Series FOURIER SERIES. Graham S McDonald. A self-contained Tutorial Module for learning the technique of Fourier series analysis

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1 Series FOURIER SERIES Graham S McDonald A self-contained Tutorial Module for learning the technique of Fourier series analysis Table of contents Begin Tutorial c 24 g.s.mcdonald@salford.ac.uk

2 1. Theory 2. Exercises 3. Answers 4. Integrals 5. Useful trig results 6. Alternative notation 7. Tips on using solutions Full worked solutions Table of contents

3 Section 1: Theory 3 1. Theory A graph of periodic function f(x) that has period L exhibits the same pattern every L units along the x-axis, so that f(x + L) = f(x) for every value of x. If we know what the function looks like over one complete period, we can thus sketch a graph of the function over a wider interval of x (that may contain many periods) f(x ) x P E R IO D = L

4 Section 1: Theory 4 This property of repetition defines a fundamental spatial frequency k = 2 L that can be used to give a first approximation to the periodic pattern f(x): f(x) c 1 sin(kx + α 1 ) = a 1 cos(kx) + b 1 sin(kx), where symbols with subscript 1 are constants that determine the amplitude and phase of this first approximation A much better approximation of the periodic pattern f(x) can be built up by adding an appropriate combination of harmonics to this fundamental (sine-wave) pattern. For example, adding c 2 sin(2kx + α 2 ) = a 2 cos(2kx) + b 2 sin(2kx) c 3 sin(3kx + α 3 ) = a 3 cos(3kx) + b 3 sin(3kx) (the 2nd harmonic) (the 3rd harmonic) Here, symbols with subscripts are constants that determine the amplitude and phase of each harmonic contribution

5 Section 1: Theory 5 One can even approximate a square-wave pattern with a suitable sum that involves a fundamental sine-wave plus a combination of harmonics of this fundamental frequency. This sum is called a Fourier series F u n d a m e n ta l F u n d a m e n ta l + 2 h a rm o n ic s x F u n d a m e n ta l + 5 h a rm o n ic s F u n d a m e n ta l + 2 h a rm o n ic s P E R IO D = L

6 Section 1: Theory 6 In this Tutorial, we consider working out Fourier series for functions f(x) with period L = 2. Their fundamental frequency is then k = 2 L = 1, and their Fourier series representations involve terms like a 1 cos x, a 2 cos 2x, a 3 cos 3x, b 1 sin x b 2 sin 2x b 3 sin 3x We also include a constant term a /2 in the Fourier series. This allows us to represent functions that are, for example, entirely above the x axis. With a sufficient number of harmonics included, our approximate series can exactly represent a given function f(x) f(x) = a /2 + a 1 cos x + a 2 cos 2x + a 3 cos 3x b 1 sin x + b 2 sin 2x + b 3 sin 3x +...

7 Section 1: Theory 7 A more compact way of writing the Fourier series of a function f(x), with period 2, uses the variable subscript n = 1, 2, 3,... f(x) = a 2 + [a n cos nx + b n sin nx] n=1 We need to work out the Fourier coefficients (a, a n and b n ) for given functions f(x). This process is broken down into three steps STEP ONE STEP TWO STEP THREE a = 1 f(x) dx 2 a n = 1 f(x) cos nx dx 2 b n = 1 f(x) sin nx dx where integrations are over a single interval in x of L = 2 2

8 Section 1: Theory 8 Finally, specifying a particular value of x = x 1 in a Fourier series, gives a series of constants that should equal f(x 1 ). However, if f(x) is discontinuous at this value of x, then the series converges to a value that is half-way between the two possible function values " V e rtic a l ju m p " /d is c o n tin u ity in th e fu n c tio n re p re s e n te d f(x ) x F o u rie r s e rie s c o n v e rg e s to h a lf-w a y p o in t

9 Section 2: Exercises 9 2. Exercises Click on Exercise links for full worked solutions (7 exercises in total). Exercise 1. Let f(x) be a function of period 2 such that { 1, < x < f(x) =, < x <. a) Sketch a graph of f(x) in the interval 2 < x < 2 b) Show that the Fourier series for f(x) in the interval < x < is [sin x + 13 sin 3x + 15 ] sin 5x +... c) By giving an appropriate value to x, show that = Theory Answers Integrals Trig Notation

10 Section 2: Exercises 1 Exercise 2. Let f(x) be a function of period 2 such that {, < x < f(x) = x, < x <. a) Sketch a graph of f(x) in the interval 3 < x < 3 b) Show that the Fourier series for f(x) in the interval < x < is 4 2 [cos x cos 3x + 15 ] 2 cos 5x [sin x 12 sin 2x + 13 ] sin 3x... c) By giving appropriate values to x, show that (i) 4 = and (ii) 8 = Theory Answers Integrals Trig Notation

