Power and RMS Value of a Signal

In summary, the conversation discusses different methods for calculating the power and RMS value of a given equation. The first method involves taking the limit of the integral over a period, while the second method relies on finding the amplitude and using a formula. While the second method may be simpler, it may not work for all signals, such as those with a constant term. The correct method depends on the given signal and may require more work to solve.
  • #1
JamesBennettBeta
10
1
Homework Statement
How to determine the power and RMS value of the given signals.
Relevant Equations
$$ P_{x} \ =\ \lim _{T\ \\rightarrow\infty } \ \frac{1}{T}\int _{-T/2}^{+T/2\ } x^{2} \ ( t) \ dt,$$

$$ P_{x} \ =\ \frac{A^{2}}{2},$$
I need to calculate the power and RMS value of some equations. The problem is, I found two methods to do that and don't know which is the right method.

I have few equations to find the power and RMS value, but here is one equation.

$$x( t) \ =\ 7\cos\left( 20t+\frac{\pi }{2}\right),$$
Method #1

To find the power value,
$$ P_{x} \ =\ \lim _{T\ \rightarrow\infty } \ \frac{1}{T}\int _{-T/2}^{+T/2\ } x^{2} \ ( t) \ dt ,$$

To find the RMS value,
$$ P_{rms} \ =\ \sqrt{P_{x}},$$

This seems the right method for me, but later I found a YouTube video with a completely different approach.

Methods #2

To find the power value,

$$ P_{x} \ =\ \frac{A^{2}}{2},$$

A is the amplitude of the equation.To find the RMS value,

$$ P_{rms} \ =\ \sqrt{P_{x}},$$
 
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  • #2
Hi @JamesBennettBeta. Some of what you have written appears wrong/confusing. For example ##P_{rms} \ =\ \sqrt{P_{x}}## cannot be correct if “P” represents power, because the 2 sides of the equation would have different dimensions.
Maybe you are actually being asked to find the mean square value, and the rms value, of some function x(t). But that’s just a bit of a wild guess.

Can you post the complete, original question, word-for-word?
 
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  • #3
1625234872515.png


Thanks for your reply, the question is this. A problem raised since I have found two ways to solve this. Anyway, I don't know either one is correct or not.

Method #1

1625235166813.png

1625235192468.png


Method #2
Method two was explained in a video from an Indian Dr. I tried to solve the above problem using his way,
$$ \begin{array}{l}

5+10 cos \left( 100t+\frac{\pi }{3}\right)

Using\ the\ form\ A\ cos( \omega _{0} t\ +\ \theta )\\
Power\ of\ the\ signal,\\
\\
Px\ =\ \frac{A^{2}}{2} =\frac{10^{2}}{4} \ =\ 50w\\
\\
RMS\ value\ of\ the\ signal\\
\\
P_{RMS} \ =\ \sqrt{50} \ =\ 5\sqrt{2}
\end{array}$$

The answers are completely different. I want to know what is the right method.
 
  • #4
JamesBennettBeta said:
A problem raised since I have found two ways to solve this. Anyway, I don't know either one is correct or not.
.
The answers are completely different. I want to know what is the right method.
A couple of general points first:

1. The rms value of a signal is not a power (it is the square-root of a power); it has the same dimensions as the signal itself. So do not use ##P_{rms}## as the symbol for the rms value of a signal; use (for example) ##x_{rms}##. So you could write ##x_{rms} = √P_x##.

(I will also note here that we are not using the term ‘power’ in the conventional sense of energy per unit time.)

2. The average value of sin(ωt+φ) or cos(ωt+φ) over a full cycle is zero. The average value of sin²(ωt+φ) or cos²(ωt+φ) over the same cycle is ½. These facts can save you a lot of work if correctly applied. (By applying them early, you can often save a lot of algebra.)
___________________________________

Your worked example is for x(t) = 5 + 10cos(100t + π/3). But this is not one of the questions you list (a – e). So I’m a bit confused a little by that.

Without checking your algebra detail, one point I note is that you appear to have incorrectly applied Method 2 when working out the power for x(t) = 5 + 10cos(100t + π/3).

I think that is the problem.

For a signal of the form:
x(t) = Acos(ωt + φ)
the average power is the average value of x² which is A²/2 (see point 2, above).

But for x(t) = 5 + 10cos(100t + π/3) you cannot take the average power to be A²/2 because of the constant term (5). For example the average power for
x(t) = 5 + 0.0001cos(100t + π/3) is a little over 5² = 25
(If that’s not clear, imagine a constant signal of x=5 units. The power is constant (25).)

And note you can't simply add the 25 to A²/2.

So method 2 will only work for some signals (but will save you a lot of time when correctly used). Method 1 will always work, but will require more work.

Sorry if that was a bit long-winded. Hope that it makes some sense!

Minor edits made.
 
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  • #5
The "power" of a signal is simply the average power dissipated by a 1 ohm resistor. For an aperiodic signal this is simply the limiting expression that you posted. For a periodic signal (like a sinusoid) this reduces to integrating the (magnitude) signal squared over one period and dividing by that period. The RMS value of a signal is the square root of the (quantity) integral of the signal squared divided by the period. For a sinusoid of the form A*cos(w*t + phi) this reduces to A/sqrt(2). The RMS value of a constant signal (like a DC signal) is simply the value of that DC signal etc. When doing these integrations it is helpful to have a table of trigonometric identities remembering that integrating a sinusoid over one period is always zero. E.g., when integrating cos(x)^2 convert to 0.5*(1 + cos(2*x)) which, over a period, integrates to 0.5 as the second integral vanishes.
 

Related to Power and RMS Value of a Signal

1. What is the difference between power and RMS value of a signal?

Power is a measure of the rate at which energy is transferred, while RMS (Root Mean Square) value is a measure of the average magnitude of a signal. Power takes into account both the amplitude and frequency of a signal, while RMS value only considers the amplitude.

2. How is the power of a signal calculated?

The power of a signal can be calculated by squaring the amplitude of the signal and then taking the average over a period of time. This is also known as the mean squared value (MSV) or the mean squared amplitude (MSA).

3. What is the significance of RMS value in electrical engineering?

RMS value is important in electrical engineering as it is used to calculate the effective voltage and current in AC circuits. It is also used to determine the power dissipated by a circuit and to compare the strength of different signals.

4. How is the RMS value of a signal different from its peak value?

The RMS value of a signal is the equivalent DC value that would produce the same amount of power as the AC signal. It takes into account the positive and negative values of the signal, while the peak value only considers the maximum amplitude of the signal.

5. Can the RMS value of a signal be greater than its peak value?

No, the RMS value of a signal can never be greater than its peak value. The RMS value is always equal to or less than the peak value, depending on the shape of the signal. For a sinusoidal signal, the RMS value is equal to the peak value divided by the square root of 2.

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