Two similar problems, but once I find out how to do the first one, I can figure out how to do the second. My signals book tells me the answers to the following "Dn"s are:
First problem: Dn = (1/∏) ∫ sin(t) * e^(-j2nt) dt = 2/(∏ (1-4n^2) )
if x(t) = rectified sin(t) wave, period T = ∏.
Second problem: Dn = ∫(t/2∏) * e^(-jn*ωnaught*t) = 1/(2∏n).
Formula for Dn = (1/period) ∫x(t)*e^(-(j*n*ωnaught*t))
See, when I pop the first one into Wolfram, I get
e^(-2*j*n*t) * (cos(t) + 2*j*n*sin(t))
/
∏(4n^2 - 1).
Since we integrate over 0 to ∏ in the first one, I understand Euler' identity is used to get the e^(-2*j*n*t) term to -1, and the cos(∏) goes away, so i'm left with 2jn / ∏(4n^2 - 1). How did they simplify that to get what I put up top as the answer?
First post so I'm a noob.
Thanks
First problem: Dn = (1/∏) ∫ sin(t) * e^(-j2nt) dt = 2/(∏ (1-4n^2) )
if x(t) = rectified sin(t) wave, period T = ∏.
Second problem: Dn = ∫(t/2∏) * e^(-jn*ωnaught*t) = 1/(2∏n).
Formula for Dn = (1/period) ∫x(t)*e^(-(j*n*ωnaught*t))
See, when I pop the first one into Wolfram, I get
e^(-2*j*n*t) * (cos(t) + 2*j*n*sin(t))
/
∏(4n^2 - 1).
Since we integrate over 0 to ∏ in the first one, I understand Euler' identity is used to get the e^(-2*j*n*t) term to -1, and the cos(∏) goes away, so i'm left with 2jn / ∏(4n^2 - 1). How did they simplify that to get what I put up top as the answer?
First post so I'm a noob.
Thanks