11 Section 2: Exercises 11 Exercise 3. Let f(x) be a function of period 2 such that { x, < x < f(x) =, < x < 2. a) Sketch a graph of f(x) in the interval 2 < x < 2 b) Show that the Fourier series for f(x) in the interval < x < 2 is [cos x cos 3x + 15 ] 2 cos 5x +... [sin x + 12 sin 2x + 13 ] sin 3x +... c) By giving appropriate values to x, show that (i) 4 = and (ii) 8 = Theory Answers Integrals Trig Notation

12 Section 2: Exercises 12 Exercise 4. Let f(x) be a function of period 2 such that f(x) = x 2 over the interval < x < 2. a) Sketch a graph of f(x) in the interval < x < 4 b) Show that the Fourier series for f(x) in the interval < x < 2 is [sin 2 x + 12 sin 2x + 13 ] sin 3x +... c) By giving an appropriate value to x, show that 4 = Theory Answers Integrals Trig Notation

13 Section 2: Exercises 13 Exercise 5. Let f(x) be a function of period 2 such that { x, < x < f(x) =, < x < 2 a) Sketch a graph of f(x) in the interval 2 < x < 2 b) Show that the Fourier series for f(x) in the interval < x < 2 is + 2 [cos x cos 3x + 15 ] 2 cos 5x sin x sin 2x sin 3x + 1 sin 4x c) By giving an appropriate value to x, show that 2 8 = Theory Answers Integrals Trig Notation

14 Section 2: Exercises 14 Exercise 6. Let f(x) be a function of period 2 such that f(x) = x in the range < x <. a) Sketch a graph of f(x) in the interval 3 < x < 3 b) Show that the Fourier series for f(x) in the interval < x < is 2 [sin x 12 sin 2x + 13 ] sin 3x... c) By giving an appropriate value to x, show that 4 = Theory Answers Integrals Trig Notation

15 Section 2: Exercises 15 Exercise 7. Let f(x) be a function of period 2 such that f(x) = x 2 over the interval < x <. a) Sketch a graph of f(x) in the interval 3 < x < 3 b) Show that the Fourier series for f(x) in the interval < x < is 2 [cos 3 4 x 12 2 cos 2x + 13 ] 2 cos 3x... c) By giving an appropriate value to x, show that 2 6 = Theory Answers Integrals Trig Notation

16 Section 3: Answers Answers The sketches asked for in part (a) of each exercise are given within the full worked solutions click on the Exercise links to see these solutions The answers below are suggested values of x to get the series of constants quoted in part (c) of each exercise 1. x = 2, 2. (i) x = 2, (ii) x =, 3. (i) x = 2, (ii) x =, 4. x = 2, 5. x =, 6. x = 2, 7. x =.

17 Section 4: Integrals Integrals Formula for integration by parts: f (x) x n 1 x b a u dv dx dx = [uv]b a b du a dx v dx f(x)dx f (x) f(x)dx xn+1 n+1 (n 1) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+1 n+1 (n 1) ln g (x) e x e x a x ax ln a (a > ) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tan x ln cos x tanh x ln cosh x cosec x ln tan x 2 cosech x ln tanh x 2 sec x ln sec x + tan x sech x 2 tan 1 e x sec 2 x tan x sech 2 x tanh x cot x ln sin x coth x ln sinh x sin 2 x sin 2x x 2 4 sinh 2 sinh 2x x 4 x 2 cos 2 x sin 2x x cosh 2 sinh 2x x 4 + x 2

18 Section 4: Integrals 18 f (x) f (x) dx f (x) f (x) dx 1 1 a 2 +x 2 a tan 1 x 1 1 a a 2 x 2 2a ln a+x a x (< x <a) 1 1 (a > ) x 2 a 2 2a ln x a x+a ( x > a>) 1 a 2 x 2 sin 1 x a 1 a 2 +x 2 ( a < x < a) 1 x2 a 2 ln ln x+ a 2 +x 2 a x+ x 2 a 2 a (a > ) (x>a>) a2 x 2 a 2 2 [ sin 1 ( ) x a a2 +x 2 a 2 2 ] x2 + x a 2 x 2 a a 2 a [ [ sinh 1 ( x a cosh 1 ( x a ) + x a 2 +x 2 a 2 ] ) + x ] x 2 a 2 a 2

19 Section 5: Useful trig results Useful trig results When calculating the Fourier coefficients a n and b n, for which n = 1, 2, 3,..., the following trig. results are useful. Each of these results, which are also true for n =, 1, 2, 3,..., can be deduced from the graph of sin x or that of cos x s in (x ) 1 sin n = 3 2 x c o s (x ) 1 cos n = ( 1) n 3 2 x 2 3 1

20 Section 5: Useful trig results 2 s in (x ) 1 c o s (x ) x x sin n, n even 2 = 1, n = 1, 5, 9,... 1, n = 3, 7, 11,... cos n 2 =, n odd 1, n =, 4, 8,... 1, n = 2, 6, 1,... Areas cancel when when integrating over whole periods sin nx dx = 2 cos nx dx = s in (x ) x

