The derivative of periodagram

Question

The periodagram is defined as $I(f)= \left |\frac{1}{N}\sum_{n=0}^{N-1} x[n]\exp(-j2\pi fn) \right |^2$. If we represent it based on sinusoidal functions we have $I(f)= [\frac{1}{N}(\sum_{n=0}^{N-1} x[n]\cos(2\pi fn))^2+(\sum_{n=0}^{N-1} x[n]\sin(2\pi fn))^2 ]$ Taking the derivative of the above with respect to $f$ yields $$\frac{\partial I(f)}{\partial f}= \frac{1}{N}[2(\sum_{n=0}^{N-1} ((x[n]\cos(2\pi fn))(-2\pi n\sin(2\pi fn))))+2(\sum_{n=0}^{N-1} ((x[n]\sin(2\pi fn))(2\pi n\cos(2\pi fn)))) ]$$

which is zero for all $f$. I think, I have made a mistake because periodagram is not a constant function for all $f$.

Answer 1

Your mistake is that: $$\frac{\partial }{ \partial f} \left(\sum_{n=0}^{N-1} \phi_n(f) \right)^2 \neq 2\left(\sum_{n=0}^{N-1}\frac{\partial \phi_n(f)}{ \partial f} \phi_n(f) \right)$$ but: $$\frac{\partial }{ \partial f} \left(\sum_{n=0}^{N-1} \phi_n(f) \right)^2 = 2\left(\sum_{n=0}^{N-1}\frac{\partial \phi_n(f)}{ \partial f}\right) \left( \sum_{n=0}^{N-1}\phi_n(f) \right)$$