I am stuck with deriving the Fourier transform of an expression mentioned below:
$X(t) = \sin(\omega_1 t)\frac{a - e^2 \sin^2( \omega_2 t)}{b - e^2 \sin^2( \omega_2 t)}$ where $a\neq b$ and $a/e^2, b/e^2 > 1$. The denominator is causing the problem during the transformation.
Looking at the expression it is quite evident that the Fourier transform will have components corresponding to $(\omega \pm \omega_1)$, $(\omega \pm \omega_2 \pm \omega_1)$ and$(\omega \pm \omega_2 \mp \omega_1)$. But I am trying to analyze the amplitude corresponding to each of the component and its variation with $e$.
Can anyone help me with that?
Thanks in advance.
Let's define $A=a/e^2$ and $B=b/e^2$, so $A>1$ and $B>1$. The first thing we want to do is to transform the sines to exponential functions with $\sin(\omega_2 t)= \frac{\exp(i\omega_2 t)-\exp(-i\omega_2 t)}{2i}$. You'll find $$ X(t)=\sin(\omega_1 t) \frac{\exp(4\omega_2 i t)+(4A-2)\exp(2\omega_2 i t)+1}{\exp(4\omega_2 i t)+(4B-2)\exp(2\omega_2 i t)+1}. $$ Let us focus on the fraction, which has the problematic denominator. We see that the denominator is quadratic in $x=\exp(2\omega_2 i t)$. If we solve for $x$ we find the solutions $x_1=1-2B+2\sqrt{B^2-B}$ and $x_2=1-2B-2\sqrt{B^2-B}=\frac{1}{x_1}$. After some calculations, you can verify that the second part can then be split in partial fractions as $$ \frac{x^2+(4A-2)x+1}{x^2+(4B-2)x+1}=1+\frac{A-B}{\sqrt{B^2-B}}\left(\frac{1}{1-\frac{x}{x_2}}-\frac{1}{1-\frac{x}{x_1}}\right). $$ From $B>1$, you can derive that $x_2<-1$ and thus $|x_1|<1$ and $|x_2|>1$. Since $|x|=1$, we can then use the formula for the geometric series to find $$ \frac{1}{1-\frac{x}{x_1}}=\frac{-\frac{x_1}{x}}{1-\frac{x_1}{x}}=-\sum_{k=1}^{\infty} \left(\frac{x_1}{x}\right)^k $$ and $$ \frac{1}{1-\frac{x}{x_2}}=\sum_{k=0}^{\infty} \left(\frac{x}{x_2}\right)^k=\sum_{k=0}^{\infty} \left(x_1 x\right)^k. $$ So \begin{align} \frac{1}{1-\frac{x}{x_2}}-\frac{1}{1-\frac{x}{x_1}}&=1+\sum_{k=1}^{\infty}x_1^k \left(\exp(2 k \omega_2 it)+\exp(-2k\omega_2 it)\right)\\ &=1+2\sum_{k=1}^{\infty}x_1^k \cos(2 k \omega_2 t). \end{align} Then we can plug everything in the first equation to find \begin{align} X(t)&= \sin(\omega_1 t) \frac{x^2+(4A-2)x+1}{x^2+(4B-2)x+1}\\ &=\left(1+\frac{A-B}{\sqrt{B^2-B}}\right)\sin(\omega_1 t)+2\frac{A-B}{\sqrt{B^2-B}}\sum_{k=1}^{\infty}x_1^k \cos(2 k \omega_2 t)\sin(\omega_1 t)\\ &=\left(1+\frac{A-B}{\sqrt{B^2-B}}\right)\sin(\omega_1 t)+\frac{A-B}{\sqrt{B^2-B}}\sum_{k=1}^{\infty}x_1^k\left(\sin((\omega_1+2 k \omega_2)t)+\sin((\omega_1-2 k \omega_2)t)\right)\\ &=\sin(\omega_1 t)+\frac{A-B}{\sqrt{B^2-B}}\sum_{k=-\infty}^{\infty}\left(1-2B+2\text{sgn}(k)\sqrt{B^2-B}\right)^k \sin((\omega_1+2 k \omega_2)t), \end{align} where the last line uses simpson's formula. If you want you can put this again into exponential form, but you can directly read the fourier coëfficients from this expression, depending on your definition of fourier series.