2
$\begingroup$

I am having difficulty with a homework problem which asks:

enter image description here

I have some ideas in mind but I have no clue as to whether they are correct or not. Below is my attempt:

enter image description here

$\endgroup$
  • $\begingroup$ This seems very much like a homework problem or self-study problem. These are on-topic here, BUT we prefer you to include your attempt (no matter whether you think it's right or not). Upload a picture of your working if that helps. I'll re-open (or another mod will) once that's done. $\endgroup$ – Peter K. Feb 23 '18 at 1:27
  • $\begingroup$ Alright, that makes sense. I have uploaded what I think the answer is. $\endgroup$ – user34067 Feb 23 '18 at 1:49
  • $\begingroup$ Not really what you're asking, but that is not a sensible definition of the center of mass of a signal. More sensible definitions would use either $|x|$ or $|x|^2$ in place of $x$. $\endgroup$ – Jazzmaniac Feb 23 '18 at 12:39
  • $\begingroup$ @user34067, I just noticed that there's a minus sign missing in the exponent in your first equation for X. $\endgroup$ – applesoup Feb 23 '18 at 12:44
  • $\begingroup$ I don't think your formula for the spectral centroid is correct. Please, see here en.wikipedia.org/wiki/Spectral_centroid. Anyway, I have a c++ implementaion and even if you don't use c++ it should be fairly easy to convert it to whatever language you're using. $\endgroup$ – dsp_user Feb 23 '18 at 18:40
1
$\begingroup$

The first thing you do is correct. You can calculate the sum of all the values of a signal evaluating its DTFT at $\omega=0$.

When you try to express the numerator with the DTFT though, you make a pretty clever thing but it is not correct, because you are evaluating the DTFT at $\omega=\ln\left( n^{\frac{1}{j2\pi n}}\right)$, which is not always real-valued. Remember that $\omega\in\mathbb{R}$.

To find how the DTFT of $x(n)$ and the numerator of $c$ are related, you can use the following property:

$$x(n)\xrightarrow{\mathscr{F}}X(\omega) \implies nx(n)\xrightarrow{\mathscr{F}}j\frac{dX(\omega)}{d\omega}$$

Using this property and your approach for the denominator (evaluating the transform at $\omega=0$), you should get the desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.