Hi, i need help with the following question.
Represent by a Fourier transform the following function;
f(x)=sin(ax)
Thanks!
Right I'm currently revising this stuff too so ill give it a go...
Is it supposed to be expressed in real form or complex?
Heres the complex form: $\displaystyle \sum_{n=-\infty}^{\infty} c_n e^{-inx}$
And heres the real form: $\displaystyle \frac{a_0}{2} + \sum_{k=1}^{\infty}(a_k cos(kx) + b_k sin(kx))$
Heres what it all means...
$\displaystyle c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} dx$
$\displaystyle a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos(kx) dx$
$\displaystyle b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin(kx) dx$
BUT... This is for when the function is $\displaystyle 2 \pi$ periodic... But yours is $\displaystyle \frac{2 \pi}{\alpha}$ periodic (i.e. $\displaystyle f(x) = f(x + \frac{2 \pi}{\alpha})$. So you must change the integral limit accordingly.
If the function is odd the a_k = 0 and if its is even then b_k = 0. So as the sine function is odd... a_k = 0!
Is this enough to start you off?