University of Connecticut Lecture Notes for ME5507 Fall 2014 Engineering Analysis I Part III: Fourier Analysis

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1 University of Connecticut Lecture Notes for ME557 Fall 24 Engineering Analysis I Part III: Fourier Analysis Xu Chen Assistant Professor United Technologies Engineering Build, Rm. 382 Department of Mechanical Engineering University of Connecticut xchen@engr.uconn.edu

2 Contents Background. Periodic functions Orthogonal decomposition Fourier series 3 2. Arbitrary period Even and odd functions Application example: ODE with special inputs Complex Fourier series 7 4 Fourier integral 8 5 Fourier transform 6 *Discrete Fourier analysis 6. Discrete Fourier series Discrete-time Fourier transform (DTFT) Discrete Fourier transform (DFT)

3 Notations: If x is complex, R (x) and I (x) denote, respectively, the real and imaginary parts of x. x, y denotes the inner product of x and y. 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.. 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. 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..2 Orthogonal decomposition In vector spaces equipped with inner products, we have the following essential theorem. Theorem 2. Suppose {e, e 2,..., e n } is an orthogonal basis of a vector space V. Then for every v V, we have v = v, e e, e e + + v, e n e n, e n e n or, by using the norm notation, v = v, e e 2 e + + v, e n e n 2 e n Proof. Let v V. Because {e, e 2,..., e n } is a basis, there exists scalars α,..., α n such that v = α e + + α n e n Taking the inner product with e j (j =,..., n) yields v, e j = α e + + α n e n, e j = α e, e j + + α n e n, e j But the basis is orthogonal, hence e i, e j = if i j. This yields α e + + α n e n, e j = α j e j, e j

4 and thus v, e j = α j e j, e j α j = v, e j e j, e j 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 and the norm of f is f, g = ˆ β α f (x) g (x) dx ˆ β f (x) = α f (x) 2 dx As a very useful special case, the trigonometric system is a function space with orthogonal basis {sin nx, cos nx, } n=. To check, we first notice that the functions are all periodic with period 2π. The inner product in this case is f, g = f (x) g (x) dx Noting that, by checking the area under the graphs and noting the shape of sine and cosine functions, we have cos nxdx =, n =, 2,... () sin nxdx =, n =, 2,... This confirms the orthogonality between and sin nx or cos nx. Furthermore, using () we immediately know cos nx, cos mx = cos nx cos mxdx = = 2 cos (n + m) x + cos (n m) x dx 2 cos (n + m) xdx + 2 = + as long as n m cos (n m) xdx namely, cos nx and cos mx are orthogonal if m n. Similarly, as long as n m (by assumption, both m and n are positive integers), sin nx and sin mx are orthogonal. This is because sin nx, sin mx = sin nx sin mxdx = 2 cos (n m) x cos (n + m) x dx = 2

5 Furthermore, regardless of the values of m and n, we have sin nx, cos mx = sin nx cos mxdx = sin (n + m) x + sin (n m) x dx = 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 sin nx 2 = sin nx, sin nx = cos nx 2 = 2 = 2 Fourier series (cos nx) 2 dx = dx = 2π (sin nx) 2 cos 2nx dx = dx = π 2 cos 2nx + dx = π 2 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 where and a n = a + a = (a n cos nx + b n sin nx) (2) n= f (x),, f (x), cos nx cos nx, cos nx = f (x) cos nxdx, b n = π = f (x) dx (3) 2π f (x), sin nx sin nx, sin nx = f (x) sin nxdx (4) 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 <, and f (x + 2π) = f (x) k if < x < π Such functions occur as external forces acting on mechanical systems, electromotive forces in electric circuits, etc. 3

6 (Solution: (sin x + 3 sin 3x + 5 sin 5x +... ) 4k π ) Theorem 4 (Existance of Fourier Series). Let f (x) be periodic with period 2π and piecewise continuous in the interval 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 x o where f (x) is discontinuous. There the sum of the series is the average of the leftand right-hand limits of f (x) at x o. Hence a periodic function f (x + 2π) with /x, x [, π does not have a Fourier series extension, as it does not have a left- or right-hand derivative at x =. A discontinuous periodic function, with period 2π and {, < x <, π > x is piecewise continuous. At x =, it has a left-hand derivative of and a right-hand derivative of. Hence the function has a Fourier series expansion. 2. 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 v = x L π You can see, that x = ±L corresponds to v = ±π. So ( ) L f π v is periodic in v with period 2π. Now forget about f (x). Focus just on the new f ( L π v) with period 2π. The Fourier series is where f ( ) L π v = a + (a n cos nv + b n sin nv) n= a = ( ) L f 2π π v dv, a n = ( ) L f π π v cos nvdv, b n = ( ) L f π π v sin nvdv (5) 4

