ENGIN 211, Engineering Math. Fourier Series and Transform

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1 ENGIN 11, Engineering Math Fourier Series and ransform 1

2 Periodic Functions and Harmonics f(t) Period: a a+ t Frequency: f = 1 Angular velocity (or angular frequency): ω = ππ = π Such a periodic function can be expressed in terms of a series of cosine and sine functions, known as the Fourier series: f t = a + a n cos nnn + b n sin nnn n=1 = a + a n cos nnn n=1 + b n sin nnn where cos nnn and sin nnn are the harmonics. 1 st (or fundamental) Harmonic: cos ωω and sin ωω, (n = 1) nd Harmonic: cos ωω and sin ωω, (n = ) 3 rd Harmonic: cos 3ωω and sin 3ωω, (n = 3).

3 Significance of the Harmonics n = 1 n = 1, n = 1,,3 n = 1,,3,4 Fourier series of a square wave containing various orders of harmonics 3

4 Orthogonality f t and g t are orthogonal to each other over the interval a t b if b f t a g t dd = In fact, the harmonics cos nnn and sin nnn (n =,1,, ) form an infinite collection of periodic functions that are mutually orthogonal on the interval / t / because / / / / cos mmm sin mmm cos nnn dd =, for m n sin nnn dd =, for m n / / cos mmm sin nnn dd = 4

5 Fourier Coefficients f t = a + a n cos nnn + b n sin nnn n=1 a n = / / f t cos nnn dd, for n =,1,, b n = f t sin nnn dd, for n = 1,,3, Please note a has been included in the first set of integrals. Example: Determine the Fourier series for the function 1 + t 1 < t < f t = < t < 1 f t + = f t 1 1 f t t 5

6 Example (Cont d) f t = a + a n cos nnn n=1 he coefficients: a n = For n = 1 1 f t + b n sin nnn cos nnn dd = = a + a n cos nnn + b n sin nnn 1 n=1 1 + t cos nnn dd, n =,1,, a = 1 + t 1 For n, we can use integral by parts dd = t + t 1 = 1 a n = 1 + t 1 cos nnn dd = nn =, n even nn 1 cos nn = 4 nn, n odd 1 + t sin nnn 1 sin nnn dd 1 6

7 Example (Cont d) Similarly 1 b n = f t 1 sin nnn dd = 1 + t 1 sin nnn dd =, n = 1,,3, nn he first few terms of the Fourier series: f t = π cos ππ cos 3ππ + 1 cos 5ππ + 5 π sin ππ + 1 sin ππ sin 3ππ + 1 sin 4ππ + 4 7

8 Odd and Even Functions Odd function: f t = f t, symmetrical about the origin. Sine function is odd. Fourier series of an odd function contains only sine terms (a n =, n =,1,, ), because Even function: f t = f t, symmetrical about the y-axis Cosine function is even. Fourier series of an even function contains only cosine terms (b n =, n = 1,,3, ), / f t cos nnn dd / = if f t is odd. / f t sin nnn dd = if f t is even. / 8

9 Example (Even Function) he function is symmetrical about the y-axis 3π/ y π/ 4 π/ 3π/ x f x = a + a n cos nn a = 1 π n=1 π f x dd π π/ = π = 4dd = 4 π π f x dd a n = 1 π π f x cos nn dd π = 8 nn sin nn = = π f x cos nn dd = π π, n = k, (n =,4,6, ) 8, n = 4k 3, (n = 1,5,9, ) nn π/ 4 cos nn dd 8, n = 4k 1, (n = 3,7,11, ) nn k = 1,,3, 9

10 Example (Odd Function) y A shift of π/ in x-axis and shift of in y-axis in the previous example change it into an odd function b n = 1 π = 4 π π π f x sin nn dd π π sin nn dd = 4 nn π π x = π f x sin nn dd π, n even 1 1 n = 8, n odd nn g x = b n sin nn n=1 Connection between the two examples f x π = 8 nn cos x π 1 3 cos 3 x π cos 5 x π 1 7 cos 7 x π + g x = 8 nn sin x sin 3x sin 5x 1 sin 7x + 7 1

11 Complex Fourier Series Euler s Identity e jj = cos θ + j sin θ and e jj = cos θ j sin θ hus, Now the Fourier series cos θ = ejj +e jj and sin θ = ejj e jj j f t = a + a n cos nnn + b n sin nnn n=1 = a + a n + j b n e jjjj + a n j b n n=1 where the complex coefficients = a + a n ejjjj + e jjjj + b n j ejjjj e jjjj n=1 ejjjj = c n n= e jjjj c n = a n j b n = 1 f t cos nnn j sin nnn dd = 1 f t e jjjj dd Obviously, c n = c n = 1 f t ejjjj dd, and c = 1 f t dd = a since b = 11

12 Example 1 (Complex Fourier) f(t) 1, / < t < a/ f t = 1, a/ < t < a/,, a/ < t < / and f t + = f(t) / a/ a/ / t f(t) = n= c n e jjjj, ω = π c n = 1 / / f(t)ejjjj dd = 1 a/ ejjjj dd a/ = nnn sin nnn = a sin nnn/ nnn/ = a = 1 a/ jjjj ejjjj a/ sin nnn/ nnn/ = a sinc nnn = 1 jjjj ejjjj/ e jjjj/, for n sinc x = sin x x sin x is undefined at x =, but lim sinc x = lim x x x = 1, If we define sin x, for x sinc x = x 1, for x = and notice c = 1 / / f(t)dd = 1 a/ a/ 1dd = a, then c n = a sinc nnn, for all n, < n < 1

