Troubleshooting Fourier Series Analysis for Periodic Functions | Help Needed

[cybernoodles]

Hi All,

I'm having a lot of trouble determining the a and b coefficients of the Fourier Series. I am given the following periodic function below and am asked to determine the coefficients. I am usually off by a factor of 2 or 1/2, if that means anything. It could be my bounds or my value for f(t). I was able to determine the function has half-wave odd symmetry and thus able to conclude that all of the a's are zero for the first four coefficients, and b=0, but I failed at finding the odd b coefficients. Here is the integral and equation I used, which are incorrect I guess:

(32/T)*(Integral_UpperB=(3T/8)_LowerB=(T/8)) of sin(nwt)dt

and this yielded the expression: (32/(2*pi*n))* (-cos(6*pi*n/8) + cos(2*pi*n/8))

Forgive me, Latex isn't working at the moment. Any help is appreciated. Been struggling with this since 10AM. Thanks!

 

[Valerie Park]

</code>For the a coefficients, we have:a_0 = 1/T * Integral_UpperB=T_LowerB=0 of (f(t)) dt a_n = 1/T * Integral_UpperB=T_LowerB=0 of (f(t) * cos(nwt)) dt For the b coefficients, we have:b_n = 1/T * Integral_UpperB=T_LowerB=0 of (f(t) * sin(nwt)) dt The period of your function is T = 8. Therefore, for the a_0 coefficient, you have:a_0 = 1/8 * Integral_UpperB=8_LowerB=0 of (f(t)) dt For the a_n coefficients, you have:a_n = 1/8 * Integral_UpperB=8_LowerB=0 of (f(t) * cos(nwt)) dt For the b_n coefficients, you have:b_n = 1/8 * Integral_UpperB=8_LowerB=0 of (f(t) * sin(nwt)) dt You can then evaluate the integrals for each of the a_n and b_n coefficients. The a_0 coefficient is simply the average value of the function f(t).Hope this helps!