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Fourier series and transforms
- 3. 3 3 11.1 Fourier Series 11.1 Fourier Series Periodic Functions A function is said to be periodic of period if for all x Example 1 cos 2 cos 2 cos 2 sin sin 2 cos Hence cos is periodic of period 2 Example 2 sin 4 2 sin 4 2 sin 4 2 sin 4 cos 2 cos 4 sin 2 sin 4 = Hence sin is periodic of period , observe that 2 is also a period of sin 4
- 4. 4 4 11.1 Fourier Series Useful Identities • sin sin cos cos sin • cos cos cos sin sin Notes • Any function can be considered periodic with period zero, this period is trivial and is not considered as a period. • If is a period of , then is a period for any integer . Proof: want: we know that 2 3 2 • If is a period then is not necessarily a period. Fundamental Period The most interesting period for a periodic function is the smallest positive period , this period is called the Fundamental Period. The fundamental period of • sin is 2 • sin 3 is
- 5. 5 5 11.1 Fourier Series Period of Multiple Functions If and are periodic of period then so is Proof Denote by want is periodic of period If is periodic of period then the graph of repeats itself every units 2 Therefore if we know the curve of a periodic function on , , then we can draw the entire graph. Exercise If is periodic of period then , 0 0,2 0,4 0,6 0,8 1 1,2 ‐2 ‐1 0 1 2 3 4 5
- 6. 6 6 11.1 Fourier Series Fourier Series Our purpose is to approximate periodic functions by sine and cosine. we define Fourier series of the periodic function f(x) by: cos sin Fourier coefficients , can be obtained by Euler formulas. Derivation: Suppose cos sin 5 * cos sin 5 2 1 2 * cos cos cos sin 5 cos 1 cos * .... 0
- 7. 7 7 11.1 Fourier Series In general Example 3 Find the Fourier series: 1 0 1 0 Solution: 1 2 0 ‐1,5 ‐1 ‐0,5 0 0,5 1 1,5 ‐4 ‐3 ‐2 ‐1 0 1 2 3 4 • When the phrase "Fourier series" is mentioned then we implicitly understand that is periodic. • If the period is not given , then we implicitly understand that its 2 1 2 1 cos 1 sin Fourier coefficients of f(x), given by the Euler formulas
- 8. 8 8 11.1 Fourier Series cos 1 1 cos 0 1 sin 1 sin sin 1 cos cos 1 1 cos cos 1 1 2 2 cos 2 1 1 0 4 Now Fourier series cos sin 4 2 1 sin 2 1
- 9. 9 9 11.1 Fourier Series Example 4 Evaluate: 2 3 cos 4 sin Solution: Denote function by * We need to find * Remember that 1 2 2 * We need to find to be able to find * However isn't in Fourier form because of " " , so we need to simplify using identity 1 2 1 cos so 2 3 cos 2 2 cos 2 0 * And now substitute to find ...
- 10. 10 10 11.1 Fourier Series Notes By a trigonometric polynomial we mean a finite part of the Fourier series. For instance: • 1 sin 3 cos 5 • 2 sin sin 2 sin 3 • 2 sin sin 2 (Trigonometric but not Fourier form) Notes • . • sin sin 0 • sin cos 0 • cos cos 0
- 11. 11 11 11.2 Functions of Any Period p = 2L 11.2 Functions of Any Period p = 2L Example 1 Find the Fourier series of , 1 1 Solution: In this example, p = 2 (period = 2 ) In this case when p = 2 L Thus in our example L = 1 1 2 1 3 cos sin 1 2 1 cos 1 sin , 2 In general
- 12. 12 12 11.2 Functions of Any Period p = 2L 1 1 sin 0 1 1 cos 1 1 cos 2 cos 2 1 1 4 1 1 3 4 1 cos cos 2 sin 2 cos Integration by parts 0
- 13. 13 13 11.6a Parseval's Identity 11.6a Parseval's Identity Consider Fourier series and expand it cos sin cos sin cos 2 sin 2 … Square it cos sin 2 cos sin 2 cos sin 2 cos cos sin 2 cos sin Integrate cos sin 2 0 2| | | | | | 1 2| | | | | | 1 Parseval's Identity Standard form General form
- 14. 14 14 11.6a Parseval's Identity Applications Example 1 From Chapter 11.1 , Example 1 ∑ sin 2 1 1 0 1 0 * L.H.S of Parseval's 2 0 0 4 2 1 16 1 2 1 *R.H.S of Parseval's 1 1 2 *Therefore 1 2 1 8 Example 2 From Chapter 11.2 Example 1 1 3 4 1 cos , Apply Parseval's 2 1 3 16 2 5 16 2 5 2 9 8 45 1 π 90
- 15. 15 15 11.6a Parseval's Identity Example 3 Find 1 Now series is given but not unlike Example 2 Solution: We need such that 1 1 We attempt with since when integrating by parts , we get in the denominator Taking = 0 0 1 sin 1 cos | 1 1 2 1 cos 1 cos Integration by parts 0