7 Changing back to the x notation, by using ( ) L f π v, v = x L π, dv = π L dx we have with a + (a n cos nπ L x + b n sin nπ ) L x n= (6) a = f (x) dx, a n = f (x) cos nπ 2L L L xdx, b n = f (x) sin nπ xdx (7) L L Example 5. Find the Fourier series of if 2 < x < k if < x <, p = 2L = 4, L = 2 if < x < 2 Answer: (cos π2 x 3 cos 3π2 x + 5 cos 5π2 x +... ) k 2 + 2k π 2.2 Even and odd functions It turns out that, if f(x) is even or odd, the Fourier series can be significantly simplified. because, if f (x) is an even function, then f (x) sin nxdx = as the integral of any odd function is zero. Hence in (7) b n = so that (6) simplifies to Furthermore, since f (x) is even, we have a + a n cos nπ L x n= This is a = 2L f (x) dx = L f (x) dx, a n = L Similarly, if f (x) is an odd function, then f (x) cos nxdx = 5 f (x) cos nπ L xdx = 2 L f (x) cos nπ L xdx

8 and b n sin nπ L x, b n = 2 L n= The following Theorem is intuitive and very useful. f (x) sin nπ L xdx Theorem 6. The Fourier series of a sum f + f 2 are the sums of the corresponding Fourier coefficients of f and f 2. 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 x + π if π < x < π and f (x + 2π) = f (x) The basic idea is to decompose the function as f (x) + f 2 (x), f (x) = x, f 2 (x) = π f (x) is an odd function; f 2 (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 m d2 dt y + b d y + ky = r (t) 2 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 { t + π if π < t < 2 r (t) =, r (t + 2π) = r (t) t + π if < t < π 2 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 the solution looks like that in Fig.. d 2 dt y +.5 d y + 25y = r (t) 2 dt 6

9 494 CHAP. Fourier Analysis y.3.2 Output t. Input.2 3 Complex Fourier series PROBLEM SET.3 Fig Input and steady-state output in Example Figure : Fig. 277 from [EK Instead of sin and cos functions, we can use {e jnx } j= as the (complex) basis for Fourier series. For. Coefficients better physical C n. Derive intuitions, the formula we for usually C n from write A n n the= program ω to Probs. 7 and with initial values of your s l. The formula for Fourier series (when it exists) is and B n. choice. 2. Change of spring and damping. In Example, what happens to the amplitudes C if we take a stiffer spring, f (x), e jω slx n say, of k 49? If we increase the damping? 3 6 e jωslx STEADY-STATE, e jωslx ejωslx (8) DAMPED OSCILLATIONS l= 3. Phase shift. Explain the role of the B n s. What happens Find the steady-state oscillations of ys cyr y r (t) if we let c :? Fact (Inner product for complex functions). with The c set and ofr (t) allas complex-valued given. Note that the continuous spring constantfunctions f (x), 4. Differentiation of input. In Example, what happens is k. Show the details. In Probs. 4 6 sketch r (t). if g we (x), replace.. r.(t) xwith its [α, derivative, β is athe complex rectangular vector wave? space undernthe usual addition of functions and multiplication What by scalars. is the ratio An of the inner new product C n to the old onones? this function 3. space r (t) is a (a n cos nt b n sin nt) n 5. Sign of coefficients. Some of the A n in Example are ˆ β positive, some negative. All B n are positive. Is this if p t g, f physically understandable? 4. = r (t) g b(x)f (x) dx and r(t 2p) r(t) α if t p 6 GENERAL SOLUTION 5. r (t) t (p 2 t 2 ) if p t p and and the norm of f is Find a general solution of the ODE ys v 2 y r (t) with r (t 2p) r (t) r (t) as given. Show the details of your work. 6. r (t) f (x) = ˆ β f, f = f (x) 2 dx 6. r (t) sin at sin bt, v 2 a 2, b 2 t if p>2 t p>2 7. r (t) sin t, v.5,.9,.,.5, e α and r(t 2p) r(t) p t if p>2 t 3p>2 8. Rectifier. Fact 8. r (t) e jωslx p/4 has ƒ cos at ƒ if norm p and r (t 2p) r (t), ƒ v ƒ, 2, 4, Á of t pt s = 2π/ω s 9. What Proof. kind By of definition solution is excluded in Prob. 8 by 7 9 RLC-CIRCUIT ƒ v ƒ, 2, 4, Á? Find the steady-state current I (t) in the RLC-circuit in. Rectifier. r (t) p/4 ƒ sin t ƒ if and r (t 2p) r (t), ƒ v ƒ, 2, 4, Á t 2p Fig. 275, where R, L H, C Fand with e jωslx = ˆ Ts/2 ˆ Ts/2 e jωslx, e jωslx = E (t) V as follows e jωslx and 2 periodic dx = with period 2p dx. Graph = or T s T if p t. r (t) b ƒ v ƒ, 3, 5, Á sketch the s/2 T first four partial sums. Note that s/2 the coefficients of the solution decrease rapidly. Hint. Remember that the if t p, ODE contains Er(t), not E (t), cf. Sec CAS Program. Write a program for solving the ODE The Fourier series expansion (8) hence simplifies to just considered and for jointly graphing input and output 7. E (t) b 5t 2 if p t of an initial value problem involving that ODE. Apply 5t 2 if t p f (x), e jω slx e jωslx T s l= 7

