I'm trying to find the fundamental frequency of
$\cos(23t + \pi/2) + \sin(5t + \pi/5)$
Now the period for both are $2\pi/23$ and $2\pi/5$ respectively. So i need to find the LCM of these two fractions.
As far as i'm aware LCM(a/b , c/d) = LCM(a,c)/HCF(b,d)
This would result in $2\pi/1 = 2\pi$.
When plugged into wolfram alpha it says the period is also 2pi. However I have an online quiz in which i entered 2pi and it says it is wrong.
Am I doing something wrong? Any help is greatly appreciated.
Start with $$ \begin{align} x(t+T) &= \Big[\cos(23t + 23T+ \frac{\pi}{2})\Big] + \Big[\sin(5t + 5T+ \frac{\pi}{5})\Big] \\ &= \Big[c(23t)c(23T)c(\frac{\pi}{2}) - s(23t)s(23T)c(\frac{\pi}{2}) -s(23t)c(23T)s(\frac{\pi}{2}) - c(23t)s(23T)s(\frac{\pi}{2}) \Big] \\ &+\Big[ s(5t)c(5T)c(\frac{\pi}{5}) + c(5t)s(5T)c(\frac{\pi}{5}) + c(5t)c(5T)s(\frac{\pi}{5})-s(5t)s(5T)s(\frac{\pi}{5}) \Big] \tag{1} \end{align} $$ Also, notice that $$ \begin{align} x(t) &= \cos(23t+\frac{\pi}{2})+\sin(5t+\frac{\pi}{5}) \\ &= c(23t)c(\frac{\pi}{2})-s(23t)s(\frac{\pi}{2}) + s(5t)c(\frac{\pi}{5})+c(5t)s(\frac{\pi}{5}) \tag{2} \end{align} $$ Now in order for this equality $x(t)=x(t+T)$ to hold, $$ \begin{align} \Big[c(23t)c(23T)c(\frac{\pi}{2}) &- s(23t)s(23T)c(\frac{\pi}{2}) -s(23t)c(23T)s(\frac{\pi}{2}) - c(23t)s(23T)s(\frac{\pi}{2}) \Big] \\ &= c(23t)c(\frac{\pi}{2})-s(23t)s(\frac{\pi}{2}) \end{align} $$ This occurs if $23T_1=2\pi \implies T_1=\frac{2\pi}{23}$ where $T_1$ is the fundamental period of $\cos(23t+\frac{\pi}{2})$. And $$ \begin{align} \Big[ s(5t)c(5T)c(\frac{\pi}{5}) &+ c(5t)s(5T)c(\frac{\pi}{5}) + c(5t)c(5T)s(\frac{\pi}{5})-s(5t)s(5T)s(\frac{\pi}{5}) \Big] \\ &= s(5t)c(\frac{\pi}{5})+c(5t)s(\frac{\pi}{5}) \end{align} $$ This occurs if $5T_2=2\pi \implies T_2=\frac{2\pi}{5}$ where $T_2$ is the fundamental period of $\sin(5t+\frac{\pi}{5})$. The ratio of these fundamental periods is $\frac{T_1}{T_2}=\frac{2\pi/23}{2\pi/5}=\frac{5}{23}$ which is a rational number, therefore, the fundamental period is $T_o=23T_1=5T_2=2\pi$ sec. We can also use the least common divisor $T_o=$LCM($T_1,T_2$). Now the fundamental frequency $f_o=\frac{1}{T_o} = \frac{1}{2\pi}$ Hz.