21 Section 6: Alternative notation Alternative notation For a waveform f(x) with period L = 2 k f(x) = a 2 + [a n cos nkx + b n sin nkx] n=1 The corresponding Fourier coefficients are STEP ONE a = 2 f(x) dx L L STEP TWO a n = 2 f(x) cos nkx dx L L STEP THREE b n = 2 f(x) sin nkx dx L L and integrations are over a single interval in x of L

22 Section 6: Alternative notation 22 For a waveform f(x) with period 2L = 2 k k = 2 2L = nx L and nkx = L f(x) = a 2 + n=1 [ a n cos nx L The corresponding Fourier coefficients are STEP ONE STEP TWO STEP THREE a = 1 f(x) dx L a n = 1 L b n = 1 L 2L 2L 2L f(x) cos nx L f(x) sin nx L, we have that + b n sin nx ] L and integrations are over a single interval in x of 2L dx dx

23 Section 6: Alternative notation 23 For a waveform f(t) with period T = 2 ω f(t) = a 2 + [a n cos nωt + b n sin nωt] n=1 The corresponding Fourier coefficients are STEP ONE a = 2 f(t) dt T T STEP TWO a n = 2 f(t) cos nωt dt T T STEP THREE b n = 2 f(t) sin nωt dt T T and integrations are over a single interval in t of T

24 Section 7: Tips on using solutions Tips on using solutions When looking at the THEORY, ANSWERS, INTEGRALS, TRIG or NOTATION pages, use the Back button (at the bottom of the page) to return to the exercises Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct Try to make less use of the full solutions as you work your way through the Tutorial

25 Solutions to exercises 25 Full worked solutions Exercise 1. f(x) = { 1, < x <, < x <, and has period 2 a) Sketch a graph of f(x) in the interval 2 < x < 2 1 f(x ) 2 2 x

26 Solutions to exercises 26 b) Fourier series representation of f(x) STEP ONE a = 1 f(x)dx = 1 f(x)dx + 1 f(x)dx = 1 1 dx + 1 dx = 1 dx = 1 [x] = 1 ( ( )) = 1 () i.e. a = 1.

27 Solutions to exercises 27 STEP TWO a n = 1 cos nx dx = f(x) 1 cos nx dx + f(x) 1 f(x) cos nx dx = 1 1 cos nx dx + 1 cos nx dx = 1 cos nx dx = 1 [ ] sin nx = 1 n n [sin nx] = 1 (sin sin( n)) n = 1 ( + sin n) n i.e. a n = 1 ( + ) =. n

28 Solutions to exercises 28 STEP THREE b n = 1 f(x) sin nx dx = 1 f(x) sin nx dx + 1 f(x) sin nx dx = 1 1 sin nx dx + 1 sin nx dx [ cos nx ] i.e. b n = 1 i.e. b n = sin nx dx = 1 = 1 n [cos nx] = 1 (cos cos( n)) n = 1 1 (1 cos n) = n n (1 ( 1)n ), see Trig { {, n even 1, n even 2 n, n odd n, since ( 1) n = 1, n odd

29 Solutions to exercises 29 We now have that f(x) = a 2 + [a n cos nx + b n sin nx] n=1 with the three steps giving {, n even a = 1, a n =, and b n =, n odd 2 n It may be helpful to construct a table of values of b n n ( 4 5 b n ( 3) 2 1 5) Substituting our results now gives the required series f(x) = [sin x + 13 sin 3x + 15 ] sin 5x +...

30 Solutions to exercises 3 c) Pick an appropriate value of x, to show that 4 = Comparing this series with f(x) = [sin x + 13 sin 3x + 15 ] sin 5x +..., we need to introduce a minus sign in front of the constants 1 3, 1 7,... So we need sin x = 1, sin 3x = 1, sin 5x = 1, sin 7x = 1, etc The first condition of sin x = 1 suggests trying x = 2. This choice gives sin sin sin sin i.e Looking at the graph of f(x), we also have that f( 2 ) =.

31 Solutions to exercises 31 Picking x = 2 thus gives [ = sin sin sin 5 2 ] sin i.e. = [ ] A little manipulation then gives a series representation of 4 2 [ ] +... = = 4. Return to Exercise 1

32 Solutions to exercises 32 Exercise 2. f(x) = {, < x < x, < x <, and has period 2 a) Sketch a graph of f(x) in the interval 3 < x < 3 f(x ) x

33 Solutions to exercises 33 b) Fourier series representation of f(x) STEP ONE a = 1 f(x)dx = 1 f(x)dx + 1 = 1 dx + 1 = 1 [ ] x 2 2 = 1 ( ) 2 2 i.e. a = 2. f(x)dx x dx

