This must be a very basic question but I am a finance student just learning some basics about Fourier Transformation to apply to time series analysis. I did a fourier transform on a function in time domain to get the following function in frequency domain:
$Y_1[\omega] = \frac{1}{1-\phi_1 e^{-jw}}$
$Y_2[\omega] = \frac{1}{1-(\phi_1 + \phi_2)e^{-jw} +\phi_1\phi_2e^{-2jw}}$
How do I find the spectrum of this function for given $\phi_1$ and $\phi_2$ coefficients and in the discretization interval $w = [-\pi:.1*\pi: \pi]$? Then, how do I find the 'magnitude' of spectrum and 'phase' of spectrum?
Best Answer
Let \begin{align*} Z(\omega) & = 1-\phi e^{-jw} = 1-\phi (\cos(\omega) - j \sin(\omega))\\ & = 1-\phi \cos(\omega) + j \phi \sin(\omega)\\ \end{align*} Hence, we get that \begin{align*} |Z(\omega)| = & \sqrt{(1-\phi (\cos(\omega))^2 + (\phi \sin(\omega))^2} = \sqrt{1 + \phi^2 - 2 \phi \cos(\omega)}\\ \angle{Z(\omega)} = & \tan^{-1}(\frac{\phi \sin(\omega)}{1-\phi \cos(\omega)}) \end{align*}
$$Y(\omega) = \frac{1}{Z(\omega)}$$ Hence, we get that \begin{align*} |Y(\omega)| = & \frac{1}{\sqrt{1 + \phi^2 - 2 \phi \cos(\omega)}}\\ \angle{Y(\omega)} = & -\tan^{-1}(\frac{\phi \sin(\omega)}{1-\phi \cos(\omega)}) \end{align*}
Finding the magnitude and phase for $|Y_2(\omega)|$ is a trivial process once you have done the previous one.
$Y_2(\omega) = \frac{1}{1-\phi_1 e^{-jw}} \times \frac{1}{1-\phi_2 e^{-jw}}$.
Hence, \begin{align*} |Y_2(\omega)| = & \frac{1}{\sqrt{1 + \phi_1^2 - 2 \phi_1 \cos(\omega)}} \times \frac{1}{\sqrt{1 + \phi_2^2 - 2 \phi_2 \cos(\omega)}}\\ \angle{Y_2(\omega)} = & -\tan^{-1}(\frac{\phi_1 \sin(\omega)}{1-\phi_1\cos(\omega)}) -\tan^{-1}(\frac{\phi_2 \sin(\omega)}{1-\phi_2\cos(\omega)}) \end{align*}