10 Example 9 (Fourier series of an impulse train). Show that l= δ (t lt s ) = T s l= e jωslt, ω s = 2π T s where δ (x) is the Kronecker-delta function satisfying δ (x) = if x = and δ (x) = otherwise. l= δ (t lt s) is periodic with period T s. You can check that e jωst also has a period of T s if ω s = 2π/T s. For the Fourier coefficients, we have ˆ Ts/2 f (t), e jω slt = δ (t) e jωslt dt = T s/2 Hence l= δ (t lt s ) = T s l= f (x), e jω slt e jωslt = T s l= e jωslt, ω s = 2π T s 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 { if < x < f(x) = (9) otherwise This is not a periodic function. Construct a rectangular wave if < x < f L (x) = if < x <, f L (x + 2L) = f L (x) if < 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 a = 2L ˆ dx = L, a n = L ˆ cos nπx L dx = 2 L ˆ cos nπx L dx = 2 sin (nπ/l) L nπ/l As L increases, the frequency of sin (nπ/l) increases. More generally, consider any periodic function f L (x) of period 2L that can be represented by f L (x) = a + [a n cos ω n x + b n sin ω n x, ω n = nπ L n= 8

11 where a = f (x) dx, a n = f (x) cos nπ 2L L L xdx, b n = f (x) sin nπ L L xdx or, in the notation of the newly introduced ω n = nπ/l: f L (x) = f L (v) dv + [ cos ω n x f L (v) cos ω n vdv + sin ω n x 2L L Notice that if we define then () is actually f L (x) = 2L + π n= f L (v) dv n= ω = ω n+ ω n = π L L = ω π f L (v) cos ω n vdv [ ω cos (ω n x) f L (v) cos ω n vdv + ω sin (ω n x) f L (v) cos ω n vdv We want to have an understanding of the case when L. Assume that lim L f L (x) is absolutely integrable, i.e. the following finite limit exists Then lim a ˆ a lim L f (x) dx + lim b ˆ b ˆ L f L (v) dv = 2L f (x) dx Let g (ω) = cos (ωx) L f L (v) cos ωvdv. The first summation in () is ωg (ω n ) = g (ω ) ω + g (ω 2 ) ω +... n= As ω = π/l if L, the last term in () thus looks like an integration [ cos ωx f (v) cos ωvdv + sin ωx f (v) cos ωvdv dω π Letting yields A (ω) = π f (v) cos ωvdv, B (w) = π [A (ω) cos ωx + B (ω) sin ωx dω f (v) sin ωvdv This is called a representation of f (x) (no longer periodic) by a Fourier integral. 9 () ()