34 Solutions to exercises 34 STEP TWO a n = 1 cos nx dx = f(x) 1 cos nx dx + f(x) 1 f(x) cos nx dx = 1 cos nx dx + 1 x cos nx dx i.e. a n = 1 x cos nx dx = 1 {[ ] sin nx } sin nx x dx n n (using integration by parts) i.e. a n = 1 {( ) sin n 1 [ cos nx ] } n n n = 1 { ( ) + 1 } n 2 [cos nx] 1 1 = {cos n cos } = n2 n 2 {( 1)n 1} {, n even i.e. a n = 2, see Trig. n, n odd 2

35 Solutions to exercises 35 STEP THREE b n = 1 f(x) sin nx dx = 1 f(x) sin nx dx + 1 f(x) sin nx dx = 1 sin nx dx + 1 x sin nx dx i.e. b n = 1 x sin nx dx = 1 { [ ( cos nx )] ( cos nx ) } x dx n n (using integration by parts) = 1 { 1 n [x cos nx] + 1 } cos nx dx n = 1 { 1 n ( cos n ) + 1 [ ] } sin nx n n = 1 n ( 1)n + 1 ( ), see Trig n2 = 1 n ( 1)n

36 Solutions to exercises 36 { 1 n, n even i.e. b n = + 1 n, n odd We now have f(x) = a 2 + [a n cos nx + b n sin nx] n=1 { where a =, n even 2, a n = 2 n, n odd, b n = 2 Constructing a table of values gives { 1 n 1 n, n even, n odd n a n b n

37 Solutions to exercises 37 This table of coefficients gives f(x) = 1 2 ( 2 ) ( 2 ) cos x + cos 2x ( 2 ) cos 3x + cos 4x ( 2 ) cos 5x sin x 1 2 sin 2x + 1 sin 3x... 3 [cos x cos 3x + 15 ] 2 cos 5x +... i.e. f(x) = [sin x 12 sin 2x + 13 ] sin 3x... and we have found the required series!

38 Solutions to exercises 38 c) Pick an appropriate value of x, to show that (i) 4 = Comparing this series with [cos x cos 3x + 15 ] 2 cos 5x +... f(x) = [sin x 12 sin 2x + 13 ] sin 3x..., the required series of constants does not involve terms like 1 3, 1 2 5, 1 2 7,... 2 So we need to pick a value of x that sets the cos nx terms to zero. The Trig section shows that cos n 2 = when n is odd, and note also that cos nx terms in the Fourier series all have odd n i.e. cos x = cos 3x = cos 5x =... = when x = 2, i.e. cos 2 = cos 3 2 = cos 5 2 =... =

39 Solutions to exercises 39 Setting x = 2 in the series for f(x) gives ( ) f = [cos cos cos 52 ] [sin 2 12 sin sin sin sin 52 ]... = 4 2 [ ] sin }{{} ( 1) 1 sin 4 }{{ 2 } + 1 (1)... 5 = = The graph of f(x) shows that f ( ) 2 = 2, so that i.e. 2 4 = =

40 Solutions to exercises 4 Pick an appropriate value of x, to show that (ii) 2 8 = Compare this series with [cos x cos 3x + 15 ] 2 cos 5x +... f(x) = [sin x 12 sin 2x + 13 ] sin 3x.... This time, we want to use the coefficients of the cos nx terms, and the same choice of x needs to set the sin nx terms to zero Picking x = gives sin x = sin 2x = sin 3x = and cos x = cos 3x = cos 5x = 1 Note also that the graph of f(x) gives f(x) = when x =

41 Solutions to exercises 41 So, picking x = gives = 4 2 [cos cos cos cos +... ] + sin sin + sin i.e. = 4 2 [ ] We then find that 2 [ ] = 4 and = Return to Exercise 2

42 Solutions to exercises 42 Exercise 3. f(x) = { x, < x <, < x < 2, and has period 2 a) Sketch a graph of f(x) in the interval 2 < x < 2 f(x ) 2 2 x

43 Solutions to exercises 43 b) Fourier series representation of f(x) STEP ONE a = 1 2 f(x)dx = 1 = 1 [ x 2 f(x)dx + 1 xdx + 1 ] [ x = = 1 ( ) = ] 2 2 dx ( 2 f(x)dx ) i.e. a = 3 2.

44 Solutions to exercises 44 STEP TWO a n = 1 2 f(x) cos nx dx 2 = 1 x cos nx dx + 1 cos nx dx [ [ = 1 ] ] sin nx sin nx x n n dx + [ sin nx n }{{} using integration by parts [ ] = 1 1 n ( ) sin n sin n [ cos nx n 2 ] ] (sin n2 sin n) n

45 Solutions to exercises 45 i.e. a n = 1 [ 1 n ( ) ( cos n + n 2 cos n 2 ) ] + 1 ( ) n = = 1 n 2 (cos n 1), 1 ( ( 1) n n 2 1 ), see Trig i.e. a n = { 2 n 2, n odd, n even.