- 16. 16 16 11.6a Parseval's Identity Now apply Parseval's 4 1 4 1 2 3 1 6 Exercise Find 1 1 Example 4 Evaluate 2 sin 5 cos 3 cos 10 Solution: Let 2 sin 3 cos 3 cos 10 Want According to Parseval's 2 2 1 1 3 We can't find any sum using this method , like ∑
- 17. 17 17 11.7 Dirichlet's Theorem 11.7 Dirichlet's Theorem If is a nice function , then lim lim 2 Suppose that is periodic of period 2 and that is piecewise continuous , that and both exist. Example 1 Suppose 2 1 sin ; Plug 0 , 0 = 0 Plug 2 2 1 sin 2 2 1 sin 2 2 , 2 1 1 2 1 sin 2 1 2 2 2 1 2 1 2 1 2 1 4 Plug 0 , lim lim 2 2 0
- 18. 18 18 11.4 Complex Fourier Series 11.4 Complex Fourier Series cos sin is called Real Fourier series Note cos sin cos sin 2 sin 2 cos cos 2 sin 2 1 2 The Complex Fourier Series of is defined to be
- 19. 19 19 11.4 Complex Fourier Series Remark 1 2 1 2 1 cos 1 sin 1 2 , 0 1 2 , 0 , 0 Example 1 Write the complex Fourier transform of 2 sin cos 10 Solution: 2 2 2 1 1 1 2 1 2 1 , 1 2 , 1 , 1 2 Example 2 Find the real Fourier series of 5 Solution: 5 sin cos sin cos 2 2
- 20. 20 20 11.4 Complex Fourier Series Example 3 Find the complex Fourier series of , Solution: 1 2 1 2 1 2 1 2 1 1 1 1 1 , 0 For 0 1 2 1 2 0 Therefore complex Fourier series is 0 1 ,
- 21. 21 21 11.4 Complex Fourier Series Note By a complex trigonometric polynomial , we mean a finite part of For example • Trig. 1 5 • Not Trig. 1 • Trig. 11 sin 11 ! Note that a complex Fourier series of a complex trigonometric polynomial is the same function. Exercise Show that 0 , 2 ,
- 22. 22 22 11.6b Parseval's Identity 11.6b Parseval's Identity Note |2 3 | 4 9 13 |3 | |2 3 | 0 3 9 | | 0 1 1 Example 1 1 lets apply Parseval's | | 1 , 1 , 2 1 2 1 1 2 3 1 6 | | 1 2 | | | | Parseval's Identity for complex Fourier series
- 23. 23 23 11.6b Parseval's Identity Example 2 Evaluate 1 3 1 cos 4 Solution: Let 1 3 1 cos 4 Want | | 2 | | 1 , 1 , 3 1 2 , 1 2 , 1 | | 2 1 1 9 1 4 1 4 2
- 24. 24 24 11.9 Fourier Transform 11.9 Fourier Transform Fourier Transform Example 1 Find the Fourier transform of 1 2 2 0 Then apply Parseval's identity and see what it gives Solution: 1 √2 1 √2 1 √2 0 0,5 1 1,5 ‐4 ‐2 0 2 4 1 √2 | | Let be defined on ∞, ∞ We define its Fourier transform by Parseval's Identity
- 25. 25 25 11.9 Fourier Transform Note here if we are asked about 0 , we take the limit 1 √2 cos 2 sin 2 cos 2 sin 2 1 √2 2 sin 2 Let's apply Parseval's 2 sin 2 4 2 sin 2 2 1) Let's play with 2 sin 2 2 2) Let 2 sin sin 2 3) Let's find sin sin 2
- 26. 26 26 11.9 Fourier Transform sin 2 sin cos 2 sin 2 2 Let 2 sin 2 4) sin Note is continuous regardless of lim ∞ 0 lim lim lim 0 sin 1 t Integration by parts 2 sin cos
- 27. 27 27 11.9 Fourier Transform Fourier Sine and Cosine Transforms Where did these equations come from? Recall Fourier transform 1 √2 If is even 1 √2 cos sin 1 √2 cos √2 sin 2 √2 cos 2 cos 2 cos 2 sin If is defined on 0, ∞ , we define its Fourier Cosine Transform by And Fourier Sine Transform
- 28. 28 28 11.9 Fourier Transform Note Practically when is even. when is odd. Note that when is defined on 0, ∞ , we can consider it even or odd. Example 2 Find and for , 0 1 0 , Solution: 2 cos 2 cos 2 sin cos 2 sin cos 1 Using limits 0 2 1 1 2 1 2 2
- 29. 29 29 11.9 Fourier Transform Inverse Fourier Transform Useful Rules 2 0 1 √2 2 cos 2 sin Fourier Inverse Transform Fourier Inverse Cosine Transform Fourier Inverse Sine Transform
- 30. 30 30 11.9 Fourier Transform Applications Example 3 Find ; Solution: Using the rules 2 0 2 2 2 ...
- 31. 31 31 11.9 Fourier Transform Example 4 You are given that 2 1 2 2 1 sin 2 sin 1 1 sin 1 2 0 1 0 ? ? ? The formula of the Fourier Inverse Sine Transform sin is true when is continuous at . Moreover , recall that is computed for odd function . If we extend to be odd , we get Not continuous at 0 when taking , so we use Dirichlet's Theorem.