I created a signal with lets say W1$\omega_1$ that is fixed at 1 bin in the fftFFT: $$A_1\cos(W_1t)$$$$A_1\cos(\omega_1t)$$
I was asked to find the closest $W_2$$\omega_2$ to $W_1$ which$\omega_1$ so that the signals $A_1\cos(W_1t)$,$A_1\cos(\omega_1t)$ and $A_2\cos(W_2t)$$A_2\cos(\omega_2t)$ are separable.
Now iI only know the ratio A1/A2$A_1/A_2$ in dbdB and W1the $\omega_1$ that I chose.
$$s(t)=A_1\cos(W_1t)+A_2\cos(W_2t)$$$$s(t)=A_1\cos(\omega_1t)+A_2\cos(\omega_2t)$$
Now my questions are:
- What is the optimum(minimal)minimum window length of the windows?
- Number of points for the DftDFT?
- What is the relationship (If existingif any exists) is the relation between the ratio $\frac{A_1}{A_2}$ ratio toand the DftDFT points?