Recent content by Benabruzzo

My final answer is then: for M odd \begin{equation} P_x=\frac{1}{M}\sum_{n=0}^{M-1}{|Acos(\frac{2\pi n}{M}+\phi)|}^2=|A|^2cos^2(\phi) \end{equation} for M even \begin{equation} P_x=\frac{1}{M}\sum_{n=0}^{M-1}{|Acos(\frac{2\pi n}{M}+\phi)|}^2=|A|^2cos^2(\phi)+\frac{|A|^2}{2}...

Done similarly, \begin{equation} (3) \frac{|A|^2}{M}\sum_{n=0}^{M-1}|-2sin(\alpha)sin(\phi)cos(\alpha)cos(\phi)| \end{equation} \begin{equation} \frac{|A|^2}{M}sin(\phi)cos(\phi)\sum_{n=0}^{M-1}|-2sin(\alpha)cos(\alpha)| \end{equation} using the same polar defintiions of sine and...

I'm not 100% sure that I have done this part correctly, my result is a two-case answer, where the period, M, is either and even or odd integer, \begin{equation} (2) \frac{|A|^2}{M}\sum_{n=0}^{M-1}|sin^2(\alpha)| \end{equation} The next term is somewhat simplified, since I already know what...

\begin{equation} (1) \frac{|A|^2}{M}\sum_{n=0}^{M-1}|cos^2(\phi)(cos^2(\alpha)-sin^2(\alpha)|, or\cdots \frac{|A|^2}{M}cos^2(\phi)\sum_{n=0}^{M-1}|(cos^2(\alpha)-sin^2(\alpha)| \end{equation} \begin{equation} (2) \frac{|A|^2}{M}\sum_{n=0}^{M-1}|sin^2(\alpha)| \end{equation} and...

(I didn't mention earlier that M is the period of the sinusoid, and the average power equation sums the sampled terms during one period only, thus M=N) Here is what I have so far: \begin{equation} x[n]=Acos(\frac{2\pi n}{M}+\phi) \end{equation} and \begin{equation}...

I think I may have accidentally caused some confusion with the title of the thread. Did my tex/mathjax script load properly on your browser? I may want to clarify for what I am looking before going too much further.. (id est, did I fail to type some sort of script bookend to make my equations...

Homework Statement Given \begin{equation} x[n]=A\cos(\frac{2\pi n}{M}+\phi) \end{equation} Evaluate(average power of a sinusoidal digitally sampled signal) : \begin{equation} P_x=\frac{1}{N}\sum_{n=0}^{N-1}{|x[n]|}^2 \end{equation}Homework Equations Trig Identities...