Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For the sum to be periodic, $f_1$ and $f_2$ must be commensurable; that is, there most be a number $f_0$ contained in each an integral number of times. Thus, if $f_0$ is the largest such number, $f_1$=$n_1 f_0$ and $f_2=n_2 f_0$ ($f_0$ is the fundamental frequency and $n_1$ and $n_2$ are integers) Which of the signals are periodic:

a) $x_1(t)=2 \cos(2t) + 3\sin(5\pi t)$

b) $x_2(t)=2 \cos(4\pi t) + 5\cos(6\pi t) + 6\sin(22\pi t)$

Find the period of those that are periodic.

How can I find $f_0$? I have an idea but I don't know if it's correct for:

a) We know that $\cos(2t)$ is $\cos wt$, where $w=2\pi f$ so $f_1$ would be $f_1=2/(2\pi)=0.318$ for $3\sin(5\pi t)$, $f_2=2.5$. But how do I find $f_0$ if I don't have $n_1$ or $n_2$??? After that I'm block... what am I missing?

Thanks for your help

share|cite|improve this question
up vote 0 down vote accepted

DON'T turn the $f_i$ into decimals. You have $$f_1={2\over2\pi},\qquad f_2={5\over2}$$ You want to know whether there are integers $n_1$ and $n_2$ such that $f_1/n_1=f_2/n_2=f_0$. That would make $f_2/f_1=n_2/n_1$, a rational number. So: is $f_2/f_1$ a rational number?

share|cite|improve this answer
if f2/f1 is a rational number, the sum of the sinusoidals are periodic. Accordingly, with your answer we will obtained the period like this, f2/f1=f0=10*pi/4 and the period is To=1/fo=4/10*pi rigth??? – user43680 Oct 31 '12 at 20:33
Also for the second signal b). How I can employ the same analysis of f1/n1=f2/n2=fo. Thanks for your help! – user43680 Oct 31 '12 at 21:34
Is $10\pi/4$ a rational number? For the second signal, can you calculate $x_2(t+1)$, and compare it to $x_2(t)$? – Gerry Myerson Oct 31 '12 at 21:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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