12 Remark. The above construction is only an intuition. Nonetheless, the conclusion is correct. Theorem (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 ˆ lim a a f (x) dx + lim b ˆ b exists, then f (x) can be represented by a Fourier integral with A (ω) = π f (x) dx [A (ω) cos ωx + B (ω) sin ωx dω (2) f (v) cos ωvdv, B (w) = π f (v) sin ωvdv (3) At a point where f (x) is discontinuous the value of the Fourier integral equals the average of the leftand right-hand limits of f (x) at that point. 5 Fourier transform With the Fourier integral formulas (2)-(3), we have Note that π = π = π [A (ω) cos ωx + B (ω) sin ωx dω f (v) [cos (ωv) cos (ωx) + sin (ωv) sin (ωx) dvdω [ f (v) cos (ωx ωv) dv dω f (v) cos (ωx ωv) dv is even in ω. Hence [ f (v) cos (ωx ωv) dv dω = 2π [ f (v) cos (ωx ωv) dv dω To simplify the equations, we add an integral term that equals : ˆ [ f (v) i sin (ωx ωv) dv dω =, i = 2π 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 [ f (v) e iω(x v) dv dω = 2π

13 or, equivalently 2π F (ω) = 2π F (ω) e iωx dω f (v) e iωv dv = 2π f (x) e iωx dx 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 (F (ω)) Remark 2. Many people prefer to use a different normalization coefficient, and write: f (x)= F (ω)= 2π F (ω) e iωx dω f (x) e iωx dx Theorem 3 (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 4. Find the Fourier transform of Solution: By definition {, x <, otherwise F (ω) = 2π ˆ e iωx dx = iω 2π ( e iω e iω) π sin ω = 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. Discrete Fourier series A periodic bounded discrete sequence x [n has a discrete Fourier series (DFS) expansion x[n = N N k= X[ke j 2π N nk (4)

14 where the unnormalized DFS coefficients are X[k = N n= x[ne j 2π N nk (5) DFS properties Let x [n and X [k be defined as in (4) and (5). Then Sequence DFS coefficient x[n X[k x[n m e j 2π N km X[k e j 2π nl x[n N X[k l x 3 [n = N m= x [m x 2 [n m X3 [k = X [k X 2 [k X [n N x[ k (duality) 6.2 Discrete-time Fourier transform (DTFT) A bounded infinite sequence x [n can be decomposed as x[n = 2π ˆπ X(e jω )e jωn dω where X(e jω ) is called the DTFT of x [n and is defined as X(e jω ) = k= x[ke jωk Let x [n denote the complex conjugate sequence of x [n. If X (e jω ) is the DFT of x [n, then the following is true Sequence DTFT coefficient x [n X (e jω ) x [ n X (e jω ) (x[n + x [n)/2 R {X(e jω )} (x[n x [n)/(2j) I {X(e jω )} e jωn δ[n n e jωn x[n X(e j(ω ω) ) x[n = a n u[n X(e jω ) = ae jω 2

15 6.3 Discrete Fourier transform (DFT) A bounded sequence x [n defined on n {,..., N } can be decomposed as N X[ke j 2π N kn, n {,..., N } N x[n = k=, n / {,..., N } where X[k is the DFT of x [n and is defined as N x[ne 2πj N kn, k {,..., N } X[k = n=, k / {,..., N } 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 let x [n N x 2 [n := N m= if x[n = x[(( n)) N = x[n n, then X[k is real DFT coefficient table: Reference x [mx 2 [(n m) mod N. Then x [n N x 2 [n X [kx 2 [k Sequence DFT coefficient x[n X[k x [n X [(( k)) N x [(( n)) N X [k x[ne j2πnl/n X[(k l) mod N x[((n l)) N X[ke j2πkl/n R(x[n) 2 N + X [(( k)) N } I(x[n) 2 N X [(( k)) N } 2 x [(( n)) N } R(X[k) 2 x [(( n)) N } ji(x[k) N 2πkn 2 n= x[n cos( 2N x[n R X[k = X [N k [EK: ERwin Kreyszig, Advanced Engineering Mathematics, th edition [AO: Alan Oppenheim et al., Discrete-time Signal Processing, 2nd edition 3