# Fourier Series in Circuit analysis

Discussion in 'Homework Help' started by Hanck, Dec 27, 2008.

1. ### Hanck Thread Starter Member

Nov 25, 2008
18
0
The question given Vs = 1/2 + (2/pi)Ʃ(1/2)cos(n*pi*t - 90°)

i cannot figure out why when nth harmonic
Vs = (2/[n*pi])cos(n*pi*t - 90°) = (2/[n*pi]) polar -90°

The lecturer told me just simply take the θ in cosine term [cos (wt-θ)] as the angle in polar coordination but i still cannot understand why...

If it is simply like that then why
given Vs = 1 + Ʃ [2*(-1)^n / (sqrt (1 + n^2))]* (cos nt - n sin nt)

for nth harmonic
Vs = [2*(-1)^n / (sqrt (1 + n^2))]* polar arctan n

why arctan n?
I appreciate any help, thanks.

2. ### DrNick Active Member

Dec 13, 2006
110
2
convert the second formula you gave into it's complex form. That will make things clear.

(complex form being f(t) = sum(Ck*exp(-j*w*n*t))...)

Also, If you do compute the complex form you will find that the solution for your n'th harmonic is incorrect.

Vs = [2*(-1)^n / (sqrt (1 + n^2))]* polar arctan n , should be

Vs = [(-1)^n]* polar arctan n

because,

when changed to complex form you get

sum(bn/2*(1-j*n)*exp(-jwnt))

and 1-j*n = sqrt(1+n^2)*exp(j*atan(n))

when you perform the multiplication

f(t) = sum(bn/(2*sqrt(1+n^2)*exp(-jwnt+atan(n)))

in your case bn = 2*(-1)^n / (sqrt (1 + n^2))

SO, f(t) = sum((-1)^n*exp(-jwnt+atan(n))). Note this is summed from -inf to +inf. this means the nth harmonic should be (as stated above) (-1)^n polar atan(n).

NOTE: please check my math...it seems right to me, but there may be some lame error. long story short, this explains the atan(n)...

Last edited: Dec 27, 2008