46 Solutions to exercises 46 STEP THREE b n = 1 2 f(x) sin nx dx = 1 x sin nx dx sin nx dx [ = 1 [ ( cos nx )] ( ) ] cos nx x dx + [ ] 2 cos nx n n n }{{} using integration by parts [ ( = 1 ) [ ] ] cos n sin nx + + n n 2 1 (cos 2n cos n) n [ = 1 ( 1) n ( ) ] sin n sin + n n 2 1 ( 1 ( 1) n ) n = 1 n ( 1)n + 1 n ( 1 ( 1) n )

47 Solutions to exercises 47 We now have i.e. b n = 1 n ( 1)n 1 n + 1 n ( 1)n i.e. b n = 1 n. f(x) = a 2 + [a n cos nx + b n sin nx] where a = 3 2, a n = n=1 { Constructing a table of values gives, n even 2 n 2, n odd, b n = 1 n n ( 4 5 a n ) ( ) b n

48 Solutions to exercises 48 This table of coefficients gives f(x) = 1 2 ( ) ( ) [ ] cos x + cos 2x cos 3x ( ) [ 1 sin x sin 2x sin 3x +... ] i.e. f(x) = [ ] cos x cos 3x cos 5x [ ] sin x sin 2x sin 3x +... and we have found the required series.

49 Solutions to exercises 49 c) Pick an appropriate value of x, to show that (i) 4 = Compare this series with f(x) = [cos x cos 3x + 15 ] 2 cos 5x +... [sin x + 12 sin 2x + 13 ] sin 3x +... Here, we want to set the cos nx terms to zero (since their coefficients 1 1 are 1, 3, 2 5,...). Since cos n 2 2 = when n is odd, we will try setting x = 2 in the series. Note also that f( 2 ) = 2 This gives 2 = [ cos cos cos ] [ sin sin sin sin sin ]

50 Solutions to exercises 5 and 2 = [ ] then [ (1) () ( 1) () (1) +...] 2 = 3 4 ( ) = = 4, as required. To show that (ii) 2 8 = , We want zero sin nx terms and to use the coefficients of cos nx

51 Solutions to exercises 51 Setting x = eliminates the sin nx terms from the series, and also gives cos x cos 3x cos 5x cos 7x +... = (i.e. the desired series). The graph of f(x) shows a discontinuity (a vertical jump ) at x = The Fourier series converges to a value that is half-way between the two values of f(x) around this discontinuity. That is the series will converge to 2 at x = i.e. 2 = [cos cos cos + 17 ] 2 cos +... [sin + 12 sin + 13 ] sin +... and 2 = [ ] [ ]

52 Solutions to exercises 52 Finally, this gives ( ) and = 2 = Return to Exercise 3

53 Solutions to exercises 53 Exercise 4. f(x) = x 2, over the interval < x < 2 and has period 2 a) Sketch a graph of f(x) in the interval < x < 4 f(x ) x

54 Solutions to exercises 54 b) Fourier series representation of f(x) STEP ONE a = 1 = 1 i.e. a =. 2 2 [ x 2 = 1 4 = 1 [ (2) 2 4 f(x) dx x 2 dx ] 2 ]

55 Solutions to exercises 55 STEP TWO a n = 1 = 1 i.e. a n =. 2 2 { [ = 1 x 2 f(x) cos nx dx x cos nx dx 2 ] 2 2 } sin nx 1 sin nx dx n n }{{} using integration by parts { ( = 1 sin n2 2 sin n ) } 1n 2 n n { } = 1 ( ) 1n 2, see Trig

56 Solutions to exercises 56 STEP THREE b n = 1 = 1 2 = f(x) sin nx dx = 1 2 ( x 2 ) sin nx dx x sin nx dx { [ ( )] 2 cos nx 2 ( ) } cos nx x dx n n }{{} using integration by parts { } = n ( 2 cos n2 + ) + 1 n, see Trig = 2 2n cos(n2) = 1 n cos(2n) i.e. b n = 1 n, since 2n is even (see Trig)

57 Solutions to exercises 57 We now have f(x) = a 2 + [a n cos nx + b n sin nx] n=1 where a =, a n =, b n = 1 n These Fourier coefficients give f(x) = 2 + ( 1n ) sin nx n=1 i.e. f(x) = 2 { sin x sin 2x sin 3x +... }.