13 Example (Complex Fourier) g(t) 1 1, < t < a g t =, a < t <, and f t + = f(t) a t g(t) = n= g n e jjjj, ω = π g n = 1 f(t)ejjjj dd = ejjjj/ nnn = 1 a ejjjj dd nnn sin = a sin nnn/ nnn/ = 1 a jjjj ejjjj e jjjj/ = a = 1 jjjj ejjjj 1 = ejjjj/ jjjj sinc nnn ejjjj/ e jjjj/ e jjπa/, Complex coefficients he same result can be obtained from Example 1 by shifting the time origin by half the pulse width, g t = f t + a = a n= sinc nnn e jjj t+a = a n= sinc nnn e jjπa/ e jjjj 13

14 Complex Spectra he Fourier coefficients of square-pulse wave all have sinc nnn like?, what does it look In example, we have c n = a sinc nnn e jjjj/ = c n e jφ n In both examples, their amplitudes c n = a sinc nnn for n, and c c n describes the spectrum of f t in the frequency domain = a c c 1 c c 1 c 14

15 Power in ime and Frequency Domains Power content of a periodic function f(t) = c n n= e jjjj is defined as the mean square value in time domain, and is related to the spectrum in the frequency domain Because, P = 1 / / f t dd = c n n= P = 1 / / n= = c n c n = n= c n e jjjj f(t)dd = c n n= / / e jjjj f(t)dd c n c n = c n c n = c n n= n= n= 15

16 Continuous Spectrum In the example of the periodic square pulse function, f t = n= c n e jjjj = n= c n e jjjjj/, and c n = a sinc nnn the spacing between two neighboring harmonics is δx = ππ. If we increase the period, this spacing δx decreases, and when, the function has a single pulse and no longer periodic, the spacing δx, the spectrum becomes continuous., 16

17 Fourier ransform We start from a periodic function f t = n= c n e jjjjj/, where c n = 1 f τ e jjjjj/ dd We can plug in c n into the series and use δδ = π f t = 1 f τ e jjjjj/ dd n= e jjjjj/ = 1 π δδ f τ e jjjjj dd n= e jjjjj Let, then δδ, thus δδ = dd, and nnn = ω, f t = 1 π n= f τ e jjj dd e jjj dd = 1 π 1 π f τ e jjj dd e jjj dd So if we introduce F ω = 1 π f t e jjt dt - Fourier transform hen f t = 1 π F ω e jjj dd - inverse Fourier ransform As, the discrete harmonic values nω = ππ/ become a continuous value ω, and the discrete spectrum c n = c n e jφ n becomes the continuous spectrum F ω = F ω e jj ω. 17

18 1 f(t) Example f t = 1, < t < a, otherwise F ω a t F ω = 1 π f t e jjj dd = 1 j πω 1 e jjj = ae jjj/ π sin ωω ωω = 1 a π e jjj dd = e jjj/ j πω ejjj/ e jjj/ = ae jjj/ π sinc ωω = 1 a j πω e jjj = e jjj/ j πω ω j sin ωω his is a continuous spectrum! 18

19 Properties of Fourier ransform Linearity F α 1 f 1 t + α f t = α 1 F 1 ω + α F ω ime scaling F f kk = 1 k F ω k ime shifting F f t t = e jjt F ω Frequency shifting F e jω t f t = F ω ω Symmetry F F t = f ω Proof: We start with the inverse Fourier transform and then employ two variable substitutions f t = 1 π F ω e jjj dd = 1 π Let t ω, then f ω = 1 π F u F u e jjj dd e jjj dd = 1 π F t e jjj dd = F F t Example: F 1 =? (hard to solve without using the symmetry) Since F δ(t) = 1 π, then using symmetry: F 1 π = δ ω = δ ω, thus F 1 = πδ ω. 19

20 Cosine and Sine ransforms F ω = 1 π f t e jjj dd If f t is even function, f t sin ωω dd = 1 π f t cos ωω + j sin ωω dd =, then the cosine transform: F c ω = π f t cos ωω dd If f t is odd function, f t cos ωω dd =, then the sine transform: F s ω = π f t sin ωω dd For those functions that are defined only for t, and the extension into t < can make them either even or odd, where cosine and sine transforms can then be used, respectively.

21 1 f(t) Example 1, < t < a Example: f t =, a t < and undefined in < t < 1 a t Cosine transform: F c ω = π f t cos ωω dd = π a cos ωω dd a a t = π sin ωω ω = asinc ωω π 1 Sine transform: a 1 a t F s ω = π = π f t sin ωω dd = π a sin ωω dd = π ωω sin ω = ωa sin ωω π ωω = ωa π sinc 1 cos ωω ω ωω 1

22 Summary Key points: Periodic functions Fourier series and coefficients Significance of harmonics Sine and cosine for odd and even functions Complex Fourier series and discrete spectrum Fourier transform and continuous spectrum Properties of Fourier transform

CHAPTER 4 FOURIER SERIES S A B A R I N A I S M A I L

CHAPTER 4 FOURIER SERIES S A B A R I N A I S M A I L CHAPTER 4 FOURIER SERIES 1 S A B A R I N A I S M A I L Outline Introduction of the Fourier series. The properties of the Fourier series. Symmetry consideration Application of the Fourier series to circuit