Sinusoidal function. Top percentage of values.

[Ry122]

Homework Statement

For a sinusoidal function, how do you determine the highest value exceeded 10% of the time?
The pink line in the attached pic indicates that value.
Just wondering how you actually determine the value for a periodic function?

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

For a sinusoidal function, how do you determine the highest value exceeded 10% of the time?
The pink line in the attached pic indicates that value.
Just wondering how you actually determine the value for a periodic function?

Homework Equations

The Attempt at a Solution

That function is not very sinusoidal. In that case it's simply a numerical problem. If you mean that you have a function like y=sin(x) that is, then just draw a graph, say over the range [0,2pi]. The highest value is 1 and the lowest value is -1, so you want to find the values of x where sin(x)=0.8. That isn't so hard, is it?

 

[Ry122]

I was just posting that graph to show what I mean by sound level exceeded 10% of the time I'm actually just trying to do it for a periodic function.
I'm not sure if what's you're saying answers the question or not. Maybe you misunderstood the question.
Someone else told me length of interval [pi/2 - pi/10, pi/2 + pi/10] is 10% of [0, 2pi]
Does this make sense to you?

 

[Dick]

I was just posting that graph to show what I mean by sound level exceeded 10% of the time I'm actually just trying to do it for a periodic function.
I'm not sure if what's you're saying answers the question or not. Maybe you misunderstood the question.
Someone else told me length of interval [pi/2 - pi/10, pi/2 + pi/10] is 10% of [0, 2pi]
Does this make sense to you?

Yes, I did misunderstand. I was thinking it was 10% of the max not 10% of the time. 10% of 2*pi is pi/5. Since the max of sin is at pi/2 then I think that "somebody else" is correct.

 

[Ry122]

Yeah thanks, that gave me the correct answer. If pi/2 - pi/10 is 10% of the time, what is it for 90% of the time?

 

[Dick]

Yeah thanks, that gave me the correct answer. If pi/2 - pi/10 is 10% of the time, what is it for 90% of the time?

Well, x is outside of the range you gave 90% of the time. And y=sin(x) is less than the value of sin(pi/2-pi/10), isn't it?

 

[Ry122]

i mean what value is exceed 90% of the time though so it's line on the above graph would be down the bottom somewhere

 

[Ray Vickson]

i mean what value is exceed 90% of the time though so it's line on the above graph would be down the bottom somewhere

If you draw a horizontal line at $y = L \; (0 < L < 1)$ it cuts the graph of $y = \sin(x)$ at two points, say $x=a$ and $x=b$, with $\sin(a) = \sin(b) = L.$ If you want $\sin(x) > L$ for 10% of the time, you want $b-a = 2 \pi/10 = \pi/5,$ so you need to solve the equation
$$\sin(a) = \sin(a + \pi/5)$$
Expanding $\sin(a + \pi/5)$ we get the equation
$$\sin(a) = \sin(a) \cos(\pi/5) + \cos(a) \sin(\pi/5)$$
Thus
$$[1-\cos(\pi/5)] \sin(a) = \sin(\pi/5) \cos(a) \Rightarrow \tan(a) = <br /> \frac{\sin(\pi/5)}{1-\cos(\pi/5)}$$
From this we can find $a$ and can get the level $L$ as
$$L = \sin(a) = \frac{\sin(\pi/5)}{\sqrt{2\left(1-\cos(\pi/5) \right)}}$$
We have $a = 1.256637061,$ and $L = .9510565160$ is the desired level.

 

[Dick]

If you draw a horizontal line at $y = L \; (0 < L < 1)$ it cuts the graph of $y = \sin(x)$ at two points, say $x=a$ and $x=b$, with $\sin(a) = \sin(b) = L.$ If you want $\sin(x) > L$ for 10% of the time, you want $b-a = 2 \pi/10 = \pi/5,$ so you need to solve the equation
$$\sin(a) = \sin(a + \pi/5)$$
Expanding $\sin(a + \pi/5)$ we get the equation
$$\sin(a) = \sin(a) \cos(\pi/5) + \cos(a) \sin(\pi/5)$$
Thus
$$[1-\cos(\pi/5)] \sin(a) = \sin(\pi/5) \cos(a) \Rightarrow \tan(a) = <br /> \frac{\sin(\pi/5)}{1-\cos(\pi/5)}$$
From this we can find $a$ and can get the level $L$ as
$$L = \sin(a) = \frac{\sin(\pi/5)}{\sqrt{2\left(1-\cos(\pi/5) \right)}}$$
We have $a = 1.256637061,$ and $L = .9510565160$ is the desired level.

a=1.256637061 is, not surprisingly, pi/2-pi/10. It's actually kind of obvious if you just look at the graph. Why are you choosing this difficult path to the result?