Engineering Mathematics II

Transcription

1 PSUT Engineering Mathematics II Fourier Series and Transforms Dr. Mohammad Sababheh 4/14/2009

2 11.1 Fourier Series 2 Fourier Series and Transforms Contents 11.1 Fourier Series... 3 Periodic Functions... 3 Fundamental Period... 4 Period of Multiple Functions... 5 Fourier Series Functions of Any Period p = 2L a Parseval's Identity Applications Dirichlet's Theorem Complex Fourier Series b Parseval's Identity Fourier Transform Fourier Transform Fourier Sine and Cosine Transforms Inverse Fourier Transform Applications

3 11.1 Fourier Series Fourier Series Periodic Functions A function is said to be periodic of period if for all x Example 1 cos 2 cos 2 cos 2sinsin2 cos Hence cos is periodic of period 2 Example 2 sin4 2 sin4 2 sin4 2 sin 4 cos 2 cos 4 sin 2 sin4 = Hence sin is periodic of period, observe that 2 is also a period of sin 4 3

4 11.1 Fourier Series 4 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 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 4

5 11.1 Fourier Series 5 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 1,2 1 0,8 0,6 0,4 0, 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, 5

6 11.1 Fourier Series 6 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 * 1 2 * cos sin 5 2 cos cos cos sin 5 cos 1 cos *

7 11.1 Fourier Series 7 In general Fourier coefficients of f(x), given by the Euler formulas cos sin Example 3 Find the Fourier series: 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 Solution: 1,5 1 0, , ,

8 11.1 Fourier Series 8 1 cos 0 1 sin cos 1 1 sin sin 1 cos cos 1 1 cos cos cos Now Fourier series 0 cos sin sin2 1 8

9 11.1 Fourier Series 9 Example 4 Evaluate: 23cos4sin Solution: Denote function by * We need to find * Remember that * 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 1 cos 2 so 2 3 cos 2 2 cos 2 * And now substitute to find

10 11.1 Fourier Series 10 Notes By a trigonometric polynomial we mean a finite part of the Fourier series. For instance: 1sin3cos5 2sinsin2sin3 2 sin sin 2 (Trigonometric but not Fourier form) Notes. 0 sin sin sin cos 0 0 cos cos 10

11 11.2 Functions of Any Period p = 2L Functions of Any Period p = 2L In general cos sin cos sin, 2 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 =

12 11.2 Functions of Any Period p = 2L sin cos 1 1 cos cos Integration by parts cos sin cos cos 12

13 11.6a Parseval's Identity a 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 Integrate 2 cos cos sin 2 cos sin cos sin 2 0 Parseval's Identity Standard form 2 General form

14 11.6a Parseval's Identity 14 Applications Example 1 From Chapter 11.1, Example sin * L.H.S of Parseval's *R.H.S of Parseval's *Therefore Example 2 From Chapter 11.2 Example cos, Apply Parseval's π

15 11.6a Parseval's Identity 15 Example 3 Find Now series is given but not unlike Example 2 Solution: We need such that We attempt with since when integrating by parts, we get in the denominator Taking = 0 0 Integration by parts 1 sin 1 cos cos cos 15

16 11.6a Parseval's Identity 16 Now apply Parseval's We can't find any sum using this method, like Exercise Find 1 1 Example 4 Evaluate 2 sin 5 cos 3 cos 10 Solution: Let 2sin 3 cos 3 cos 10 Want According to Parseval's

17 11.7 Dirichlet's Theorem 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 21 Plug 0, 0 = 0 sin ; Plug 21 sin 2 2, 21 sin sin Plug 0, lim lim

18 11.4 Complex Fourier Series Complex Fourier Series cos sin is called Real Fourier series The Complex Fourier Series of is defined to be 1 2 Note cossin cossin 2sin 2cos cos 2 sin 2 18

19 11.4 Complex Fourier Series 19 Remark cos 1 1 2, 0 sin 1 2, 0, 0 Example 1 Write the complex Fourier transform of 2 sin cos 10 Solution: , 1 2, 1, 1 2 Example 2 Find the real Fourier series of 5 Solution: 5sincossincos2 2 19

20 11.4 Complex Fourier Series 20 Example 3 Find the complex Fourier series of, Solution: , 0 For Therefore complex Fourier series is 0 1, 20

21 11.4 Complex Fourier Series 21 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, 21

22 11.6b Parseval's Identity b Parseval's Identity Parseval's Identity for complex Fourier series 1 2 Note Example 1 1 lets apply Parseval's 1,,

23 11.6b Parseval's Identity 23 Example 2 Evaluate cos4 Solution: Let cos4 Want 2 1, 1, 3 1 2, 1 2,

24 11.9 Fourier Transform Fourier Transform Fourier Transform Let be defined on, We define its Fourier transform by 1 2 Parseval's Identity Example 1 Find the Fourier transform of Then apply Parseval's identity and see what it gives Solution: ,5 1 0,

25 11.9 Fourier Transform 25 1 cos 2 sin 2 cos2 sin sin 2 2 Note here if we are asked about 0, we take the limit Let's apply Parseval's 2 2 sin 2 2 sin 2 4 1) Let's play with 2 2 sin 2 2) Let 2 sin sin 2 3) Let's find sin sin 2 25

26 11.9 Fourier Transform 26 Integration by parts sin 2 sin 2 2 sin cos 2 sin 1 t 2sincos Let 2 sin 2 4) sin Note is continuous regardless of lim 0 lim lim lim 0 26

27 11.9 Fourier Transform 27 Fourier Sine and Cosine Transforms If is defined on 0,, we define its Fourier Cosine Transform by 2 cos And Fourier Sine Transform 2 sin Where did these equations come from? Recall Fourier transform If is even cos sin 1 2 cos 2 sin 2 2 cos 2 cos 27

28 11.9 Fourier Transform 28 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 Solution:, 0 1 0, 2 cos 2 cos 2 sin cos 2 sin cos 1 Using limits

29 11.9 Fourier Transform 29 Inverse Fourier Transform Fourier Inverse Transform 1 2 Fourier Inverse Cosine Transform 2 cos Fourier Inverse Sine Transform 2 sin Useful Rules

30 11.9 Fourier Transform 30 Applications Example 3 Find ; Solution: Using the rules

31 11.9 Fourier Transform 31 Example 4 You are given that sin 1 2 sin 1 1 sin ??? The formula of the Fourier Inverse Sine Transform sin continuous at. Moreover, recall that is computed for odd function. is true when is If we extend to be odd, we get Not continuous at 0 when taking, so we use Dirichlet's Theorem. 31