MIT2_017JF09_p03

MIT2_017JF09_p03 - 3 FOURIER SERIES 3 4 Fourier Series...

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Unformatted text preview: 3 FOURIER SERIES 3 4 Fourier Series Compute the Fourier series co efficients A0 , An , and Bn for the following signals on the interval t = [0, 2π ]: 1. f (t) = 4 sin(t + π /3) + cos(3t) First, write this in a fully expanded form: y (t) = 4 sin(t) cos(π/3) + 4 cos(t) sin(π/3) + cos(3t). Then it is obvious that A0 A1 B1 A3 = = = = 0 (the mean) 4 sin(π/3) 4 cos(π/3) 1, and all other terms are zero, due to orthogonality. � t, t < T /2 (biased sawtooth) t − T /2, t ≥ T /2 A0 = π /2, the mean value of the function. Let’s next do the An ’s: � 1 2π An = cos(nt)f (t)dt π �0 � 1π 1 2π = cos(nt)tdt + cos(nt)(t − π )dt π �0 ππ � 2π 1 2π = cos(nt)tdt − cos(nt)dt π � 0 π ��2π 1 cos(nt) t sin(nt) � � � −0 = + � π n2 n 0 =0 2. f (t) = This makes sense intuitively because the cosines are symmetric functions around zero (even), whereas f (t) is not. The signal’s information is carried in the sine terms: � 1 2π sin(nt)f (t)dt Bn = π �0 � 1π 1 2π = sin(nt)tdt + sin(nt)(t − π )dt π �0 ππ 2π � 2π 1 = sin(nt)tdt − sin(nt)dt π0 π Now the second integral is −2/n for n o dd, and zero otherwise. Let’s call it q (n). Then continuing we see � Bn 1 sin(nt) t cos(nt) = − π n2 n = −2/n − q (n). ��2π � � � � − q (n) 0 Hence Bn = −2/n for even n, and zero otherwise. Try it out by making a plot! MIT OpenCourseWare http://ocw.mit.edu 2.017J Design of Electromechanical Robotic Systems Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . ...
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MIT2_017JF09_p03 - 3 FOURIER SERIES 3 4 Fourier Series...

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