58 Solutions to exercises 58 c) Pick an appropriate value of x, to show that Setting x = 2 gives f(x) = [ ]... 4 = and = 2 [ ] = 2 [ ] = 4 i.e = 4. Return to Exercise 4

59 Solutions to exercises 59 Exercise 5. f(x) = { x, < x <, < x < 2, and has period 2 a) Sketch a graph of f(x) in the interval 2 < x < 2 f(x ) 2 2 x

60 Solutions to exercises 6 b) Fourier series representation of f(x) STEP ONE a = 1 i.e. a = 2. 2 f(x) dx = 1 ( x) dx + 1 = 1 [ x 1 ] 2 x2 + = 1 ] [ dx

61 Solutions to exercises 61 STEP TWO a n = 1 2 f(x) cos nx dx 2 = 1 ( x) cos nx dx + 1 dx i.e. a n = 1 {[ ] sin nx } sin nx ( x) ( 1) dx + n n }{{} { ( ) + = 1 = 1 [ cos nx n n using integration by parts } ] sin nx n dx = 1 (cos n cos ) n2 i.e. a n = 1 n 2 (( 1)n 1), see Trig, see Trig

62 Solutions to exercises 62 STEP THREE b n = 1 = 1 i.e. b n = 1 n. 2 i.e. a n = = 1 { [ ( x) f(x) sin nx dx, n even 2 n 2 ( x) sin nx dx + 2 ( cos nx )] n, n odd dx = 1 { ( ( )) 1 } n n, see Trig ( cos nx ) } ( 1) dx + n

63 Solutions to exercises 63 In summary, a = 2 and a table of other Fourier cofficients is n a n = 2 n 2 (when n is odd) b n = 1 n f(x) = a 2 + [a n cos nx + b n sin nx] n=1 = cos x + 2 i.e. f(x) = cos 3x cos 5x sin x sin 2x sin 3x + 1 sin 4x [cos x cos 3x + 15 ] 2 cos 5x sin x sin 2x sin 3x + 1 sin 4x

64 Solutions to exercises 64 c) To show that 2 8 = , note that, as x, the series converges to the half-way value of 2, and then 2 = (cos cos cos +... ) + sin sin sin +... giving = ( ) = 2 ( ) = Return to Exercise 5

65 Solutions to exercises 65 Exercise 6. f(x) = x, over the interval < x < and has period 2 a) Sketch a graph of f(x) in the interval 3 < x < 3 f(x ) 3 2 x 2 3

66 Solutions to exercises 66 b) Fourier series representation of f(x) STEP ONE a = 1 f(x) dx = 1 x dx [ ] x 2 i.e. a =. = 1 2 = 1 ( )

67 Solutions to exercises 67 STEP TWO a n = 1 f(x) cos nx dx = 1 x cos nx dx { [ = 1 ] sin nx ( ) } sin nx x dx n n }{{} using integration by parts i.e. a n = 1 { 1 n ( sin n ( ) sin( n)) 1 } sin nx dx n = 1 { 1 n ( ) 1 } n, since sin n = and sin nx dx =, i.e. a n =. 2

68 Solutions to exercises 68 STEP THREE b n = 1 f(x) sin nx dx = 1 x sin nx dx { [ x = 1 ] cos nx ( ) } cos nx dx n n = 1 { 1 n [x cos nx] + 1 } cos nx dx n = 1 { 1n ( cos n ( ) cos( n)) + 1n } = (cos n + cos n) n = 1 (2 cos n) n i.e. b n = 2 n ( 1)n.

69 Solutions to exercises 69 We thus have f(x) = a 2 + [ ] a n cos nx + b n sin nx with n=1 a =, a n =, b n = 2 n ( 1)n and n Therefore b n f(x) = b 1 sin x + b 2 sin 2x + b 3 sin 3x +... i.e. f(x) = 2 [sin x 12 sin 2x + 13 ] sin 3x... and we have found the required Fourier series.

70 Solutions to exercises 7 c) Pick an appropriate value of x, to show that 4 = Setting x = 2 gives f(x) = 2 and [ = 2 sin sin sin sin sin 52 ]... This gives 2 2 i.e. 4 = 2 [ ( 1) + 15 (1) + 17 ] ( 1) +... = 2 [ ] +... = Return to Exercise 6

71 Solutions to exercises 71 Exercise 7. f(x) = x 2, over the interval < x < and has period 2 a) Sketch a graph of f(x) in the interval 3 < x < 3 2 f(x ) x

72 Solutions to exercises 72 b) Fourier series representation of f(x) STEP ONE a = 1 f(x)dx = 1 x 2 dx [ ] x 3 = 1 3 = 1 ( 3 3 ( ) 2 3 ( 3 3 )) = 1 3 i.e. a = 22 3.

73 Solutions to exercises 73 STEP TWO a n = 1 f(x) cos nx dx = 1 x 2 cos nx dx { [ = 1 ] 2 sin nx ( ) } sin nx x 2x dx n n }{{} using integration by parts { = 1 1 ( 2 sin n 2 sin( n) ) 2 } x sin nx dx n n { = 1 1 n ( ) 2 } x sin nx dx, see Trig n = 2 n x sin nx dx

74 Solutions to exercises 74 i.e. a n = 2 n = 2 n = 2 n = 2 n = 2 n { [ ( )] cos nx ( ) } cos nx x dx n n }{{} { using integration by parts again 1 n [x cos nx] + 1 cos nx dx n { 1 ( ) } cos n ( ) cos( n) + 1n n { 1 ) (( 1) } n + ( 1) n n { } 2 n ( 1)n }

75 Solutions to exercises 75 i.e. a n = 2 n { 2n ( 1)n } = +4 n 2 ( 1)n = 4 n 2 ( 1)n i.e. a n = { 4 n 2, n even 4 n 2, n odd.

