Suppose two phase-shifted cosine waves (lets say they are almost in antiphase and have a phase shift of 170 deg). Their sum produces a new sinusoid. What is the integral over the amplitude of the sum wave? Background: I need to know for two discrete signals what their average/sum amplitude is in dependence on their phase shift and variances.
[Math] Integral of sum of two sine waves
trigonometry
Related Solutions
Start with a "triangle" function of period $2 \pi/n$, minimum value $0$ and maximum value $\pi/n$: a suitable choice is $T_n(t) = \frac{1}{n} \arccos(\cos(n t))$. A regular $n$-gon of inradius $r_0$ is given, in polar coordinates, by $r = r_0/\cos(T_n(\theta - \theta_0))$. Its $y$ coordinate is then $$y = \dfrac{r_0 \sin(\theta)}{\cos(T_n(\theta - \theta_0))}$$
Here is a similar animation using a pentagon ($n=5$):
I have confirmed the envelope function to be
$$ e(t) = X_1+X_2 \cos((\omega_1-\omega_2) t+(\varphi_1-\varphi_2)) $$
based on two waves interfering. I tried it on a bunch of wave combinations and it seems to work reasonably well.
Best Answer
$y=\sin \left( x \right)$ and $y=\sin \left( x-\frac{170\pi }{180} \right)$ are the two waves with the latter shifted; phase=170ยบ.
Now the sum is: $$y=\sin \left( x \right)+\sin \left( x-\frac{17\pi }{18} \right)$$
Let's find two adjacent x intercepts to integrate:
$\sin \left( x \right)+\sin \left( x-\frac{17\pi }{18} \right)=0$
$x=\frac{17\pi }{18}-x+2\pi k,\; k\; belongs\; to\; Z$
$x=\frac{17\pi }{36}+\pi k$
I'll take k=0 and k=1:
$$x=\frac{17\pi }{36},\frac{53\pi }{36}$$
Now we integrate:
$$\int_{\frac{17\pi }{36}}^{\frac{53\pi }{36}}{\sin \left( x \right)+\sin \left( x-\frac{17\pi }{18} \right)dx=2\left( \cos \left( \frac{17\pi }{36} \right)-\cos \left( \frac{19\pi }{36} \right) \right)}$$
which is approximately 0.348623
Here's an image of the shaded area: