Average and rms values of a given signal

In summary, the conversation was about finding the average and rms values of a given signal in an Intro to EE class. The formulas for average power and rms were provided, but the student was unsure how to apply them to the given signal. Through the help of the expert, the student was able to solve for the average and rms values, with the final result for the rms value being confirmed by Wolfram Alpha.
  • #1
hdp12
68
2

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



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)

The Attempt at a Solution



I don't even know where to start.
 
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  • #2
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
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
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
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
hdp12 said:
I got xav = -4
Can you show your work? Where did the negative sign come from?
 
  • #7
ImageUploadedByPhysics Forums1447879175.478820.jpg
 
  • #8
Why does the last term have a minus sign?
 
  • #9
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
yeah its the other way around, y(2π) - y(0)

so the answer I get now is xav = 4

is that correct?
 
  • #11
hdp12 said:
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...
 
  • #12
hdp12 said:
is that correct?
Yes. :smile:
 
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  • #13
awesome! Thanks

now for the second part, how do I determine the rms?
 
  • #14
hdp12 said:
now for the second part, how do I determine the rms?
That's for you to figure out! :smile:

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
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
hdp12 said:
I found a formula that says
xrms = x/√2
so... xrms = ## \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... :smile:
 
  • #17
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
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
ImageUploadedByPhysics Forums1447882540.489669.jpg


This feels wrong
 
  • #21
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
I didn't check the math in detail, but it looks like you were doing the right things. Sure turned out messy, though...
 
  • #23
That's what I was thinking too, should I use theta instead of 7 omega t ?
 
  • #24
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
ImageUploadedByPhysics Forums1447884719.450146.jpg


This looks about right..
 
  • #26
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? :smile:
 
  • #27
berkeman said:
Did you double-check the answer with Wolfram Alpha yet? :smile:

I sure did √√√
thanks for everything
 
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Related to Average and rms values of a given signal

1. What is the difference between average and rms values?

The average value of a signal is the arithmetic mean of all the values in the signal. It represents the overall magnitude of the signal. The rms (root mean square) value, on the other hand, takes into account both the magnitude and the duration of the signal, giving a more accurate measure of the signal's strength.

2. How are average and rms values calculated?

To calculate the average value, you add all the values in the signal and divide by the number of values. To calculate the rms value, you square each value, add them up, divide by the number of values, and take the square root of the result.

3. Why are average and rms values important in signal processing?

Average and rms values are important because they provide a way to quantify the strength of a signal. They are used in various applications such as audio and video processing, power analysis, and data analysis.

4. Can the average value of a signal be higher than the rms value?

Yes, the average value of a signal can be higher than the rms value. This can occur when the signal has a large peak or spike that contributes significantly to the average but does not affect the rms value as much.

5. How are average and rms values used in AC circuits?

In AC circuits, the average value is used to calculate the average power, while the rms value is used to calculate the root mean square voltage or current. This is important in determining the power dissipation and the effective voltage or current in the circuit.

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