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I take that as an yes :D

Hmm I would probably use the voltage divider Could I not do that with the transfor function?

we have mostly been working with phasors, which is those tools i am using. I know that the following. v(t)=Re{Vejωt}=VAcos(ωt+Φ) and V=VAejΦ=VA∠Φ somehow I can't seem to get further then this. I am considering to use the transfer function, and use that to solve for X

Homework Statement Find X [/B] Homework Equations V1(t)=8cos(12000t) V0(t)=Xcos(12000t+θ) The Attempt at a Solution [/B] I've tried quiet a lot actually, but all in vain. Seems like I am either misunderstanding something, or i am just too tired to think at the moment. Either way, I am...

so when taking the interval into account i get the result ##a_k=\frac{-sin(k\pi)}{k}## which is wroooong

yeah sorry! Actually that is what confuses me. if ##\sink\pi=0## then there is only ##\frac{cos(x)sin(kx)}{k}## left? Does that mean that this is my ##a_k##??

I need it for a Fourier series where the inteval is from [0, ##\pi## ]. f(x)=cos(x) so it is an even function which means i need to calculate ##a_0## and ##a_k##. It is ##a_k## who i am trying to calculate right now by parts, but i am stuck at it.

k is just an integer that's about it. I'll remember that next time.

Homework Statement Hello! I am having some trouble solving this integral by parts. I hope someone can help me. ##\int \cos(x)cos(kx) dx## It is need for a Fourier seriesHomework Equations I am using this definition: ##\int f(x)g(x) dx = f(x)G(x)-\int f'(x)G(x) dx## since its an even...

Perfekt. Thank you :)

I came to this result after integrating by parts several times. I hope anyone can confirm this. ##\left ( \frac{12}{\left ( \pi n\right )^{3}}-\frac{2}{\pi n} \right ) \left ( -1 \right )^{n}##

I don't think i need to evaluate the integrals, I just want to know exactly how it works. Trying to understand the mechanics in how to solve such a problem.

I will try giving it a shot, but you are more then correct about me being puzzled right now. I will report back if i gives me a problem

I understand what you are saying, but isn't what I'm already trying to do? By finding an expression for ##b_n## or ##c_n## in your definition. The only difference i guess, is that it is ## \pi## and not ##2 \pi##??

I am really lost here, hope someone can help me!