� � � � � � � � � � � � 3 FOURIER SERIES 4 3 Fourier Series Compute the Fourier series coefficients 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 = 0 (the mean) A1 =4 sin(π/3) B1 =4 cos(π/3) A3 =1, and all other terms are zero, due to orthogonality. t, t < T/2 2. f(t)= (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 1 π 1 2π = cos(nt)tdt + cos(nt)(t − π)dt π 0 π π 1 � 2π 2π = cos(nt)tdt − cos(nt)dt π 0 π � ��2π1 cos(nt) t sin(nt) � = + � − 0 π n2 n 0 =0 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 Bn = sin(nt)f(t)dt π 0 1 � π 1 2π = sin(nt)tdt + sin(nt)(t − π)dt π 0 π π 1 2π 2π = sin(nt)tdt − sin(nt)dt π 0 π Now the second integral is −2/n for n odd, and zero otherwise. Let’s call it q(n). Then continuing we see � ��2π1 sin(nt) t cos(nt)Bn = 2 − �� − q(n)π n n 0 = −2/n − q(n). Hence Bn = −2/n for even n, and zero otherwise. Try it out by making a plot!MIT OpenCourseWarehttp://ocw.mit.edu 2.017J Design of Electromechanical Robotic Systems Fall 2009 For information about citing these materials or our Terms of Use, visit:
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