76 Solutions to exercises 76 STEP THREE b n = 1 f(x) sin nx dx = 1 x 2 sin nx dx { [ = 1 ( )] cos nx ( ) } cos nx x 2 2x dx n n }{{} using integration by parts { = 1 1 [ x 2 cos nx ] n + 2 } x cos nx dx n { = 1 1 ( 2 cos n 2 cos( n) ) + 2 } x cos nx dx n n { = 1 1 ( 2 cos n 2 cos(n) ) + 2 } x cos nx dx n }{{} n = = 2 x cos nx dx n

77 Solutions to exercises 77 { [ i.e. b n = 2 x n i.e. b n =. ] } sin nx sin nx dx n n }{{} using integration by parts { = 2 1 n n ( sin n ( ) sin( n)) 1 sin nx dx n { = 2 1 n n ( + ) 1 } sin nx dx n = 2 n 2 sin nx dx }

78 Solutions to exercises 78 where f(x) = a 2 + [a n cos nx + b n sin nx] n=1 { 4 a = 22 3, a n = n, n even 4 2 n, n odd, b n = 2 i.e. f(x) = 1 2 n a n 4(1) 4 ( ) ( ) ( ) ( ) [ 4 cos x 12 2 cos 2x cos 3x 14 ] 2 cos 4x... + [ ] i.e. f(x) = [cos x 12 2 cos 2x cos 3x 14 ] 2 cos 4x +....

79 Solutions to exercises 79 c) To show that 2 6 = , 2 { 1, n even use the fact that cos n = 1, n odd i.e. cos x cos 2x cos 3x cos 4x +... with x = gives cos cos cos cos i.e. ( 1) (1) ( 1) (1) +... i.e = 1 ( ) }{{} (the desired series)

80 Solutions to exercises 8 The graph of f(x) gives that f() = 2 and the series converges to this value. Setting x = in the Fourier series thus gives 2 = (cos cos cos 3 14 ) 2 cos = ( ) = ( ) = 4 ( ) i.e. 2 6 = Return to Exercise 7

81 EE 12 spring Handout #23 Lecture 11 The Fourier transform definition examples the Fourier transform of a unit step the Fourier transform of a periodic signal properties the inverse Fourier transform 11 1

82 The Fourier transform we ll be interested in signals defined for all t the Fourier transform of a signal f is the function F (ω) = f(t)e jωt dt F is a function of a real variable ω; thefunction value F (ω) is (in general) a complex number F (ω) = f(t)cosωt dt j f(t)sinωt dt F(ω) is called the amplitude spectrum of f; F (ω) is the phase spectrum of f notation: F = F(f) means F is the Fourier transform of f; asfor Laplace transforms we usually use uppercase letters for the transforms (e.g., x(t) and X(ω), h(t) and H(ω), etc.) The Fourier transform 11 2

83 Fourier transform and Laplace transform Laplace transform of f F (s) = f(t)e st dt Fourier transform of f G(ω) = f(t)e jωt dt very similar definitions, with two differences: Laplace transform integral is over t< ; Fouriertransform integral is over <t< Laplace transform: s can be any complex number in the region of convergence (ROC); Fourier transform: jω lies on the imaginary axis The Fourier transform 11 3

84 therefore, if f(t) =for t<, if the imaginary axis lies in the ROC of L(f), then G(ω) =F (jω), i.e., thefourier transform is the Laplace transform evaluated on the imaginary axis if the imaginary axis is not in the ROC of L(f), thenthe Fourier transform doesn t exist, but the Laplace transform does (at least, for all s in the ROC) if f(t) for t<, thenthe Fourier and Laplace transforms can be very different The Fourier transform 11 4

85 examples one-sided decaying exponential f(t) = { t< e t t Laplace transform: F (s) =1/(s +1)with ROC {s Rs> 1} Fourier transform is one-sided growing exponential f(t)e jωt dt = 1 jω +1 = F (jω) f(t) = { t< e t t Laplace transform: F (s) =1/(s 1) with ROC {s Rs>1} f doesn t have a Fourier transform The Fourier transform 11 5

86 Examples double-sided exponential: f(t) =e a t (with a>) F (ω) = e a t e jωt dt = = = e at e jωt dt + 1 a jω + 1 a + jω 2a a 2 + ω 2 e at e jωt dt 1 2/a f(t) F (ω) 1/a t 1/a a ω a The Fourier transform 11 6

87 { 1 T t T rectangular pulse: f(t) = t >T F (ω) = T T e jωt dt = 1 jω ( e jωt e jωt) = 2sinωT ω 1 2T f(t) F (ω) T unit impulse: f(t) =δ(t) t T /T /T ω F (ω) = δ(t)e jωt dt =1 The Fourier transform 11 7

