer periodic) by a Fourier integral.

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Remark 10. The above construction is only an intuition. Nonetheless, the conclusion is correct. Theorem 11 (Existance of Fourier Integral). If f (x) is piecewise continuous in every ﬁnite interval and has a right hand derivative and a left-hand derivative at every point and if the integral Z 0 Z b lim |f (x)| dx + lim |f (x)| dx a→−∞ a b→∞ 0 exists, then f (x) can be represented by a Fourier integral

Z ∞ f (x) = [A (ω) cos ωx + B (ω) sin ωx] dω (12) 0 with 1 Z ∞ 1 Z ∞ A (ω) = f (v) cos ωvdv, B (w) = f (v) sin ωvdv (13) π −∞ π −∞ At a point where f (x) is discontinuous the value of the Fourier integral equals the average of the left- and right-hand limits of f (x) at that point.

5 Fourier transform

With the Fourier integral formulas (12)-(13), we have Z ∞ f (x) = [A (ω) cos ωx + B (ω) sin ωx] dω 0 1 Z ∞ Z ∞ = f (v) [cos (ωv) cos (ωx) + sin (ωv) sin (ωx)] dvdω π 0 −∞ 1 Z ∞ Z ∞ = f (v) cos (ωx − ωv) dv dω π 0 −∞ R ∞ Note that −∞ f (v) cos (ωx − ωv) dv is even in ω. Hence 1 Z ∞ Z ∞ 1 Z ∞ Z ∞ f (v) cos (ωx − ωv) dv dω = f (v) cos (ωx − ωv) dv dω π 0 −∞ 2π −∞ −∞ To simplify the equations, we add an integral term that equals 0: 1 Z ∞ Z ∞ √ f (v) i sin (ωx − ωv) dv dω = 0, i = −1 2π −∞ −∞ R ∞ where the equality comes from the fact that −∞ f (v) sin (ωx − ωv) dv is an odd function of ω. Adding up the last two equations gives 1 Z ∞ Z ∞ f (x) = f (v) eiω(x−v)dv dω = 0 2π −∞ −∞

10 X. Chen Fourier Analysis September 29, 2019 or, equivalently 1 Z ∞ f (x) = √ F (ω) eiωxdω 2π −∞ 1 Z ∞ 1 Z ∞ F (ω) = √ f (v) e−iωvdv = √ f (x) e−iωxdx 2π −∞ 2π −∞ F (ω) is called the Fourier transform of f (x); f (x) is the inverse Fourier transform of F (ω). The above are often denoted as:

F (ω) = F (f (x)) , f (x) = F −1 (F (ω))

Remark 12. Many people prefer to use a diﬀerent normalization coeﬃcient, and write: Z ∞ f (x)= F (ω) eiωxdω −∞ 1 Z ∞ F (ω)= f (x) e−iωxdx 2π −∞

Theorem 13 (Existance of Fourier transform). If f (x) is absolutely integrable on the x-axis and piecewise continuous on every ﬁnite interval, then the Fourier transform of f (x) exists. Example 14. Find the Fourier transform of ( 1, |x| < 1 f (x) = 0, otherwise

Solution: By deﬁnition 1 Z 1 1 rπ sin ω F (ω) = √ e−iωxdx = √ e−iω − eiω = 2π −1 −iω 2π 2 ω

6 *Discrete Fourier analysis

Fourier series (FS) and Fourier transforms (FT) apply only to continuous-time functions. Discrete Fourier series (DFS) and discrete Fourier transform (DFT) are their discrete versions when the function f (x) is defined at finitely many points. The main equations are summarized below. You can find the strong analogy between FS and DFS; as well as FT and DFT. For more details, see the reference [AO].

6.1 Discrete Fourier series A periodic bounded discrete sequence x˜ [n] has a discrete Fourier series (DFS) expansion

N−1 1 X j 2π nk x˜[n] = X˜[k]e N (14) N k=0

11 X. Chen Fourier Analysis September 29, 2019 where the unnormalized DFS coeﬃcients are

N−1 X −j 2π nk X˜[k] = x˜[n]e N (15) n=0

DFS properties Let x˜ [n] and X˜ [k] be deﬁned as in (14) and (15). Then

Sequence DFS coeﬃcient x˜[n] X˜[k] −j 2π km x˜[n − m] e N X˜[k] j 2π nl e N x˜[n] X˜[k − l] PN−1 ˜ ˜ ˜ x˜3[n] = m=0 x˜1[m]˜x2[n − m] X3[k] = X1[k] · X2[k] X˜ [n] Nx˜[−k] (duality)

6.2 Discrete-time Fourier transform (DTFT) A bounded inﬁnite sequence x [n] can be decomposed as

π 1 Z x[n] = X(ejω)ejωndω 2π −π where X(ejω) is called the DTFT of x [n] and is deﬁned as

∞ X X(ejω) = x[k]e−jωk k=−∞

Let x∗ [n] denote the complex conjugate sequence of x [n]. If X (ejω) is the DFT of x [n], then the following is true

Sequence DTFT coeﬃcient x∗[n] X∗(e−jω) x∗[−n] X∗(ejω) (x[n] + x∗[n])/2 < {X(ejω)} (x[n] − x∗[n])/(2j) = {X(ejω)} −jωn0 e δ[n − n0] ejω0nx[n] X(ej(ω−ω0)) n jω 1 x[n] = a u[n] X(e ) = 1−ae−jω

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6.3 Discrete Fourier transform (DFT) A bounded sequence x [n] deﬁned on n ∈ {0,...,N − 1} can be decomposed as

 N−1 1 2π P j N kn  N X[k]e , n ∈ {0,...,N − 1} x[n] = k=0 0 , n∈ / {0,...,N − 1} where X[k] is the DFT of x [n] and is deﬁned as

N−1 − 2πj kn  P x[n]e N , k ∈ {0,...,N − 1} X[k] = n=0 0 , k∈ / {0,...,N − 1}

DFT properties • the DFT pair x[n] ←→ X[k] (ﬁnite-length x [n]) is analogous to the DFS pair x˜[n] ←→ X˜[k] (inﬁnite-length periodic x˜ [n]) • if x[n] is real, then X(k) = X∗[N − k] • if N ≥ L, then DFT are samples of DTFT

N−1 P • let x1 [n] N x2 [n] := x1[m]x2[(n − m) mod N]. Then x1 [n] N x2[n] ←→ X1[k]X2[k] m=0

• if x[n] = x[((−n))N ] = x[N − n], then X[k] is real • DFT coeﬃcient table: Sequence DFT coeﬃcient x[n] X[k] ∗ ∗ x [n] X [((−k))N ] ∗ ∗ x [((−n))N ] X [k] x[n]ej2πn`/N X[(k − `) mod N] −j2πk`/N x[((n − `))N ] X[k]e 1 ∗ <(x[n]) 2 {X[((k))N ] + X [((−k))N ]} 1 ∗ =(x[n]) 2 {X[((k))N ] − X [((−k))N ]} 1 ∗ 2 {x[n] + x [((−n))N ]} <(X[k]) 1 ∗ 2 {x[n] − x [((−n))N ]} j=(X[k]) 1 PN−1 2πkn 2 {x[n] + x[−n]} n=0 x[n] cos( 2N−1 ) x[n] ∈ R X[k] = X∗[N − k] Reference

[EK]: ERwin Kreyszig, Advanced Engineering Mathematics, 10th edition [AO]: Alan Oppenheim et al., Discrete-time Signal Processing, 2nd edition