Question #217902
Using the 6 ordinate scheme analyze harmonically the data to third harmonics
X 0 π/3 2π/3 π 4π/3 5π/3 2π
Y 10 12 15. 20. 17. 11. 10
1
Expert's answer
2021-07-19T16:23:22-0400

k=17yk=95\sum_{k=1}^7y_k=95

k=17ykcos(xk)=4.5\sum_{k=1}^7y_kcos(x_k)=-4.5

k=17yksin(xk)=0.866\sum_{k=1}^7y_ksin(x_k)=-0.866

k=17ykcoc(2xk)=12.5\sum_{k=1}^7y_kcoc(2x_k)=12.5

k=17yksin(2xk)=2.598\sum_{k=1}^7y_ksin(2x_k)=2.598

we can get the values of aia_i and bib_i as follows:

a0=17×95=13.57a_0={1\over 7}\times95=13.57

a1=27×(4.5)=1.29a_1={2\over 7}\times(-4.5)=-1.29

a2=27×12.5=3.57a_2={2\over 7}\times 12.5=3.57

b1=27×(0.866)=0.25b_1={2\over 7}\times(-0.866)=-0.25

b2=27×2.598=0.74b_2={2\over 7}\times 2.598=0.74

Substituting those values into Fourier series

f(x)=a0+n=1(ancos(nx)+bnsin(nx))f(x)=a_0+\sum_{n=1}^\infin(a_ncos(nx)+b_nsin(nx)) gives

f(x)=a0+a1cos(x)+a2cos(2x)+b1sin(x)+b2sin(2x)f(x)=a_0+a_1cos(x)+a_2cos(2x)+b_1sin(x)+b_2sin(2x)

y=13.571.29cos(x)+3.75cos(2x)0.25sin(x)+0.74sin(2x)\therefore y=13.57-1.29cos(x)+3.75cos(2x)-0.25sin(x)+0.74sin(2x)



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