Average and rms values of a given signal

[hdp12]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I don't even know where to start.

 

[gneill]

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 $\omega t$.

 

[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)
--> P_av = -8π

I feel as if that is wrong

 

[gneill]

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 $x_{av}$.

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

 

[hdp12]

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

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

 

[berkeman]

I got xav = -4

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

 

[hdp12]

 

[berkeman]

Why does the last term have a minus sign?

 

[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

 

[hdp12]

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

so the answer I get now is x_av = 4

is that correct?

 

[berkeman]

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

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...

 

[berkeman]

is that correct?

Yes.

 

[hdp12]

awesome! Thanks

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

 

[berkeman]

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

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.

 

[hdp12]

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

yes?

 

[berkeman]

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

yes?

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...

 

[hdp12]

I can only find V_rms and I_rms 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

 

[berkeman]

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:

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

 

[hdp12]

This feels wrong

 

[berkeman]

As described at wikipedia: https://en.wikipedia.org/wiki/Root_mean_square

 

[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

 

[berkeman]

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

 

[hdp12]

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

 

[berkeman]

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...?

 

[hdp12]

This looks about right..

 

[berkeman]

Yeah! I would guess a number a bit over 4, so that looks about right. Good job!

Did you double-check the answer with Wolfram Alpha yet?

 

[hdp12]

Did you double-check the answer with Wolfram Alpha yet?

I sure did √√√
thanks for everything