Fourier Series for Periodic Functions Lecture #8 5CT3,4,6,7 BME 333 Biomedical Signals and Systems 2 - J.Schesser Fourier Series for Periodic Functions • Up to now we have solved the problem of approximating a function f(t) by fa(t) within an interval T. • However, if f(t) is periodic with period T, i.e., f(t)=f(t+T), then the approximation is true for all t. • And if we represent a periodic function in terms of an infinite Fourier series, such that the frequencies are all integral multiples of the frequency 1/T, where k=1 corresponds to the fundamental frequency of the function and the remainder are its harmonics. Tjkt 2 1 2 T aftedtk T () T 2 j22kt j kt j 2 kt j 2 kt TT T T ft() a00 [aakk e * e ] a k e a2Re[ a k e ] kkk11 jkt2 j k ftak( ) a0 C cos( kkk ), where 2aa Ce and 00 C k1 T BME 333 Biomedical Signals and Systems 3 - J.Schesser Another Form for the Fourier Series 2 kt ft() C0 Ckk cos( ) k1 T 22kt kt AA0 kkcos( ) B sin( ) k1 TT where Akk CBCACcos k and k - k sin k and 00 BME 333 Biomedical Signals and Systems 4 - J.Schesser Fourier Series Theorem • Any periodic function f (t) with period T which is integrable ( f ( t ) dt ) can be represented by an infinite Fourier Series • If [f (t)]2 is also integrable, then the series converges to the value of f (t) at every point where f(t) is continuous and to the average value at any discontinuity. BME 333 Biomedical Signals and Systems 5 - J.Schesser Properties of Fourier Series • Symmetries – If f (t) is even, f (t)=f (-t), then the Fourier Series contains only cosine terms – If f (t) is odd, f (t)=-f (-t), then the Fourier Series contains only sine terms – If f (t) has half-wave symmetry, f (t) = -f (t+T/2), then the Fourier Series will only have odd harmonics – If f (t) has half-wave symmetry and is even, even quarter-wave, then the Fourier Series will only have odd harmonics and cosine terms – If f (t) has half-wave symmetry and is odd, odd quarter- wave,then the Fourier Series will only have odd harmonics and sine terms BME 333 Biomedical Signals and Systems 6 - J.Schesser Properties of FS Continued • Superposition holds, if f(t) and g(t) have coefficients fk and gk, respectively, then Af(t)+Bg(t)=>Afk+Bgk jkt2 T • Time Shifting: ft() aek k j22ktjkt2211 jkt j kt TT TT ft() t1 aekk e ae e kk jkt2 1 T ()aaekkdelayed () original • Differentiation and Integration jk2 ()aa () kkderivativeT original T ()aa () kkintegraljk2 original BME 333 Biomedical Signals and Systems 7 - J.Schesser FS Coefficients Calculation Example f (t) 0, 2 t 0 E f (t) E, 0 t 1 f (t) 0, 1 t 2 -4 -3 -2 -1 1234 f (t) f (t 4) 1 01jkt222 jkt 2 jkt a[0 e444 dt Ee dt 0 e dt ] Since 2aCkk , then k 4 20 1 k jkt2 sin 1 E 1 E Ektk2 e 4 ft()4 cos( ) |0 4 jk2 42k1 k 4 4 4 4 jk jk jk 2 EEeee jk 44 4 [1][e 2 ] 1.5 jk22 k j k 1 sin( ) jk E 4 ek4 , 0 0.5 4 k 4 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 EE1 a dt -0.5 0 44 8 0 BME 333 Biomedical Signals and Systems - J.Schesser Example Continued 2 1 term 1.5 1 0.5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.5 2 1.5 10 terms 1 0.5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.5 BME 333 Biomedical Signals and Systems 9 - J.Schesser Frequency Spectrum of the Pulse Function • In the preceding example, the coefficients for each of the cosine terms was proportional to sin(k 4) (k 4) • We call the function Sa(x) sin x x the Sampling Function 1.2 1 • If we plot these coefficients along the 0.8 0.6 frequency axis we have 0.4 0.2 the frequency spectrum of 0 -30 -20 -10 0 10 20 30 -0.2 f(t) -0.4 BME 333 Biomedical Signals and Systems 10 - J.Schesser Frequency Spectrum Note that the periodic signal in the time domain exhibits a 1.5 discrete spectrum (i.e., in the frequency domain) 1 0.5 0 -15 -10 -5 0 5 10 15 -0.5 Frequency (k/4) BME 333 Biomedical Signals and Systems 11 - J.Schesser Another Example V This signal is odd, H/W symmetrical, and mean=0; this means no cosine terms, odd T harmonics and mean=0. 2V 2kt jkt22 T jkt f (t) 2 cos( 90 ) 0 1 TT2 k T aVedtVedt[]T k odd k T 0 2 4V 2kt jkt22 jktT sin( ) 0 V 11TT2 k T [ eeT kodd ||0 T jk222 jk 2 TT 1.5 VV2 1 [1 cos k ] for k odd, 0.5 jk jk 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 -0.5 0 for k even. -1 2V -1.5 90 for k odd. k BME 333 Biomedical Signals and Systems 12 - J.Schesser Fourier Series of an Impulse Train or Sampling Function xt() ( t nT ) n T jkt2 1 2 T axtedtk T () T 2 T jkt2 1 2 T T ()te dt T 2 1 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 for all k T 1 jkt2 xt() e T T k BME 333 Biomedical Signals and Systems 13 - J.Schesser Fourier Series for Discrete Periodic Functions j2kn j2kn j2kn T T T x[n] a0 ak e ak e ak e k 1 k 1 k T 2 j2kn 1 T ak x[n]e Since x[n] is discrete, this T nT 2 becomes a summation Notice the similarities to the previous example. BME 333 Biomedical Signals and Systems 14 - J.Schesser Homework • Problem (1) – Compute the Fourier Series for the periodic functions a) f(t) = 1 for 0<t<π, f(t)=0 for π<t<2π b) f(t) = t for 0<t<3 • Problem (2) – Compute the Fourier series of the following Periodic Functions: • f(t)= t, 2nπ < t < (2n+1)π for n ≥ 0 = 0, (2n+1)π < t < (2n+2)π for n ≥ 0 -t/π • f(t)= e , 2nπ < t < (2n+2)π for n ≥ 0 Use Matlab to plot f(t)usingak for maximum number of components, N=5,10, 100, and1000. Show your code. • Problem (3) – Problems: 4.1/3 Find the Fourier series of the following waveforms (choose tc=To/4): 1.5 1 0.5 0 012345 -0.5 -To -To/2 -tc tc To/2 To -1 T -t T + t -(To+ tc) - (To-tc) o c o c -1.5 BME 333 Biomedical Signals and Systems 15 - J.Schesser Homework • Problem (4) – Deduce the Fourier series for the functions shown (hint: deduce the second one using superposition): -π+a -a π+a 2π-a -π -2π/3 -π/3 4π/3 5π/3 -π a π-a π 2π π/3 2π/3 π 2π • 5CT.7.1 BME 333 Biomedical Signals and Systems 16 - J.Schesser.