88 shifted rectangular pulse: f(t) = { 1 1 T t 1+T t<1 T or t>1+t F (ω) = 1+T 1 T e jωt dt = 1 jω (e jω(1+t ) jω(1 T e )) 2T = e jω jω = 2sinωT ω ( e jωt e jωt) e jω (phase assuming T =1) F (ω) F (ω) /T /T ω /T /T ω The Fourier transform 11 8

89 Step functions and constant signals by allowing impulses in F(f) we can define the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f(t) = { t< 1 t? the Laplace transform is 1/s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1/jω in fact, the integral f(t)e jωt dt = e jωt dt = cos ωt dt j sin ωt dt is not defined The Fourier transform 11 9

90 however, we can interpret f as the limit for α of a one-sided decaying exponential { e αt t g α (t) = t<, (α >), which has Fourier transform G α (ω) = 1 a + jω = a jω a 2 + ω 2 = a a 2 + ω 2 jω a 2 + ω 2 as α, a jω a 2 δ(ω), + ω2 a 2 + ω 2 1 jω let s therefore define the Fourier transform of the unit step as F (ω) = e jωt dt = δ(ω)+ 1 jω The Fourier transform 11 1

91 negative time unit step f(t) = { 1 t t> F (ω) = e jωt dt = e jωt dt = δ(ω) 1 jω constant signals: f(t) =1 f is the sum of a unit step and a negative time unit step: F (ω) = e jωt dt = e jωt dt + e jωt dt =2δ(ω) The Fourier transform 11 11

92 Fourier transform of periodic signals similarly, by allowing impulses in F(f), wecandefinethefourier transform of a periodic signal sinusoidal signals: Fourier transform of f(t) =cosω t F (ω) = 1 2 ( e jω t + e jω t ) e jωt dt = 1 e j(ω ω)t dt = δ(ω ω )+δ(ω + ω ) e j(ω+ω )t dt F (ω) ω ω ω The Fourier transform 11 12

93 Fourier transform of f(t) =sinω t F (ω) = 1 2j = 1 2j ( e jω t e jω t ) e jωt dt e j(ω ω )t dt + 1 2j = jδ(ω ω )+jδ(ω + ω ) e j(ω +ω)t dt F (ω) j ω ω ω j The Fourier transform 11 13

94 periodic signal f(t) with fundamental frequency ω express f as Fourier series f(t) = k= a k e jkω t F (ω) = a k k= e j(kω ω)t dt =2 F (ω) 2a k= a k δ(ω kω ) 2a 1 2a 1 2a 3 2a 2 2a 2 2a 3 3ω 2ω ω ω 2ω 3ω ω The Fourier transform 11 14

95 Properties of the Fourier transform linearity af(t)+bg(t) af (ω)+bg(ω) time scaling f(at) 1 a F (ω a ) time shift f(t T ) e jωt F (ω) differentiation df (t) dt d k f(t) dt k jωf(ω) (jω) k F (ω) integration t f(τ)dτ F (ω) jω + F()δ(ω) multiplication with t t k f(t) j k d k F (ω) convolution f(τ)g(t τ) dτ F dω k (ω)g(ω) multiplication f(t)g(t) 1 2 F ( ω)g(ω ω) d ω The Fourier transform 11 15

96 Examples sign function: f(t) = { 1 t 1 t< write f as f(t) = 1+2g(t), whereg is a unit step at t =,andapply linearity F (ω) = 2δ(ω)+2δ(ω)+ 2 jω = 2 jω sinusoidal signal: f(t) =cos(ω t + φ) write f as f(t) =cos(ω (t + φ/ω )) and apply time shift property: F (ω) =e jωφ/ω (δ(ω ω )+δ(ω + ω )) The Fourier transform 11 16

97 pulsed cosine: f(t) = { t > 1 cos t 1 t f(t) write f as a product f(t) =g(t)cost where g is a rectangular pulse of width 2 (see page 12-7) t F(cos t) =δ(ω 1) + δ(ω +1), F(g(t)) = 2sin1ω ω The Fourier transform 11 17

98 now apply multiplication property F (jω) = = sin 1 ω ω sin(1(ω 1)) ω 1 (δ(ω ω 1) + δ(ω ω +1))d ω + sin(1(ω +1)) ω F (ω) ω The Fourier transform 11 18

99 The inverse Fourier transform if F (ω) is the Fourier transform of f(t), i.e., F (ω) = f(t)e jωt dt then let s check 1 2 ω= f(t) = 1 2 F (ω)e jωt dω = 1 2 = 1 2 = ω= τ= = f(t) F (ω)e jωt dω ( τ= ( f(τ) f(τ)δ(τ t)dτ f(τ)e jωτ ) e jωt dω ω= ) e jω(τ t) dω dτ The Fourier transform 11 19