# Homework Help: Average and rms values of a given signal

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1. Nov 18, 2015

### hdp12

1. The problem statement, all variables and given/known data
I am in an Intro to EE class and we were given a homework assignment including the following question:

Find the average and rms values of the signal:
x(t)=3cos(7ωt)+4
I have formulas for average power and rms but they include V & I, however the signal was given as x(t). I understand that this is arbitrary.. or something but basically I'm extremely lost and need help.

Thanks

2. Relevant equations

Pav = 1/T ∫_0^T p(t)dt = 1/T ∫_0^T ½VI cos(θ)dt + 1/T ∫_0^T ½VI cos(2ωt-θ)dt
Pav = ½VI cos(θ)

3. The attempt at a solution

I don't even know where to start.

2. Nov 18, 2015

### Staff: Mentor

The problem wants the average and rms values for the given signal. Power doesn't enter into the problem. Just apply the definitions of average and rms to the given function.

Hint: Both definitions expect you to integrate over one period of a repeating function. You can go to the trouble of working out the time interval that represents one period and integrate over that, dividing by that time intervaland so on, or, since your function is based on a trig function you can integrate over one angular period instead of time. That is, the period is $2 \pi$ and you integrate over the angle from $0$ to $2 \pi$. No need to muss around with $\omega t$.

3. Nov 18, 2015

### hdp12

so I did
∫_0^2π 3cos(7x)+4 dx = 3/7 sin(7x) + 4x = 3/7sin(7*0) + 4(0) - 3/7 sin(7*2π) - 4(2π)
--> Pav = -8π

I feel as if that is wrong

4. Nov 18, 2015

### Staff: Mentor

The argument of the cosine should just be $\theta$. Not 7x. And don't forget to divide by the period. And it's not a power that you're finding, its the average of signal x(t), so maybe call it $x_{av}$.

$$x_{av} = \frac{1}{T} \int_0^T \{ 3 cos(7 \omega t)+4 \} ~dt ~~→~~ \frac{1}{2 \pi} \int_0^{2 \pi} \{ 3 cos(\theta)+4 \} ~d \theta$$

5. Nov 18, 2015

### hdp12

okay so after using the formula you gave me I got xav = -4

that feels weird too.. is there supposed to be a constant after I integrate?

6. Nov 18, 2015

### Staff: Mentor

Can you show your work? Where did the negative sign come from?

7. Nov 18, 2015

### hdp12

8. Nov 18, 2015

### Staff: Mentor

Why does the last term have a minus sign?

9. Nov 18, 2015

### hdp12

I may have finished the integral incorrectly... I haven't done integrals intensely since calc 2 (2 years ago) ...

I did ...
say y(t) is the integral I determined... the 3/2π sin ... etc
to solve a definite integral... y(0) - y(2π)

is that incorrect? I'm going to look it up right now

10. Nov 18, 2015

### hdp12

yeah its the other way around, y(2π) - y(0)

so the answer I get now is xav = 4

is that correct?

11. Nov 18, 2015

### Staff: Mentor

Most of it's right except for how you handled the constant at the end. You should end up with 4*2π/2π - 0*2π/2π...

12. Nov 18, 2015

### Staff: Mentor

Yes.

13. Nov 18, 2015

### hdp12

awesome! Thanks

now for the second part, how do I determine the rms?

14. Nov 18, 2015

### Staff: Mentor

That's for you to figure out!

What is the equation definition of the RMS value of a function? Your textbook should show it, or use wikipedia to look it up.

15. Nov 18, 2015

### hdp12

I found a formula that says
xrms = x/√2
so... xrms = $\frac{3}{\sqrt 2} cos{(7 \omega t)} + \frac{4}{\sqrt 2}$

yes?

16. Nov 18, 2015

### Staff: Mentor

It might be right, but I don't think so. If it were just the cos() term, then yes, the RMS value is the peak value divided by √2. But having the constant term there means you need to use the full equation for calculating the RMS of the addition of two signals. Just look for the equation that has the square root of the sum of the squares in it...

17. Nov 18, 2015

### hdp12

I can only find Vrms and Irms equations

I also found $\sqrt{\frac{1}{T} \int_0^T{A^2 sin^2(\omega t) dt}}$

If I use 2π for T like I did in the first part, does this formula work?

Also for A do I use 3 since it's in front of the cos() in the given signal equation? What do I do about the 4 at the end of that?

Thank you

18. Nov 18, 2015

### Staff: Mentor

To calculate the RMS value of a function, you need to square the function, average it over a period, and then take the square root. So more like this:

$\sqrt{\frac{1}{T} \int_0^T{(3cos(7ωt)+4)^2 dt}}$

19. Nov 18, 2015

### hdp12

This feels wrong

20. Nov 18, 2015

### Staff: Mentor

21. Nov 18, 2015

### hdp12

I have the wikipedia article open in another tab, and that's actually where I got the formula. Same one you gave me three posts back.

Is the work I just posted correct? I worked out the formula you & wikipedia gave me

22. Nov 18, 2015

### Staff: Mentor

I didn't check the math in detail, but it looks like you were doing the right things. Sure turned out messy, though...

23. Nov 18, 2015

### hdp12

That's what I was thinking too, should I use theta instead of 7 omega t ?

24. Nov 18, 2015

### Staff: Mentor

The significance of the 7ωt is in relation to the period that you integrate over. The period of 7ωt is not 2π. What is the period in terms of t, since that is what your are integrating over...?

25. Nov 18, 2015

### hdp12

This looks about right..