I am having a problem in finding the fundamental period of the signal $x(t)$ given below: \begin{align} x(t) &= 2\cos\left(\frac 45 \pi t\right)\sin^2\left(\frac{16}{3} t\right)\\ &= 2\cos\left(\frac 45 \pi t\right)\cdot \frac 12\left[1-\cos\left(\frac{32}{3}t\right)\right]\\ &= \cos\left(\frac 45 \pi t\right) - \cos\left(\frac 45 \pi t\right)\cos\left(\frac{32}{3} t\right)\\ & = \cos\left(\frac 45 \pi t\right)-\frac 12\bigg\{\cos\left[\left(\frac 45 \pi -\frac{32}{3}\right)t\right]+\cos\left[\left(\frac 45 \pi -\frac{32}{3}\right)t\right]\bigg\} \end{align}

  • hay Ahmad, we know you're new here. and welcome. can you please learn to use $\LaTeX$ and spell your equations out like Dan did below? i realize it may be easier for you to just take a snapshot of your notes, but it's harder for us to read that. and you're asking us for help. so help us help you and save the photographs for other functions than displaying math. – robert bristow-johnson Sep 13 at 22:42

If your top equation is really $$ x(t) = 2\cos\left(\frac 45 \pi t\right)\sin^2\left(\frac{16}{3} t\right)\tag{1} $$ You gonna have a hard time getting the fundamental period/frequency as the there isn't an exact integer relating the two periods/frequencies. You can make approximations of the $\pi$ multiplier/divisor but the errors accumulates and this doesn't cut it. However if your equation is: $$ x(t) = 2\cos\left(\frac 45 \pi t\right)\sin^2\left(\frac{16}{3} \color{magenta}\pi t\right)\tag{2} $$ You could then easily extract the fundamental frequency $f_0$, and simply get the inverse $1/f_0$ to get the fundamental period $T_0$. For a signal $x(t)$

$$ x(t) = \sum_{k=1}^N a_k\cos\left(2\pi f_k t+ \phi_k\right) $$

The fundamental frequency is the greatest common divisor (GCD) of all the frequencies; or $$\mathrm{GCD}\left\{f_k\right\}\quad\big\vert \quad k = 1, \ldots, N$$

With non-integer values of $f_k$, you get the frequencies to the same common denominator or their least common multiple (LCM) and $f_0$ is the GCD of the numerators over this LCM, their common denominator. In your case, you have the form: $$ x(t) = \prod_{k=1}^N a_k\cos\left(2\pi f_k t+ \phi_k\right) $$ But the same principle applies. So, looking at the second-last equation you have two distinct frequencies: $f_1 = \frac 25$ and $f_2 =\frac{16}{3}$.

So in the end, you get $f_0 = \frac{2}{15}\implies T_0 = \frac 1f_0 = 7.5$.

Below is a plot illustrating this with the four waveforms below:

\begin{align} \color{cyan}{y_1} &= 2\cos\left(\frac 45 \pi t\right)\\ \color{green}{y_2} &= \sin^2\left(\frac{16}{3}\pi t\right)\\ \color{blue}{y_3} &= 2\cos\left(\frac 45\pi t\right)\cdot \sin^2\left(\frac{16}{3}\pi t\right)\\ \color{red}{y_4} &= 2\cos\left(2\pi\frac{2}{15}t\right)\\ \end{align} enter image description here

You see that the fundamental period of your signal $y_3$ ($\color{blue}{\rm blue}$) is at exactly $7.5 \ \rm seconds$ in the plot, matching exactly that of a pure sinusoidal wave $y_4$ ($\color{red}{\rm red}$) at $f_0 = \frac{2}{15}$.

Determine the repetition rate in time of both functions independently instead of multiplying it out, and then determine how many integer cycles of each where that time would be equal. Given that each function cyclically repeats (and that repetition rate is easily determined by observation), the product can only cyclically repeat when their cycles align (otherwise it would not be the same waveform repeating!).

Other approaches involve solving the auto-correlation function as that would readily show correlation peaks at the repetition period, but I think in this case the above approach would be very straightforward.

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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